# Math riddle?

A kid has to buy a toy of 100 Rupees, he takes Rs.50 from his father and 50 from his mother. Later when he goes to the store he finds that the toy is only 97 rupees, he buys the toy and 3 rupees are left, still with me? Now he goes to his mother and father and out of those 3 rupees gives one rupee to his father, one rupee to his mother and keeps the remaining 1 rupee aside (not for himself).

Now here is the question, as he has already given 1 rupee to his mother and father each, he has remaining 49 rupees to pay each one, so, my friend asks that add 49 to 49, 98, and then add to it the remaining 1 rupee he had kept aside, 99. Where did the remaining 1 rupee go?

I just can't understand what is he trying to ask, the calculation he just did, I think, has nothing to do with amount the kid has to pay or paid.

active| newest | oldestLet us see it from the kid's point of view:

+ 50 (from father)

+ 50 (from mother)

- 97 (for toy)

- 1 (father)

- 1 (mother)

total: 1 rupee he kept aside.

What the riddle is trying to do is:

+ 49 (50-1 from father)

+ 49 (50-1 from mother)

+ 1 he kept

As you can see two of the three loose rupees are considered negative but one is considered positive. This is where the riddle tries to trick you. If you had consider all three rupees negative you end up with 97, which is what the toy really costs.

Lets solve this riddle in reverse direction...

You kept 1$. It means you have to 99$, and you bought the shirt for 97$ it means you have left with 2$ (and from those 2$ you returned 1$ to your mother and 1$ to your father).

Cheers

Start on one hand and count down, 10,9,8,7,6 then go to the other hand and count back up, 7,8,9,10,11 so suddenly you have 11 fingers.