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?

Step-by-step explanation: The most logical thing to do is to solve it backwards, that is, from what he had at the end of the third day up till the beginning of the first day.

On the third day he ate one-third and had 8 sweets left over. To determine how many he started with on the third day, let the total on day three be called a. If one-third of a is eaten, then the left over which is two-thirds is 8. That is;

8/a = 2/3

By cross multiplication we now have

8 x 3 = 2a

24/2 = a

a = 12

Let the number of sweets he had on day two be called b. If he ate one-third of b and he had 12 left over, then the two-thirds left over is 12 and we now have;

12/b = 2/3

By cross multiplication we now have

12 x 3 = 2b

36 = 2b

36/2 = b

b = 18

Let the number of sweets he had on day one be called x. If he ate one-third of x and he had 18 left over, then the two-thirds left over is 18, and we now have;

18/x = 2/3

By cross multiplication we now have

18 x 3 = 2x

54 = 2x

x = 27

Therefore Tim was given 27 sweets at the beginning.

adioabiola adioabiola · Answer 2 · 2021-11-30T12:22:49+01:00

The total number of sweets Tim was given in the beginning is 27 sweets

Given:

let

total number of sweets = x

Day 1

Quality eaten = 1/3 of x

= 1/3x

Quantity remaining = x - 1/3x

= (3x - x) / 3

= 2/3x

Day 2:

Quality eaten = 1/3 of 2/3x

= 1/3 × 2/3x

= 2/9x

Quantity remaining = 2/3x - 2/9x

= (6x-2x) / 9

= 4/9x

Day 3:

Quality eaten = 1/3 of 4/9x

= 1/3 × 4/9x

= 4/27x

Quantity remaining = 4/9x - 4/27x

= (12x-4x) / 27

= 8/27x

Quantity left = 8

So,

8/27x = 8

divide both sides by 8/27

x = 8 ÷ 8/27

x = 8 × 27/8

x = 27 sweets