Mathematics puzzle questions are often asked in competitive exams and entrance tests. These are mostly arithmetic or algebra puzzles that test you on the basics of time, speed & distance, probability, clocks, etc. Here, we have compiled a variety of mathematics puzzles in three groups: easy, medium and difficult puzzles. Each article contains a set of 10 questions with solutions. In this article, you will come across difficult maths puzzles with answers and explanations.

Solve the given questions and check your mental aptitude:

Q.1. The house number in Raja Karorimal lane starts with natural numbers, i.e. 1,2,3….. Sum of all the house numbers on the left side of Raja's residence in that lane is equal to the sum of all the house on its right side. There are more but less than 3000 houses on the right side of Raja's residence.

Q. What is my house number?

204

6929

288

7014

Suppose there are n houses in the left side and the total number of houses in the lane is m. Then Raja's house will be (n+1). Now n(n+1)/2= m(m+1)/2- (n+1)(n+2)/2 (n+1)2 = m(m+1)/2 There are many pairs of n and m that satisfy the equation, but n=6929 and m= 9801 is the only pair from which we shall get more than 100 but less than 3000 houses on the right side of Raja's house. So answer is (b) option.

Q.2. The house number in Raja Karorimal's lane starts with natural numbers, i.e. 1,2,3….. Sum of all the house numbers on the left side of Raja's house is equal to the sum of all the house on its right side. There are more but less than 3000 houses on the right side of Raja's house.

How many houses are there in the lane?

9885

288

9801

Data inadequate

Suppose there are n houses in the left side and the total number of houses in the lane is m. Then Raja's house will be (n+1). Now n(n+1)/2= m(m+1)/2- (n+1)(n+2)/2 (n+1)2 = m(m+1)/2 There are many pairs of n and m that satisfy the equation, but n=6929 and m= 9801 is the only pair from which we shall get more than 100 but less than 3000 houses on the right side of Raja's house. So answer is C option.

Q.3. If ABC are distinct single digit natural numbers satisfying.

ABC

+ ACB

_______

BCA

Find the Values of A, B, C.

B<= 9 & A <=4

B+C>= C. This is only possible if B is 9 and there is a carry of one from unit place.

So B=9.

2A + 1 = B, So A =4.

C + B = A, So C+9 = 4 ⇒ C=5

Q.4. Seth SS and the Seth JS were partners, in a business for many years. JS's share in the business was three-fifths as large as that of SS. As both were aging. They agreed to take in as a partner SS's son-in-law, Chottu. Chottu agreed to pay them Rs. 8,00,000 with the condition that after he became a partner, everybody should own exactly one-third of the business.

"I suppose the bulk of this amount will go to my father-in-law” said Chottu. "Why will it not be the whole amount?" exclaimed SS. "What do you mean?" retorted JS, "it has to be divided in the proportion of our share of profit."

How should this amount of Rs. 8,00,000 be shared?

As Chottu paid Rs. 8,00,000 to own one-third of the business, the whole business is worth Rs. 24,00,000 out of which Rs. 15,00,000 is SS’s share and the balance of Rs. 9,00,000 that of JS. After Chottu joins in, each is to own exactly one third of the business and so the old partners should with draw their investments over and above Rs. 8,00,000. Thus SS gets Rs. 7,00,000 and Rs. 1,00,000 only is JS's rightful share.

Q.5. "Today my age is equal to the last two digits of my year of birth." said a boy to his grandfather in 1942. To this, his grandfather, with a chuckle, said," it is true in my case also." The boy was trying to figure out how. Will you explain?

To calculate the year of birth of the boy, we divide the last two digits of the year of birth (1942) by 2, that is 42/2 = 21. So he was born in 1921 and he was 21 years old in 1942. We have divided the number of years elapsed in that century by 2. If we take the previous century also into consideration, the number of years elapsed would be 142 which when divided by 2 gives 71. This number 71 must be the last two digits of the year of birth of grandfather. So he was born in 1871 AD and he was 71 years old in 1942 AD.

Q.6. Peter and his son Repeter go around a track formed by an equilateral triangle. Peter's running speed is 5 mph and his walking speed is 1mph. Repeter' s walking and running rates are four times that of his father. Both start together from one vertex of the triangle, the son going clockwise & the father going anti-clockwise. Initially Peter runs and Repeter walks but thereafter, as soon as the father starts walking, the son starts running and both complete the course in 80 minutes.

For how long does Peter run? Where do the two cross each other?

This problem can be solved using elementary Algebra. Assuming that the first phase, during which Peter runs and Repeter walks, lasts ‘x’ minutes, and the second phase ‘y’ minutes, we get two question: x + y = 80 and 5x + y = 4x + 20 y
Solving the above equations we find that ‘x’ equals 38 minutes and ‘y’ equals 2 minutes. Thus Peter runs for 38 minutes. As the ratio of their speeds in the first phase.is 5:4 the two meet along the side opposite the start point after Peter has covered two thirds of that side and Repeter one-third.

Q.7. A diamond worker has been contracting for 15 months with a diamond merchant. The merchant told him that he would be getting one diamond each month from a diamond chain with 15 diamonds . However, he can break only three links in the first month. He cannot break any more links after the first month. Which three links should he break up in order to be able to get one diamond each month?

Break the link between second and third diamond, the link between third and fourth diamond and the link between seventh and eighth diamonds so that there will be 4 parts: one single diamond, 2 diamonds, 4 diamonds and 8 diamonds. He can take the single diamond at first month. He can get the double diamonds and give back the single diamond at second month. He can get the single diamond at third month. He can get the 4 diamonds and give back the single and the double diamonds and so on.

Q.8. If only 2 couples are existing today. Suppose each of these couple give birth to 4 children & the 8 children form 4 couples & each couple in turn give birth to 4 children, & those 16 form 8 couples & give birth to 4 children each & so on, can you tell exactly the number of persons existing after 10 generations, excluding the two initial couples?

We start out with 2 couples, four people or 2^{2} people who increase their progeny as follows: 2^{3}+2^{4}+2^{5}+2^{6}+2^{7}+2^{8}+2^{9}+2^{10}+2^{11}+2^{12}= 8184

Q.9. Out of two coins, one is fair (the probability of getting head is 1/2) and the other is biased with probability of getting a tail 1/3. One of the coins is tossed once, resulting in heads. The other is tossed three times, resulting in two heads. Which coin is more likely to be the biased one?

Let's find the probability of the given outcomes in two different cases: the first coin being the fair one, and the second-the biased one, and vice versa.

If we assume that the first coin is fair, then the probability of the heads is 1/2. The second coin must be the biased one, and the probability of it coming up with 2 heads and 1 tail in three tosses is 3*2/3*2/3*1/3 = 4/9. Note that there are three ways to get 2 heads: HHT, HTH, THH, the probability of each is being 4/27. Thus, the probability of both coins coming up with the given results is 2/9.

If, on the other hand, the first coin is the biased one, and the second coin is fair the probability of them resulting in the combination given in the problem is 3 * 2/3 * 1/2 * 1/2 * 1/2 = 1/4, or 2/8 which is greater than the 2/9. Therefore it is more probable that the first coin is the biased one.

Q.10. A clock gains 2% time during the first week and then loses 2% time during the next one week. If the clock was set right at 12 noon on a Wednesday, what will be the time that the clock will show exactly 14 days from the time it was set right?

The clock gains 2% time during the first week. In a day there are 24 hours and in a week there are 7 days. Therefore, there are 7 * 24 = 168 hours in a week.

If the clock gains 2% time during the first week, then it will show a time which is 2% of 168 hours more than 12 Noon on next Wednesday (end of first week) = 3.36 hours more.

Subsequently, the clock loses 2% during the next week. The second week again has 168 hours and the clock loses 2% time = 2% of 168 hours = 3.36 hours less than the actual time.

As it gained 3.36 hours during the first week and then lost 3.36 hours during the next week, the net result will be a -3.36 + 3.36 = 0 hour net gain in time.

So the clock will show a time, which is exactly 12 Noon on Wednesday two weeks from the time it was set right.