\[ If \frac{a}{b}=-\frac{2}{3},\ then \frac{b-a}{a} \]

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VJesus12: Legendary Member; Posts: 2276; Joined: Sat Oct 14, 2017 6:10 am; Followed by:3 members

\[ If \frac{a}{b}=-\frac{2}{3},\ then \frac{b-a}{a} \]

by VJesus12 » Mon May 21, 2018 7:52 am

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$$If\ \ \ \frac{a}{b}=-\frac{2}{3},\ then\ \ \frac{b-a}{a}=\ ?$$ (A) âˆ’ 5/2
(B) âˆ’ 5/3
(C) âˆ’ 1/3
(D) 0
(E) 7

The OA is the option A .

Could someone explain this PS question to me? I don't know how to solve it without finding the values of a and b.

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Source: Beat The GMAT — Problem Solving | Mon May 21, 2018 7:52 am

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Sionainn@PrincetonReview: GMAT Instructor; Posts: 41; Joined: Mon Mar 12, 2018 9:54 am; Followed by:1 members

by Sionainn@PrincetonReview » Mon May 21, 2018 10:28 am

This is a great one for plugging in values. While technically a and b can be a range of values since the first equation just gives you the ratio of a to b, you can plug in one possible value that makes the equation true. For example if a = -2 and b = 3 the equation is true. Then plug these values into the expression (b - a)/a = (3-(-2))/-2 = 5/-2= -5/2. So the answer is choice A.

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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pm; Thanked: 6 times; Followed by:5 members

by Vincen » Mon May 21, 2018 11:58 pm

VJesus12 wrote:$$If\ \ \ \frac{a}{b}=-\frac{2}{3},\ then\ \ \frac{b-a}{a}=\ ?$$ (A) âˆ’ 5/2
(B) âˆ’ 5/3
(C) âˆ’ 1/3
(D) 0
(E) 7

The OA is the option A .

Could someone explain this PS question to me? I don't know how to solve it without finding the values of a and b.

Hello Vjesus12.

It is not necessary to find the values of "a" and "b". Let's see how can we solve it.

We just have to rewrite the given expression and use that a/b = -2/3 as follows: $$\frac{b-a}{a}=\frac{b}{a}-\frac{a}{a}=\frac{1}{\frac{a}{b}}-1=\frac{1}{-\frac{2}{3}}-1=-\frac{3}{2}-1=-\frac{5}{2}.$$ Hence, we get the that the correct answer is the option A.

I hope it can help you.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members