3 boys shared a box of chocolate. The first boy got 3/8 of the total and the 2nd boy got 2/5 of the remainder and the 3rd boy got 9. How many chocolates were in the box and how many did the boy with the smallest share get?​

1. there were 24 chocolates in the box at first.

2. the boy with the smallest share received 6 chocolates.

Explanation:

let's represent the number of chocolates in the box with 8x.

the first boy got ⅜ × 8x  = 3x chocolates. now, there are 8x - 3x = 5x chocolates remaining.

the second boy got ⅖ × 5x = 2x chocolates. now, there are 5x - 2x = 3x chocolates remaining.

3x = 9 chocolates

x = 3 chocolates

since the original number of chocolates was represented by 8x, we now know that there were 8 × 3 = 24 chocolates in the box at first.

the first boy got 9 chocolates, the second boy got 6 chocolates and the last boy got 9 chocolates. thus, the boy with the smallest share got 6 chocolates.

i hope this helps! :D

answer: first option i.e. 'a' is correct.

step-by-step explanation:

since we have given that

number of minutes to wrap a small gift box = 4 minutes

we need to find the number of gifts to be wrapped in 5 minutes

since there is a direct variation, as the number of minutes will increase with   number of gifts to be wrapped.

so, let the number of gifts to be wrapped in 5 minutes be x.

so, it becomes,

hence, first option 'a'   is correct.

hmu if ur willing to do a algebra a online class i’ll pay 600 hmu on the gram profitpreme