Mathematical riddles are commonly asked in aptitude tests and other competitive exams. These are based on various concepts of number system, algebra, clocks, probability, etc. Here is a compilation of variety of math riddles in the order of difficulty - easy, medium and difficult. Each article consists of 10 maths number puzzles with answers and explanations. Solve these medium-level maths puzzle questions and check your level of preparation:

Q.1. Santa was copying down his Mathematics Homework the other day, but because he was in a hurry he copied it down incorrectly. They had been learning about multiplication and had five of them to do at home.Santa, however, copied every number down backwards:

Instead of writing 14 x 82, Santa wrote 41 x 28

Instead of writing 13 x 62, Santa wrote 31 x 26

Fortunately for Santa all his answers were still correct!

What were the other three questions?

Instead of writing 14 x 82, boy wrote 41x 28.

Both give the answer 1148 because 1 x 8 is the same as 4 x 2.

Instead of writing 13 x 62, he wrote 31 x 26.

Both give the answer 806 because 1 x 6 is the same as 3 x 2.

We also know that:

1 x 4 is the same as 2 x 2,

1 x 8 is the same as 2 x 4,

and 1 x 6 is the same as 2 x 3.

So the other three questions were 12 x 42, 12 x 84 and 12 x 63.

Q.2. There are 17 beads in a necklace with the largest and most valuable in the middle.

Starting from one end, each successive beads is worth Rs.100 more than the one before (up to the middle one), but starting from the other end each beads is worth Rs.50 more than the one before, up to the big bead. The whole necklace is worth Rs.15,000. What is the value of the middle pearl?

The value of the central pearl is Rs.1200.

The pearl at one end (from which they increased in value by Rs.100): Rs.400.

The pearl at the other end: Rs.800.

Starting from the central pearl, two A.P.’s are formed and sum of those 2 progressions will be Rs.15000. Moving to one side, we have a = x, d =100 and n=9. While on the other side, we have a = x-50, d = -50 and n=8

Q.3. At 8 AM, a clock ticks 8 times. The time between first and last ticks is 35 seconds. How long does it tick at 11 Noon?

It is given that the time between first and last ticks at 8'o clock is 35 seconds.

Total time gaps between first and last ticks at 8'o clock= 7

Now, total time gaps between first and last ticks at 11'o clock = 10

Therefore time taken for 11 ticks = 10 * 5 = 50 seconds .

Q.4. A truck driver, once in a week, leaves his house and drives to the river dock to pick up supplies. At 6:05 PM, one-fifth of the way to the dock, he passes the Temple. At 6:15 PM, one-third of the way, he passes the Amritsari Dhaba. At what time does he reached the dock?

At 6:05 PM, the wagon driver passes the temple, one-fifth of the way to the dock.Also, at 6:15 PM, he passes the Amritsari Dhaba, one-third of the way. Thus, he travels 2/15 (1/3 - 1/5) of the distance in 10 minutes.

At 6:15 PM, he has already travelled 1/3 of the distance. Thus 2/3 of the way is remaining, which can be travelled in = ((2/3) * 10) / (2/15) = 50 minutes.

At 4:15, he was at Amritsari Dhaba and remaining way will take 50 more minutes. Hence, the driver will reach at 7:05 PM to the dock.

Q.5. Sami decided to give Kachoris to her friends on her Birthday. If she gives 3 Kachoris to each friend, one friend will get only 1 Kachoris. Also, if she gives 2 Kachoris to each friends, she will be left with 13 Kachoris. How many Kachoris Sami got on her Birthday? And how many friends are there?

Let's assume that there are total C kachoris and F friends

According to first case, if she gives 3 Kachoris to each friend, one friend will get only 2 Kachoris.

3*(F - 1) + 1 = C

Similarly, if she gives 2 Kachoris to each friend, she will leave with 15 Kachoris

2*F + 13 = C

Solving above 2 equations, F = 15 and C = 43. Hence, Sami got 43 Kachoris and 15 friends

Q.6. In the equation shown, P, Q, R, S, and T are 5 consecutive positive integers. What are they?

P^{2} + Q^{2 }+ R^{2} = S^{2} + T^{2}

Squares of 5 consecutive integers P=10, Q=11, R=12, S=13, T=14
i.e. 10x10 + 11x11 + 12x12 = 13x13 + 14x14

Q.7. Neeti puts into the basket one banana when ordered ‘one’, one kiwi when ordered ‘Two’, one litchi when ordered ‘Three’, and is asked to take out from the basket one banana and kiwi when asked Four.

A sequence of order is given as: 12342231124124213

How many total fruits will be in the basket at the end of the above order sequence?

A. 9

B. 8

C. 11

D. 10

Since no. of fruits will be total no. of 1's,2's and3's and subtracting twice the no. of 4's.

Number of fruits at the end= Number of (1’s + 2’ s + 3’s ) – 2*Number of 4’s = 14 - 6 = 8. Hence option B is the correct option.

Q.8. 3 shots can be fired from an air gun at a time. If the probabilities of the first, second and the last shot hitting the enemy helicopter are 0.6, 0.5 and 0.4, what is the probability that three shots aimed at the enemy helicopter will bring the helicopter down?

The enemy helicopter will be brought down even if one of the three shots hits the helicopter. The opposite of this situation is that none of the three shots hit the helicopter. The probability that none of the 3 shots hit the helicopter is given by:

(1-0.6)*(1-0.5)*(1-0.4) = 0.4*0.5*0.6 = 0.12.

So, the probability that at least one of the 3 shots hit the helicopter = 1 – 0.12 = 0.88

Q.9. The price of a commodity increases by 25% every odd year and reduces by 20% every even year. By how much percentage, the prices would have risen or fallen after exactly 8 years?

Let the price at the beginning of the year is 100.

There will be four pairs of multiplication with 5/4 and 4/5.

Thus, it will remain unchanged. So there will be no change after 8 years.

Q.10. Find the next 2 numbers in the series: 1, 8, 11, 69, 88, 96, 101, 111, 181, ...

The next 2 numbers in the series will be 609 and 619. All numbers are the same when flipped upside-down (known as invertible numbers). Only 1, 6, 8, 9 and 0 represent digitss which when inverted produce the same digit or another digit (as in case of 6 and 9).