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Alfred saves $X$ dollars more than Barry. If together they saved $Y$ d
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23 Mar 2025, 10:40
23 Mar 2025, 10:40
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Alfred saves $\(X\)$ dollars more than Barry. If together they saved $\(Y\)$ dollars, then how many dollars did Alfred save in terms of $\(X\)$ and $\(Y\)$ ?
(A) $\(\frac{X-Y}{2}\)$
(B) $\(\frac{X}{2}\)$
(C) $\(\frac{Y}{2}\)$
(D) $\(\frac{X+Y}{2}\)$
(E) $\(X+\frac{Y}{2}\)$
(A) $\(\frac{X-Y}{2}\)$
(B) $\(\frac{X}{2}\)$
(C) $\(\frac{Y}{2}\)$
(D) $\(\frac{X+Y}{2}\)$
(E) $\(X+\frac{Y}{2}\)$
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Re: Alfred saves $X$ dollars more than Barry. If together they saved $Y$ d
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26 Mar 2025, 04:00
26 Mar 2025, 04:00
Expert Reply
We know that Alfred saves X dollars more than Barry and together they saved Y dollars; we need to find the amount that Alfred save in terms of X and Y .
Let the amount saved by Alfred and Barry be A \& B respectively.
As we know that Alfred saves X dollars more than Barry and together they saved Y dollars, we get $\(\mathrm{A}=\mathrm{B}+\mathrm{X} \& \mathrm{~A}+\mathrm{B}=\mathrm{Y}\)$
$\(\mathrm{A}=\mathrm{B}+\mathrm{X} \Rightarrow \mathrm{B}=\mathrm{A}-\mathrm{X}\)$ substituting the value of B in terms of $\(\mathrm{A} \& \mathrm{X}\)$ in second equation i.e. in $\(A+B=Y\)$, we get $\(A+(A-X)=Y \Rightarrow A=\frac{X+Y}{2}\)$
Hence the savings of $\(A\)$ in terms of $\(X \& Y\)$ is $\(\frac{X+Y}{2}\)$, so the answer is (D).
Let the amount saved by Alfred and Barry be A \& B respectively.
As we know that Alfred saves X dollars more than Barry and together they saved Y dollars, we get $\(\mathrm{A}=\mathrm{B}+\mathrm{X} \& \mathrm{~A}+\mathrm{B}=\mathrm{Y}\)$
$\(\mathrm{A}=\mathrm{B}+\mathrm{X} \Rightarrow \mathrm{B}=\mathrm{A}-\mathrm{X}\)$ substituting the value of B in terms of $\(\mathrm{A} \& \mathrm{X}\)$ in second equation i.e. in $\(A+B=Y\)$, we get $\(A+(A-X)=Y \Rightarrow A=\frac{X+Y}{2}\)$
Hence the savings of $\(A\)$ in terms of $\(X \& Y\)$ is $\(\frac{X+Y}{2}\)$, so the answer is (D).
gmatclubot
Re: Alfred saves $X$ dollars more than Barry. If together they saved $Y$ d [#permalink]
26 Mar 2025, 04:00
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