He bought chocolates as follows:

10 pieces at Rs. 5 each  = 50/-
2 pieces at Rs. 3 each = 6/-
88 pieces at Rs, 0.5 each = 44/-

Total No. of pieces = 10 + 2 + 88 = 100; Total Amount = 50 + 6 + 44 = 100/-

There is no single solution that fits all. A somewhat generic form could be the N-digit number such that the number of digits equals ((number of digits of the multiple of 9) + 1), wherein the first and last digits are 1 and the intermediate digits are 0s.

e.g. When the multiplier is 9: No. of digits  of multiplier = 1, hence the required number is the two-digit (1 + 1) number with first & last digits as 1, i.e., 11. (9 x 11= 99)

When Multiplier is 18, 27… 99, (No. of digits = 2) the required number is three-digit (2 + 1) number with 1st & last digits as 1 and remaining 0., i.e,, the number 101.

101 x 18 = 1818
101 x 27 = 2727
101 x 36 = 3636 and so on.

101 x 99 = 9999.

When the  multiplier is more than 99: (i.e., 108, 117, 126…) No. of digits = 3, and the required number four-digit (3+1) number 1001.

101 x 108 = 108108
101 x 117 = 117117
101 x 126 = 126126 and so on.

It’s “hidden” among the stripes on the tiger’s body. 🙂

Coffee Break?

CHINA

The question seems faulty. Instead of “Every word…”, it seems that it should be “Every LETTER…” Otherwise, there can be multiple solutions. e.g. the word “SEVEN” may stand for digit 1, “SIX” may stand for digit 2 and “TWENTY” may stand for digit 4, so that the equation becomes 1 + 1 + 2 = 4. Similarly, “SEVEN” may stand for 2, “SIX”, for 3 and “TWENTY” for 7 => 2 + 2 + 3 = 7, and so on.

One possible solution:

(Note: The question does NOT require that the final should be a closed figure.)

That’s one very poor show indeed…! Just three animals, one of each kind…!!

If I already know that all three labels are wrong (before proceeding to check), checking only ONE box should suffice.

There are only TWO possible combinations of labels such that they are all wrong, as shown in the picture below.

In either case, once I open one box, say, Box 1, I can know on which box each label should go.

e.g. Suppose the scenario is Case 1 (150/200/100). I open box 1 (Labelled ‘150’) and sees that it contains Rs. 100 (2 x 50) ,  I know for sure that the box labelled ‘200’ actually contains Rs. 150 and hence the label should go to that box (Box 2). Moreover, since I have already seen that Box 1 contains Rs. 100, I know that I should remove the label that says ‘100’ (which is Box 3) and place it on the Box 1.  Hence solved. Likewise in the second case as well.