Among 12 identical looking golf balls there is one that is defective in weight. It is either heavier or lighter than the standard one. You have a balance. You can only weigh 3 times to find out which one is defective and whether it is heavier or lighter.
- 5 Answers
- 13774 Views
- I also Faced
- E-Mail Answers
Answers were Sorted based on User's Feedback
Answer / swati
In order to explain better, give each ball a number, said from 1 to 12. We also use ball zero to indicate a standard ball if that ball is identified as a standard ball.
Step 1. Take 8 balls to weight, ball 1, 2, 3, and 4 on the left, ball 5, 6, 7, and 8 on the right.
If they are balanced, go to Step 2-1.
If they are not balanced, go to Step 2-2.
Step 2-1.
At this time, you know the ball 1 to 8 are OK. Take ball 9, 10 and 11, plus a standard ball - ball zero. Just assume you have ball zero and 9 on the left and ball 10 and ball 11 on the right.
If they are balanced, go to Step 3-1.
If left side is heavier, go to step 3-2.
If right side is heavier, go to Step 3-3.
Step 2-2.
At this time, you know the ball 9 to ball 12 are standard balls. Assume the left side is heavier. Now you need to take 6 balls to weight the second time. You need to choose a standard ball - ball zero, ball 6 which might be lighter, and ball 3 which might be heavier, put them on the left side. Also you choose ball 2 which might be heavier, ball 5 and 7 which might be lighter. Notice that we change sides for ball 2 and ball 6.
If they are balanced, go to step 3-4.
If the left side is heavier, go to step 3-5.
If the right side is heavier, go to step 3-6.
Step 3-1.
From step 2-1, we know ball 9, 10 and 11 are standard balls. We just need to determine the ball 12 is heavier or lighter. We put a standard ball on the left, put ball 12 on the right.
If right side is heavier, the ball 12 is heavier.
If the right side is lighter, the ball 12 is lighter
Step 3-2.
From step 2-1, you know that either ball 9 is heavier, or one of statements should be true: the ball 10 is lighter or the ball 11 is lighter. We can put the ball 10 on the left, the ball 11 on the right.
If they are balanced, they should be OK, the ball 9 is heavier.
If the ball 10 is heavier, the ball 10 is Ok, ball 11 is lighter.
If the ball 10 is lighter, then the ball 11 is OK.
Step 3-3.
From Step 2-1, you know that either ball 9 is lighter, or one of statements should be true: the ball 10 is heavier or the ball 11 is heavier. We can put the ball 10 on the left, the ball 11 on the right.
If they are balanced, they should be OK, the ball 9 is lighter.
If the ball 10 is lighter, the ball 10 is Ok, ball 11 is Heavier.
If the ball 10 is heavier, then the ball 11 is OK.
Step 3-4.
From Step 2-2, you know ball 2, 3, 5, 6, 7 are standard balls. You can determine the ball 1 or ball 4 is heavier, or ball 8 is lighter. We can put ball 1 on the left and the ball 4 on the right.
If they are balanced, we are sure the ball 8 is lighter.
If the ball 1 is lighter, the ball 1 is Ok and ball 4 is heavier.
If the ball 4 is lighter, the ball 4 is OK, the ball 1 is heavier.
Step 3-5.
From Step 2-2, you know ball 2 and ball 6 are OK, because the change side does not affect the consequence. You can determine the ball 3 is heavier, or one of the statements is true: ball 5 is lighter or the ball 7 is lighter. We can put ball 5 on the left and the ball 7 on the right.
If they are balanced, we are sure the ball 3 is heavier.
If the ball 5 is heavier, the ball 5 is Ok and ball 7 is lighter.
If the ball 7 is heavier, the ball 7 is OK, the ball 5 is lighter.
Step 3-6.
From step 2-2, you know that the weighting result is changed because the ball 2 and ball 6 are exchanged. It really means that either ball 2 is heavier or ball 6 is lighter. You still have one more weight, I am sure you know how to do it!
| Is This Answer Correct ? | 4 Yes | 3 No |
Answer / amit
Make 4 groups of 3 balls each (G1, G2, G3, G4. Picturing a decision tree structure may help as I explain below.
1st Weighing: Place G1 and G2
If Equal then: defective ball in G3 or G4 (Consequently G1 & G2 contain balls all having the same weight i.e. normal balls or control group)
If Not Equal then: defective ball in G1 or G2 (Consequently G3 & G4 contain balls all having the same weight i.e. normal balls or control group)
Assume 1st Weighing is Equal.
2nd Weighing: Place G3 (test) and G1 (control)on pans
If Equal then: defective ball is in G4
If Not Equal then: defective ball in G3. Now if G3 is heavier (pan goes down) that means the abnormal ball is heavier than all others. If G3 is lighter (pan goes up) then that means the abnormal ball is lighter than all others.
Assume 2nd Weighing is Not Equal and G3 group is heavier.
Now we name the 3 balls in G3 group as G3-1, G3-2, G3-3.
3rd Weighing: Place G3-1 and G3-2 on pans.
If Equal then: The G3-3 is the defective (and heavier ball).
If Not Equal then: Then ball in the pan that goes down (i.e. heavier ball) is the defective one.
So 1st Weighing - Narrowing the test group
2nd Weighing - Determining heavy/light status
3rd Weighing - Determining abnormal ball
At any place above depending upon the outcome (Equal / Not Equal) you will have to change the Group in the next weighing accordingly. (E.g. Assume 2nd Weighing was equal then defective ball in G4 - so 3rd weighing you follow same procedure described above with the G4 balls).
I was stumped when I first came across this. And I found no perfect answers online. But some attempts by others helped me figure out part of the solution. Finally I found the missing part (2nd weighing finding heavy/light is crucial and 3rd Weighing with individual balls).
Since many people may be struggling with this - I thought I should help by posting the solution. Test it and vote up if its right!
Cheers!
| Is This Answer Correct ? | 3 Yes | 4 No |
1. take and put 6 ball on each side of the scale
2. One side will be heavier Try #1
3. Take the heavier side and put 3 balls on each side of
side of the scale. one side is heavier Try #2
4. you now have 3 balls. Take and place one ball on each
side of the scale. If one is the heavier ball the scale
will show it. If the scale measures the same then to ball
in your hand is the heavier. Try #3
| Is This Answer Correct ? | 4 Yes | 7 No |
Answer / anand
first take 3 balls on one side of the scale and another 3
balls on the other side of the scale. If they don’t weigh
equal then the defective ball is here in this bunch.if they
weigh equal then defective ball is in the set of other 6
balls.Thus we have 6 balls in which we need to find out the
defective one.
now the bunch of other six balls are equal weighted..take 3
balls of from them and weigh them with the bunch of 3 balls
of the above weighted, if they are equal then defective ball
is in the other bunch of 3 balls and if they are not equal
defective ball is in this 3 ball set.(if this set weighs
heavier than the ideal group of 3 balls {that we have
selected from the remaining set of 6 balls} than defective
ball is heavier and vice versa.
now we have 3 balls..take any 2 and weigh them with the
other equal set of 2 balls..if they weigh equal then the
remaining ball is defective..and if they don’t weigh equal
then the ball which is heavier(as we have assumed the set if
3 is heavier than ideal set in the second step.) is
defective in the scale..
| Is This Answer Correct ? | 9 Yes | 20 No |
Answer / neelesh patil
First take 6 balls on one side of the scale and another 6
balls on the other side of the scale. The defective ball will be there in one of this bunch. Consider the bunch which is Imbalanced. Again weight that 6 balls in the Bunch of 3 balls on one side of the scale and another 3 balls on the other side of the scale. Again Considering the imbalanced weight.weight the ball Single-Single if it is balanced then defective is remaining ball.
in these way we can complete in 3 steps.
| Is This Answer Correct ? | 4 Yes | 22 No |
Implement a multiple-reader-single-writer lock given a compare-and-swap instruction. Readers cannot overtake waiting writers.
How will you differentiate a 5 rupee(indian currency)or any five currency coin?Answer me how will you differentiate????? This question is asked in a college appitute test.Any one know help me plzzzzz!!!!!!!!!!!!!
If you started a business in which you earned Rs.1 on the first day, Rs.3 on the second day, Rs.5 on the third day, Rs.7 on the fourth day, & so on. How much would you have earned with this business after 50 years (assuming there are exactly 365 days in every year)?
A, B and C are three points on a straight line, not necessarily equidistant with B being between A and C. Three semicircles are drawn on the same side of the line with AB, BC and AC as the diameters. BD is perpendicular to the line ABC, and D lies on the semicircle AC. If the funny shaped diagram between the three semicircles has an area of 1000 square cms, find the length of BD.
500 men are arranged in an array of 10 rows and 50 columns according to their heights. Tallest among each row of all are asked to come out. And the shortest among them is A. Similarly after resuming them to their original positions, the shortest among each column are asked to come out. And the tallest among them is B. Now who is taller A or B ?
There are 2 items combined cost is 110.. one item cost is 100 rupees more than the cost of other. So tell me indiviual cost of items......
What is the smallest number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, when divided by 8 leaves a remainder of 7, when divided by 7 leaves a remainder of 6 and so on until when divided by 2 leaves a remainder of 1?
Once a week a wagon driver leaves his hut and drives his wagon to the river dock to pick up supplies for his town. At 4:05 PM, one-fifth of the way to the dock, he passes the Temple. At 4:15 PM, one-third of the way, he passes the Preetam-Da-Dhabaa. At what time does he reached the dock?
Jim lies a lot. He tells the truth on only one day in a week. One day he said: "I lie on Mondays and Tuesdays." The next day he said: "Today is either Sunday, Saturday or Thursday." The next day he said: "I lie on Fridays and Wednesdays." On which day of the week does Jim tell the truth?
If A * [(B + C)(D - E) - F(G*H) ] / J = 10 What number is ABCDEFGHJ where each letter is a digit?
A soldier looses his way in a thick jungle. At random he walks from his camp but mathematically in an interesting fashion. First he walks one mile East then half mile to North. Then 1/4 mile to West, then 1/8 mile to South and so on making a loop. Finally how far he is from his camp and in which direction?
Given an N × N array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array. Input Format: First line contains the size of matrix. Followed by n lines and each line contain n integers separated by space. Output format: Single integer which represents maximum sum of rectangle. Sample Input: 4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2 Sample Output: 15
English 
