Dear Reader, below are three puzzle type aptitude questions that you can expect for placement papers of companies like Infosys, Syntel etc.

Question 1

Ramakanth and his son went to a ball shop. There were 37 balls in the shop. Out of these one ball alone was weighing 6 gms whereas the other balls weighed 7 gms. How many minimum weighings are required to find out the 6 gm ball.( You are given a balance with two pans for weighing.)

a) 3 b) 4 c) 5 d) not possible to determine.

Answer : b) 4.

Solution :

Below let us discuss how we can handle the problem assuming three possible situations that can occur.

SITUATION ONE

1. Divide balls into three groups of 12, 12 and 13 balls each.Put one set of 12 balls into one pan and another set of 12 balls in another pan.Assume one pan shows lesser weight than the other.(a pan will go up and the other down).Discard the balls in the pan weighing more and the13 balls. .The ball weighing 6 gms has to be one among the 12 balls that were there in the pan that weighed less.
2.Now divide the 12 balls into 4,4 and 4. Put one set of 4 balls and another set of 4 balls in each pan. In case one of them weighs less –
3.Divide 4 into 2 and 2. Put them in two pans with 2 balls each from this lot. One pan will weight less.
4.Take the balls from the pan weighing less and then weigh one in each pan. The ball in the pan going up is the lesser weighing ball.(6 gm). So, in this case minimum 4 weighings are required.

SITUATION TWO

1. Divide balls into 12, 12 and 13. Put one set of 12 balls into one pan and another set of 12 balls in another pan.Consider both weigh same. NOW the ball weighing 6 gms has to be within 13 balls.
2. Divide 13 balls as 4,4 and 5. Put 4 balls each in the two pans.
3. Assume one pan goes down and the other goes up. Then the lighter ball has to be in the pan going up. Divide 4 into 2 and 2. Put them in two pans with 2 balls each from this lot. One pan will weight less.
4. Take the balls from the pan weighing less and then weigh one in each pan and the ball in the pan going up is the lesser weighing ball.(6 gm). So minimum 4 weighings are required.

SITUATION THREE

1 Divide balls into 12, 12 and 13. Put one set of 12 balls into one pan and another set of 12 balls in another pan. Consider both weigh same. NOW the ball weighing 6 gms has to be within 13 balls.
2. Divide 13 balls into 4,4 and 5. Put 4 balls each in the two pans. Assume both the pans weigh same. Then the lighter ball has to be one among the 5 balls.
3. Divide these 5 balls into 2,2 and 1. Weigh 2 and 2. If they weigh same then lighter ball is the fifth ball.
4. If one pan goes up and the other down take the two balls in the pan going up and then weigh as 1 and 1. Totally 4 weighings are required.

RESULT : In each of the above situations, the weighings required was 4. Hence 4 is the answer.

Question 2

England cricket captain and Indian cricket team captain went to a ball shop in Chennai. There they were told that there are 22 balls and out of them one weighs 15 gms whereas other balls weigh 16 gms. They were given a balance with two pans for weighing. How many minimum weighings are required by them to find out the ball weighing 15 gms.

a) 3 b) 4 c) 2 d) none of these.

Answer : a) 3

Solution :

Below let us discuss how we can handle the problem assuming three possible situations that can occur.

SITUATION ONE

1. Divide the balls into groups of 7, 7 and 8 balls each. Put the 7 balls in one pan and another set of 7 balls in another pan. Assume one pan goes up indicating that there is a ball which weighs 15 gms.(lesser weighing ball)
2. Discard the balls in the pan weighing more. Also discard the group of 8 balls set aside. Divide 7 balls in lesser weighing pan into 2,2 and 3. From these put 2 balls in one pan and another set of 2 balls in another pan.
3. Assume both weigh same. Then the lighter ball has to be one among the 3 balls set aside. So in third weighing put one ball in one pan and another ball from these in another pan. If one pan goes up then the ball in that pan is weighing 15 gms. If they weigh same then the third ball in this lot is the lighter ball.

SITUATION TWO

1. Divide the balls into 7, 7 and 8. Put the 7 balls in one pan and another set of 7 balls in another pan. Assume both them weigh same. Then the lighter ball has to be one among the 8 balls.
2. Divide 8 balls into 3,3 and 2. Weigh 3 balls in one pan and another set of 3 balls in another pan. Imagine they weigh same. Then the lighter ball has to be one among the two.
3. Do a third weighing putting one in each pan. The ball in the pan going up is the lighter ball.

SITUATION THREE

1. Divide the balls into 7, 7 and 8. Put the 7 balls in one pan and another set of 7 balls in another pan. Assume one pan goes up. Then the lighter ball has to be one among the balls in that pan.
2. Divide 7 balls into 2,2 and 3. Weigh 2 balls in one pan and 2 balls in another pan. Assume they weigh the same.
3. The lighter ball has to be one among the 3. Weigh 1 ball in one pan and another 1 ball in another pan. The ball in the pan going up is lighter one. If they weigh same then the third ball among these is the lighter ball.

RESULT : In each of the above situations, the weighings required was 3. Hence 3 is the answer.

Question 3

Ramesh Khanna wanted to buy 87 balls of the same weight – say 12 gms each. In a shop that he visited they were showing 88 balls. Out of them one weighed 14 gms and all others weighed 12 gms. How many minimum weighings are required to find out the heavier ball?

a) 4 b) 3 c) 5 d) none of these.

Answer : c) 5.

Solution :

Now in this problem we are going to use a shortcut which we haven't used in the above two problems. We could had said this earlier, but knowing the formula after knowing the actual solution would make you more knowledgeable and equipped to solve these kinds of problems.
If the number of objects (balls in our example) is > 3^n-1 and <= 3^n, then n weighings would be required to find the one odd object (ball) that will be weighing lesser or heavier than the rest.

Based on the above formula, for determining lighter or heavier one ball among 9 balls (3^2 balls) – 2 weighings are required. For determining lighter or heavier one ball from 10 to 27 balls (3^3 balls) – 3 weighings are required. From 28 to 81 – 4 weighings are required. From 82 to 243 balls 5 weighings are required. (follow the same principle of dividing into three parts.)