Paul had 2/5 as many VCDs as Peter.

When **Peter threw away 210 VCDs** that are not working,

the ratio of Peter's VCDs to Paul's VCDs became 10:11.

(a) How many VCDs has Peter at first ?

(b) What fraction of Peter's VCDs was thrown away?

Guest Feb 20, 2022

#1**+1 **

Write it as an equation: \({2x\over5 x-210 } = {10\over11}\)

Cross Multiply: \(22x = 50x-2100\)

Subtract 50x from both sides: \(-28x = -2100\)

Divide by -28: \(x = 75\)

Because the original ratio was: \(2x \over 5x\), Peter originally had \(\color{brown}\boxed{375}\) VCDs.

Because Peter threw away 210, he threw away \({210 \over 375 }= \color{brown}\boxed{14\over25}\) of his VCDs.

BuilderBoi Feb 20, 2022

#3**0 **

I can't delete my answer, but this is answer is incorrect. I misread the problem. 🤦♂️

BuilderBoi
Feb 20, 2022

#2**+1 **

Paul = 2/5 Peter

Paul/Peter = 2/5

2x/(5x - 210) = 11/10

20x = 55x - 2310

35x = 2310

x = 66

(a) Peter = 5x = 330

(b) Peter throw away = 210/330

= 7/11

Slimesewer Feb 20, 2022