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
Are You Up For the Challenge: 700 Level QuestionsA'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
C
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