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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?

 Feb 20, 2022
 #1
avatar+2455 
-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.

 Feb 20, 2022
 #3
avatar+2455 
-1

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

BuilderBoi  Feb 20, 2022
 #2
avatar+231 
+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

 Feb 20, 2022

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