Bat and Ball
A bat costs Rs 100 more than a ball. The two together cost Rs 150. What is the cost of the ball?
Ans: It's not Rs 50! Let the cost of the ball be X. Then the bat costs Rs 100 more, so it costs 100+X. The total cost, which is X+(100+X)=100+2X, equals 150, or the cost of the ball is X=(150-100)/2 = Rs 25.

Weigh it!
You have a balance scale and 12 coins, 11 of which are genuine and identical in weight; but one is counterfeit, and is either lighter or heavier than the others. Can you determine, in three weighings with a balance scale, which coin is counterfeit and whether it is heavier or lighter than the rest?
Ans: In order to optimise the result, so that we can have an answer in 4 weighings, we can take 4 of the coins and weigh them against 4 others, leaving 4 over. Let us say coins ABCD are first weighed against coins EFGH. Now there are two possibilities. If they balance, then the culprit is in the remaining four coins, i.e., one of IJKL. If they don't then, any of these eight can be wrong. Let us consider the simpler option first.
Suppose they balance. Then all 8 coins, A,B,...,H, are all genuine. And one of IJKL is counterfeit. In the second round, weigh any three of these--say ABC--against three of the remaining, say IJK. If they balance, the bad coin is L, and by weighing L against any of the remaining good coins (third weighing), we can tell if it is lighter or heavier than the others. So in this case, we are done in 3 weighings.
What if ABC and IJK don't balance? Then one of the I,J,K is a bad coin. Since ABC are good coins, we will also already know whether the bad coin is lighter or heavier. To find out which one it is, weigh I against J (third weighing). If they balance, then K is the bad coin. If they don't balance, use the information on whether the bad coin is lighter or heavier to tell which of I or J is the bad coin, and we are done.
Now we consider the case of the first weighing itself being unbalanced. Then we know that I,J,K,L are good coins. Suppose the initial weighing tilts left. (You can easily work out the case if the initial weighing tilts right). Then either one of ABCD is heavy or one of EFGH is light. To find out, re-weigh the coins by mixing them up. Weigh (say) ABE against CDF. Remember AB and CD were on the same side in the first weighing; EF were on the other side, and GH have been left out.
If they balance, the culprit is G or H. Since ABCD was the heavier one and EFGH were lighter, we also know that G or H are lighter than the other coins which are good ones. In the third weighing, we therefore test G (say) against any one of the genuine I,J,K,L. If they are equal, H is the bad one, otherwise it is G.
Suppose ABE does not balance CDF in the second weighing. Then the problem is with one of these coins. Let us say they tilt left. Since ABCD were heavier than EFGH, there are two options: the bad coin is one of ABCD (and is heavier) or the bad coin is one of EFGH (and is lighter). Since we are weighing ABE against CDF, and it is tilted to the left, we are left with either A or B heavy or F light. To find out which is true, we use the third weighing to put A and F against two genuine coins, say K and L. If they weigh the same, then AF are good coins and B is the bad (heavier one). If AF weigh less than KL, then F is the bad (lighter) coin. If AF weigh more than KL, then A is the bad (heavier) coin! We are now completely done with all possibilities, and in all cases, in three weighings!
Congratulations to the reader who figured out these complex possibilities. This puzzle is an example of one where there is no specific answer, but only a procedure outlined by which the answer can be found with the given constraints (three weighings only, in this case).
Source: Puzzle Book from math.dartmouth.edu