## Purvik brings a certain number of sweets in a box to his class on his birthday. He distributes 1 sweet less

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:- Purvik brings a certain number of sweets in a box to his class on his birthday. He distributes 1 sweet less than half the number of sweets in the box in the 1st period. Then in the 2nd period he distributes 2 sweets less than one-third of the remaining and then, in the 3rd period he distributes 3 sweets less than one fourth of the remaining. If there are still 36 sweets left in the box, what was the initial number of sweets in the box ?complete question:-SolutionLet us assume that, the initial number of sweets in the box was

x.so,

→ in 1st period he distributes = 1 sweet less than half the number of sweets in the box = {(x/2) – 1} .

then,

→ sweets left after 1st period = x – {(x/2) – 1} = (x – x/2) + 1 = {(x/2) + 1} .

now,

→ in 2nd period he distributes = 2 sweets less than one-third of the remaining = (1/3){(x/2) + 1} – 2 = (x/6 + 1/3) – 2 = (x/6) + (1/3 – 2) = (x/6) + {(1 – 6)/3} = (x/6) + (-5/3) = {(x/6) – (5/3)}

then,

→ sweets left after 2nd period = {(x/2) + 1} – {(x/6) – (5/3)} = {(x/2) – (x/6)} + {1 + (5/3)} = {(3x-x)/6} + (8/3) = {(2x/6)} + (8/3) = {(x/3) + (8/3)} .

now,

→ in 3rd period he distributes = 3 sweets less than one fourth of the remaining. = (1/4){(x/3) + (8/3)} – 3 = {(x/12) + (2/3)} – 3 = (x/12) + (2/3 – 3) = (x/12) + (2 – 9)/3 = {(x/12) – (7/3)}.

then,

→ sweets left after 3rd period = {(x/3) + (8/3)} – {(x/12) – (7/3)} = {(x/3) – (x/12)} + {(8/3) + (7/3)} = {(4x – x)/12} + (15/3) = (3x/12) + 5 = (x/4) + 5 .

A/q,

→ (x/4) + 5 = 36

→ (x/4) = 36 – 5

→ (x/4) = 31

→ x = 31 * 4

→ x =

124 sweets. (Ans.)Hence, the initial number of sweets in the box was 124.Learn more:-There are some cows, hens and goats in Baburam’s fam His 8 year-old son, Vistinu, is fond of counting One day, he counts…

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