(Puzzles) Challenging Mathematical Puzzles (23)

Puzzles : Challenging Mathematical Puzzles

66. 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 X minutes.
If the tank is empty initially and if all the three pipes A, B and C are opened simultaneously, the tank will NEVER be full. Give the maximal possible value of X.

Answer : -

The maximum possible value of X is 13 minutes 20 seconds.

In one minute,
pipe A can fill 1/30 part of the tank.
pipe B can fill 1/24 part of the tank.

Thus, the net water level increase in one minute is
= 1/30 + 1/24
= 3/40 part of the tank

In order to keep the tank always empty, outlet pipe C should empty at least 3/40 part of the tank in one minute. Thus, pipe C can empty the full tank in 40/3 i.e. 13 minutes 20 seconds.

 

67. 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?

Answer : -

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

 

68. 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)

Answer : -


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 

 

69. At the Party:
1. There were 9 men and children.
2. There were 2 more women than children.
3. 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:
4. There were 4 of one group.
5. There were 6 of one group.
6. 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?

Answer : -

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.

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