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A and B brought a radio each for the same price, $z. A and B sold thei

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New post 07 Nov 2019, 03:04
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A and B brought a radio each for the same price, $z. A and B sold their radio for $x and $y respectively. If x>y>z, A’s profit is what percent greater than B’s?

(1) x – y = 20
(2) x – z = 40


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A and B brought a radio each for the same price, $z. A and B sold thei  [#permalink]

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New post 07 Nov 2019, 21:14
Bunuel wrote:
A and B brought a radio each for the same price, $z. A and B sold their radio for $x and $y respectively. If x>y>z, A’s profit is what percent greater than B’s?

(1) x – y = 20
(2) x – z = 40


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A's profit = \(\frac{x-z}{z}\), and B's profit = \(\frac{y-z}{z}\)

% change of A over B = \(\frac{A's.profit-B's.profit}{b"s.profit}*100...........\frac{\frac{x-z}{z}-\frac{y-z}{z}}{\frac{y-z}{z}}*100=\frac{\frac{x-y}{z}}{\frac{y-z}{z}}*100=\frac{x-y}{y-z}*100\)...

Hence, we require values of x, y and z OR values of x-y and y-z

We have some fractions for A’s and B’s profit. When we substitute it in the formula
\(\frac{\frac{(x-z)}{z}-\frac{(y-z)}{z}}{(\frac{y-z}{z})}=\frac{\frac{(x-z-(y-z))}{z}}{\frac{((y-z)}{z)}}=\frac{(x-z-y+z)}{y-z}=\frac{(x-y)}{(y-z)}\)
Now we know x-y and y-z can be found from the two statements.
Subtract statement I from ii
\((x-z)-(x-y)=40-20=20....y-z=20\)

Statement I and II give us these values when combined
Answer = \(\frac{20}{20}*100=100%\)
y-z can be found by subtracting Statement I from statement II

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Re: A and B brought a radio each for the same price, $z. A and B sold thei  [#permalink]

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New post 25 Nov 2019, 22:17
chetan2u : I have a doubt, the stmt. 2 says "x-z = 40", I'm wondering how was it substituted into "y-z", please help me with this as I'm stuck at this point when I solved it myself.
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A and B brought a radio each for the same price, $z. A and B sold thei  [#permalink]

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New post 29 Nov 2019, 09:18
Not sure if I agree with the answer;

I think it should be a ratio for percent of A/ B. Not percent change formula as we are not dealing with one item and seeing how they change over time.

Also, I set up the equations like this: PROFIT (A)= REVENUE- COST = X-Z / PROFIT (B) = Y-Z

SO (X-Z)/ (Y-Z). This is the equation to use for both statements.

1) SInce Z is the same and based on profit formula can we conclude that x is 20 % more than y. Sufficeint.

2) 40/ (y-z). Not Suff.

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A and B brought a radio each for the same price, $z. A and B sold thei  [#permalink]

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New post 29 Nov 2019, 09:57
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pkky wrote:
chetan2u : I have a doubt, the stmt. 2 says "x-z = 40", I'm wondering how was it substituted into "y-z", please help me with this as I'm stuck at this point when I solved it myself.



Hi

We have some fractions for A’s and B’s profit. When we substitute it in the formula
\(\frac{\frac{(x-z)}{z}-\frac{(y-z)}{z}}{(\frac{y-z}{z})}=\frac{\frac{(x-z-(y-z))}{z}}{\frac{((y-z)}{z)}}=\frac{(x-z-y+z)}{y-z}=\frac{(x-y)}{(y-z)}\)
Now we know x-y and y-z can be found from the two statements.
Subtract statement I from ii
\((x-z)-(x-y)=40-20=20....y-z=20\)
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A and B brought a radio each for the same price, $z. A and B sold thei   [#permalink] 29 Nov 2019, 09:57
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