(Puzzles) Challenging Mathematical Puzzles (21)

Puzzles : Challenging Mathematical Puzzles

60. 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?

Answer : -

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


61. There are four people in a room (not including you). Exactly two of these four always tell the truth. The other two always lie.

You have to figure out who is who IN ONLY 2 QUESTIONS. Your questions have to be YES or NO questions and can only be answered by one person. (If you ask the same question to two different people then that counts as two questions). Keep in mind that all four know each other's characteristics whether they lie or not.

What questions would you ask to figure out who is who? Remember that you can ask only 2 questions.

You have 3 baskets, & each one contains exactly 4 balls, each of which is of the same size. Each ball is either red, black, white, or purple, & there is one of each color in each basket.

If you were blindfolded, & lightly shook each basket so that the balls would be randomly distributed, & then took 1 ball from each basket, what chance is there that you would have exactly 2 red balls?


There are 64 different possible outcomes, & in 9 of these, exactly 2 of the balls will be red. There is thus a slightly better than 14% chance [(9/64)*100] that exactly 2 balls will be red.

A much faster way to solve the problem is to look at it this way. There are 3 scenarios where exactly 3 balls are red:

1 2 3 
R R X 
R X R 
X R R 

X is any ball that is not red.

There is a 4.6875% chance that each of these situations will occur.

Take the first one, for example: 25% chance the first ball is red, multiplied by a 25% chance the second ball is red, multiplied by a 75% chance the third ball is not red.

Because there are 3 scenarios where this outcome occurs, you multiply the 4.6875% chance of any one occurring by 3, & you get 14.0625%


62. Consider a state lottery where you get to choose 8 numbers from 1 to 80, no repetition allowed. The Lottery Commission chooses 11 from those 80 numbers, again no repetition. You win the lottery if at least 7 of your numbers are there in the 11 chosen by the Lottery Commission.

What is the probability of winning the lottery?

Answer : -

The probability of winning the lottery is two in one billion i.e. only two person can win from one billion !!!

Let's find out sample space first. The Lottery Commission chooses 11 numbers from the 80. Hence, the 11 numbers from the 80 can be selected in 80C11 ways which is very very high and is equal to 1.04776 * 1013

Now, you have to select 8 numbers from 80 which can be selected in 80C8 ways. But we are interested in only those numbers which are in 11 numbers selected by the Lottery Commission. There are 2 cases.

  • You might select 8 numbers which all are there in 11 numbers choosen by the Lottery Commission. So there are 11C8 ways.
  • Another case is you might select 7 lucky numbers and 1 non-lucky number from the remaining 69 numbers. There are ( 11C7 ) * ( 69C1 ) ways to do that.

So total lucky ways are
= ( 11C8 ) + ( 11C7 ) * ( 69C1 )
= (165) + (330) * (69)
= 165 + 22770
= 22935

Hence, the probability of the winning lottery is
= (Total lucky ways) / (Total Sample space)
= (22935) / ( 1.04776 * 1013)
= 2.1889 * 10-9
i.e. 2 in a billion.