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\(\dfrac{2\frac35-1\frac23}{\frac23-\frac35}\)

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Re: \(\dfrac{2\frac35-1\frac23}{\frac23-\frac35}\)

by deloitte247 » Sat Sep 26, 2020 6:18 pm
$$\frac{2\frac{3}{5}-1\frac{2}{3}}{\frac{2}{3}-\frac{3}{5}}=\frac{\frac{13}{5}-\frac{5}{3}}{\frac{2}{3}-\frac{3}{5}}$$
$$=\frac{\frac{39-25}{15}}{\frac{10-9}{15}}$$
$$=\frac{\frac{14}{15}}{\frac{1}{15}}$$
$$=\frac{14}{15}\cdot\frac{15}{1}=14$$
Answer = option B
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Re: \(\dfrac{2\frac35-1\frac23}{\frac23-\frac35}\)

by Scott@TargetTestPrep » Thu Oct 01, 2020 10:01 am
Gmat_mission wrote: ↑
Thu Sep 24, 2020 2:24 am
 \(\dfrac{2\frac35-1\frac23}{\frac23-\frac35}\)

(A) 16
(B) 14
(C) 3
(D) 1
(E) -1

Answer: B

Source: Official Guide
Solution:

We want to rid both the numerator and denominator of fractions. Multiplying the expression by 15/15, we have:

[(30 + 9) - (15 + 10)] / (10 - 9) = (39 - 25)/1 = 14

Answer: B

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