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
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

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