Question

Question 7.
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?​

Answer 1

Answer:

124

Step-by-step explanation:

Let 'x' be the total number of sweets purvik had.

Now,

He distributes 1 sweet less than half the number of sweets in the box in the 1st period.Let's assume it as 'a'.

a= a=(x/2)-1

Then in the 2nd period he distributes 2 sweets less than one-third of

the remaining.Let's assume it as 'b'.

b=((1/3)*(x-a))-2

In the 3rd period he distributes 3 sweets less than one fourth of the remaining.Let's assume it as 'c'.

There are still 36 sweets left in the box after all distribution.Therefore,we can conclude that subtracting the sum of a,b,c from 'x' gives us 36.

Therefore, x-(a+b+c)=36

If we substitute the options, we could come up with answer as '124'.

But,this is a very lengthy computation.I solved it using python program.

That's the only answer I could come up with for now. But, If I get any shortcut, then I will post it in the future.

              "Any shortcuts or modifications are appreciated."

Answer 2

Solution :-

Let 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.

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