For questions 1-10 : Find the odd one
1)
1)518 2)448 3)426 4)574
2)
1)4 2)28 3)48 4)100
3)
1)AON 2)EWR 3)IEV 4)OLB
4)
1)Mouse 2)Monitor 3)Keyboard 4)Joy stick
5)
1).jpeg 2).exe 3).jpg 4).gif
6)
1)Scissors 2)Knives 3)Trousers 4)Knickers
7)
1)Kill 2)Will 3)Fill 4)Bill
8)
1)Xylophone 2)Telephone 3)Violin 4)Clarinet
9)
1)Nucles 2)Mitochondria 3)Celloplasm 4)Tissue
10)
1)RAM 2)Cache 3)Hard disk 4)Compact Disc
Direction for questions 11 to 15: Each question contains five or six statements followed by four sets of three statements each.Choose the set in which the statements are logically related.
11)
(A)No teacher is patient.
(B)Some teacher are not patients.
(C)No teachers is a doctor.
(D)Some nurses are not teachers.
(E)Some nurses are doctors.
(F)All patients are doctors.
1)CAF 2)CED 3)BCF 4)All the three
12)
(A)Some bowlers are not cricketers.
(B)All batsmen are bowlers.
(C)All cricketers are not batsmen.
(D)All bowlers are cricketers.
(E)Some cricketers are bowlers.
(F)Some bowlers are not batsmen.
1)ECF 2)ABC 3)CDF 4)None of these
13)
(A)Some SMS are not GMS.
(B)Each SMS is TMS.
(C)Some SMS are MMS.
(D)No MMS is GMS.
(E)Most TMS is MMS.
(F)No GMS is TMS.
1)ADC 2)DEF 3)BDF 4)None of these
14)
(A)No CPI is BJP.
(B)No BJP is Congress.
(C)Some BJP's are SAD
(D)No Congress is CPI
(E)Some SAD's are not Congress.
(F)Some SAD's are CPI's.
1)ABD 2)ECB 3)DFC 4)None of these
15)
(A)Every credit card is not a green card.
(B)Some credit cards are direct credit cards.
(C)Some direct credit cards are not debit cards.
(D)No credit card is a debit card.
(E)Some green cards are debit cards.
(F)No credit card is a direct credit card.
1)CDF 2)AED 3)BCD 4)None of these
Directions for questions 16 to 20: Read the following passage, and from the statements that follow each passage choose the one that can be concluded from the passage.
16)Those who can speak English. Some Marathi speaking people cannot speak English. All Bengali speaking people are Indians.All Indians can speak Hindi.
1)Some Bengali speaking people cannot speak Marathi.
2)Some English speaking people are not Indians.
3)Some Marathi speaking people cannot speak Bengali.
4)None of these
17)Some non vegetarians are animals.Vegetarian eat only vegatables.All non-vegetarians are vegetarians. Those who eat only vegetables do not eat meat.No animal is a human.
1)Some, who do not eat meat,are human.
2)No animal is vegetarian.
3)Human do not eat meat.
4)None of these.
18)Norah Jones is a singer.Those who don't know how to play violin,know how to play guitar.Those who sing songs,don't know how to play violin.Every singer sings songs.
1)Singers know how to play guitar.
2)All singers don't know how to play violine.
3)Norah Jones knows how to play guitar.
4)None of these.
19)Coal is black.Somethings which glitter are diamonds.No glitter is gold.Diamond is coal.
1)Some gold is coal.
2)Somethings, which are gold are not black.
3)Somethings, which are black are not gold.
4)None of these
20)Leaders are assertive people.Trainees are developers.Those who take classes are not Doctors.Doctors are not leaders.Some developers take classes.
1)Some assertive people are not trainees.
2)All trainees are not assertive people.
3)Some leaders are not trainees.
4)None of these.
Click here for the answers
Showing posts with label puzzles. Show all posts
Showing posts with label puzzles. Show all posts
Logical Puzzles -2
For questions 1 to 10 :Complete the following series.
1)2,3,11,38,102,__
1)200 2)208 3)227 4)236
2)6,27,138,831,5820,__
1)50523 2)46563 3)41203 4)43123
3)2,8,13,41,79,239,__
1)475 2)470 3)467 4)462
4)25,62,123,214,341,__
1)488 2)510 3)496 4)574
5)11,48,927,25125,__
1)49343 2)81729 3)1001000 4)64512
6)3,4,9,32,155,__
1)729 2)924 3)876 4)961
7)29,27,31,29,37,31,41,__
1)30 2)33 3)35 4)37
8)8,9,16,51,200,__
1)995 2)805 3)795 4)1005
9)15,66,132,363,726,__
1)1353 2)1303 3)1293 4)1403
10)5040,2520,3360,2800,3200,__
1)3000 2)2880 3)2960 4)3020
For questions 11-25:Choose the correct alternative from the given choices
11)In a certain code language, if the word PRINTER is coded as EINPRRT and the word SERVICE is coded as EINPRRT and the word SERVICE is coded as CEEIRSV, then how is the word SCANNING coded in that language?
1)ACIGSNNN 2)ACGNNNSI 3)ACGINNNS 4)ACIGNNNS
12)In a certain code language, if the word MATERIAL is coded as LAIRETAM, then how is the word FOUNDATION coded in that language?
1)NOITADNOF 2)NOITDANUOF 3)NOITANDUOF 4)NIOTADNUOF
13)In a certain code language, if the word PROJECT is coded as QTRNJIA, then how is the word PACKAGE coded in the language?
1)QCFOFML 2)QCFOEML 3)QCFOFNK 4)QCFOFMK
14)In a certain code language, if the word KIDNAP is coded as 30, and the word TERRORIST as 72, then how is the word MISCHIEF coded in that language?
1)80 2)56 3)72 4)42
15)In a certain code language, if the word MADRID is coded as 49, and the word TOKYO as 86, then how is the word RANGOON coded in that language?
1)83 2)85 3)84 4)86
16)Raju decided to go to Ravi's house.So he started out by his bike.Initially he travelled 10 km towards east and then he turned to his left and travelled 5 km.Later he travelled 8 km more to his right before turning to his left to travel 3 km more.Finally he travelled 3 km to his left to reach Raju's house.What is the distance between Raju's starting point and Ravi's house and in which direction is Ravi's house and in which direction is Ravi's house with respect to Raju's starting point?
1)17 km,South-east 2)23 km,North-east 3)17 km,North-east 4)23 km,North-west
17)One evening,Srinivas and Balaji were standing in a park,facing each other.It was observed that Balaji's shadow was falling to his right.Then which direction was Srinivas facing?
1)East 2)West 3)North 4)South
18)If 2*5=17 and 3*4=19 then 5*6=?
1)41 2)30 3)11 4)31
19)If 3#4=7 and 7#11=72 then 6#10=?
1)60 2)61 3)16 4)64
20)In a certain code language '+' means '*','*' means '-','-' means '/' and '/' means '+',then what is the simplified value for the following expression in which the mathematical operators are written according to the above code?
12-4+15*4+6/10
1)31 2)256 3)68.6 4)53
Click here to for the questions
1)2,3,11,38,102,__
1)200 2)208 3)227 4)236
2)6,27,138,831,5820,__
1)50523 2)46563 3)41203 4)43123
3)2,8,13,41,79,239,__
1)475 2)470 3)467 4)462
4)25,62,123,214,341,__
1)488 2)510 3)496 4)574
5)11,48,927,25125,__
1)49343 2)81729 3)1001000 4)64512
6)3,4,9,32,155,__
1)729 2)924 3)876 4)961
7)29,27,31,29,37,31,41,__
1)30 2)33 3)35 4)37
8)8,9,16,51,200,__
1)995 2)805 3)795 4)1005
9)15,66,132,363,726,__
1)1353 2)1303 3)1293 4)1403
10)5040,2520,3360,2800,3200,__
1)3000 2)2880 3)2960 4)3020
For questions 11-25:Choose the correct alternative from the given choices
11)In a certain code language, if the word PRINTER is coded as EINPRRT and the word SERVICE is coded as EINPRRT and the word SERVICE is coded as CEEIRSV, then how is the word SCANNING coded in that language?
1)ACIGSNNN 2)ACGNNNSI 3)ACGINNNS 4)ACIGNNNS
12)In a certain code language, if the word MATERIAL is coded as LAIRETAM, then how is the word FOUNDATION coded in that language?
1)NOITADNOF 2)NOITDANUOF 3)NOITANDUOF 4)NIOTADNUOF
13)In a certain code language, if the word PROJECT is coded as QTRNJIA, then how is the word PACKAGE coded in the language?
1)QCFOFML 2)QCFOEML 3)QCFOFNK 4)QCFOFMK
14)In a certain code language, if the word KIDNAP is coded as 30, and the word TERRORIST as 72, then how is the word MISCHIEF coded in that language?
1)80 2)56 3)72 4)42
15)In a certain code language, if the word MADRID is coded as 49, and the word TOKYO as 86, then how is the word RANGOON coded in that language?
1)83 2)85 3)84 4)86
16)Raju decided to go to Ravi's house.So he started out by his bike.Initially he travelled 10 km towards east and then he turned to his left and travelled 5 km.Later he travelled 8 km more to his right before turning to his left to travel 3 km more.Finally he travelled 3 km to his left to reach Raju's house.What is the distance between Raju's starting point and Ravi's house and in which direction is Ravi's house and in which direction is Ravi's house with respect to Raju's starting point?
1)17 km,South-east 2)23 km,North-east 3)17 km,North-east 4)23 km,North-west
17)One evening,Srinivas and Balaji were standing in a park,facing each other.It was observed that Balaji's shadow was falling to his right.Then which direction was Srinivas facing?
1)East 2)West 3)North 4)South
18)If 2*5=17 and 3*4=19 then 5*6=?
1)41 2)30 3)11 4)31
19)If 3#4=7 and 7#11=72 then 6#10=?
1)60 2)61 3)16 4)64
20)In a certain code language '+' means '*','*' means '-','-' means '/' and '/' means '+',then what is the simplified value for the following expression in which the mathematical operators are written according to the above code?
12-4+15*4+6/10
1)31 2)256 3)68.6 4)53
Click here to for the questions
Logical Puzzles-1
1)What is the least number of cuts required to cut a cube into 504 identical pieces?
1)24 2)20 3)21 4)22
2)What is the maximum number of identical pieces obtained when a cube is cut by 15 cuts?
1)125 2)216 3)210 4)64
3)What is the maximum number of identical pieces a cube can produce when cut into by 20 cuts?
1)441 2)343 3)512 4)448
4)What is the least number of cuts required to cut a cube into 343 identical pieces?
1)18 2)21 3)24 4)20
5)A cube is decorated in such a way that, a diamond is placed at each corner, six diamonds at each edge and a diamond at the centre of every face.How many diamonds are there on the cube?
1)74 2)70 3)62 4)64
The next questions 6-14 are based on the following information:
A large cube is taken and one pair of opposite faces is painted in blue, one pair of adjacent faces is painted in green and among the remaining faces, one is painted in red and the other in blue. Now , the cube is cut into 125 small and identical pieces.
6)How many smaller cubes have all the three colours?
1)0 2)1 3)2 4)4
7)How many smaller cubes have only one colour on them?
1)54 2)57 3)60 4)63
8)How many smaller cubes have exactly one face painted?
1)54 2)96 3)63 4)75
9)How many smaller cubes have exactly two colours on them?
1)27 2)33 3)36 4)30
10)How many smaller cubes have exactly two painted faces?
1)30 2)36 3)33 4)27
11)How many smaller cubes have exactly two painted faces in exactly two colours?
1)24 2)27 3)30 4)33
12)How many smaller cubes have no blue colour on them?
1)45 2)57 3)65 4)60
13)How many smaller cubes have only blue and green on them?
1)19 2)15 3)23 4)27
14)How many smaller cubes have red or green but not blue?
1)36 2)30 3)60 4)33
The next 15-18 questions are based on the following information:
216 small but identical cubes are put together to form a large cube. This large cube is now painted on all six faces.
15)How many of the smaller cubes have no face painted at all?
1)125 2)27 3)64 4)49
16)How many of the smaller cubes have exactly one face painted?
1)96 2)16 3)120 4)112
17)How many of the smaller cubes have exactly two faces painted?
1)56 2)60 3)64 4)48
18)How many of the smaller cubes have exactly three faces painted?
1)8 2)9 3)16 4)27
The next 19-20 questions are based on the following information:
In a class of 150 students, 40 passed in Social Studies, 90 passed in Science and 30 failed in both the subjects.
19)How many students passed in at most one subject among Science and Social Science?
1)100 2)110 3)140 4)150
20)How many students passed in Science but failed in Social Studies?
1)70 2)10 3)90 4)80
Click here for the answers of all the 20 questions
1)24 2)20 3)21 4)22
2)What is the maximum number of identical pieces obtained when a cube is cut by 15 cuts?
1)125 2)216 3)210 4)64
3)What is the maximum number of identical pieces a cube can produce when cut into by 20 cuts?
1)441 2)343 3)512 4)448
4)What is the least number of cuts required to cut a cube into 343 identical pieces?
1)18 2)21 3)24 4)20
5)A cube is decorated in such a way that, a diamond is placed at each corner, six diamonds at each edge and a diamond at the centre of every face.How many diamonds are there on the cube?
1)74 2)70 3)62 4)64
The next questions 6-14 are based on the following information:
A large cube is taken and one pair of opposite faces is painted in blue, one pair of adjacent faces is painted in green and among the remaining faces, one is painted in red and the other in blue. Now , the cube is cut into 125 small and identical pieces.
6)How many smaller cubes have all the three colours?
1)0 2)1 3)2 4)4
7)How many smaller cubes have only one colour on them?
1)54 2)57 3)60 4)63
8)How many smaller cubes have exactly one face painted?
1)54 2)96 3)63 4)75
9)How many smaller cubes have exactly two colours on them?
1)27 2)33 3)36 4)30
10)How many smaller cubes have exactly two painted faces?
1)30 2)36 3)33 4)27
11)How many smaller cubes have exactly two painted faces in exactly two colours?
1)24 2)27 3)30 4)33
12)How many smaller cubes have no blue colour on them?
1)45 2)57 3)65 4)60
13)How many smaller cubes have only blue and green on them?
1)19 2)15 3)23 4)27
14)How many smaller cubes have red or green but not blue?
1)36 2)30 3)60 4)33
The next 15-18 questions are based on the following information:
216 small but identical cubes are put together to form a large cube. This large cube is now painted on all six faces.
15)How many of the smaller cubes have no face painted at all?
1)125 2)27 3)64 4)49
16)How many of the smaller cubes have exactly one face painted?
1)96 2)16 3)120 4)112
17)How many of the smaller cubes have exactly two faces painted?
1)56 2)60 3)64 4)48
18)How many of the smaller cubes have exactly three faces painted?
1)8 2)9 3)16 4)27
The next 19-20 questions are based on the following information:
In a class of 150 students, 40 passed in Social Studies, 90 passed in Science and 30 failed in both the subjects.
19)How many students passed in at most one subject among Science and Social Science?
1)100 2)110 3)140 4)150
20)How many students passed in Science but failed in Social Studies?
1)70 2)10 3)90 4)80
Click here for the answers of all the 20 questions
Puzzles
puzzles are very important for getting jobs,i am putting some important puzzles and others puzzles which covers almost all the models those asking in interviews.
Thanks,
All The Best
Answers for the PRACTICE PUZZLES
Thanks,
All The Best
Answers For Practice Puzzles - 4
1) Start with ZBG and ZBGJ. It should be either "the/then" or "you/your" combination as they appear more.
B R W Q H L F K W H J K Q I B W K
o b s t a c l e s a r e t h o s e
Q I C E D W Z B G W K K M I K E
t h i n g s y o u s e e w h e n
Z B G Q H S K Z B G J K Z K W
y o u t a k e y o u r e y e s
B U U Z B G J D B H F W.
o f f y o u r g o a l s.
2) It is obvious that between 5 O'clock and 6 O'clock the hands will not be exactly opposite to each other. It is also obvious that the hands will be opposite to each other just before 5 O'clock. Now to find exact time:
The hour hand moves 1 degree for every 12 degrees that the minute hand moves. Let the hour hand be X degree away from 5 O'clock. Therefore the minute hand is 12X degree away from 12 O'clock.
Therefore solving for X
Angle between minute hand and 12 O'clock + Angle between 12 O'clock and 4 O'clock + Angle between 4 O'clock and hour hand = 180
12X + 120 + (30-X) = 180
11X = 30
Hence X = 30/11 degrees
(hour hand is X degree away from 5 O'clock)
Now each degree the hour hand moves is 2 minutes.
Therefore minutes are
= 2 * 30/11
= 60/11
= 5.45 (means 5 minutes 27.16 seconds)
Therefore the exact time at which the hands are opposite to each other is
= 4 hrs. 54 min. 32.74 seconds
3) They took their percentages from 40 and not from 39, so they got more than their share.
The oldest son got 1/2 of 40 = 20 which is 0.5 more
The second son got 1/4 of 40 = 10 which is 0.25 more
The third son got 1/8 of 40 = 5 which is 0.125 more
The youngest son got 1/10 of 40 = 4 which is 0.1 more
And the stranger got 1/40 of 40 = 1 which is 0.025 more (As he is not supposed to get anything)
All these fractions add to = 0.5 + 0.25 + 0.125 + 0.1 + 0.025 = 1 which stranger took away.
4) There are total 2 couples and a son. Grandfather and Grand mother, their son and his wife and again their son. So total 5 people.
Grandfather, Grandmother
|
|
Son, wife
|
|
Son
5) He lost Rs.600
First time restaurant has given food worth Rs.105 and Rs. 395 change. Similarly second time, food worth Rs.80 and Rs.20 change. Here, we are not considering food restaurant profits.
6) 3651 represents LENS.
Let's assign possible values to each letter and then use trial-n-error.
S must be 1.
Then D (under L) must be greater than 5. If D is 6, then L is 0. But then A must be 0 or 1 which is impossible. Hence, the possible values of D are 7, 8 or 9.
N must be E + 1. Also, D must be A + 5 as the possible values of D are 7, 8 or 9, D can not be (10+A) + 5.
Now using trial-n-error, we get S=1, I=2, L=3, A=4, N=5, E=6 and D=9
S L I D E 1 3 2 9 6
- D E A N - 9 6 4 5
----- -----
3 6 5 1 L E N S
Hence, 3651 represents LENS.
7) Clark is the landlord.
Assume each statement true, one at a time and see that no other statement is contradicted.
Let's assume that Statement (1) is true. Then, Adam is the landlord and lives in apartment 1. Also, other three's apartments will be on the right of his apartment - which contradicts Statement (4) i.e. If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin. Thus, Adam is not the landlord.
Let's assume that Statement (2) is true. Then, Edmund is the landlord and lives in apartment 4. Also, other three's apartments will be on the left of his apartment - which again contradicts Statement (4) i.e. If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin. Thus, Edmund is not the landlord either.
Let's assume that Statement (3) is true. Then, Clark is the landlord and lives in apartment 3. It satisfies all the statements for
(1) Adam - (2) Edmund - (3) Clark - (4) Burzin
Hence, Clark is the landlord.
Similarly, you can assume Statement (4) true and find out that it also contradicts.
8) J is the married man.
Note that a person's sister-in-law may be the wife of that person's brother or the sister of that person's spouse.
There are 2 cases:
If B's legal spouse is J, then J's sibling must be P and P's sister-in-law must be B.
If B's legal spouse is P, then P's sister-in-law must be J and J's sibling must be B.
It is given that B's legal spouse and J's sibling are of the same sex. Also, it is obvious that P's sister-in-law is female. Then, B's legal spouse and J's sibling both must be males.
B's spouse J's sibling P's sister-in-law
(male) (male) (female)
Case I J P B
Case II P B J
Case II is not possible as B & P are married to each other and both are male. Hence, J is the married man.
9) 44
36 of the cubes have EXACTLY 2 of their sides painted black, but because a cube with 3 of its sides painted black has 2 of its sides painted black, you must also include the corner cubes. This was a trick question, but hopefully the title of the puzzle tipped you off to this.
10) t takes 5 moves to make the triangle with 5 rows point the other way.
0 = a coin that has not been moved.
X = the old position of the moved coin
8 = the new position of the moved coin.
________X
_______X X
____8 0 0 0 8
_____0 0 0 0
____X 0 0 0 X
_______8 8
________8
For traingle of any number of rows, the optimal number of moves can be achieved by moving the vertically symmetrical coins i.e. by moving same number of coins from bottom left and right, and remaining coins from the top.
For a triangle with an odd number of rows, the total moves require are :
(N^2/4) - (N-4) Where N = 4, 6, 8, 10, ...
For a triangle with even number of rows, the total moves require are :
((N^2-1)/4) - (N-4) Where N = 5, 7, 9, 11, ...
11) Each envelope contains the money equal to the 2 raised to the envelope number minus 1. The sentence "Each envelope contains the least number of bills possible of any available US currency" is only to misguide you. This is always possible for any amount !!!
One more thing to notice here is that the man must have placed money in envelopes in such a way that if he bids for any amount less than $25000, he should be able to pick them in terms of envelopes.
First envelope contains, 2^0 = $1
Second envelope contains, 2^1 = $2
Third envelope contains, 2^2 = $4
Fourth envelope contains, 2^3 = $8 and so on...
Hence the amount in envelopes are $1, $2, $4, $8, $16, $32, $64, $128, $256, $512, $1024, $2048, $4096, $8192, $8617
Last envelope (No. 15) contains only $8617 as total amount is only $25000.
Now as he bids for $8322 and gives envelope number 2, 8 and 14 which contains $2, $128 and $8192 respectively.
Envelope No 2 conrains one $2 bill
Envelope No 8 conrains one $100 bill, one $20 bill, one $5 bill, one $2 bill and one $1 bill
Envelope No 14 conrains eighty-one $100 bill, one $50 bill, four $10 bill and one $2 bill
12) The minute and the hour hand meet 11 times in 12 hours in normal watch i.e. they meet after every
= (12 * 60) / 11 minutes
= 65.45 minutes
= 65 minutes 27.16 seconds
But in our case they meet after every 65 minutes means the watch is gaining 27.16 seconds.
13) The number is 45, simply because
45 = 5 * (4 + 5)
How does one find this number?
Let T be the digit in the tens place and U be the digit in the units place. Then, the number is 10*T + U, and the sum of its digits is T + U.
The following equation can be readily written:
10*T + U = 5*(T + U) or
10*T + U = 5*T + 5*U or
5*T = 4*U
Thus, T / U = 4 / 5
Since T and U are digits, T must be 4 and U must be 5.
14) Total no of balls = 5 + 7 + 14 + 16 + 18 + 29 = 89
Total number of balls are odd. Also, same number of red balls and blue balls are left out after selling one box. So it is obvious that the box with odd number of balls in it is sold out i.e. 5, 7 or 29.
Now using trial and error method,
(89-29) /2 = 60/2 = 30 and
14 + 16 = 5 + 7 + 18 = 30
So box with 29 balls is sold out.
15) 47 Chocolates and 16 Friends
Let's assume that there are total C chocolates and F friends.
According to first case, if she gives 3 chocolates to each friend, one friend will get only 2 chocolates.
3*(F - 1) + 2 = C
Similarly, if she gives 2 chocolates to each friends, she will left with 15 chocolates.
2*F + 15 = C
Solving above 2 equations, F = 16 and C = 47. Hence, Ekta got 47 chocolates and 16 friends
16) Esha's age is 45 years.
Assume that Esha's age is 10X+Y years. Hence, her hunsbands age is (10Y + X) years.
It is given that difference between their age is 1/11th of the sum of their age. Hence,
[(10Y + X) - (10X + Y)] = (1/11)[(10Y + X) + (10X + Y)]
(9Y - 9X) = (1/11)(11X + 11Y)
9Y - 9X = X + Y
8Y = 10X
4Y = 5X
Hence, the possible values are X=4, Y=5 and Esha's age is 45 years.
17) The fish is 128 inches long.
It is obvious that the lenght of the fish is the summation of lenghts of the head, the body and the tail. Hence,
Fish (F) = Head (H) + Body (B) + Tail (T)
But it is given that the lenght of the head is 4 inches i.e. H = 4. The body is three-quarters of its total length i.e. B = (3/4)*F. And the tail is its head plus a quarter the lenght of its body i.e. T = H + B/4. Thus, the equation is
F = H + B + T
F = 4 + (3/4)*F + H + B/4
F = 4 + (3/4)*F + 4 + (1/4)*(3/4)*F
F = 8 + (15/16)*F
(1/16)*F = 8
F = 128 inches
Thus, the fish is 128 inches long.
18) Everybody in the World will get to know the scandal in 8 hours.
You came to know of a scandal and you passed it on to 4 persons in 30 minutes. So total (1+4) 5 persons would know about it in 30 minutes.
By the end of one hour, 16 more persons would know about it. So total of (1+4+16) 21 persons would know about it in one hour.
Similarly, the other (1+4+16+64) persons would have know about it in one and a half hours. (1+4+16+64+256) persons would have know about it in two hours and so on...
It can be deduced that the terms of the above series are the power of 4 i.e. 4^0, 4^1, 4^2, 4^3 and so on upto (2N+1) terms. Also, the last term would be 4^2N where N is the number of hours.
Sum of the above mentioned series = [4^(2N+1)-1]/3
***END OF ANSWER SESSION FOR PRACTISE PUZZLES _3 ******
Thanks,
All The Best
B R W Q H L F K W H J K Q I B W K
o b s t a c l e s a r e t h o s e
Q I C E D W Z B G W K K M I K E
t h i n g s y o u s e e w h e n
Z B G Q H S K Z B G J K Z K W
y o u t a k e y o u r e y e s
B U U Z B G J D B H F W.
o f f y o u r g o a l s.
2) It is obvious that between 5 O'clock and 6 O'clock the hands will not be exactly opposite to each other. It is also obvious that the hands will be opposite to each other just before 5 O'clock. Now to find exact time:
The hour hand moves 1 degree for every 12 degrees that the minute hand moves. Let the hour hand be X degree away from 5 O'clock. Therefore the minute hand is 12X degree away from 12 O'clock.
Therefore solving for X
Angle between minute hand and 12 O'clock + Angle between 12 O'clock and 4 O'clock + Angle between 4 O'clock and hour hand = 180
12X + 120 + (30-X) = 180
11X = 30
Hence X = 30/11 degrees
(hour hand is X degree away from 5 O'clock)
Now each degree the hour hand moves is 2 minutes.
Therefore minutes are
= 2 * 30/11
= 60/11
= 5.45 (means 5 minutes 27.16 seconds)
Therefore the exact time at which the hands are opposite to each other is
= 4 hrs. 54 min. 32.74 seconds
3) They took their percentages from 40 and not from 39, so they got more than their share.
The oldest son got 1/2 of 40 = 20 which is 0.5 more
The second son got 1/4 of 40 = 10 which is 0.25 more
The third son got 1/8 of 40 = 5 which is 0.125 more
The youngest son got 1/10 of 40 = 4 which is 0.1 more
And the stranger got 1/40 of 40 = 1 which is 0.025 more (As he is not supposed to get anything)
All these fractions add to = 0.5 + 0.25 + 0.125 + 0.1 + 0.025 = 1 which stranger took away.
4) There are total 2 couples and a son. Grandfather and Grand mother, their son and his wife and again their son. So total 5 people.
Grandfather, Grandmother
|
|
Son, wife
|
|
Son
5) He lost Rs.600
First time restaurant has given food worth Rs.105 and Rs. 395 change. Similarly second time, food worth Rs.80 and Rs.20 change. Here, we are not considering food restaurant profits.
6) 3651 represents LENS.
Let's assign possible values to each letter and then use trial-n-error.
S must be 1.
Then D (under L) must be greater than 5. If D is 6, then L is 0. But then A must be 0 or 1 which is impossible. Hence, the possible values of D are 7, 8 or 9.
N must be E + 1. Also, D must be A + 5 as the possible values of D are 7, 8 or 9, D can not be (10+A) + 5.
Now using trial-n-error, we get S=1, I=2, L=3, A=4, N=5, E=6 and D=9
S L I D E 1 3 2 9 6
- D E A N - 9 6 4 5
----- -----
3 6 5 1 L E N S
Hence, 3651 represents LENS.
7) Clark is the landlord.
Assume each statement true, one at a time and see that no other statement is contradicted.
Let's assume that Statement (1) is true. Then, Adam is the landlord and lives in apartment 1. Also, other three's apartments will be on the right of his apartment - which contradicts Statement (4) i.e. If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin. Thus, Adam is not the landlord.
Let's assume that Statement (2) is true. Then, Edmund is the landlord and lives in apartment 4. Also, other three's apartments will be on the left of his apartment - which again contradicts Statement (4) i.e. If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin. Thus, Edmund is not the landlord either.
Let's assume that Statement (3) is true. Then, Clark is the landlord and lives in apartment 3. It satisfies all the statements for
(1) Adam - (2) Edmund - (3) Clark - (4) Burzin
Hence, Clark is the landlord.
Similarly, you can assume Statement (4) true and find out that it also contradicts.
8) J is the married man.
Note that a person's sister-in-law may be the wife of that person's brother or the sister of that person's spouse.
There are 2 cases:
If B's legal spouse is J, then J's sibling must be P and P's sister-in-law must be B.
If B's legal spouse is P, then P's sister-in-law must be J and J's sibling must be B.
It is given that B's legal spouse and J's sibling are of the same sex. Also, it is obvious that P's sister-in-law is female. Then, B's legal spouse and J's sibling both must be males.
B's spouse J's sibling P's sister-in-law
(male) (male) (female)
Case I J P B
Case II P B J
Case II is not possible as B & P are married to each other and both are male. Hence, J is the married man.
9) 44
36 of the cubes have EXACTLY 2 of their sides painted black, but because a cube with 3 of its sides painted black has 2 of its sides painted black, you must also include the corner cubes. This was a trick question, but hopefully the title of the puzzle tipped you off to this.
10) t takes 5 moves to make the triangle with 5 rows point the other way.
0 = a coin that has not been moved.
X = the old position of the moved coin
8 = the new position of the moved coin.
________X
_______X X
____8 0 0 0 8
_____0 0 0 0
____X 0 0 0 X
_______8 8
________8
For traingle of any number of rows, the optimal number of moves can be achieved by moving the vertically symmetrical coins i.e. by moving same number of coins from bottom left and right, and remaining coins from the top.
For a triangle with an odd number of rows, the total moves require are :
(N^2/4) - (N-4) Where N = 4, 6, 8, 10, ...
For a triangle with even number of rows, the total moves require are :
((N^2-1)/4) - (N-4) Where N = 5, 7, 9, 11, ...
11) Each envelope contains the money equal to the 2 raised to the envelope number minus 1. The sentence "Each envelope contains the least number of bills possible of any available US currency" is only to misguide you. This is always possible for any amount !!!
One more thing to notice here is that the man must have placed money in envelopes in such a way that if he bids for any amount less than $25000, he should be able to pick them in terms of envelopes.
First envelope contains, 2^0 = $1
Second envelope contains, 2^1 = $2
Third envelope contains, 2^2 = $4
Fourth envelope contains, 2^3 = $8 and so on...
Hence the amount in envelopes are $1, $2, $4, $8, $16, $32, $64, $128, $256, $512, $1024, $2048, $4096, $8192, $8617
Last envelope (No. 15) contains only $8617 as total amount is only $25000.
Now as he bids for $8322 and gives envelope number 2, 8 and 14 which contains $2, $128 and $8192 respectively.
Envelope No 2 conrains one $2 bill
Envelope No 8 conrains one $100 bill, one $20 bill, one $5 bill, one $2 bill and one $1 bill
Envelope No 14 conrains eighty-one $100 bill, one $50 bill, four $10 bill and one $2 bill
12) The minute and the hour hand meet 11 times in 12 hours in normal watch i.e. they meet after every
= (12 * 60) / 11 minutes
= 65.45 minutes
= 65 minutes 27.16 seconds
But in our case they meet after every 65 minutes means the watch is gaining 27.16 seconds.
13) The number is 45, simply because
45 = 5 * (4 + 5)
How does one find this number?
Let T be the digit in the tens place and U be the digit in the units place. Then, the number is 10*T + U, and the sum of its digits is T + U.
The following equation can be readily written:
10*T + U = 5*(T + U) or
10*T + U = 5*T + 5*U or
5*T = 4*U
Thus, T / U = 4 / 5
Since T and U are digits, T must be 4 and U must be 5.
14) Total no of balls = 5 + 7 + 14 + 16 + 18 + 29 = 89
Total number of balls are odd. Also, same number of red balls and blue balls are left out after selling one box. So it is obvious that the box with odd number of balls in it is sold out i.e. 5, 7 or 29.
Now using trial and error method,
(89-29) /2 = 60/2 = 30 and
14 + 16 = 5 + 7 + 18 = 30
So box with 29 balls is sold out.
15) 47 Chocolates and 16 Friends
Let's assume that there are total C chocolates and F friends.
According to first case, if she gives 3 chocolates to each friend, one friend will get only 2 chocolates.
3*(F - 1) + 2 = C
Similarly, if she gives 2 chocolates to each friends, she will left with 15 chocolates.
2*F + 15 = C
Solving above 2 equations, F = 16 and C = 47. Hence, Ekta got 47 chocolates and 16 friends
16) Esha's age is 45 years.
Assume that Esha's age is 10X+Y years. Hence, her hunsbands age is (10Y + X) years.
It is given that difference between their age is 1/11th of the sum of their age. Hence,
[(10Y + X) - (10X + Y)] = (1/11)[(10Y + X) + (10X + Y)]
(9Y - 9X) = (1/11)(11X + 11Y)
9Y - 9X = X + Y
8Y = 10X
4Y = 5X
Hence, the possible values are X=4, Y=5 and Esha's age is 45 years.
17) The fish is 128 inches long.
It is obvious that the lenght of the fish is the summation of lenghts of the head, the body and the tail. Hence,
Fish (F) = Head (H) + Body (B) + Tail (T)
But it is given that the lenght of the head is 4 inches i.e. H = 4. The body is three-quarters of its total length i.e. B = (3/4)*F. And the tail is its head plus a quarter the lenght of its body i.e. T = H + B/4. Thus, the equation is
F = H + B + T
F = 4 + (3/4)*F + H + B/4
F = 4 + (3/4)*F + 4 + (1/4)*(3/4)*F
F = 8 + (15/16)*F
(1/16)*F = 8
F = 128 inches
Thus, the fish is 128 inches long.
18) Everybody in the World will get to know the scandal in 8 hours.
You came to know of a scandal and you passed it on to 4 persons in 30 minutes. So total (1+4) 5 persons would know about it in 30 minutes.
By the end of one hour, 16 more persons would know about it. So total of (1+4+16) 21 persons would know about it in one hour.
Similarly, the other (1+4+16+64) persons would have know about it in one and a half hours. (1+4+16+64+256) persons would have know about it in two hours and so on...
It can be deduced that the terms of the above series are the power of 4 i.e. 4^0, 4^1, 4^2, 4^3 and so on upto (2N+1) terms. Also, the last term would be 4^2N where N is the number of hours.
Sum of the above mentioned series = [4^(2N+1)-1]/3
***END OF ANSWER SESSION FOR PRACTISE PUZZLES _3 ******
Thanks,
All The Best
Practise Puzzles - 4
11) A man is going to an Antique Car auction. All purchases must be paid for in cash. He goes to the bank and draws out $25,000.
Since the man does not want to be seen carrying that much money, he places it in 15 evelopes numbered 1 through 15. Each envelope contains the least number of bills possible of any available US currency (i.e. no two tens in place of a twenty).
At the auction he makes a successful bid of $8322 for a car. He hands the auctioneer envelopes number(s) 2, 8, and 14. After opening the envelopes the auctioneer finds exactly the right amount.
How many ones did the auctioneer find in the envelopes?
12) The minute and the hour hand of a watch meet every 65 minutes.
How much does the watch lose or gain time and by how much?
13) There is a number that is 5 times the sum of its digits. What is this number? Answer is not 0.
14) There are six boxes containing 5, 7, 14, 16, 18, 29 balls of either red or blue in colour. Some boxes contain only red balls and others contain only blue.
One sales man sold one box out of them and then he says, "I have the same number of red balls left out as that of blue."
Which box is the one he solds out?
15) Ekta got chocolates to give her friends on her Birthday. If she gives 3 chocolates to each friend, one friend will get only 2 chocolates. Also, if she gives 2 chocolates to each friends, she will left with 15 chocolates.
How many chocolates Ekta got on her Birthday? and how many friends are there?
16) Pooja and Esha met each other after long time. In the course of their conversation, Pooja asked Esha her age. Esha replied, "If you reverse my age, you will get my husbund's age. He is of course older than me. Also, the difference between our age is 1/11th of the sum of our age."
Can you help out Pooja in finding Esha's age?
17) A fish had a tail as long as its head plus a quarter the lenght of its body. Its body was three-quarters of its total length. Its head was 4 inches long.
What was the length of the fish?
18) Assume that you have just heard of a scandal and you are the first one to know. You pass it on to four person in a matter of 30 minutes. Each of these four in turn passes it to four other persons in the next 30 minutes and so on.
How long it will take for everybody in the World to get to know the scandal?
Assume that nobody hears it more than once and the population of the World is approximately 5.6 billions.
<<< Previous
Since the man does not want to be seen carrying that much money, he places it in 15 evelopes numbered 1 through 15. Each envelope contains the least number of bills possible of any available US currency (i.e. no two tens in place of a twenty).
At the auction he makes a successful bid of $8322 for a car. He hands the auctioneer envelopes number(s) 2, 8, and 14. After opening the envelopes the auctioneer finds exactly the right amount.
How many ones did the auctioneer find in the envelopes?
12) The minute and the hour hand of a watch meet every 65 minutes.
How much does the watch lose or gain time and by how much?
13) There is a number that is 5 times the sum of its digits. What is this number? Answer is not 0.
14) There are six boxes containing 5, 7, 14, 16, 18, 29 balls of either red or blue in colour. Some boxes contain only red balls and others contain only blue.
One sales man sold one box out of them and then he says, "I have the same number of red balls left out as that of blue."
Which box is the one he solds out?
15) Ekta got chocolates to give her friends on her Birthday. If she gives 3 chocolates to each friend, one friend will get only 2 chocolates. Also, if she gives 2 chocolates to each friends, she will left with 15 chocolates.
How many chocolates Ekta got on her Birthday? and how many friends are there?
16) Pooja and Esha met each other after long time. In the course of their conversation, Pooja asked Esha her age. Esha replied, "If you reverse my age, you will get my husbund's age. He is of course older than me. Also, the difference between our age is 1/11th of the sum of our age."
Can you help out Pooja in finding Esha's age?
17) A fish had a tail as long as its head plus a quarter the lenght of its body. Its body was three-quarters of its total length. Its head was 4 inches long.
What was the length of the fish?
18) Assume that you have just heard of a scandal and you are the first one to know. You pass it on to four person in a matter of 30 minutes. Each of these four in turn passes it to four other persons in the next 30 minutes and so on.
How long it will take for everybody in the World to get to know the scandal?
Assume that nobody hears it more than once and the population of the World is approximately 5.6 billions.
<<< Previous
Practise Puzzles - 4
1) Decipher this sentence.
B R W Q H L F K W H J K Q I B W K
Q I C E D W Z B G W K K M I K E
Z B G Q H S K Z B G J K Z K W
B U U Z B G J D B H F W.
2) At what time immediately prior to Six O'clock the hands of the clock are exactly opposite to each other. Give the exact time in hours, minutes and seconds
3) Ali Baba had four sons, to whom he bequeathed his 39 camels, with the proviso that the legacy be divided in the following way :
The oldest son was to receive one half the property, the next a quarter, the third an eighth and the youngest one tenth. The four brothers were at a loss as how to divide the inheritance among themselves without cutting up a camel, until a stranger appeared upon the scene.
Dismounting from his camel, he asked if he might help, for he knew just what to do. The brothers gratefully accepted his offer.
Adding his own camel to Ali Baba's 39, he divided the 40 as per the will. The oldest son received 20, the next 10, the third 5 and the youngest 4. One camel remained : this was his, which he mounted and rode away.
Scratching their heads in amazement, they started calculating. The oldest thought : is not 20 greater than the half of 39? Someone must have received less than his proper share ! But each brother discovered that he had received more than his due. How is it possible?
4) There is a family party consisting of two fathers, two mothers, two sons, one father-in-law, one mother-in-law, one daughter-in-law, one grandfather, one grandmother and one grandson.
What is the minimum number of persons required so that this is possible?
5) A man went into a fast food restaurant and ate a meal costing Rs. 105, giving the accountant a Rs. 500 note. He kept the change, came back a few minutes later and had some food packed for his girl friend. He gave the accountant a Rs. 100 note and received Rs. 20 in change. Later the bank told the accountant that both the Rs. 500 and the Rs. 100 notes were counterfeit.
How much money did the restaurant lose? Ignore the profit of the food restaurant.
6)
S L I D E
- D E A N
3 6 5 1
Each of seven digits from 0-9 are represented by a different letter above such that the subtraction is true.
What word represents 3651?
7) Adam, Burzin, Clark and Edmund each live in an apartment. Their apartments are arranged in a row numbered 1 to 4 from left to right. Also, one of them is the landlord.
If Clark's apartment is not next to Burzin's apartment, then the landlord is Adam and lives in apartment 1.
If Adam's apartment is right of Clark's apartment, then the landlord is Edmund and lives in apartment 4.
If Burzin's apartment is not next to Edmund's apartment, then the landlord is Clark and lives in apartment 3.
If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin and lives in apartment 2.
Who is the landlord?
8) B, J and P are related to each other.
Among the three are B's legal spouse, J's sibling and P's sister-in-law.
B's legal spouse and J's sibling are of the same sex.
Who is the married man?
9) A cube is made of a white material, but the exterior is painted black.
If the cube is cut into 125 smaller cubes of exactly the same size, how many of the cubes will have atleast 2 of their sides painted black?
10) Imagine a triangle of coins on a table so that the first row has one coin in it and the second row has two coins in it and so on. If you can only move one coin at a time, how many moves does it take to make the triangle point the other way?
For a triangle with two row it is one, for a triangle with three rows it is two, for a triangle with four rows it is three.
For a traingle with five rows is it four?
>>> NEXT
B R W Q H L F K W H J K Q I B W K
Q I C E D W Z B G W K K M I K E
Z B G Q H S K Z B G J K Z K W
B U U Z B G J D B H F W.
2) At what time immediately prior to Six O'clock the hands of the clock are exactly opposite to each other. Give the exact time in hours, minutes and seconds
3) Ali Baba had four sons, to whom he bequeathed his 39 camels, with the proviso that the legacy be divided in the following way :
The oldest son was to receive one half the property, the next a quarter, the third an eighth and the youngest one tenth. The four brothers were at a loss as how to divide the inheritance among themselves without cutting up a camel, until a stranger appeared upon the scene.
Dismounting from his camel, he asked if he might help, for he knew just what to do. The brothers gratefully accepted his offer.
Adding his own camel to Ali Baba's 39, he divided the 40 as per the will. The oldest son received 20, the next 10, the third 5 and the youngest 4. One camel remained : this was his, which he mounted and rode away.
Scratching their heads in amazement, they started calculating. The oldest thought : is not 20 greater than the half of 39? Someone must have received less than his proper share ! But each brother discovered that he had received more than his due. How is it possible?
4) There is a family party consisting of two fathers, two mothers, two sons, one father-in-law, one mother-in-law, one daughter-in-law, one grandfather, one grandmother and one grandson.
What is the minimum number of persons required so that this is possible?
5) A man went into a fast food restaurant and ate a meal costing Rs. 105, giving the accountant a Rs. 500 note. He kept the change, came back a few minutes later and had some food packed for his girl friend. He gave the accountant a Rs. 100 note and received Rs. 20 in change. Later the bank told the accountant that both the Rs. 500 and the Rs. 100 notes were counterfeit.
How much money did the restaurant lose? Ignore the profit of the food restaurant.
6)
S L I D E
- D E A N
3 6 5 1
Each of seven digits from 0-9 are represented by a different letter above such that the subtraction is true.
What word represents 3651?
7) Adam, Burzin, Clark and Edmund each live in an apartment. Their apartments are arranged in a row numbered 1 to 4 from left to right. Also, one of them is the landlord.
If Clark's apartment is not next to Burzin's apartment, then the landlord is Adam and lives in apartment 1.
If Adam's apartment is right of Clark's apartment, then the landlord is Edmund and lives in apartment 4.
If Burzin's apartment is not next to Edmund's apartment, then the landlord is Clark and lives in apartment 3.
If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin and lives in apartment 2.
Who is the landlord?
8) B, J and P are related to each other.
Among the three are B's legal spouse, J's sibling and P's sister-in-law.
B's legal spouse and J's sibling are of the same sex.
Who is the married man?
9) A cube is made of a white material, but the exterior is painted black.
If the cube is cut into 125 smaller cubes of exactly the same size, how many of the cubes will have atleast 2 of their sides painted black?
10) Imagine a triangle of coins on a table so that the first row has one coin in it and the second row has two coins in it and so on. If you can only move one coin at a time, how many moves does it take to make the triangle point the other way?
For a triangle with two row it is one, for a triangle with three rows it is two, for a triangle with four rows it is three.
For a traingle with five rows is it four?
>>> NEXT
Answers for Practise Puzzles - 3
1)
Assume that contents of the container is X
On the first day 1/3rd is evaporated.
(1 - 1/3) of X is remaining i.e. (2/3)X
On the Second day 3/4th is evaporated. Hence,
(1- 3/4) of (2/3)X is remaining
i.e. (1/4)(2/3)X = (1/6) X
Hence 1/6th of the contents of the container is remaining
2)
Vipul studied for 3 hours in candle light.
Assume that the initial lenght of both the candle was L and Vipul studied for X hours.
In X hours, total thick candle burnt = XL/6
In X hours, total thin candle burnt = XL/4
After X hours, total thick candle remaining = L - XL/6
After X hours, total thin candle remaining = L - XL/4
Also, it is given that the thick candle was twice as long as the thin one when he finally went to sleep.
(L - XL/6) = 2(L - XL/4)
(6 - X)/6 = (4 - X)/2
(6 - X) = 3*(4 - X)
6 - X = 12 - 3X
2X = 6
X = 3
Hence, Vipul studied for 3 hours i.e. 180 minutes in candle light.
3)
Rs.333,062,500
To begin with, you want to know the total number of days: 365 x 50 = 18250.
By experimentation, the following formula can be discovered, & used to determine the amount earned for any particular day: 1 + 2(x-1), with x being the number of the day. Take half of the 18250 days, & pair them up with the other half in the following way: day 1 with day 18250, day 2 with day 18249, & so on, & you will see that if you add these pairs together, they always equal Rs.36500.
Multiply this number by the total number of pairs (9125), & you have the amount you would have earned in 50 years.
4)
Rs.22176
Assume his salary was Rs. X
He earns 5% raise. So his salary is (105*X)/100
A year later he receives 2.5% cut. So his salary is ((105*X)/100)*(97.5/100) which is Rs. 22702.68
Hence, solving equation ((105*X)/100)*(97.5/100) = 22702.68
X = 22176
5)
66 seconds
It is given that the time between first and last ticks at 6'o is 30 seconds.
Total time gaps between first and last ticks at 6'o = 5
(i.e. between 1 & 2, 2 & 3, 3 & 4, 4 & 5 and 5 & 6)
So time gap between two ticks = 30/5 = 6 seconds.
Now, total time gaps between first and last ticks at 12'o = 11
Therefore time taken for 12 ticks = 11 * 6 = 66 seconds (and not 60 seconds)
6)
There are 7 members in Mr. Mehta's family. Mother & Father of Mr. Mehta, Mr. & Mrs. Mehta, his son and two daughters.
Mother & Father of Mr. Mehta
|
Mr. & Mrs. Mehta
|
One Son & Two Daughters
7)
The soldier said, "You will put me to the sword."
The soldier has to say a Paradox to save himself. If his statement is true, he will be hanged, which is not the sword and hence false. If his statement is false, he will be put to the sword, which will make it true. A Paradox !!!
8)
As given, the person wanted to withdraw 100X + Y paise.
But he got 100Y + X paise.
After spending 20 paise, he has double the amount he wanted to withdraw. Hence, the equation is
2 * (100X + Y) = 100Y + X - 20
200X + 2Y = 100Y +X - 20
199X - 98Y = -20
98Y - 199X = 20
Now, we got one equation; but there are 2 variables. We have to apply little bit of logic over here. We know that if we interchange X & Y, amount gets double. So Y should be twice of X or one more than twice of X i.e. Y = 2X or Y = 2X+1
Case I : Y=2X
Solving two equations simultaneously
98Y - 199X = 20
Y - 2X = 0
We get X = - 20/3 & Y = - 40/2
Case II : Y=2X+1
Solving two equations simultaneously
98Y - 199X = 20
Y - 2X = 1
We get X = 26 & Y = 53
Now, its obvious that he wanted to withdraw Rs. 26.53
9)
Statement (4) is false. There are 3 men, 8 women and 6 children.
Assume that Statements (4), (5) and (6) are all true. Then, Statement (1) is false. But then Statement (2) and (3) both can not be true. Thus, contradictory to the fact that exactly one statement is false.
So Statement (4) or Statement (5) or Statement (6) is false. Also, Statements (1), (2) and (3) all are true.
From (1) and (2), there are 11 men and women. Then from (3), there are 2 possible cases - either there are 8 men and 3 women or there are 3 men and 8 women.
If there are 8 men and 3 women, then there is 1 child. Then Statements (4) and (5) both are false, which is not possible.
Hence, there are 3 men, 8 women and 6 children. Statement (4) is false.
10)
42% houses do not have tubelight, bulb and fan.
Let's assume that there are 100 houses. Hence, there should be total 300 items i.e. 100 tubelights, 100 bulbs and 100 fans.
From the given data, we know that there is shortage of atleast (67+83+73) 223 items in every 100 houses.
Also, exactly 19 houses do not have just one item. It means that remaining 81 houses should account for the shortage of remaining (223-19) 204 items. If those remaining 81 houses do not have 2 items each, there would be a shortage of 162 items. But total of 204 items are short. Hence, atleast (204-162) 42 houses do not have all 3 items - tubelight, bulb and fan.
Thus, 42% houses do not have tubelight, bulb and fan.
11)
The individual student should pay Rs. 50 for their journey.
Note that 3 persons are travelling between Bandra and Colaba.
The entire trip costs Rs. 300 to Mr. Subramanian. Hence, half of the trip costs Rs. 150.
For Andheri-Bandra-Andheri, only one person i.e. Mr. Subramaniam is travelling. Hence, he would pay Rs. 150.
For Bandra-Colaba-Bandra, three persons i.e Mr. Subramaniam and two students, are travelling. Hence, each student would pay Rs. 50.
12)
You are able to swim downstream at 3 miles an hour, & upstream at 2 miles an hour. There is a difference of 1 mile an hour, which is the river helping you in 1 direction, & slowing you in the other direction.
Average the 2 rates, & you have the rate that you can swim in still water, which is 2.5 miles an hour.
You can thus swim a mile in still water in 24 minutes.
13)
There are 1024 different ways.
This is the type of Brain Teaser that can be solved using the method of induction.
If there is just a one person and one seat, that person has only one option.
If there are two persons and two seats, it can be accomplished in 2 different ways.
If there are three persons and three seats, it can be accomplished in 4 different ways. Remember that no person can take a seat that separates him/her from at least one other person.
Similarly, four persons and four seats produce 8 different ways. And five persons with five seats produce 16 different ways.
It can be seen that with each additional person and seat, the different ways increase by the power of two. For six persons with six seats, there are 32 different ways.
For any number N, the different possible ways are 2 ^ (N-1)
Thus, for 11 persons and 11 seats, total different ways are 2 ^ 10 i.e. 1024
14)
Assume there are 6X boys and 4X Girls
Total Students taking lunch in canteen
= (6X)(60/100) + (4X)(40/100)
= 36(X/10) + 16(X/10)
= 52(X/10)
Total students are = 6X + 4X = 10X
% of class taking lunch in canteen
= ((52X/10) * 100 ) / 10X
= 52 %
15)
The Club originally had 24 members.
Assume that there were initially N members.
As 4 members resigned and remaining members paid Rs 26 each, it means that total amount of 4 members is equal to Rs 26 each from remaining (N-4) members. Thus,
4 * (3120 / N) = 26 * (N - 4)
12480 = 26N2 - 104N
26N2 - 104N - 12480 = 0
Solving the quadratic equation we get N=24.
Hence, the Club originally had 24 members.
16)
The tank will be full in 16 minutes.
In one minute,
pipe A can fill 1/30 part of the tank.
pipe B can fill 1/24 part of the tank.
pipe C can empty 1/80 part of the tank.
Thus, the net water level in one minute is
= 1/30 + 1/24 - 1/80
= 15/240 part of the tank
Hence, the tank will be full in 240/15 i.e. 16 minutes.
17)
The old man temporarily added his camel to the 17, making a total of 18 camels.
First son got 1/2 of it = 9
Second son got 1/3 of it = 6
Third son got 1/9 of it = 2
For a total of 17. He then takes his camel back and rides away......
18)
Blue box contains the maximum amount Rs. 40000
As it is given that only one of the given 3 statements is true; assume in turn, each statement to be true & the other 2 false and check whether the corresponding box contains the maximum amount.
Let's assume that the statement on the Blue box is true. Thus, the given 3 statements can be interpreted as
* Atmost one, a red box or a blue box contains Rs. 10000.
* Atmost one, a green box or a red box contains Rs. 25000.
* Both, a blue box and a green box contain Rs. 15000 each.
Going through all possible combinations, we can conclude that
Red Box : Rs. 10000 + Rs. 25000 = Rs. 35000
Green Box : Rs. 10000 + Rs. 15000 = Rs. 25000
Blue Box : Rs. 15000 + Rs. 25000 = Rs. 40000
You can test out for other two statements i.e. assuming Red box statement true and then Green box statement true. In both the cases, other statements will contradict the true statement.
19)
achin, Dravid and Ganguly scored 75, 87 and 78 respectively.
Sachin's score must be less than 86, otherwise Dravid's score would be more than 99. Also, he must have scored atleast 42 - incase Dravid and Ganguly scored 99 each.
Also, as none of them scored more than 99 and the total runs scored by them is 240; their individual scores must be around 80.
20)
A tough one.
It is obvious that C=1. Also, the maximum possible value of E is 7. Now, start putting possible values of D, E and R as they occure frequently and use trial-n-error.
W O R L D 5 3 6 8 4
+ T R A D E + 7 6 0 4 2
--- ---
C E N T E R 1 2 9 7 2 6
21)
7.5 degrees
At 3:15 minute hand will be perfactly horizontal pointing towards 3. Whereas hour hand will be towards 4. Also, hour hand must have covered 1/4 of angle between 3 and 4.
The angle between two adjacent digits is 360/12 = 30 degrees.
Hence 1/4 of it is 7.5 degrees.
22)
1, 2, 4, 8, 16, 32, 64, 128, 256, 489
Let's start from scratch.
The apple vandor can ask for only 1 apple, so one box must contain 1 apple.
He can ask for 2 apples, so one box must contain 2 apples.
He can ask for 3 apples, in that case box one and box two will add up to 3.
He can ask for 4 apples, so one box i.e. third box must contain 4 apples.
Now using box number one, two and three containing 1, 2 and 4 apples respectively, his son can give upto 7 apples. Hence, forth box must contain 8 apples.
Similarly, using first four boxes containing 1, 2, 4 and 8 apples, his son can give upto 15 apples. Hence fifth box must contain 16 apples.
You must have noticed one thing till now that each box till now contains power of 2 apples. Hence the answer is 1, 2, 4, 8, 16, 32, 64, 128, 256, 489. This is true for any number of apples, here in our case only upto 1000.
23)
The sequence of letters from the lowest value to the highest value is TUSQRPV.
From (3), Q is the middle term.
___ ___ ___ _Q_ ___ ___ ___
From (4), there must be exactly 2 numbers between P and S which gives two possible positions.
[1] ___ _S_ ___ _Q_ _P_ ___ ___
[2] ___ ___ _S_ _Q_ ___ _P_ ___
From (1), the number of letters between U and Q must be same as the number of letters between S and R. Also, the number of letters between them can be 1, 2 or 3.
Using trial and error, it can be found that there must be 2 letters between them. Also, it is possible only in option [2] above.
[2] ___ _U_ _S_ _Q_ _R_ _P_ ___
From (2) V must be the highest and the remaining T must be the lowest number.
_T_ _U_ _S_ _Q_ _R_ _P_ _V_
Thus, the sequence of letters from the lowest value to the highest value is TUSQRPV.
A contractor had employed 100 labourers for a flyover construction task. He did not allow any woman to work without her husband. Also, atleast half the men working came with their wives.
he sequence of letters from the lowest value to the highest value?
24)
16 men, 12 women and 72 children were working with the constructor.
Let's assume that there were X men, Y women and Z children working with the constructor. Hence,
X + Y + Z = 100
5X + 4Y + Z = 200
Eliminating X and Y in turn from these equations, we get
X = 3Z - 200
Y = 300 - 4Z
As if woman works, her husband also works and atleast half the men working came with their wives; the value of Y lies between X and X/2. Substituting these limiting values in equations, we get
if Y = X,
300 - 4Z = 3Z - 200
7Z = 500
Z = 500/7 i.e. 71.428
if Y = X/2,
300 - 4Z = (3Z - 200)/2
600 - 8Z = 3Z - 200
11Z = 800
Z = 800/11 i.e. 72.727
But Z must be an integer, hence Z=72. Also, X=16 and Y=12
There were 16 men, 12 women and 72 children working with the constructor.
25)13 not 12
He makes 9 originals from the 27 butts he found, and after he smokes them he has 9 butts left for another 3 cigars. And then he has 3 butts for another cigar.
So 9+3+1=13
26)
Cruz is Tanya's date.
As no two of them have the same number of preferred traits - from (1), exactly one of them has none of the preferred traits and exactly one of them has all the preferred traits.
From (4) and (5), there are only two possibilities:
* Adam & Cruz both are tall and Bond & Dumbo both are fair.
* Adam & Cruz both are short and Bond & Dumbo both are dark.
But from (2), second possibility is impossible. So the first one is the correct possibility i.e. Adam & Cruz both are tall and Bond & Dumbo both are fair.
Then from (3), Bond is short and handsome.
Also, from (1) and (2), Adam is tall and fair. Also, Dumbo is the person without any preferred traits. Cruz is Dark. Adam and Cruz are handsome. Thus, following are the individual preferred traits:
Cruz - Tall, Dark and Handsome
Adam - Tall and Handsome
Bond - Handsome
Dumbo - None :-(
Hence, Cruz is Tanya's date.
Consider a game of Tower of Hanoi
27)
There are number of ways to find the answer.
To move the largest disc (at level N) from one tower to the other, it requires 2(N-1) moves. Thus, to move N discs from one tower to the other, the number of moves required is
= 2^(N-1) + 2^(N-2) + 2^(N-3) + ..... + 2^2 + 2^1 + 2^0
= 2^N - 1
For N discs, the number of moves is one more than two times the number of moves for N-1 discs. Thus, the recursive function is
F(1) = 1
F(N) = 2*[F(N-1)] + 1
where N is the total number of discs
Also, one can arrive at the answer by finding the number of moves for smaller number of discs and then derive the pattern.
For 1 disc, number of moves = 1
For 2 discs, number of moves = 3
For 3 discs, number of moves = 7
For 4 discs, number of moves = 15
For 5 discs, number of moves = 31
Thus, the pattern is 2^N – 1
A boy found that he had a 48 inch strip of paper. He could cut an inch off every second.
28)
47 seconds.
To get 48 pieces, the boy have to put only 47 cuts. i.e. he can cut 46 pieces in 46 seconds. After getting 46 pieces, he will have a 2 inches long piece. He can cut it into two with just a one cut in 1 second. Hence, total of 47 seconds.
29)
A simple one. Use the given facts and put down all the players in order. The order is as follow with Harbhajan, the highest scorer and Laxman, the lowest scorer.
Harbhajan
Ganguly
Dravid
Badani, Agarkar
Sachin
Laxman
Also, as the lowest score was 10 runs. Laxman must have scored 10, Sachin 20, Badani & Agarkar 30 and so on.
Harbhajan - 60 runs
Ganguly - 50 runs
Dravid - 40 runs
Badani, Agarkar - 30 runs each
Sachin - 20 runs
Laxman - 10 runs
*******END OF ANSWER SESSION FOR PRACTISE PUZZLES - 3 *******
Thanks,
All the Best
Assume that contents of the container is X
On the first day 1/3rd is evaporated.
(1 - 1/3) of X is remaining i.e. (2/3)X
On the Second day 3/4th is evaporated. Hence,
(1- 3/4) of (2/3)X is remaining
i.e. (1/4)(2/3)X = (1/6) X
Hence 1/6th of the contents of the container is remaining
2)
Vipul studied for 3 hours in candle light.
Assume that the initial lenght of both the candle was L and Vipul studied for X hours.
In X hours, total thick candle burnt = XL/6
In X hours, total thin candle burnt = XL/4
After X hours, total thick candle remaining = L - XL/6
After X hours, total thin candle remaining = L - XL/4
Also, it is given that the thick candle was twice as long as the thin one when he finally went to sleep.
(L - XL/6) = 2(L - XL/4)
(6 - X)/6 = (4 - X)/2
(6 - X) = 3*(4 - X)
6 - X = 12 - 3X
2X = 6
X = 3
Hence, Vipul studied for 3 hours i.e. 180 minutes in candle light.
3)
Rs.333,062,500
To begin with, you want to know the total number of days: 365 x 50 = 18250.
By experimentation, the following formula can be discovered, & used to determine the amount earned for any particular day: 1 + 2(x-1), with x being the number of the day. Take half of the 18250 days, & pair them up with the other half in the following way: day 1 with day 18250, day 2 with day 18249, & so on, & you will see that if you add these pairs together, they always equal Rs.36500.
Multiply this number by the total number of pairs (9125), & you have the amount you would have earned in 50 years.
4)
Rs.22176
Assume his salary was Rs. X
He earns 5% raise. So his salary is (105*X)/100
A year later he receives 2.5% cut. So his salary is ((105*X)/100)*(97.5/100) which is Rs. 22702.68
Hence, solving equation ((105*X)/100)*(97.5/100) = 22702.68
X = 22176
5)
66 seconds
It is given that the time between first and last ticks at 6'o is 30 seconds.
Total time gaps between first and last ticks at 6'o = 5
(i.e. between 1 & 2, 2 & 3, 3 & 4, 4 & 5 and 5 & 6)
So time gap between two ticks = 30/5 = 6 seconds.
Now, total time gaps between first and last ticks at 12'o = 11
Therefore time taken for 12 ticks = 11 * 6 = 66 seconds (and not 60 seconds)
6)
There are 7 members in Mr. Mehta's family. Mother & Father of Mr. Mehta, Mr. & Mrs. Mehta, his son and two daughters.
Mother & Father of Mr. Mehta
|
Mr. & Mrs. Mehta
|
One Son & Two Daughters
7)
The soldier said, "You will put me to the sword."
The soldier has to say a Paradox to save himself. If his statement is true, he will be hanged, which is not the sword and hence false. If his statement is false, he will be put to the sword, which will make it true. A Paradox !!!
8)
As given, the person wanted to withdraw 100X + Y paise.
But he got 100Y + X paise.
After spending 20 paise, he has double the amount he wanted to withdraw. Hence, the equation is
2 * (100X + Y) = 100Y + X - 20
200X + 2Y = 100Y +X - 20
199X - 98Y = -20
98Y - 199X = 20
Now, we got one equation; but there are 2 variables. We have to apply little bit of logic over here. We know that if we interchange X & Y, amount gets double. So Y should be twice of X or one more than twice of X i.e. Y = 2X or Y = 2X+1
Case I : Y=2X
Solving two equations simultaneously
98Y - 199X = 20
Y - 2X = 0
We get X = - 20/3 & Y = - 40/2
Case II : Y=2X+1
Solving two equations simultaneously
98Y - 199X = 20
Y - 2X = 1
We get X = 26 & Y = 53
Now, its obvious that he wanted to withdraw Rs. 26.53
9)
Statement (4) is false. There are 3 men, 8 women and 6 children.
Assume that Statements (4), (5) and (6) are all true. Then, Statement (1) is false. But then Statement (2) and (3) both can not be true. Thus, contradictory to the fact that exactly one statement is false.
So Statement (4) or Statement (5) or Statement (6) is false. Also, Statements (1), (2) and (3) all are true.
From (1) and (2), there are 11 men and women. Then from (3), there are 2 possible cases - either there are 8 men and 3 women or there are 3 men and 8 women.
If there are 8 men and 3 women, then there is 1 child. Then Statements (4) and (5) both are false, which is not possible.
Hence, there are 3 men, 8 women and 6 children. Statement (4) is false.
10)
42% houses do not have tubelight, bulb and fan.
Let's assume that there are 100 houses. Hence, there should be total 300 items i.e. 100 tubelights, 100 bulbs and 100 fans.
From the given data, we know that there is shortage of atleast (67+83+73) 223 items in every 100 houses.
Also, exactly 19 houses do not have just one item. It means that remaining 81 houses should account for the shortage of remaining (223-19) 204 items. If those remaining 81 houses do not have 2 items each, there would be a shortage of 162 items. But total of 204 items are short. Hence, atleast (204-162) 42 houses do not have all 3 items - tubelight, bulb and fan.
Thus, 42% houses do not have tubelight, bulb and fan.
11)
The individual student should pay Rs. 50 for their journey.
Note that 3 persons are travelling between Bandra and Colaba.
The entire trip costs Rs. 300 to Mr. Subramanian. Hence, half of the trip costs Rs. 150.
For Andheri-Bandra-Andheri, only one person i.e. Mr. Subramaniam is travelling. Hence, he would pay Rs. 150.
For Bandra-Colaba-Bandra, three persons i.e Mr. Subramaniam and two students, are travelling. Hence, each student would pay Rs. 50.
12)
You are able to swim downstream at 3 miles an hour, & upstream at 2 miles an hour. There is a difference of 1 mile an hour, which is the river helping you in 1 direction, & slowing you in the other direction.
Average the 2 rates, & you have the rate that you can swim in still water, which is 2.5 miles an hour.
You can thus swim a mile in still water in 24 minutes.
13)
There are 1024 different ways.
This is the type of Brain Teaser that can be solved using the method of induction.
If there is just a one person and one seat, that person has only one option.
If there are two persons and two seats, it can be accomplished in 2 different ways.
If there are three persons and three seats, it can be accomplished in 4 different ways. Remember that no person can take a seat that separates him/her from at least one other person.
Similarly, four persons and four seats produce 8 different ways. And five persons with five seats produce 16 different ways.
It can be seen that with each additional person and seat, the different ways increase by the power of two. For six persons with six seats, there are 32 different ways.
For any number N, the different possible ways are 2 ^ (N-1)
Thus, for 11 persons and 11 seats, total different ways are 2 ^ 10 i.e. 1024
14)
Assume there are 6X boys and 4X Girls
Total Students taking lunch in canteen
= (6X)(60/100) + (4X)(40/100)
= 36(X/10) + 16(X/10)
= 52(X/10)
Total students are = 6X + 4X = 10X
% of class taking lunch in canteen
= ((52X/10) * 100 ) / 10X
= 52 %
15)
The Club originally had 24 members.
Assume that there were initially N members.
As 4 members resigned and remaining members paid Rs 26 each, it means that total amount of 4 members is equal to Rs 26 each from remaining (N-4) members. Thus,
4 * (3120 / N) = 26 * (N - 4)
12480 = 26N2 - 104N
26N2 - 104N - 12480 = 0
Solving the quadratic equation we get N=24.
Hence, the Club originally had 24 members.
16)
The tank will be full in 16 minutes.
In one minute,
pipe A can fill 1/30 part of the tank.
pipe B can fill 1/24 part of the tank.
pipe C can empty 1/80 part of the tank.
Thus, the net water level in one minute is
= 1/30 + 1/24 - 1/80
= 15/240 part of the tank
Hence, the tank will be full in 240/15 i.e. 16 minutes.
17)
The old man temporarily added his camel to the 17, making a total of 18 camels.
First son got 1/2 of it = 9
Second son got 1/3 of it = 6
Third son got 1/9 of it = 2
For a total of 17. He then takes his camel back and rides away......
18)
Blue box contains the maximum amount Rs. 40000
As it is given that only one of the given 3 statements is true; assume in turn, each statement to be true & the other 2 false and check whether the corresponding box contains the maximum amount.
Let's assume that the statement on the Blue box is true. Thus, the given 3 statements can be interpreted as
* Atmost one, a red box or a blue box contains Rs. 10000.
* Atmost one, a green box or a red box contains Rs. 25000.
* Both, a blue box and a green box contain Rs. 15000 each.
Going through all possible combinations, we can conclude that
Red Box : Rs. 10000 + Rs. 25000 = Rs. 35000
Green Box : Rs. 10000 + Rs. 15000 = Rs. 25000
Blue Box : Rs. 15000 + Rs. 25000 = Rs. 40000
You can test out for other two statements i.e. assuming Red box statement true and then Green box statement true. In both the cases, other statements will contradict the true statement.
19)
achin, Dravid and Ganguly scored 75, 87 and 78 respectively.
Sachin's score must be less than 86, otherwise Dravid's score would be more than 99. Also, he must have scored atleast 42 - incase Dravid and Ganguly scored 99 each.
Also, as none of them scored more than 99 and the total runs scored by them is 240; their individual scores must be around 80.
20)
A tough one.
It is obvious that C=1. Also, the maximum possible value of E is 7. Now, start putting possible values of D, E and R as they occure frequently and use trial-n-error.
W O R L D 5 3 6 8 4
+ T R A D E + 7 6 0 4 2
--- ---
C E N T E R 1 2 9 7 2 6
21)
7.5 degrees
At 3:15 minute hand will be perfactly horizontal pointing towards 3. Whereas hour hand will be towards 4. Also, hour hand must have covered 1/4 of angle between 3 and 4.
The angle between two adjacent digits is 360/12 = 30 degrees.
Hence 1/4 of it is 7.5 degrees.
22)
1, 2, 4, 8, 16, 32, 64, 128, 256, 489
Let's start from scratch.
The apple vandor can ask for only 1 apple, so one box must contain 1 apple.
He can ask for 2 apples, so one box must contain 2 apples.
He can ask for 3 apples, in that case box one and box two will add up to 3.
He can ask for 4 apples, so one box i.e. third box must contain 4 apples.
Now using box number one, two and three containing 1, 2 and 4 apples respectively, his son can give upto 7 apples. Hence, forth box must contain 8 apples.
Similarly, using first four boxes containing 1, 2, 4 and 8 apples, his son can give upto 15 apples. Hence fifth box must contain 16 apples.
You must have noticed one thing till now that each box till now contains power of 2 apples. Hence the answer is 1, 2, 4, 8, 16, 32, 64, 128, 256, 489. This is true for any number of apples, here in our case only upto 1000.
23)
The sequence of letters from the lowest value to the highest value is TUSQRPV.
From (3), Q is the middle term.
___ ___ ___ _Q_ ___ ___ ___
From (4), there must be exactly 2 numbers between P and S which gives two possible positions.
[1] ___ _S_ ___ _Q_ _P_ ___ ___
[2] ___ ___ _S_ _Q_ ___ _P_ ___
From (1), the number of letters between U and Q must be same as the number of letters between S and R. Also, the number of letters between them can be 1, 2 or 3.
Using trial and error, it can be found that there must be 2 letters between them. Also, it is possible only in option [2] above.
[2] ___ _U_ _S_ _Q_ _R_ _P_ ___
From (2) V must be the highest and the remaining T must be the lowest number.
_T_ _U_ _S_ _Q_ _R_ _P_ _V_
Thus, the sequence of letters from the lowest value to the highest value is TUSQRPV.
A contractor had employed 100 labourers for a flyover construction task. He did not allow any woman to work without her husband. Also, atleast half the men working came with their wives.
he sequence of letters from the lowest value to the highest value?
24)
16 men, 12 women and 72 children were working with the constructor.
Let's assume that there were X men, Y women and Z children working with the constructor. Hence,
X + Y + Z = 100
5X + 4Y + Z = 200
Eliminating X and Y in turn from these equations, we get
X = 3Z - 200
Y = 300 - 4Z
As if woman works, her husband also works and atleast half the men working came with their wives; the value of Y lies between X and X/2. Substituting these limiting values in equations, we get
if Y = X,
300 - 4Z = 3Z - 200
7Z = 500
Z = 500/7 i.e. 71.428
if Y = X/2,
300 - 4Z = (3Z - 200)/2
600 - 8Z = 3Z - 200
11Z = 800
Z = 800/11 i.e. 72.727
But Z must be an integer, hence Z=72. Also, X=16 and Y=12
There were 16 men, 12 women and 72 children working with the constructor.
25)13 not 12
He makes 9 originals from the 27 butts he found, and after he smokes them he has 9 butts left for another 3 cigars. And then he has 3 butts for another cigar.
So 9+3+1=13
26)
Cruz is Tanya's date.
As no two of them have the same number of preferred traits - from (1), exactly one of them has none of the preferred traits and exactly one of them has all the preferred traits.
From (4) and (5), there are only two possibilities:
* Adam & Cruz both are tall and Bond & Dumbo both are fair.
* Adam & Cruz both are short and Bond & Dumbo both are dark.
But from (2), second possibility is impossible. So the first one is the correct possibility i.e. Adam & Cruz both are tall and Bond & Dumbo both are fair.
Then from (3), Bond is short and handsome.
Also, from (1) and (2), Adam is tall and fair. Also, Dumbo is the person without any preferred traits. Cruz is Dark. Adam and Cruz are handsome. Thus, following are the individual preferred traits:
Cruz - Tall, Dark and Handsome
Adam - Tall and Handsome
Bond - Handsome
Dumbo - None :-(
Hence, Cruz is Tanya's date.
Consider a game of Tower of Hanoi
27)
There are number of ways to find the answer.
To move the largest disc (at level N) from one tower to the other, it requires 2(N-1) moves. Thus, to move N discs from one tower to the other, the number of moves required is
= 2^(N-1) + 2^(N-2) + 2^(N-3) + ..... + 2^2 + 2^1 + 2^0
= 2^N - 1
For N discs, the number of moves is one more than two times the number of moves for N-1 discs. Thus, the recursive function is
F(1) = 1
F(N) = 2*[F(N-1)] + 1
where N is the total number of discs
Also, one can arrive at the answer by finding the number of moves for smaller number of discs and then derive the pattern.
For 1 disc, number of moves = 1
For 2 discs, number of moves = 3
For 3 discs, number of moves = 7
For 4 discs, number of moves = 15
For 5 discs, number of moves = 31
Thus, the pattern is 2^N – 1
A boy found that he had a 48 inch strip of paper. He could cut an inch off every second.
28)
47 seconds.
To get 48 pieces, the boy have to put only 47 cuts. i.e. he can cut 46 pieces in 46 seconds. After getting 46 pieces, he will have a 2 inches long piece. He can cut it into two with just a one cut in 1 second. Hence, total of 47 seconds.
29)
A simple one. Use the given facts and put down all the players in order. The order is as follow with Harbhajan, the highest scorer and Laxman, the lowest scorer.
Harbhajan
Ganguly
Dravid
Badani, Agarkar
Sachin
Laxman
Also, as the lowest score was 10 runs. Laxman must have scored 10, Sachin 20, Badani & Agarkar 30 and so on.
Harbhajan - 60 runs
Ganguly - 50 runs
Dravid - 40 runs
Badani, Agarkar - 30 runs each
Sachin - 20 runs
Laxman - 10 runs
*******END OF ANSWER SESSION FOR PRACTISE PUZZLES - 3 *******
Thanks,
All the Best
Practise Puzzles - 3
21) If you look at a clock and the time is 3:15.
What is the angle between the hour and the minute hands? ( The answer to this is not zero!)
22) An apple vendor has 1000 apples and 10 empty boxes. He asks his son to place all the 1000 apples in all the 10 boxes in such a manner that if he asks for any number of apples from 1 to 1000, his son should be able to pick them in terms of boxes.
How did the son place all the apples among the 10 boxes, given that any number of apples can be put in one box.
23) The letters P, Q, R, S, T, U and V, not necessarily in that order represents seven consecutive integers from 22 to 33.
U is as much less than Q as R is greater than S.
V is greater than U.
Q is the middle term.
P is 3 greater than S.
Can you find the sequence of letters from the lowest value to the highest value?
24) He paid five rupees per day to each man, four ruppes to each woman and one rupee to each child. He gave out 200 rupees every evening.
How many men, women and children were working with the constructor?
25) Because cigars cannot be entirely smoked, a Bobo who collects cigar butts can make a cigar to smoke out of every 3 butts that he finds.
Today, he has collected 27 cigar butts. How many cigars will he be able to smoke?
26) anya wants to go on a date and prefers her date to be tall, dark and handsome.
Of the preferred traits - tall, dark and handsome - no two of Adam, Bond, Cruz and Dumbo have the same number.
Only Adam or Dumbo is tall and fair.
Only Bond or Cruz is short and handsome.
Adam and Cruz are either both tall or both short.
Bond and Dumbo are either both dark or both fair.
Who is Tanya's date?
27)If the tower has 2 discs, the least possible moves with which you can move the entire tower to another peg is 3.
If the tower has 3 discs, the least possible moves with which you can move the entire tower to another peg is 7.
What is the least possible moves with which you can move the entire tower to another peg if the tower has N discs?
28) How long would it take for him to cut 48 pieces? He can not fold the strip and also, can not stack two or more strips and cut them together.
29) The cricket match between India and Pakistan was over.
Harbhajan scored more runs than Ganguly.
Sachin scored more runs than Laxman but less than Dravid
Badani scored as much runs as Agarkar but less than Dravid and more than Sachin.
Ganguly scored more runs than either Agarkar or Dravid.
Each batsman scored 10 runs more than his immediate batsman. The lowest score was 10 runs. How much did each one of them score
Previous
What is the angle between the hour and the minute hands? ( The answer to this is not zero!)
22) An apple vendor has 1000 apples and 10 empty boxes. He asks his son to place all the 1000 apples in all the 10 boxes in such a manner that if he asks for any number of apples from 1 to 1000, his son should be able to pick them in terms of boxes.
How did the son place all the apples among the 10 boxes, given that any number of apples can be put in one box.
23) The letters P, Q, R, S, T, U and V, not necessarily in that order represents seven consecutive integers from 22 to 33.
U is as much less than Q as R is greater than S.
V is greater than U.
Q is the middle term.
P is 3 greater than S.
Can you find the sequence of letters from the lowest value to the highest value?
24) He paid five rupees per day to each man, four ruppes to each woman and one rupee to each child. He gave out 200 rupees every evening.
How many men, women and children were working with the constructor?
25) Because cigars cannot be entirely smoked, a Bobo who collects cigar butts can make a cigar to smoke out of every 3 butts that he finds.
Today, he has collected 27 cigar butts. How many cigars will he be able to smoke?
26) anya wants to go on a date and prefers her date to be tall, dark and handsome.
Of the preferred traits - tall, dark and handsome - no two of Adam, Bond, Cruz and Dumbo have the same number.
Only Adam or Dumbo is tall and fair.
Only Bond or Cruz is short and handsome.
Adam and Cruz are either both tall or both short.
Bond and Dumbo are either both dark or both fair.
Who is Tanya's date?
27)If the tower has 2 discs, the least possible moves with which you can move the entire tower to another peg is 3.
If the tower has 3 discs, the least possible moves with which you can move the entire tower to another peg is 7.
What is the least possible moves with which you can move the entire tower to another peg if the tower has N discs?
28) How long would it take for him to cut 48 pieces? He can not fold the strip and also, can not stack two or more strips and cut them together.
29) The cricket match between India and Pakistan was over.
Harbhajan scored more runs than Ganguly.
Sachin scored more runs than Laxman but less than Dravid
Badani scored as much runs as Agarkar but less than Dravid and more than Sachin.
Ganguly scored more runs than either Agarkar or Dravid.
Each batsman scored 10 runs more than his immediate batsman. The lowest score was 10 runs. How much did each one of them score
Previous
Practise Puzzles - 3
11) Mr. Subramaniam rents a private car for Andheri-Colaba-Andheri trip. It costs him Rs. 300 everyday.
One day the car driver informed Mr. Subramaniam that there were two students from Bandra who wished to go from Bandra to Colaba and back to Bandra. Bandra is halfway between Andheri and Colaba. Mr. Subramaniam asked the driver to let the students travel with him.
On the first day when they came, Mr. Subramaniam said, "If you tell me the mathematically correct price you should pay individually for your portion of the trip, I will let you travel for free."
How much should the individual student pay for their journey?
12) In training for a competition, you find that swimming downstream (with the current) in a river, you can swim 2 miles in 40 minutes, & upstream (against the current), you can swim 2 miles in 60 minutes.
How long would it take you to swim a mile in still water?
13) How many different ways can this be accomplished? Note that the first person can choose any of the 11 seats.
(question is not completly avilable..will be updated asap)
14) The ratio of Boys to Girls is 6:4. 60% of the boys and 40% of the girls take lunch in the canteen. What % of class takes lunch in canteen?
15) In the General meeting of "Friends Club", Sameer said, "The repairs to the Club will come to a total of Rs 3120 and I propose that this amount should be met by the members, each paying an equal amount."
The proposal was immediately agreed. However, four members of the Club chose to resign, leaving the remaining members to pay an extra Rs 26 each.
How many members did the Club originally have?
16) A tank can be filled by pipe A in 30 minutes and by pipe B in 24 minutes. Outlet pipe C can empty the full tank in one hour and twenty minutes.
If the tank is empty initially and if all the three pipes A, B and C are opened simultaneously, in how much time will the tank be full?
17) A rich old Arab has three sons. When he died, he willed his 17 camels to the sons, to be divided as follows:
First Son to get 1/2 of the camels Second Son to get 1/3rd of the camels Third Son to get 1/9th of the camels.
The sons are sitting there trying to figure out how this can possibly be done, when a very old wise man goes riding by. They stop him and ask him to help them solve their problem. Without hesitation he divides the camels properly and continues riding on his way.
How did he do it?
18) There are 3 colored boxes - Red, Green and Blue. Each box contains 2 envelopes. Each envelope contains money - two of them contain Rs. 25000 each, two of them contain Rs. 15000 each and remaining two contain Rs. 10000 each.
There is one statement written on the cover of each box.
* Red Box: Both, a red box and a blue box contain Rs. 10000 each.
* Green Box: Both, a green box and a red box contain Rs. 25000 each.
* Blue Box: Both, a blue box and a green box contain Rs. 15000 each.
Only one of the above 3 statements is true and the corresponding box contains the maximum amount.
Can you tell which box contains the maximum amount and how much?
19) Sachin, Dravid and Ganguly played in a Cricket match between India and England.
None of them scored more than 99 runs.
If you add the digits of the runs scored by Sachin to his own score, you will get the runs scored by Dravid.
If you reverse the digits of the runs scored by Dravid, you will get the runs scored by Ganguly.
The total runs scored by them is 240.
Can you figure out their individual scores?
20) Substitute digits for the letters to make the following relation true.
W O R L D
+ T R A D E
----
C E N T E R
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter W, no other letter can be 3 and all other W in the puzzle must be 3.
<<< Previous >>> NExt
One day the car driver informed Mr. Subramaniam that there were two students from Bandra who wished to go from Bandra to Colaba and back to Bandra. Bandra is halfway between Andheri and Colaba. Mr. Subramaniam asked the driver to let the students travel with him.
On the first day when they came, Mr. Subramaniam said, "If you tell me the mathematically correct price you should pay individually for your portion of the trip, I will let you travel for free."
How much should the individual student pay for their journey?
12) In training for a competition, you find that swimming downstream (with the current) in a river, you can swim 2 miles in 40 minutes, & upstream (against the current), you can swim 2 miles in 60 minutes.
How long would it take you to swim a mile in still water?
13) How many different ways can this be accomplished? Note that the first person can choose any of the 11 seats.
(question is not completly avilable..will be updated asap)
14) The ratio of Boys to Girls is 6:4. 60% of the boys and 40% of the girls take lunch in the canteen. What % of class takes lunch in canteen?
15) In the General meeting of "Friends Club", Sameer said, "The repairs to the Club will come to a total of Rs 3120 and I propose that this amount should be met by the members, each paying an equal amount."
The proposal was immediately agreed. However, four members of the Club chose to resign, leaving the remaining members to pay an extra Rs 26 each.
How many members did the Club originally have?
16) A tank can be filled by pipe A in 30 minutes and by pipe B in 24 minutes. Outlet pipe C can empty the full tank in one hour and twenty minutes.
If the tank is empty initially and if all the three pipes A, B and C are opened simultaneously, in how much time will the tank be full?
17) A rich old Arab has three sons. When he died, he willed his 17 camels to the sons, to be divided as follows:
First Son to get 1/2 of the camels Second Son to get 1/3rd of the camels Third Son to get 1/9th of the camels.
The sons are sitting there trying to figure out how this can possibly be done, when a very old wise man goes riding by. They stop him and ask him to help them solve their problem. Without hesitation he divides the camels properly and continues riding on his way.
How did he do it?
18) There are 3 colored boxes - Red, Green and Blue. Each box contains 2 envelopes. Each envelope contains money - two of them contain Rs. 25000 each, two of them contain Rs. 15000 each and remaining two contain Rs. 10000 each.
There is one statement written on the cover of each box.
* Red Box: Both, a red box and a blue box contain Rs. 10000 each.
* Green Box: Both, a green box and a red box contain Rs. 25000 each.
* Blue Box: Both, a blue box and a green box contain Rs. 15000 each.
Only one of the above 3 statements is true and the corresponding box contains the maximum amount.
Can you tell which box contains the maximum amount and how much?
19) Sachin, Dravid and Ganguly played in a Cricket match between India and England.
None of them scored more than 99 runs.
If you add the digits of the runs scored by Sachin to his own score, you will get the runs scored by Dravid.
If you reverse the digits of the runs scored by Dravid, you will get the runs scored by Ganguly.
The total runs scored by them is 240.
Can you figure out their individual scores?
20) Substitute digits for the letters to make the following relation true.
W O R L D
+ T R A D E
----
C E N T E R
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter W, no other letter can be 3 and all other W in the puzzle must be 3.
<<< Previous >>> NExt
Practise Puzzles - 3
1) 1/3 rd of the contents of a container evaporated on the 1st day. 3/4th of the remaining contents of the container evaporated on the second day. What part of the contents of the container is left at the end of the second day?
2) Vipul was studying for his examinations and the lights went off. It was around 1:00 AM. He lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last six hours and the thin one two hours less. When he finally went to sleep, the thick candle was twice as long as the thin one.
For how long did Vipul study in candle light?
3) If you started a business in which you earned Rs.1 on the first day, Rs.3 on the second day, Rs.5 on the third day, Rs.7 on the fourth day, & so on.
How much would you have earned with this business after 50 years (assuming there are exactly 365 days in every year)?
Math gurus may use series formula to solve it.(series: 1,3,5,7,9,11.....upto 18250 terms)
4) A worker earns a 5% raise. A year later, the worker receives a 2.5% cut in pay, & now his salary is Rs. 22702.68
What was his salary to begin with?
5) At 6'o a clock ticks 6 times. The time between first and last ticks is 30 seconds. How long does it tick at 12'o.
6) In Mr. Mehta's family, there are one grandfather, one grandmother, two fathers, two mothers, one father-in-law, one mother-in-law, four children, three grandchildren, one brother, two sisters, two sons, two daughters and one daughter-in-law. How many members are there in Mr. Mehta's family? Give minimal possible answer.
7) When Alexander the Great attacked the forces of Porus, an Indian soldier was captured by the Greeks. He had displayed such bravery in battle, however, that the enemy offered to let him choose how he wanted to be killed. They told him, "If you tell a lie, you will put to the sword, and if you tell the truth you will be hanged."
The soldier could make only one statement. He made that statement and went free. What did he say?
8) A person wanted to withdraw X rupees and Y paise from the bank. But cashier made a mistake and gave him Y rupees and X paise. Neither the person nor the cashier noticed that.
After spending 20 paise, the person counts the money. And to his surprise, he has double the amount he wanted to withdraw.
Find X and Y. (1 Rupee = 100 Paise)
9) At the Party:
There were 9 men and children.
There were 2 more women than children.
The number of different man-woman couples possible was 24. Note that if there were 7 men and 5 women, then there would have been 35 man-woman couples possible.
Also, of the three groups - men, women and children - at the party:
There were 4 of one group.
There were 6 of one group.
There were 8 of one group.
Exactly one of the above 6 statements is false.
Can you tell which one is false? Also, how many men, women and children are there at the party
10) There is a shortage of tube lights , bulbs and fans in a village - Kharghar. It is found that
All houses do not have either tubelight or bulb or fan.
exactly 19% of houses do not have just one of these.
atleast 67% of houses do not have tubelights.
atleast 83% of houses do not have bulbs.
atleast 73% of houses do not have fans.
What percentage of houses do not have tubelight, bulb and fan?
>>> Next
2) Vipul was studying for his examinations and the lights went off. It was around 1:00 AM. He lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last six hours and the thin one two hours less. When he finally went to sleep, the thick candle was twice as long as the thin one.
For how long did Vipul study in candle light?
3) If you started a business in which you earned Rs.1 on the first day, Rs.3 on the second day, Rs.5 on the third day, Rs.7 on the fourth day, & so on.
How much would you have earned with this business after 50 years (assuming there are exactly 365 days in every year)?
Math gurus may use series formula to solve it.(series: 1,3,5,7,9,11.....upto 18250 terms)
4) A worker earns a 5% raise. A year later, the worker receives a 2.5% cut in pay, & now his salary is Rs. 22702.68
What was his salary to begin with?
5) At 6'o a clock ticks 6 times. The time between first and last ticks is 30 seconds. How long does it tick at 12'o.
6) In Mr. Mehta's family, there are one grandfather, one grandmother, two fathers, two mothers, one father-in-law, one mother-in-law, four children, three grandchildren, one brother, two sisters, two sons, two daughters and one daughter-in-law. How many members are there in Mr. Mehta's family? Give minimal possible answer.
7) When Alexander the Great attacked the forces of Porus, an Indian soldier was captured by the Greeks. He had displayed such bravery in battle, however, that the enemy offered to let him choose how he wanted to be killed. They told him, "If you tell a lie, you will put to the sword, and if you tell the truth you will be hanged."
The soldier could make only one statement. He made that statement and went free. What did he say?
8) A person wanted to withdraw X rupees and Y paise from the bank. But cashier made a mistake and gave him Y rupees and X paise. Neither the person nor the cashier noticed that.
After spending 20 paise, the person counts the money. And to his surprise, he has double the amount he wanted to withdraw.
Find X and Y. (1 Rupee = 100 Paise)
9) At the Party:
There were 9 men and children.
There were 2 more women than children.
The number of different man-woman couples possible was 24. Note that if there were 7 men and 5 women, then there would have been 35 man-woman couples possible.
Also, of the three groups - men, women and children - at the party:
There were 4 of one group.
There were 6 of one group.
There were 8 of one group.
Exactly one of the above 6 statements is false.
Can you tell which one is false? Also, how many men, women and children are there at the party
10) There is a shortage of tube lights , bulbs and fans in a village - Kharghar. It is found that
All houses do not have either tubelight or bulb or fan.
exactly 19% of houses do not have just one of these.
atleast 67% of houses do not have tubelights.
atleast 83% of houses do not have bulbs.
atleast 73% of houses do not have fans.
What percentage of houses do not have tubelight, bulb and fan?
>>> Next
Answers for Practise Puzzles - 2
1)
The required number is 236 and the sum is 11.
It is given that the first two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, first two digits must be either 2 or 3 i.e. 22, 23, 32 or 33 which means that there are four possible numbers - 224, 236, 326 and 339.
Now, it is also given that - the difference between it's reverse and itself is 396. Only 236 satisfies this condition. Hence, the sum of the three digits is 11.
2)
A=2, B=1, C=9, D=4, E=3, F=8, G=6, H=5, I=7
Let's start with GHI = 9 * IE. Note that I appears on both the side. Also, after multiplying IE by 9 the answer should have I at the unit's place. The possible values of IE are 19, 28, 37, 46, 55, 64, 73, 82 and 91; out of which only 64, 73 and 82 satisfies the condition. (as all alphabet should represent different digits)
Now, consider DEF = 6 * IE. Out of three short-listed values, only 73 satisfies the equation. Also, ABC = 3 * IE is satisfied by 73.
Hence, A=2, B=1, C=9, D=4, E=3, F=8, G=6, H=5, I=7
219 438 657
--- = 73 --- = 73 --- = 73
3 6
3)
A, B & D are males; C is female. B is C's only son. A & D are C's brothers.
A(male) --- C(female) --- D(male)
|
|
B(male)
Work out which relation can hold and discard the contradictory options.
From (2) and (4), D can not be a only daughter and have a sister (B or C). Hence, D is A's brother i.e. D is a Male.
From (4), let's say that B is D's sister i.e. B is Female.
From (3), A is C's only son i.e. A is Male.
But D is A's brother which means that A is not C's only son. Hence, our assumption was wrong.
Thus, C is D's sister i.e. C is Female. And B must be C's only son.
Now it is clear that D & B are Males and C is Female. A must be a Male as only one of them is of opposite sex from each of the other three. And he is C & D's brother.How are they related to each other?
4)
32 minutes 43.6 seconds
In twelve hours, the minute hand and the hour hand are together for 11 times. It means that after every 12/11 hours, both the hands are together.
Similarly in twelve hours, the minute hand and the hour hand are exactly opposite to each other for 11 times. It means that after every 12/11 hours, both the hands are opposite.
Now, let's take an example. We know that at 12 both the hands are together and at 6 both the hands are exactly opposite to each other.
After 6, both the hands are in opposition at [6+(12/11)] hours, [6+2*(12/11)] hours, [6+3*(12/11)] hours and so on. The sixth such time is [6+6*(12/11)] hours which is the first time after 12. Thus after 12, both the hands are opposite to each other at 12:32:43.6
Hence, Dr. DoLittle takes 32 minutes and 43.6 seconds to reach home from the clinic.
5)
Mr. Haani crossed 7 SlowRun Expresses during his journey.
Let's say that Mr. Haani travelled by SlowRun Express on Wednesday 10:00PM from Mumbai. The first train he would have crossed is the one scheduled to arrive at Mumbai at 11:30 PM the same day i.e. the one that left Bangalore at 10:00 PM on last Sunday.
Also, he would have crossed the last train just before reaching Bangalore on Saturday.
6)
The cabins from left to right (1-6) are of Mr. Solanki, Mr. Sinha, Mr. Shaan, Mr. Sharma, Miss Shudha and Miss Shalaka.
From (2), cabin number 5 is assigned to Miss Shudha.
As Miss Shudha is allergic to smoke and Mr. Sinha, Mr. Shaan & Mr. Solanki all smoke, they must be in cabin numbers 1, 2 and 3 not necessarily in the same order. Also, Miss Shalaka and Mr. Sharma must be in cabin 4 and 6.
From (3), Mr. Shaan must be in cabin 3 and Mr. Sharma must be in cabin 4. Thus, Miss Shalaka is in cabin 6.
As Mr. Solanki needs silence during work and Mr. Shaan is in cabin 3 who often talks to Mr. Sharma during work, Mr. Solanki must be in cabin 1. Hence, Mr. Sinha is in cabin 2.
Thus, the cabins numbers are
1# Mr. Solanki,
2# Mr. Sinha,
3# Mr. Shaan,
4# Mr. Sharma,
5# Miss Shudha,
6# Miss Shalaka
7)
It is given that
2 wixsomes = 3 changs
8 wixsomes = 12 changs ----- (I)
Also, given that
4 changs = 1 plut
12 changs = 3 plutes
8 wixsomes = 3 plutes ----- From (I)
Therefore,
6 plutes = 16 wixsomes
8)
The required number is 236 and the sum is 11.
It is given that the first two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, first two digits must be either 2 or 3 i.e. 22, 23, 32 or 33 which means that there are four possible numbers - 224, 236, 326 and 339.
Now, it is also given that - the difference between it's reverse and itself is 396. Only 236 satisfies this condition. Hence, the sum of the three digits is 11.
9)
A simple one.
As it is given that only one of the four statement is correct, the correct number can not appear in more than one statement. If it appears in more than one statement, then more than one statement will be correct.
Hence, there are 4 marbles under each mug.
10)
aswer is 7
11)
The cicumference of the earth is
= 2 * PI * r
= 2 * PI * 6400 km
= 2 * PI * 6400 * 1000 m
= 2 * PI * 6400 * 1000 * 100 cm
= 1280000000 * PI cm
where r = radius of the earth, PI = 3.141592654
Hence, the length of the thread is = 1280000000 * PI cm
Now length of the thread is increasd by 12 cm. So the new length is = (1280000000 * PI) + 12 cm
This thread will make one concentric circle with the earth which is slightly away from the earth. The circumfernce of that circle is nothing but (1280000000 * PI) + 12 cm
Assume that radius of the outer circle is R cm
Therefore,
2 * PI * R = (1280000000 * PI) + 12 cm
Solving above equation, R = 640000001.908 cm
Radius of the earth is r = 640000000 cm
Hence, the thread will be separatedfrom the earth by
= R - r cm
= 640000001.908 - 640000000
= 1.908 cm
12)
At the beginning of the 11th year, there would be 1,024,000 rabbits.
At the beginning, there were 1000 rabbits. Also, there were 4000 rabbits at the beginning of third year which is equal to 2828Z. Thus, Z = 4000/2828 i.e. 1.414 (the square root of 2)
Note that 2828Z can be represented as 2000*Z*Z (Z=1.414), which can be further simplified as 1000*Z*Z*Z*Z
Also, it is given that at the end of 6 months, there were 1000Z rabbits.
It is clear that the population growth is 1.414 times every six months i.e. 2 times every year. After N years, the population would be 1000*(Z^(2N)) i.e. 1000*(2^N)
Thus, at the beginning of the 11th year (i.e. after 10 years), there would be 1000*(2^10) i.e. 1,024,000 rabbits.
13)
The man drunk 190oz of water.
It is given that the man has 20oz bottle of sprite. Also, he will drink 1oz on the first day and refill the bottle with water, will drink 2oz on the second day and refill the bottle, will drink 3oz on the third day and refill the bottle, and so on till 20th day. Thus at the end of 20 days, he must have drunk (1 + 2 + 3 + 4 + ..... +18 + 19 + 20) = 210oz of liquid.
Out of that 210oz, 20oz is the sprite which he had initially. Hence, he must have drunk 190oz of water.ed
14)
There are 5 such expressions.
99 + (9/9) = 100
(99/.99) = 100
(9/.9) * (9/.9) = 100
((9*9) + 9)/.9 = 100
(99-9)/.9 = 100
15)
answer is 7/9
Assume that number of boys graduated from City High School = B
Therefore, number of girls graduated from City High School = 2*B
It is given that 3/4 of the girls and 5/6 of the boys went to college immediately after graduation.
Hence, total students went to college
= (3/4)(2*B) + (5/6)(B)
= B * (3/2 + 5/6)
= (7/3)B
Fraction of the graduates that year went to college immediately after graduation
= [(7/3)B] / [3*B]
= 7/9
Therefore, the answer is 7/9
16)
The mule was carrying 5 sacks and the donkey was carrying 7 sacks.
Let's assume that the mule was carrying M sacks and the donkey was carrying D sacks.
As the donkey told the mule, "If you gave me one of your sacks I'd have double what you have."
D + 1 = 2 * (M-1)
D + 1 = 2M - 2
D = 2M - 3
The donkey also said, "If I give you one of my sacks we'd have an even amount."
D - 1 = M + 1
D = M + 2
Comparing both the equations,
2M - 3 = M + 2
M = 5
Substituting M=5 in any of above equation, we get D=7
Hence, the mule was carrying 5 sacks and the donkey was carrying 7 sacks.
17) Person A came in first.
Let's assume that the distance between start and the point is D miles.
Total time taken by Person A to finish
= (D/20) + (D/20)
= D/10
= 0.1D
Total time taken by Person B to finish
= (D/10) + (D/30)
= 2D/15
= 0.1333D
Thus, Person A is the Winner.
18)
Assume that initial there were 3*X bullets.
So they got X bullets each after division.
All of them shot 4 bullets. So now they have (X - 4) bullets each.
But it is given that, after they shot 4 bullets each, total number of bullets remaining is equal to the bullets each had after division i.e. X
Therefore, the equation is
3 * (X - 4) = X
3 * X - 12 = X
2 * X = 12
X = 6
Therefore the total bullets before division is = 3 * X = 18
19)The sum of the digits of D is 1.
Let E = sum of digits of D.
It follows from the hint that A = E (mod 9)
consider, (" ^ " means power)
A = 1999 ^ 1999
< 2000 ^ 2000
= 2 ^ 2000 * 1000 ^ 2000
= 1024 ^ 200 * 10 ^ 6000
< 10 ^ 800 * 10 ^ 6000
= 10 ^ 6800
i.e. A < 106800
i.e. B < 6800 * 9 = 61200
i.e. C < 5 * 9 = 45
i.e. D < 2 * 9 = 18
i.e. E <= 9
i.e. E is a single digit number.
Also,
1999 = 1 (mod 9)
so 19991999 = 1 (mod 9)
Therefore we conclude that E=1.
20)
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backword - are equal.
Let's assume that when the last person reached the first person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total distance covered by the last person is 100 meters, as he ran the total length of the platoon (50 meters) twice. TRUE, but that's the relative distance covered by the last person i.e. assuming that the platoon is stationary.
21) answer is 9
To verify this, you can make a drawing of a cube, & number each of its 12 edges. Then, always starting from 1 corner & 1 edge, you can determine all of the possible combinations for tracing along the edges of a cube.
There is no need to start from other corners or edges of the cube, as you will only be repeating the same combinations. The process is a little more involved than this, but
is useful for solving many types of spatial puzzles.
22)
Mr. Bajaj is the Engineer and either his wife or his son is the Doctor.
Mr. Bajaj's wife and mother are not blood relatives. So from 3, if the Engineer is a female, the Doctor is a male. But from 1, if the Doctor is a male, then the Engineer is a male. Thus, there is a contradiction, if the Engineer is a female. Hence, either Mr. Bajaj or his son is the Engineer.
Mr. Bajaj's son is the youngest of all four and is blood relative of each of them. So from 2, Mr. Bajaj's son is not the Engineer. Hence, Mr. Bajaj is the Engineer.
Now from 2, Mr. Bajaj's mother can not be the Doctor. So the Doctor is either his wife or his son . It is not possible to determine anything further.
23)
Carrie is married to Cam.
"Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge."
It means that Sam is not married to either Billy or Carrie. Thus, Sam is married to Tina.
As Cam does not play bridge, Billy's husband must be Laurie.
Hence, Carrie is married to Cam
24)
One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Backtracing.
It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X & Y as many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48)
Similarly, before Y gave tractors to X & Z, they had 6 & 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24)
Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)
Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.
25)
The sign-maker will need 192 zeroes.
Divide 1000 building numbers into groups of 100 each as follow:
(1..100), (101..200), (201..300), ....... (901..1000)
For the first group, sign-maker will need 11 zeroes.
For group numbers 2 to 9, he will require 20 zeroes each.
And for group number 10, he will require 21 zeroes.
The total numbers of zeroes required are
= 11 + 8*20 + 21
= 11 + 160 + 21
= 192
26)
It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.
Take 8 coins and weigh 4 against 4.
If both are not equal, goto step 2
If both are equal, goto step 3
One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.
If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.
If both are equal, L4 is the odd coin and is lighter.
If L2 is light, L2 is the odd coin and is lighter.
If L3 is light, L3 is the odd coin and is lighter.
If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2
If both are equal, there is some error.
If H1 is heavy, H1 is the odd coin and is heavier.
If H2 is heavy, H2 is the odd coin and is heavier.
If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4
If both are equal, L1 is the odd coin and is lighter.
If H3 is heavy, H3 is the odd coin and is heavier.
If H4 is heavy, H4 is the odd coin and is heavier.
The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.
If both are equal, there is some error.
If X is heavy, X is the odd coin and is heavier.
If X is light, X is the odd coin and is lighter.
27)
Total 36 medals were awarded and the contest was for 6 days.
On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals
On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals
On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals
On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals
On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals
On day 6: Medals awarded 6
I got this answer by writing small program. If anyone know any other simpler method, do submit it.
28)
The answer is 381654729
One way to solve it is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.
The other way to solve this problem is by writing a computer program that systematically tries all possibilities
***********************************ENF OF ANSWER SESSION FOR PRACTISE PUZZLES - 2 ***************************
Thanks,
All The Best
The required number is 236 and the sum is 11.
It is given that the first two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, first two digits must be either 2 or 3 i.e. 22, 23, 32 or 33 which means that there are four possible numbers - 224, 236, 326 and 339.
Now, it is also given that - the difference between it's reverse and itself is 396. Only 236 satisfies this condition. Hence, the sum of the three digits is 11.
2)
A=2, B=1, C=9, D=4, E=3, F=8, G=6, H=5, I=7
Let's start with GHI = 9 * IE. Note that I appears on both the side. Also, after multiplying IE by 9 the answer should have I at the unit's place. The possible values of IE are 19, 28, 37, 46, 55, 64, 73, 82 and 91; out of which only 64, 73 and 82 satisfies the condition. (as all alphabet should represent different digits)
Now, consider DEF = 6 * IE. Out of three short-listed values, only 73 satisfies the equation. Also, ABC = 3 * IE is satisfied by 73.
Hence, A=2, B=1, C=9, D=4, E=3, F=8, G=6, H=5, I=7
219 438 657
--- = 73 --- = 73 --- = 73
3 6
3)
A, B & D are males; C is female. B is C's only son. A & D are C's brothers.
A(male) --- C(female) --- D(male)
|
|
B(male)
Work out which relation can hold and discard the contradictory options.
From (2) and (4), D can not be a only daughter and have a sister (B or C). Hence, D is A's brother i.e. D is a Male.
From (4), let's say that B is D's sister i.e. B is Female.
From (3), A is C's only son i.e. A is Male.
But D is A's brother which means that A is not C's only son. Hence, our assumption was wrong.
Thus, C is D's sister i.e. C is Female. And B must be C's only son.
Now it is clear that D & B are Males and C is Female. A must be a Male as only one of them is of opposite sex from each of the other three. And he is C & D's brother.How are they related to each other?
4)
32 minutes 43.6 seconds
In twelve hours, the minute hand and the hour hand are together for 11 times. It means that after every 12/11 hours, both the hands are together.
Similarly in twelve hours, the minute hand and the hour hand are exactly opposite to each other for 11 times. It means that after every 12/11 hours, both the hands are opposite.
Now, let's take an example. We know that at 12 both the hands are together and at 6 both the hands are exactly opposite to each other.
After 6, both the hands are in opposition at [6+(12/11)] hours, [6+2*(12/11)] hours, [6+3*(12/11)] hours and so on. The sixth such time is [6+6*(12/11)] hours which is the first time after 12. Thus after 12, both the hands are opposite to each other at 12:32:43.6
Hence, Dr. DoLittle takes 32 minutes and 43.6 seconds to reach home from the clinic.
5)
Mr. Haani crossed 7 SlowRun Expresses during his journey.
Let's say that Mr. Haani travelled by SlowRun Express on Wednesday 10:00PM from Mumbai. The first train he would have crossed is the one scheduled to arrive at Mumbai at 11:30 PM the same day i.e. the one that left Bangalore at 10:00 PM on last Sunday.
Also, he would have crossed the last train just before reaching Bangalore on Saturday.
6)
The cabins from left to right (1-6) are of Mr. Solanki, Mr. Sinha, Mr. Shaan, Mr. Sharma, Miss Shudha and Miss Shalaka.
From (2), cabin number 5 is assigned to Miss Shudha.
As Miss Shudha is allergic to smoke and Mr. Sinha, Mr. Shaan & Mr. Solanki all smoke, they must be in cabin numbers 1, 2 and 3 not necessarily in the same order. Also, Miss Shalaka and Mr. Sharma must be in cabin 4 and 6.
From (3), Mr. Shaan must be in cabin 3 and Mr. Sharma must be in cabin 4. Thus, Miss Shalaka is in cabin 6.
As Mr. Solanki needs silence during work and Mr. Shaan is in cabin 3 who often talks to Mr. Sharma during work, Mr. Solanki must be in cabin 1. Hence, Mr. Sinha is in cabin 2.
Thus, the cabins numbers are
1# Mr. Solanki,
2# Mr. Sinha,
3# Mr. Shaan,
4# Mr. Sharma,
5# Miss Shudha,
6# Miss Shalaka
7)
It is given that
2 wixsomes = 3 changs
8 wixsomes = 12 changs ----- (I)
Also, given that
4 changs = 1 plut
12 changs = 3 plutes
8 wixsomes = 3 plutes ----- From (I)
Therefore,
6 plutes = 16 wixsomes
8)
The required number is 236 and the sum is 11.
It is given that the first two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, first two digits must be either 2 or 3 i.e. 22, 23, 32 or 33 which means that there are four possible numbers - 224, 236, 326 and 339.
Now, it is also given that - the difference between it's reverse and itself is 396. Only 236 satisfies this condition. Hence, the sum of the three digits is 11.
9)
A simple one.
As it is given that only one of the four statement is correct, the correct number can not appear in more than one statement. If it appears in more than one statement, then more than one statement will be correct.
Hence, there are 4 marbles under each mug.
10)
aswer is 7
11)
The cicumference of the earth is
= 2 * PI * r
= 2 * PI * 6400 km
= 2 * PI * 6400 * 1000 m
= 2 * PI * 6400 * 1000 * 100 cm
= 1280000000 * PI cm
where r = radius of the earth, PI = 3.141592654
Hence, the length of the thread is = 1280000000 * PI cm
Now length of the thread is increasd by 12 cm. So the new length is = (1280000000 * PI) + 12 cm
This thread will make one concentric circle with the earth which is slightly away from the earth. The circumfernce of that circle is nothing but (1280000000 * PI) + 12 cm
Assume that radius of the outer circle is R cm
Therefore,
2 * PI * R = (1280000000 * PI) + 12 cm
Solving above equation, R = 640000001.908 cm
Radius of the earth is r = 640000000 cm
Hence, the thread will be separatedfrom the earth by
= R - r cm
= 640000001.908 - 640000000
= 1.908 cm
12)
At the beginning of the 11th year, there would be 1,024,000 rabbits.
At the beginning, there were 1000 rabbits. Also, there were 4000 rabbits at the beginning of third year which is equal to 2828Z. Thus, Z = 4000/2828 i.e. 1.414 (the square root of 2)
Note that 2828Z can be represented as 2000*Z*Z (Z=1.414), which can be further simplified as 1000*Z*Z*Z*Z
Also, it is given that at the end of 6 months, there were 1000Z rabbits.
It is clear that the population growth is 1.414 times every six months i.e. 2 times every year. After N years, the population would be 1000*(Z^(2N)) i.e. 1000*(2^N)
Thus, at the beginning of the 11th year (i.e. after 10 years), there would be 1000*(2^10) i.e. 1,024,000 rabbits.
13)
The man drunk 190oz of water.
It is given that the man has 20oz bottle of sprite. Also, he will drink 1oz on the first day and refill the bottle with water, will drink 2oz on the second day and refill the bottle, will drink 3oz on the third day and refill the bottle, and so on till 20th day. Thus at the end of 20 days, he must have drunk (1 + 2 + 3 + 4 + ..... +18 + 19 + 20) = 210oz of liquid.
Out of that 210oz, 20oz is the sprite which he had initially. Hence, he must have drunk 190oz of water.ed
14)
There are 5 such expressions.
99 + (9/9) = 100
(99/.99) = 100
(9/.9) * (9/.9) = 100
((9*9) + 9)/.9 = 100
(99-9)/.9 = 100
15)
answer is 7/9
Assume that number of boys graduated from City High School = B
Therefore, number of girls graduated from City High School = 2*B
It is given that 3/4 of the girls and 5/6 of the boys went to college immediately after graduation.
Hence, total students went to college
= (3/4)(2*B) + (5/6)(B)
= B * (3/2 + 5/6)
= (7/3)B
Fraction of the graduates that year went to college immediately after graduation
= [(7/3)B] / [3*B]
= 7/9
Therefore, the answer is 7/9
16)
The mule was carrying 5 sacks and the donkey was carrying 7 sacks.
Let's assume that the mule was carrying M sacks and the donkey was carrying D sacks.
As the donkey told the mule, "If you gave me one of your sacks I'd have double what you have."
D + 1 = 2 * (M-1)
D + 1 = 2M - 2
D = 2M - 3
The donkey also said, "If I give you one of my sacks we'd have an even amount."
D - 1 = M + 1
D = M + 2
Comparing both the equations,
2M - 3 = M + 2
M = 5
Substituting M=5 in any of above equation, we get D=7
Hence, the mule was carrying 5 sacks and the donkey was carrying 7 sacks.
17) Person A came in first.
Let's assume that the distance between start and the point is D miles.
Total time taken by Person A to finish
= (D/20) + (D/20)
= D/10
= 0.1D
Total time taken by Person B to finish
= (D/10) + (D/30)
= 2D/15
= 0.1333D
Thus, Person A is the Winner.
18)
Assume that initial there were 3*X bullets.
So they got X bullets each after division.
All of them shot 4 bullets. So now they have (X - 4) bullets each.
But it is given that, after they shot 4 bullets each, total number of bullets remaining is equal to the bullets each had after division i.e. X
Therefore, the equation is
3 * (X - 4) = X
3 * X - 12 = X
2 * X = 12
X = 6
Therefore the total bullets before division is = 3 * X = 18
19)The sum of the digits of D is 1.
Let E = sum of digits of D.
It follows from the hint that A = E (mod 9)
consider, (" ^ " means power)
A = 1999 ^ 1999
< 2000 ^ 2000
= 2 ^ 2000 * 1000 ^ 2000
= 1024 ^ 200 * 10 ^ 6000
< 10 ^ 800 * 10 ^ 6000
= 10 ^ 6800
i.e. A < 106800
i.e. B < 6800 * 9 = 61200
i.e. C < 5 * 9 = 45
i.e. D < 2 * 9 = 18
i.e. E <= 9
i.e. E is a single digit number.
Also,
1999 = 1 (mod 9)
so 19991999 = 1 (mod 9)
Therefore we conclude that E=1.
20)
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backword - are equal.
Let's assume that when the last person reached the first person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total distance covered by the last person is 100 meters, as he ran the total length of the platoon (50 meters) twice. TRUE, but that's the relative distance covered by the last person i.e. assuming that the platoon is stationary.
21) answer is 9
To verify this, you can make a drawing of a cube, & number each of its 12 edges. Then, always starting from 1 corner & 1 edge, you can determine all of the possible combinations for tracing along the edges of a cube.
There is no need to start from other corners or edges of the cube, as you will only be repeating the same combinations. The process is a little more involved than this, but
is useful for solving many types of spatial puzzles.
22)
Mr. Bajaj is the Engineer and either his wife or his son is the Doctor.
Mr. Bajaj's wife and mother are not blood relatives. So from 3, if the Engineer is a female, the Doctor is a male. But from 1, if the Doctor is a male, then the Engineer is a male. Thus, there is a contradiction, if the Engineer is a female. Hence, either Mr. Bajaj or his son is the Engineer.
Mr. Bajaj's son is the youngest of all four and is blood relative of each of them. So from 2, Mr. Bajaj's son is not the Engineer. Hence, Mr. Bajaj is the Engineer.
Now from 2, Mr. Bajaj's mother can not be the Doctor. So the Doctor is either his wife or his son . It is not possible to determine anything further.
23)
Carrie is married to Cam.
"Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge."
It means that Sam is not married to either Billy or Carrie. Thus, Sam is married to Tina.
As Cam does not play bridge, Billy's husband must be Laurie.
Hence, Carrie is married to Cam
24)
One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Backtracing.
It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X & Y as many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48)
Similarly, before Y gave tractors to X & Z, they had 6 & 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24)
Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)
Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.
25)
The sign-maker will need 192 zeroes.
Divide 1000 building numbers into groups of 100 each as follow:
(1..100), (101..200), (201..300), ....... (901..1000)
For the first group, sign-maker will need 11 zeroes.
For group numbers 2 to 9, he will require 20 zeroes each.
And for group number 10, he will require 21 zeroes.
The total numbers of zeroes required are
= 11 + 8*20 + 21
= 11 + 160 + 21
= 192
26)
It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.
Take 8 coins and weigh 4 against 4.
If both are not equal, goto step 2
If both are equal, goto step 3
One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.
If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.
If both are equal, L4 is the odd coin and is lighter.
If L2 is light, L2 is the odd coin and is lighter.
If L3 is light, L3 is the odd coin and is lighter.
If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2
If both are equal, there is some error.
If H1 is heavy, H1 is the odd coin and is heavier.
If H2 is heavy, H2 is the odd coin and is heavier.
If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4
If both are equal, L1 is the odd coin and is lighter.
If H3 is heavy, H3 is the odd coin and is heavier.
If H4 is heavy, H4 is the odd coin and is heavier.
The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.
If both are equal, there is some error.
If X is heavy, X is the odd coin and is heavier.
If X is light, X is the odd coin and is lighter.
27)
Total 36 medals were awarded and the contest was for 6 days.
On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals
On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals
On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals
On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals
On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals
On day 6: Medals awarded 6
I got this answer by writing small program. If anyone know any other simpler method, do submit it.
28)
The answer is 381654729
One way to solve it is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.
The other way to solve this problem is by writing a computer program that systematically tries all possibilities
***********************************ENF OF ANSWER SESSION FOR PRACTISE PUZZLES - 2 ***************************
Thanks,
All The Best
Answers For Practise Puzzles - 1
1)
6,48,64,800 ways
There are total 16 pieces which can be arranged on 16 places in 16P16 = 16! ways.
(16! = 16 * 15 * 14 * 13 * 12 * ..... * 3 * 2 * 1)
But, there are some duplicate combinations because of identical pieces.
There are 8 identical pawn, which can be arranged in 8P8 = 8! ways.
Similarly there are 2 identical rooks, 2 identical knights and 2 identical bishops. Each can be arranged in 2P2 = 2! ways.
Hence, the require answer is
= (16!) / (8! * 2! * 2! * 2!)
= 6,48,64,800
2)
Rs. 250/-
Assume that initially he had Rs. X
He spent 1/3 for cloths =. (1/3) * X
Remaining money = (2/3) * X
He spent 1/5 of remaining money for food = (1/5) * (2/3) * X = (2/15) * X
Remaining money = (2/3) * X - (2/15) * X = (8/15) * X
Again, he spent 1/4 of remaining maoney for travel = (1/4) * (8/15) * X = (2/15) * X
Remaining money = (8/15) * X - (2/15) * X = (6/15) * X
But after spending for travel he is left with Rs. 100/- So
(6/15) * X = 100
X = 250
3)
20 cows
g - grass at the beginning
r - rate at which grass grows, per day
y - rate at which one cow eats grass, per day
n - no of cows to eat the grass in 96 days
From given data,
g + 24*r = 70 * 24 * y - A
g + 60*r = 30 * 60 * y - B
g + 96*r = n * 96 * y - C
Solving for (B-A),
(60 * r) - (24 * r) = (30 * 60 * y) - (70 * 24 * y)
36 * r = 120 * y - D
Solving for (C-B),
(96 * r) - (60 * r) = (n * 96 * y) - (30 * 60 * y)
36 * r = (n * 96 - 30 * 60) * y
120 * y = (n * 96 - 30 * 60) * y [From D]
120 = (n * 96 - 1800)
n = 20
Hence, 20 cows are needed to eat the grass in 96 days.
4)
65292
As per given conditions, there are three possible combinations for 2nd, 3rd and 4th digits. They are (3, 0, 7) or (4, 1, 8) or (5, 2, 9)
It is given that there are 3 pairs whose sum is 11. All possible pairs are (2, 9), (3, 8), (4, 7), (5, 6). Now required number is 5 digit number and it contains 3 pairs of 11. So it must not be having 0 and 1 in it. Hence, the only possible combination for 2nd, 3rd and 4th digits is (5, 2, 9)
Also, 1st digit is thrice the last digit. The possible combinations are (3, 1), (6, 2) and (9, 3), out of which only (6, 2) with (5, 2, 9) gives 3 pairs of 11. Hence, the answer is 65292.
5)
The answer is 285714.
If its rightmost digit is placed at its left end, then new number is 428571 which is 50% larger than the original number 285714
The simplest way is to write a small program. And the other way is trial and error !!!
6)
52 / 2703
There are two cases to be considered.
CASE 1 : King of Hearts is drawn from Pack A and shuffled with Pack B
Probability of drawing King of Hearts from Pack A = 1/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 2/53
So total probability of case 1 = (1/51) * (2/53) = 2 / (51 * 53)
CASE 2 : King of Hearts is not drawn from Pack A
Probability of not drawing King of Hearts from Pack A = 50/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 1/53
So total probability of case 2 = (50/51) * (1/53) = 50 / (51 * 53)
Now adding both the probability, the required probability is
= 2 / (51 * 53) + 50 / (51 * 53)
= 52 / (51 * 53)
= 52 / 2703
= 0.0192378
7)
The maximum number of attempts required are 16,22,400
There are 52 possible letters - a to z and A to Z, and 10 possible numbers - 0 to 9. Now, 4 characters - 2 letters and 2 numbers, can be selected in 52*52*10*10 ways. These 4 characters can be arranged in 4C2 i.e. 6 different ways - the number of unique patterns that can be formed by lining up 4 objects of which 2 are distinguished one way (i.e. they must be letters) and the other 2 are distinguished another way (i.e. they must be numbers).
Consider an example : Let's assume that @ represents letter and # represents number. the 6 possible ways of arranging them are : @@##, @#@#, @##@, #@@#, #@#@, ##@@
Hence, the required answer is
= 52*52*10*10*6
= 16,22,400 attempts
= 1.6 million approx.
8)
There are 4.3252 * 10^19 possible combinations for 3x3x3 Rubics and 7.4012 * 10^45 possible combinations for 4x4x4 Rubics.
Let's consider 3x3x3 Rubics first.
There are 8 corner cubes, which can be arranged in 8! ways.
Each of these 8 cubes can be turned in 3 different directions, so there are 3^8 orientations altogether. But if you get all but one of the corner cube into chosen positions and orientations, only one of 3 orientations of the final corner cube is possible. Thus, total ways corner cubes can be placed = (8!) * (3^8)/8 = (8!) * (3^7)
Similarly, 12 edge cubes can be arranged in 12! ways.
Each of these 12 cubes can be turned in 2 different directions, so there are 2^12 orientations altogether. But if you get all but one of the edge cube into chosen positions and orientations, only one of 2 orientations of the final edge cube is possible. Thus, total ways edge cubes can be placed = (12!) * (2^12)/2 = (12!) * (2^11)
Here, we have essentially pulled the cubes apart and stuck cubes back in place wherever we please. In reality, we can only move cubes around by turning the faces of the cubes. It turns out that you can't turn the faces in such a way as to switch the positions of two cubes while returning all the others to their original positions. Thus if you get all but two cubes in place, there is only one attainable choice for them (not 2!). Hence, we must divide by 2.
Total different possible combinations are
= [(8!) * (3^7)] * [(12!) * (2^11)] / 2
= (8!) * (3^7) * (12!) * (2^10)
= 4.3252 * 10^19
Similarly, for 4x4x4 Rubics total different possible combinations are
= [(8!) * (3^7)] * [(24!)] * [(24!) / (4!^6)] / 24
= 7.4011968 * 10^45
Note that there are 24 edge cubes, which you can not turn in 2 orientations (hence no 2^24 / 2). Also, there are 4 center cubes per face i.e. (24!) / (4!^6). You can switch 2 cubes without affecting the rest of the combination as 4*4*4 has even dimensions (hence no division by 2). But pattern on one side is rotated in 4 directions over 6 faces, hence divide by 24.
9)
Answer
A tough one!!!
Since R + E + E = 10 + E, it is clear that R + E = 10 and neither R nor E is equal to 0 or 5. This is the only entry point to
solve it. Now use trial-n-error method.
N E V E R 2 1 4 1 9
L E A V E 3 1 5 4 1
+ M E + 6 1
-------- --------
A L O N E 5 3 0 2 1
10)
Cindy is the Singer. Mr. Clinton or Monika is the Dancer.
From (1) and (3), the singer and the dancer, both can not be a man. From (3) and (4), if the singer is a man, then the dancer must be a man. Hence, the singer must be a woman.
CASE I : Singer is a woman and Dancer is also a woman
Then, the dancer is Monika and the singer is Cindy.
CASE II : Singer is a woman and Dancer is also a man
Then, the dancer is Mr. Clinton and the singer is Cindy.
In both the cases, we know that Cindy is the Singer. And either Mr. Clinton or Monika is the Dancer.
11)
There are total 450 rooms.
Out of which 299 room number starts with either 1, 2 or 3. (as room number 100 is not there) Now out of those 299 rooms only 90 room numbers end with 4, 5 or 6
So the probability is 90/450 i.e. 1/5 or 0.20
Draw 9 dots on a page, in the shape of three rows of three dots to form a square. Now place your pen on the page, draw 4 straight lines and try and cover all the dots.
You're not allowed to lift your pen.
Note: Don't be confined by the dimensions of the square.
12)
One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Backtracing.
It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X & Y as many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48)
Similarly, before Y gave tractors to X & Z, they had 6 & 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24)
Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)
Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.
13)
None
Frank gave you a pear in exchange of the apple which Marge gave you. And you gave that pear to Bret in exchange for an apple. All the others exchanges involved apples and/or organges.
Four couples are going to the movie. Each row holds eight seats. Betty and Jim don't want to sit next to Alice and Tom. Alice and Tom don't want to sit next to Gertrude and Bill. On the otherhand, Sally and Bob don't want to sit next to Betty and Jim.
14)
Let before adding X stations, total number of tickets
t = N(N-1)
After adding X stations total number of tickets are
t + 46 = (N+X)(N+X-1)
Subtracting 1st from 2nd
46 = (N+X)(N+X-1) - N(N-1)
46 = N2 + NX - N + NX + X2 - X - N2 + N
46 = 2NX + X2 - X
46 = (2N - 1)X + X2
X2 + (2N - 1)X - 46 = 0
Now there are only two possible factors of 46. They are (46,1) and (23,2)
Case I: (46,1)
2N - 1 = 45
2N = 46
N = 23
And X = 1
Case II: (23,2)
2N - 1 = 21
2N = 22
N = 11
And X = 2
Hence, there are 2 possible answers.
15)
75 miles per hour
While going to the destination, the vehicle travels 10 mils at the speed of 50 miles per hour. So the time taken to travel 10 miles is
= (60 * 10) / 50
= 12 minutes
Now it's given that round-trip travel time is 20 minutes. So the vehicle should complete its return trip of 10 miles in 8 minutes. So the speed of the vehicle must
= (60 * 10) / 8
= 75 miles per hour
16)
70 students are majoring in both, psychology & business
If 73% of the students are psychology majors, we know that 27% are not psychology majors. By the same reasoning, 38% are not business majors, because 62% of the students do major in business. So: 27 + 38 = 65
65% of the students are not majoring in both psychology & business, so 35% are double majors, a total of 70 students.
17)
Initially, there were 42 or 36 matches in the match-box.
There are 42 matches in the box with which he could form a triangle 20, 15, 7, with an area of 42 square inches. After 6 matches had been used, the remaining 36 matches would form a triangle 17, 10, 9, with an area of 36 square inches. After using another 6 matches, the remaining 30 matches would form a triangle 13, 12, 5, with an area of 30 square inches. After using another 6, the 24 remaining would form a triangle 10, 8, 6, with an area of 24 square inches.
Thus, there are two possible answers. There were either 42 or 36 matches in the match-box.
18)
Mr. Grey is wearing Black suit.
Mr. White is wearing Grey suit.
Mr. Black is wearing White suit.
Mr. Grey must not be wearing grey suit as that is the same colour as his name. Also, he was not wearing white suit as the person wearing white suit responded to his comment. So Mr Grey must be wearing a black suit.
Similarly, Mr. White must be wearing either black suit or grey suit. But Mr. Grey is wearing a black suit. Hence, Mr. White must be wearing a grey suit.
And, Mr. Black must be wearing white suit
******************************END OF ANSWERS FOR FIRST PRACTISE EXERCISE***********************************
thanks,
ALL THE BEST
6,48,64,800 ways
There are total 16 pieces which can be arranged on 16 places in 16P16 = 16! ways.
(16! = 16 * 15 * 14 * 13 * 12 * ..... * 3 * 2 * 1)
But, there are some duplicate combinations because of identical pieces.
There are 8 identical pawn, which can be arranged in 8P8 = 8! ways.
Similarly there are 2 identical rooks, 2 identical knights and 2 identical bishops. Each can be arranged in 2P2 = 2! ways.
Hence, the require answer is
= (16!) / (8! * 2! * 2! * 2!)
= 6,48,64,800
2)
Rs. 250/-
Assume that initially he had Rs. X
He spent 1/3 for cloths =. (1/3) * X
Remaining money = (2/3) * X
He spent 1/5 of remaining money for food = (1/5) * (2/3) * X = (2/15) * X
Remaining money = (2/3) * X - (2/15) * X = (8/15) * X
Again, he spent 1/4 of remaining maoney for travel = (1/4) * (8/15) * X = (2/15) * X
Remaining money = (8/15) * X - (2/15) * X = (6/15) * X
But after spending for travel he is left with Rs. 100/- So
(6/15) * X = 100
X = 250
3)
20 cows
g - grass at the beginning
r - rate at which grass grows, per day
y - rate at which one cow eats grass, per day
n - no of cows to eat the grass in 96 days
From given data,
g + 24*r = 70 * 24 * y - A
g + 60*r = 30 * 60 * y - B
g + 96*r = n * 96 * y - C
Solving for (B-A),
(60 * r) - (24 * r) = (30 * 60 * y) - (70 * 24 * y)
36 * r = 120 * y - D
Solving for (C-B),
(96 * r) - (60 * r) = (n * 96 * y) - (30 * 60 * y)
36 * r = (n * 96 - 30 * 60) * y
120 * y = (n * 96 - 30 * 60) * y [From D]
120 = (n * 96 - 1800)
n = 20
Hence, 20 cows are needed to eat the grass in 96 days.
4)
65292
As per given conditions, there are three possible combinations for 2nd, 3rd and 4th digits. They are (3, 0, 7) or (4, 1, 8) or (5, 2, 9)
It is given that there are 3 pairs whose sum is 11. All possible pairs are (2, 9), (3, 8), (4, 7), (5, 6). Now required number is 5 digit number and it contains 3 pairs of 11. So it must not be having 0 and 1 in it. Hence, the only possible combination for 2nd, 3rd and 4th digits is (5, 2, 9)
Also, 1st digit is thrice the last digit. The possible combinations are (3, 1), (6, 2) and (9, 3), out of which only (6, 2) with (5, 2, 9) gives 3 pairs of 11. Hence, the answer is 65292.
5)
The answer is 285714.
If its rightmost digit is placed at its left end, then new number is 428571 which is 50% larger than the original number 285714
The simplest way is to write a small program. And the other way is trial and error !!!
6)
52 / 2703
There are two cases to be considered.
CASE 1 : King of Hearts is drawn from Pack A and shuffled with Pack B
Probability of drawing King of Hearts from Pack A = 1/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 2/53
So total probability of case 1 = (1/51) * (2/53) = 2 / (51 * 53)
CASE 2 : King of Hearts is not drawn from Pack A
Probability of not drawing King of Hearts from Pack A = 50/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 1/53
So total probability of case 2 = (50/51) * (1/53) = 50 / (51 * 53)
Now adding both the probability, the required probability is
= 2 / (51 * 53) + 50 / (51 * 53)
= 52 / (51 * 53)
= 52 / 2703
= 0.0192378
7)
The maximum number of attempts required are 16,22,400
There are 52 possible letters - a to z and A to Z, and 10 possible numbers - 0 to 9. Now, 4 characters - 2 letters and 2 numbers, can be selected in 52*52*10*10 ways. These 4 characters can be arranged in 4C2 i.e. 6 different ways - the number of unique patterns that can be formed by lining up 4 objects of which 2 are distinguished one way (i.e. they must be letters) and the other 2 are distinguished another way (i.e. they must be numbers).
Consider an example : Let's assume that @ represents letter and # represents number. the 6 possible ways of arranging them are : @@##, @#@#, @##@, #@@#, #@#@, ##@@
Hence, the required answer is
= 52*52*10*10*6
= 16,22,400 attempts
= 1.6 million approx.
8)
There are 4.3252 * 10^19 possible combinations for 3x3x3 Rubics and 7.4012 * 10^45 possible combinations for 4x4x4 Rubics.
Let's consider 3x3x3 Rubics first.
There are 8 corner cubes, which can be arranged in 8! ways.
Each of these 8 cubes can be turned in 3 different directions, so there are 3^8 orientations altogether. But if you get all but one of the corner cube into chosen positions and orientations, only one of 3 orientations of the final corner cube is possible. Thus, total ways corner cubes can be placed = (8!) * (3^8)/8 = (8!) * (3^7)
Similarly, 12 edge cubes can be arranged in 12! ways.
Each of these 12 cubes can be turned in 2 different directions, so there are 2^12 orientations altogether. But if you get all but one of the edge cube into chosen positions and orientations, only one of 2 orientations of the final edge cube is possible. Thus, total ways edge cubes can be placed = (12!) * (2^12)/2 = (12!) * (2^11)
Here, we have essentially pulled the cubes apart and stuck cubes back in place wherever we please. In reality, we can only move cubes around by turning the faces of the cubes. It turns out that you can't turn the faces in such a way as to switch the positions of two cubes while returning all the others to their original positions. Thus if you get all but two cubes in place, there is only one attainable choice for them (not 2!). Hence, we must divide by 2.
Total different possible combinations are
= [(8!) * (3^7)] * [(12!) * (2^11)] / 2
= (8!) * (3^7) * (12!) * (2^10)
= 4.3252 * 10^19
Similarly, for 4x4x4 Rubics total different possible combinations are
= [(8!) * (3^7)] * [(24!)] * [(24!) / (4!^6)] / 24
= 7.4011968 * 10^45
Note that there are 24 edge cubes, which you can not turn in 2 orientations (hence no 2^24 / 2). Also, there are 4 center cubes per face i.e. (24!) / (4!^6). You can switch 2 cubes without affecting the rest of the combination as 4*4*4 has even dimensions (hence no division by 2). But pattern on one side is rotated in 4 directions over 6 faces, hence divide by 24.
9)
Answer
A tough one!!!
Since R + E + E = 10 + E, it is clear that R + E = 10 and neither R nor E is equal to 0 or 5. This is the only entry point to
solve it. Now use trial-n-error method.
N E V E R 2 1 4 1 9
L E A V E 3 1 5 4 1
+ M E + 6 1
-------- --------
A L O N E 5 3 0 2 1
10)
Cindy is the Singer. Mr. Clinton or Monika is the Dancer.
From (1) and (3), the singer and the dancer, both can not be a man. From (3) and (4), if the singer is a man, then the dancer must be a man. Hence, the singer must be a woman.
CASE I : Singer is a woman and Dancer is also a woman
Then, the dancer is Monika and the singer is Cindy.
CASE II : Singer is a woman and Dancer is also a man
Then, the dancer is Mr. Clinton and the singer is Cindy.
In both the cases, we know that Cindy is the Singer. And either Mr. Clinton or Monika is the Dancer.
11)
There are total 450 rooms.
Out of which 299 room number starts with either 1, 2 or 3. (as room number 100 is not there) Now out of those 299 rooms only 90 room numbers end with 4, 5 or 6
So the probability is 90/450 i.e. 1/5 or 0.20
Draw 9 dots on a page, in the shape of three rows of three dots to form a square. Now place your pen on the page, draw 4 straight lines and try and cover all the dots.
You're not allowed to lift your pen.
Note: Don't be confined by the dimensions of the square.
12)
One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Backtracing.
It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X & Y as many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48)
Similarly, before Y gave tractors to X & Z, they had 6 & 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24)
Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)
Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.
13)
None
Frank gave you a pear in exchange of the apple which Marge gave you. And you gave that pear to Bret in exchange for an apple. All the others exchanges involved apples and/or organges.
Four couples are going to the movie. Each row holds eight seats. Betty and Jim don't want to sit next to Alice and Tom. Alice and Tom don't want to sit next to Gertrude and Bill. On the otherhand, Sally and Bob don't want to sit next to Betty and Jim.
14)
Let before adding X stations, total number of tickets
t = N(N-1)
After adding X stations total number of tickets are
t + 46 = (N+X)(N+X-1)
Subtracting 1st from 2nd
46 = (N+X)(N+X-1) - N(N-1)
46 = N2 + NX - N + NX + X2 - X - N2 + N
46 = 2NX + X2 - X
46 = (2N - 1)X + X2
X2 + (2N - 1)X - 46 = 0
Now there are only two possible factors of 46. They are (46,1) and (23,2)
Case I: (46,1)
2N - 1 = 45
2N = 46
N = 23
And X = 1
Case II: (23,2)
2N - 1 = 21
2N = 22
N = 11
And X = 2
Hence, there are 2 possible answers.
15)
75 miles per hour
While going to the destination, the vehicle travels 10 mils at the speed of 50 miles per hour. So the time taken to travel 10 miles is
= (60 * 10) / 50
= 12 minutes
Now it's given that round-trip travel time is 20 minutes. So the vehicle should complete its return trip of 10 miles in 8 minutes. So the speed of the vehicle must
= (60 * 10) / 8
= 75 miles per hour
16)
70 students are majoring in both, psychology & business
If 73% of the students are psychology majors, we know that 27% are not psychology majors. By the same reasoning, 38% are not business majors, because 62% of the students do major in business. So: 27 + 38 = 65
65% of the students are not majoring in both psychology & business, so 35% are double majors, a total of 70 students.
17)
Initially, there were 42 or 36 matches in the match-box.
There are 42 matches in the box with which he could form a triangle 20, 15, 7, with an area of 42 square inches. After 6 matches had been used, the remaining 36 matches would form a triangle 17, 10, 9, with an area of 36 square inches. After using another 6 matches, the remaining 30 matches would form a triangle 13, 12, 5, with an area of 30 square inches. After using another 6, the 24 remaining would form a triangle 10, 8, 6, with an area of 24 square inches.
Thus, there are two possible answers. There were either 42 or 36 matches in the match-box.
18)
Mr. Grey is wearing Black suit.
Mr. White is wearing Grey suit.
Mr. Black is wearing White suit.
Mr. Grey must not be wearing grey suit as that is the same colour as his name. Also, he was not wearing white suit as the person wearing white suit responded to his comment. So Mr Grey must be wearing a black suit.
Similarly, Mr. White must be wearing either black suit or grey suit. But Mr. Grey is wearing a black suit. Hence, Mr. White must be wearing a grey suit.
And, Mr. Black must be wearing white suit
******************************END OF ANSWERS FOR FIRST PRACTISE EXERCISE***********************************
thanks,
ALL THE BEST
Practise Puzzles - 2
21) If you take a marker & start from a corner on a cube, what is the maximum number of edges you can trace across if you never trace across the same edge twice, never remove the marker from the cube, & never trace anywhere on the cube, except for the corners & edges?
22) One of Mr. Bajaj, his wife, their son and Mr. Bajaj's mother is an Engineer and another is a Doctor.
If the Doctor is a male, then the Engineer is a male.
If the Engineer is younger than the Doctor, then the Engineer and the Doctor are not blood relatives.
If the Engineer is a female, then she and the Doctor are blood relatives.
Can you tell who is the Doctor and the Engineer?
23)Three men - Sam, Cam and Laurie - are married to Carrie, Billy and Tina, but not necessarily in the same order.
Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge. No wife partners her husband and Cam does not play bridge.
Who is married to Cam?
24) There are 3 persons X, Y and Z. On some day, X lent tractors to Y and Z as many as they had. After a month Y gave as many tractors to X and Z as many as they have. After a month Z did the same thing. At the end of this transaction each one of them had 24.
Find the tractors each originally had?
25) A certain street has 1000 buildings. A sign-maker is contracted to number the houses from 1 to 1000. How many zeroes will he need?
26) There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?
27) In a sports contest there were m medals awarded on n successive days (n > 1).
On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.
On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.
On the nth and last day, the remaining n medals were awarded.
How many days did the contest last, and how many medals were awarded altogether?
28) A number of 9 digits has the following properties:
The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2<=n<=9.
Each digit in the number is different i.e. no digits are repeated.
The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.
Find the number.
<<< Previous
22) One of Mr. Bajaj, his wife, their son and Mr. Bajaj's mother is an Engineer and another is a Doctor.
If the Doctor is a male, then the Engineer is a male.
If the Engineer is younger than the Doctor, then the Engineer and the Doctor are not blood relatives.
If the Engineer is a female, then she and the Doctor are blood relatives.
Can you tell who is the Doctor and the Engineer?
23)Three men - Sam, Cam and Laurie - are married to Carrie, Billy and Tina, but not necessarily in the same order.
Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge. No wife partners her husband and Cam does not play bridge.
Who is married to Cam?
24) There are 3 persons X, Y and Z. On some day, X lent tractors to Y and Z as many as they had. After a month Y gave as many tractors to X and Z as many as they have. After a month Z did the same thing. At the end of this transaction each one of them had 24.
Find the tractors each originally had?
25) A certain street has 1000 buildings. A sign-maker is contracted to number the houses from 1 to 1000. How many zeroes will he need?
26) There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?
27) In a sports contest there were m medals awarded on n successive days (n > 1).
On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.
On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.
On the nth and last day, the remaining n medals were awarded.
How many days did the contest last, and how many medals were awarded altogether?
28) A number of 9 digits has the following properties:
The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2<=n<=9.
Each digit in the number is different i.e. no digits are repeated.
The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.
Find the number.
<<< Previous
Practise Puzzles - 2
11) Assume for a moment that the earth is a perfectly uniform sphere of radius 6400 km. Suppose a thread equal to the length of the circumference of the earth was placed along the equator, and drawn to a tight fit.
Now suppose that the length of the thread is increased by 12 cm, and that it is pulled away uniformly in all directions.
By how many cm. will the thread be separated from the earth's surface?
12) Scientist decided to do a study on the population growth of rabbits. Inside a controlled environment, 1000 rabbits were placed.
Six months later, there were 1000Z rabbits. At the beginning of the 3rd year, there were roughly 2828Z rabbits, which was 4 times what the scientists placed in there at the beginning of the 1st year.
If Z is a positive variable, how many rabbits would be there at the beginning of the 11th year?
13) A man is stranded on a desert island. All he has to drink is a 20oz bottle of sprite.
To conserve his drink he decides that on the first day he will drink one oz and the refill the bottle back up with water. On the 2nd day he will drink 2oz and refill the bottle. On the 3rd day he will drink 3oz and so on...
By the time all the sprite is gone, how much water has he drunk?
14) You have four 9's and you may use any of the (+, -, /, *) as many times as you like. I want to see a mathematical expression which uses the four 9's to = 100
How many such expressions can you make?
15) In a certain year, the number of girls who graduated from City High School was twice the number of boys. If 3/4 of the girls and 5/6 of the boys went to college immediately after graduation, what fraction of the graduates that year went to college immediately after graduation?
16) A mule and a donkey were carrying full sacks on their backs.
The mule started complaining that his load was too heavy. The donkey said to him "Why are you complaining? If you gave me one of your sacks I'd have double what you have and if I give you one of my sacks we'd have an even amount."
How many sacks were each of them carrying? Give the minimal possible answer
17) Two people enter a race in whick you run to a point and back. Person A runs 20 mph to and from the point. Person B runs to the point going 10 mph and 30 mph going back.
Who came in first?
18) Three friends divided some bullets equally. After all of them shot 4 bullets the total number of bullets remaining is equal to the bullets each had after division. Find the original number divided.
19) Find sum of digits of D.
Let A= 1999 (power) 1999
B = sum of digits of A, C = sum of digits of B, D = sum of digits of C. (HINT: A = B = C = D (mod 9))
20) There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.In the mean time the whole platoon has moved ahead by 50m.The question is how much distance did the last person cover in that time. Assuming that he ran the whole distance with uniform speed.
<<< Previous >>>Next
Now suppose that the length of the thread is increased by 12 cm, and that it is pulled away uniformly in all directions.
By how many cm. will the thread be separated from the earth's surface?
12) Scientist decided to do a study on the population growth of rabbits. Inside a controlled environment, 1000 rabbits were placed.
Six months later, there were 1000Z rabbits. At the beginning of the 3rd year, there were roughly 2828Z rabbits, which was 4 times what the scientists placed in there at the beginning of the 1st year.
If Z is a positive variable, how many rabbits would be there at the beginning of the 11th year?
13) A man is stranded on a desert island. All he has to drink is a 20oz bottle of sprite.
To conserve his drink he decides that on the first day he will drink one oz and the refill the bottle back up with water. On the 2nd day he will drink 2oz and refill the bottle. On the 3rd day he will drink 3oz and so on...
By the time all the sprite is gone, how much water has he drunk?
14) You have four 9's and you may use any of the (+, -, /, *) as many times as you like. I want to see a mathematical expression which uses the four 9's to = 100
How many such expressions can you make?
15) In a certain year, the number of girls who graduated from City High School was twice the number of boys. If 3/4 of the girls and 5/6 of the boys went to college immediately after graduation, what fraction of the graduates that year went to college immediately after graduation?
16) A mule and a donkey were carrying full sacks on their backs.
The mule started complaining that his load was too heavy. The donkey said to him "Why are you complaining? If you gave me one of your sacks I'd have double what you have and if I give you one of my sacks we'd have an even amount."
How many sacks were each of them carrying? Give the minimal possible answer
17) Two people enter a race in whick you run to a point and back. Person A runs 20 mph to and from the point. Person B runs to the point going 10 mph and 30 mph going back.
Who came in first?
18) Three friends divided some bullets equally. After all of them shot 4 bullets the total number of bullets remaining is equal to the bullets each had after division. Find the original number divided.
19) Find sum of digits of D.
Let A= 1999 (power) 1999
B = sum of digits of A, C = sum of digits of B, D = sum of digits of C. (HINT: A = B = C = D (mod 9))
20) There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.In the mean time the whole platoon has moved ahead by 50m.The question is how much distance did the last person cover in that time. Assuming that he ran the whole distance with uniform speed.
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Practise Puzzles - 2
1) A 3 digit number is such that it's unit digit is equal to the product of the other two digits which are prime. Also, the difference between it's reverse and itself is 396.
What is the sum of the three digits?
2) Substitute numbers for the letters so that the following mathematical expressions are correct.
ABC DEF GHI
--- = IE --- = IE --- = IE
3 6 9
Note that the same number must be used for the same letter whenever it appears.
3) A, B, C and D are related to each other.
One of the four is the opposite sex from each of the other three.
D is A's brother or only daughter.
A or B is C's only son.
B or C is D's sister.
4) Dr. DoLittle always goes walking to the clinic and takes the same time while going and while coming back. One day he noticed something.
When he left the home, the hour hand and the minute hand were exactly opposite to each other and when he reached the clinic, they were together.
Similarly, when he left the clinic, the hour hand and the minute hand were together and when he reached the home, they were exactly opposite to each other.
How much time does Dr. DoLittle take to reach home from the clinic? Give the minimal possible answer.
5) SlowRun Express runs between Bangalore and Mumbai, For the up as well as the down journey, the train leaves the starting station at 10:00 PM everyday and reaches the destination at 11:30 PM after three days.
Mr. Haani once travelled by SlowRun Express from Mumbai to Bangalore. How many SlowRun Express did he cross during his journey?
6) Thus, Mr. Haani must have crossed 7 SlowRun Expresses during his journey.
Six cabins numbered 1-6 consecutively, are arranged in a row and are separated by thin dividers. These cabins must be assigned to six staff members based on following facts.
Miss Shalaka's work requires her to speak on the phone frequently throughout the day.
Miss Shudha prefers cabin number 5 as 5 is her lucky number.
Mr. Shaan and Mr. Sharma often talk to each other during their work and prefers to have adjacent cabins.
Mr. Sinha, Mr. Shaan and Mr. Solanki all smoke. Miss Shudha is allergic to smoke and must have non-smokers adjacent to her.
Mr. Solanki needs silence during work.
Can you tell the cabin numbers of each of them?
7) SkyFi city is served by 6 subway lines - A, E, I, O, U and Z.
When it snows, morning service on line E is delayed.
When it rains or snows, service on the lines A, U and Z is delayed both morning and afternoon.
When the temperature drops below 20 C, afternoon service is cancelled on either line A or line O, but not both.
When the temperature rises above 40 C, afternoon service is cancelled on either line I or line Z, but not both.
When service on line A is delayed or cancelled, service on line I is also delayed.
When service on line Z is delayed or cancelled, service on line E is also delayed.
On February 10, it snows all day with the temperature at 18C. On how many lines service will be delayed or cancelled, including both morning and afternoon?
SkyFi city is served by 6 subway lines - A, E, I, O, U and Z.
When it snows, morning service on line E is delayed.
When it rains or snows, service on the lines A, U and Z is delayed both morning and afternoon.
When the temperature drops below 20 C, afternoon service is cancelled on either line A or line O, but not both.
When the temperature rises above 40 C, afternoon service is cancelled on either line I or line Z, but not both.
When service on line A is delayed or cancelled, service on line I is also delayed.
When service on line Z is delayed or cancelled, service on line E is also delayed.
On February 10, it snows all day with the temperature at 18C. On how many lines service will be delayed or cancelled, including both morning and afternoon?
In a certain game, if 2 wixsomes are worth 3 changs, and 4 changs are worth 1 plut, then 6 plutes are worth how many wixsomes?
8) A 3 digit number is such that it's unit digit is equal to the product of the other two digits which are prime. Also, the difference between it's reverse and itself is 396.
What is the sum of the three digits?
9) There are 4 mugs placed upturned on the table. Each mug have the same number of marbles and a statement about the number of marbles in it. The statements are: Two or Three, One or Four, Three or One, One or Two.
Only one of the statement is correct. How many marbles are there under each mug?
10) At University of Probability, there are 375 freshmen, 293 sophomores, 187 juniors, & 126 seniors. One student will randomly be chosen to receive an award.
What percent chance is there that it will be a junior? Round to the nearest whole percent.
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What is the sum of the three digits?
2) Substitute numbers for the letters so that the following mathematical expressions are correct.
ABC DEF GHI
--- = IE --- = IE --- = IE
3 6 9
Note that the same number must be used for the same letter whenever it appears.
3) A, B, C and D are related to each other.
One of the four is the opposite sex from each of the other three.
D is A's brother or only daughter.
A or B is C's only son.
B or C is D's sister.
4) Dr. DoLittle always goes walking to the clinic and takes the same time while going and while coming back. One day he noticed something.
When he left the home, the hour hand and the minute hand were exactly opposite to each other and when he reached the clinic, they were together.
Similarly, when he left the clinic, the hour hand and the minute hand were together and when he reached the home, they were exactly opposite to each other.
How much time does Dr. DoLittle take to reach home from the clinic? Give the minimal possible answer.
5) SlowRun Express runs between Bangalore and Mumbai, For the up as well as the down journey, the train leaves the starting station at 10:00 PM everyday and reaches the destination at 11:30 PM after three days.
Mr. Haani once travelled by SlowRun Express from Mumbai to Bangalore. How many SlowRun Express did he cross during his journey?
6) Thus, Mr. Haani must have crossed 7 SlowRun Expresses during his journey.
Six cabins numbered 1-6 consecutively, are arranged in a row and are separated by thin dividers. These cabins must be assigned to six staff members based on following facts.
Miss Shalaka's work requires her to speak on the phone frequently throughout the day.
Miss Shudha prefers cabin number 5 as 5 is her lucky number.
Mr. Shaan and Mr. Sharma often talk to each other during their work and prefers to have adjacent cabins.
Mr. Sinha, Mr. Shaan and Mr. Solanki all smoke. Miss Shudha is allergic to smoke and must have non-smokers adjacent to her.
Mr. Solanki needs silence during work.
Can you tell the cabin numbers of each of them?
7) SkyFi city is served by 6 subway lines - A, E, I, O, U and Z.
When it snows, morning service on line E is delayed.
When it rains or snows, service on the lines A, U and Z is delayed both morning and afternoon.
When the temperature drops below 20 C, afternoon service is cancelled on either line A or line O, but not both.
When the temperature rises above 40 C, afternoon service is cancelled on either line I or line Z, but not both.
When service on line A is delayed or cancelled, service on line I is also delayed.
When service on line Z is delayed or cancelled, service on line E is also delayed.
On February 10, it snows all day with the temperature at 18C. On how many lines service will be delayed or cancelled, including both morning and afternoon?
SkyFi city is served by 6 subway lines - A, E, I, O, U and Z.
When it snows, morning service on line E is delayed.
When it rains or snows, service on the lines A, U and Z is delayed both morning and afternoon.
When the temperature drops below 20 C, afternoon service is cancelled on either line A or line O, but not both.
When the temperature rises above 40 C, afternoon service is cancelled on either line I or line Z, but not both.
When service on line A is delayed or cancelled, service on line I is also delayed.
When service on line Z is delayed or cancelled, service on line E is also delayed.
On February 10, it snows all day with the temperature at 18C. On how many lines service will be delayed or cancelled, including both morning and afternoon?
In a certain game, if 2 wixsomes are worth 3 changs, and 4 changs are worth 1 plut, then 6 plutes are worth how many wixsomes?
8) A 3 digit number is such that it's unit digit is equal to the product of the other two digits which are prime. Also, the difference between it's reverse and itself is 396.
What is the sum of the three digits?
9) There are 4 mugs placed upturned on the table. Each mug have the same number of marbles and a statement about the number of marbles in it. The statements are: Two or Three, One or Four, Three or One, One or Two.
Only one of the statement is correct. How many marbles are there under each mug?
10) At University of Probability, there are 375 freshmen, 293 sophomores, 187 juniors, & 126 seniors. One student will randomly be chosen to receive an award.
What percent chance is there that it will be a junior? Round to the nearest whole percent.
>> NEXT
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