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TEA
12 days ago
8

Tim was given a large bag of sweets and ate one third of the sweets before stopping as he was feeling sick. The next day he ate

one third of the remaining sweets and the following day he ate one third of the remainder, before counting the sweets he had left which totaled eight. How many sweets was he given in the beginning?
Mathematics
1 answer:
Leona [4.1K]12 days ago
3 0

Answer: Initially, he had 27 sweets.

Step-by-step explanation: The most logical approach is to work backwards from what remained after the third day to the start of the first day.

On the third day, he consumed one-third of his sweets and was left with 8. If we let the total sweets on day three be denoted as a, then one-third of a equals what he ate and the two-thirds left equals 8, giving us:

8/a = 2/3

By cross-multiplying, we find:

8 x 3 = 2a

Therefore, 24 = 2a

This leads to a = 12.

Let the sweets on day two be represented as b. If he consumed one-third of b and was left with 12, we have the same structure; hence:

12/b = 2/3

Cross-multiplying gives:

12 x 3 = 2b

So, 36 = 2b, leading to b = 18.

Denote the number of sweets on day one as x. If one-third of x was eaten and 18 remained, we can set up the equation:

18/x = 2/3

Again, cross-multiplying results in:

18 x 3 = 2x

Which simplifies to 54 = 2x, yielding x = 27.

Thus, Tim received 27 sweets at the start.

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