Each of four boys P, Q, R and S, had a few chocolates with him. P first gave 33.33% of the chocolates with him to Q, Q gave 25% of what he then had to R and R gave 20% of what he then had to S. Finally, all the four boys had an equal number of chocolates. If P initially had 160 chocolates more than the number of chocolates that Q had initially. Then, find the difference between the numbers of chocolates that R and S initially had?

Each of four boys P, Q, R and S, had a few chocolates with him. P first gave 33.33% of the chocolates with him to Q, Q gave 25% of what he then had to R and R gave 20% of what he then had to S. Finally, all the four boys had an equal number of chocolates. If P initially had 160 chocolates more than the number of chocolates that Q had initially. Then, find the difference between the numbers of chocolates that R and S initially had? Correct Answer 40

CALCULATION   : 

Let’s calculate this from the last.

According to question –

Finally, all the four boys had an equal number of chocolates.

So after ‘R’ gave 20% = (1/5) of her share to S he is left with ‘x’.

(R – R/5) = x

⇒ R = 5x/4

⇒ Before R gave to S, he had 5x/4.

So, this is the situation before R gave to S or after Q gave to R.

So, Q is left with ‘x’ after she gave 25% = (1/4^th) of his chocolates to R.

(Q – Q/4) = x

⇒ Q = 4x/3

∴ Before Q gave to R, he had 4x/3.

This is the situation after P gave 33.33% = (1/3)^rd of his share to Q.

(P –P/3) = x

⇒ P = 3x/2

∴ Initially ‘P’ had 3x/2.

Given, P – Q = 160

⇒ (3x/2 – 5x/6) = 160

⇒ (9x – 5x)/6 = 160

⇒ 4x/6 = 160

⇒ x = (160 × 6)/4

⇒ x = 240

∴ The difference between the numbers of chocolates that R and S initially had = (11x/12 – 3x/4) = (11x – 9x)/12 = 2x/12 = x/6

⇒ 240/6

⇒ 40