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?
+2668 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.
+2668 I can't delete my answer, but this is answer is incorrect. I misread the problem. 🤦♂️
+237 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
