5 jars of pills

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.
1. Mark the jars with numbers 1, 2, 3, 4, and 5.
2. Take 1 pill from jar 1, take 2 pills from jar 2, take 3 pills from jar 3, take 4 pills from jar 4 and take 5 pills from jar 5.
3. Put all of them on the scale at once and take the measurement.
4. Now, subtract the measurment from 150 ( 1*10 + 2*10 + 3*10 + 4*10 + 5*10)
5. The result will give you the jar number which has contaminated pill.

8 Identical Balls Problem

Q. You have 8 balls. One of them is defective and weighs less than others. You have a balance to measure balls against each other. In 2 weighings how do you find the defective one?


A. weigh three balls against another three balls. if both weigh the same , then just weighing the remain two (one against one) will show the lighter ball. if the sets of three do not weigh equal, then weigh any two balls in the lighter set, one against the other . the balance will show if the lighter one is on the balance,if not the remaining one is the lighter one.
8= (3 + 3 ) + 2
(the numbers in the brackets are balls on either side of the balance)
if both are equal, then
2= (1 + 1) done.
else, from the lighter set of 3
3= (1 + 1) + 1 done.

There are 3 baskets. one of them have apples, one has oranges only and the other has mixture of apples and oranges. The labels on their baskets always lie. (i.e. if the label says oranges, you are sure that it doesn’t have oranges only,it could be a mixture) The task is to pick one basket and pick only one fruit from it and then correctly label all the three baskets.

HINT. There are only two combinations of distributions in which ALL the baskets have wrong labels. By picking a fruit from the one labeled MIXTURE, it is possible to tell what the other two baskets have.

in a race u drove 1st lap with 40kmph and in the second lap at what speed u must drive so that ur average speed must be 80kmph.

Its impossible! if u drove the first lap in 40 kmph, its impossible that the 
average speed of both the laps is 80kmph.
for eg. consider one lap distance = 80km.
time req. to cover 1 lap = 80km/40kmph = 2 hrs.

 if the avg. speed is 80kmph, then the total time would have taken = 160kms/80kmph = 2 hrs.
same is the case with any other distance u consider. so the avg to be 80kmph is impossible

Can u make 120 with 5 zeros?

Factorial (factorial (0)+factorial (0)+factorial (0)+factorial (0)+factorial (0)) = 120