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25 Apr 2019, 01:26
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
80% (01:43) correct
20%
(02:15)
wrong
based on 84
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[GMAT math practice question]
Suppose \(x = \frac{a}{b} (b≠0)\). Which of following is equivalent to \(\frac{(a^2+ab)}{(a^2+b^2)}\)?
\(A. \frac{(x^2+x)}{(x^2+1)}\)
\(B. \frac{1}{(x^2+1)}\)
\(C. 1\)
\(D. x^2+1\)
\(E. x^2+x\)
Suppose \(x = \frac{a}{b} (b≠0)\). Which of following is equivalent to \(\frac{(a^2+ab)}{(a^2+b^2)}\)?
\(A. \frac{(x^2+x)}{(x^2+1)}\)
\(B. \frac{1}{(x^2+1)}\)
\(C. 1\)
\(D. x^2+1\)
\(E. x^2+x\)
25 Apr 2019, 02:39
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Here is how I solved this one in ~3minutes
As I lack the superpowers in algebra, The only thing I've done is \(\frac{a^2}{a^2+b^2}\)+\(\frac{ab}{a^2+b^2}\) ; I realized I cant simplify the stem any further.
I had to put values and see which answer choice satisfies
A: is too complicated, let's begin with simpler answer choice
B: 1/\(\frac{a^2}{b^2}\)+1=\(\frac{b^2}{a^2+b^2}\) I love it, dismiss
C: no, as you can find many values that do not satisfy
D: \(\frac{a^2}{b^2}\)+1; no
E: \(\frac{a^2}{b^2}\)+\(\frac{a}{b}\)=\(\frac{a^2+ab}{b^2}\); nope
I was tempted to smash A, as I have excluded all others, but decided to spend additional time to prove it
\(\frac{a^2}{b^2}\)+\(\frac{a}{b}\) / \(\frac{a^2}{b^2}\)+1=(\(\frac{a^2+ab}{b^2}\))*\(\frac{b^2}{a^2+b^2}\)
IMO
Ans: A
As I lack the superpowers in algebra, The only thing I've done is \(\frac{a^2}{a^2+b^2}\)+\(\frac{ab}{a^2+b^2}\) ; I realized I cant simplify the stem any further.
I had to put values and see which answer choice satisfies
A: is too complicated, let's begin with simpler answer choice
B: 1/\(\frac{a^2}{b^2}\)+1=\(\frac{b^2}{a^2+b^2}\) I love it, dismiss
C: no, as you can find many values that do not satisfy
D: \(\frac{a^2}{b^2}\)+1; no
E: \(\frac{a^2}{b^2}\)+\(\frac{a}{b}\)=\(\frac{a^2+ab}{b^2}\); nope
I was tempted to smash A, as I have excluded all others, but decided to spend additional time to prove it
\(\frac{a^2}{b^2}\)+\(\frac{a}{b}\) / \(\frac{a^2}{b^2}\)+1=(\(\frac{a^2+ab}{b^2}\))*\(\frac{b^2}{a^2+b^2}\)
IMO
Ans: A
