1. Puzzles

1) Which Pill is contaminated
You have 5 jars of pills. Each pill weighs 10 gram, except for contaminated pills contained in one jar, where each pill weighs 9 gm. Given a scale, how could you tell which jar had the contaminated pills in just one measurement?

Ans. Take 1 pill from Jar1, 2 from Jar2, 3 from 3, 4 from 4th and 5 from 5th.
Now, if none of the pills were contaminated, total weight would be (1+2+3+4+5)*10 =150.
As one of jar would be contaminated, weight would be less. If its less by 1 gm, jar is jar 1 and so on.

 2) How would you measure 4 quarts
If you had an infinite supply of water and a 5 quart and 3 quart pail, how would you measure exactly 4 quarts?

Ans. follow below
               5quart           3quart
step1:        0                  3
step2:        3                  0
step3         3                  3
step4:        5                  1 (poured water from 3 quart to 5, we will have 1 quart left)
step5         1                  0
step6:        1                  3
step7:        4                  0


3. Pick 2 of same color
You have a bucket of jelly beans. Some are red, some are blue, and some green. With your eyes closed, pick out 2 of a like color. How many do you have to grab to be sure you have 2 of the same

Ans. You have to consider worst case scenario, i.e. what ever you pick are all different. So if you pick 4, it will have at least 2 of same color.


4. Defective Ball
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

Ans. Note the cache here is, we don't know whether defective ball is heavy or light.
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.

5. 12 ounces of water
There are 3 glasses. The biggest one can hold 24 ounces. The medium one can hold 11 ounces and the smallest one can hold 5 ounces. Now you have 24 ounces of soft drink in the largest glass. Can you use just these 3 glasses to make the largest glass contain 12 ounces of soft drink by pouring soft drink from one glass to another?

Ans. Follow these steps
      24_B     11_B   5_B
1)   24          0          0
2)   13         11         0
3)   13         6           5
4)   18         6           0
5)   18         1           5
6)   23         1           0          
7)   23         0           1
8)   12        11          1


continue puzzling your brain with next post