#161 - How Many Chocolates Maths Puzzle

In a certain offer, you can exchange 5 wrappers of chocolate to get an actual chocolate. After utilizing two weeks, a man manages to collect 77 wrappers.

Can you find out how many chocolates will he be able to get in return?

View Answer Discuss

He can get 19 chocolates.

77 wrappers will give him 15 chocolates and with 2 wrappers still left. The wrappers form the 15 chocolates he has just got combined with 2 left out wrappers will give him 3 more chocolates and 2 wrappers. The 3 wrappers from these chocolates with 2 left out wrappers will give him another chocolate.

15 + 3 + 1 = 19

#162 - How Many Chocolates Maths Problem

You have two jars of chocolates labelled as P and Q. If you move one chocolate from P to Q, the number of chocolates on B will become twice the number of chocolates in A. If you move one chocolate from Q to P, the number of chocolates in both the jars will become equal.

Can you find out how many chocolates are there in P and Q respectively?

View Answer Discuss

P has 5 chocolates and Q has 7 chocolates.

#163 - Complete The Equation Puzzle

Below, you can see three complete equations and one incomplete. Based on the three complete ones, can you complete the incomplete one?

5 $ 4 $ 3 $ 9 = 4215
6 $ 9 $ 2 $ 6 = 3816
4 $ 7 $ 3 $ 3 = 1122
7 $ 2 $ 7 $ 4 = ____

View Answer Discuss

2418 is the answer.

The algo that has been used here is

A $ B $ C $ D = [first digit of (A * D)] [last digit of (B * C)] [first digit of (B * C)] [last digit of (A * D)]

Using the same algo,
7 $ 2 $ 7 $ 4 = [first digit of (7 * 4)] [last digit of (2 * 7)] [first digit of (2 * 7)] [last digit of (7 * 4)]
= 2418

#164 - Number Guessing Riddle

Lets play a guessing game. Have selected a number in my mind and you have to find it out. The hint for you is:
The tens digit is 7 more than ones digit and the hundreds digit is 6 less than the tens digit.

Can you find the number now?

View Answer Discuss

The number that you have selected in your mind is 281.

#165 - Find Number Of Days Problem

A few friends decided to go on a vacation to a resort. It was raining heavily and it kept raining for thirteen days.

When it rained in the morning, the afternoon was beautiful and when it rained in the afternoon, the day was blessed with a clear morning.

Overall, the friends experienced eleven nice mornings and twelve nice afternoons. Can you find out the number of days they spent in the vacations ?

View Answer Discuss

18

Let us solve this question by denoting variables.

Nice afternoons = P
Nice mornings = Q
No rain = R

Now, the number of days with rain = P + Q = 13
Days with clear mornings = Q + R = 11
Days with clear afternoons = P + R = 12

If we solve these three equations, we will get:
P = 7
Q = 6
R = 5

The total number of days = P + Q + R = 18

#166 - Hardest Counting Squares Puzzle

A network of 20 x 10 squares is given to you.

Can you calculate how many unique squares and rectangles can be formed combining two or more individual squares ?

View Answer Discuss

11550

Total number of rectangles and squares will be equal to: (Summation of row numbers) * (Summation of column numbers)
= (1 + 2 + 3 +......+ 19 + 20) * (1 + 2 + 3 +......+ 9 + 10)
= 210 * 55
= 11550

#167 - Count Chocolates Maths Problem

A and B have a certain number of chocolates with them. If B gives one chocolate to A, they will have equal number of chocolates. But if A gives one chocolate to B, then A will be left with half the number of chocolates that B has.

Can you find out the number of chocolates they have right now?

View Answer Discuss

Let us say that A has P no. of chocolates and B has Q no. of chocolates right now.
Now according to the question:
P + 1 = Q - 1
=> Q - P = 2 ------ (i)

Also, P - 1 = (Q + 1) / 2
=> 2P - Q = 3 ------ (ii)

Equating (i) and (ii)
P = 5 and Q = 7

Therefore, A has 5 chocolates and B has 7 chocolates right now.

#168 - Hard Tricky Maths Trivia Question

Can you think of three numbers that gives a prime number as their product and also the difference between the second and first is equal to the difference between third and second?

View Answer Discuss

The three numbers are: -3, -1 and 1.
Remember that we never asked for positive numbers.

#169 - Counting Apples Basket Problem

In a basket of apples,
when counted in twos, there was one extra
when counted in threes, there were two extra when counted in fours, there were three extra
when counted in fives, there were four extra
when counted in sixes, there were five extra.

However, if the apples were counted in sevens, no extra apple was left. Can you calculate the minimum number of apples that were present in the basket ?

View Answer Discuss

119

Let the number of apples = X
X / 2 => Remainder = 1
X / 3 => Remainder = 2
X / 4 => Remainder = 3
X / 5 => Remainder = 4
X / 6 => Remainder = 5
X / 7 => Remainder = 0

Therefore, X is divisible by 7.

Whenever X is divided by any number less than 7, the remainder is 1 less than the divisor.
=> X + 1 is the LCM of 2, 3, 4, 5 and 6.

Now LCM of 2, 3, 4, 5 and 6 = 60
But 60 - 1 = 59 is not divisible by 7

60 * 2 = 120
120 - 1 = 119 which is divisible by 7

This means that X + 1 = 120 or X = 119

Number of minimum number of apples in the basket = 119

#170 - Confusing Probability Puzzle

You along with your friend are standing in front of two houses. Each of those houses inhabits a family with two children.

Your friend tells you the below two facts:
1) On your left is a family that has a boy who likes accounts but the other child loves science.
2) On the right is a family with a seven year old boy and a new born baby.

You ask him, "Does either of the family have a girl?"

To this he replies, "I am not quite sure. But can you guess that? If you are right, I will give you $200."

Which family do you think is likely to have a girl ?

View Answer Discuss

1/2

In the house on the left, there are three possibilities:
Younger Older
Girl Boy
Boy Girl
Boy Boy

We cant have a girl-girl option because it has been mentioned that there is at least one boy in that house. Now all these outcomes are equally likely, and we have two events with girls, the chances of having a girl in the left house is 2/3.

In the house on the right, there are two possibilities because we already know that the older child is a boy:
Younger Older
Girl Boy
Boy Boy
Here the chances are 1/2.

Thus you must choose the house on the left for better chances at winning.