# (Puzzles) Challenging Mathematical Puzzles (18)

**Puzzles : Challenging Mathematical Puzzles**

**49. Kate, Demi, Madona, Sharon, Britney and Nicole decided to lunch together in a restaurant. The waiter led them to a round table with six chairs.
How many different ways can they seat? **

**Answer:**There are 120 different possible seating arrangements.

Note that on a round table ABCDEF and BCDEFA is the same.

The first person can sit on any one of the seats. Now, for the second person there are 5 options, for the third person there are 4 options, for the forth person there are 3 options, for the fifth person there are 2 options and for the last person there is just one option.

Thus, total different possible seating arrangements are

= 5 * 4 * 3 * 2 * 1

= 120

**50. 3 blocks are chosen randomly on a chessboard. What is the probability that they are in the same diagonal?**

**Answer : -**

There are total of 64 blocks on a chessboard. So 3 blocks can be chosen out of 64 in 64C3 ways.

So the sample space is = 41664

There are 2 diagonal on chessboard each one having 8 blocks. Consider one of them.

3 blocks out of 8 blocks in diagonal can be chosen in 8C3 ways.

But there are 2 such diagonals, hence favourables = 2 * 8C3 = 2 * 56 = 112

The require probability is

= 112 / 41664

= 1 / 372

= 0.002688

**51. What is the area of the triangle ABC with A(e,p) B(2e,3p) and
C(3e,5p)?
where p = PI (3.141592654)**

**Answer: -**

A tricky ONE.

Given 3 points are colinear. Hence, it is a straight line.

Hence area of triangle is 0.

**52. Silu and Meenu were walking on the road.
Silu said, "I weigh 51 Kgs. How much do you weigh?"
Meenu replied that she wouldn't reveal her weight directly as she is overweight. But she said, "I weigh 29 Kgs plus half of my weight."
How much does Meenu weigh?**

**Answer :-**

Meenu weighs 58 Kgs.

It is given that Meenu weighs 29 Kgs plus half of her own weight. It means that 29 Kgs is the other half. So she weighs 58 Kgs.

Solving mathematically, let's assume that her weight is X Kgs.

X = 29 + X/2

2*X = 58 + X

X = 58 Kgs

**53. Consider the sum: ABC + DEF + GHI = JJJ
If different letters represent different digits, and there are no leading zeros, what does J represent?
Answer : **The value of J must be 9.

Since there are no leading zeros, J must be 7, 8, or 9. (JJJ = ABC + DEF + GHI = 14? + 25? + 36? = 7??)

Now, the remainder left after dividing any number by 9 is the same as the remainder left after dividing the sum of the digits of that number by 9. Also, note that 0 + 1 + ... + 9 has a remainder of 0 after dividing by 9 and JJJ has a remainder of 0, 3, or 6.

The number 9 is the only number from 7, 8 and 9 that leaves a remainder of 0, 3, or 6 if you remove it from the sum 0 + 1 + ... + 9. Hence, it follows that J must be 9.