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

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