If a certain toy store's revenue in November was $2/5$ of

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AAPL: Moderator; Posts: 2599; Joined: Sun Oct 29, 2017 2:08 pm; Followed by:2 members

If a certain toy store's revenue in November was $2/5$ of

by AAPL » Thu Aug 08, 2019 3:24 am

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Official Guide

If a certain toy store's revenue in November was $2/5$ of its revenue in December and its revenue in January was $1/4$ of its revenue in November, then the store's revenue in December was how many times the average (arithmetic mean) of its revenues in November and January?

A. $1/4$
B. $1/2$
C. $2/3$
D. $2$
E. $4$

OA E

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Source: Beat The GMAT — Problem Solving | Thu Aug 08, 2019 3:24 am

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 am; Location: Grand Central / New York; Thanked: 470 times; Followed by:34 members

by Jay@ManhattanReview » Fri Aug 09, 2019 12:51 am

AAPL wrote:Official Guide

If a certain toy store's revenue in November was $2/5$ of its revenue in December and its revenue in January was $1/4$ of its revenue in November, then the store's revenue in December was how many times the average (arithmetic mean) of its revenues in November and January?

A. $1/4$
B. $1/2$
C. $2/3$
D. $2$
E. $4$

OA E

Since we have to deal with fractions and denominators of 2/5 and 1/4 are 5 and 4, let's assume that December revenue = 5*4 = $20;

Thus, November revenue = 2/5 of 20 = $8;

And, January revenue = 1/4 of 8 = $2;

Average of November and January revenues = (8 + 2)/2 = $5

Thus, December revenue ($20) is 20/5 = 4 times of average revenue of November and January ($5).

The correct answer: E

Hope this helps!

-Jay
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deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pm; Followed by:5 members

by deloitte247 » Sat Aug 10, 2019 12:42 am

Let the revenue in November = n
Let the revenue in December = d
Let the revenue in January = j
$$Given\ that;\ n=\frac{2}{5}\cdot d$$
$$J=\frac{1}{4}\cdot\ n==>\ \frac{1}{4}\cdot\frac{2}{5}\cdot d$$
$$J=\frac{2}{20}\cdot d=\frac{1}{10}d$$
$$d=\frac{x\left(n+j\right)}{2}\ \ =>\ from\ the\ question$$
$$d=\frac{x\left(\frac{2}{5}\right)d+\left(\frac{1}{10}\right)d}{2}$$
$$d=\frac{x\left[\left(\frac{2}{5}+\frac{1}{10}\right)d\right]}{2}$$
$$d=\frac{x\left(\frac{5}{10}\right)d}{2}$$
$$2d=x\left(\frac{5}{10}\right)d$$
Divide both sides by d, we have;
$$2\cdot10=5x$$
$$x=\frac{20}{5}=4$$

Therefore, we got option E as the correct answer. Thanks

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8088; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Sun Aug 11, 2019 6:29 pm

AAPL wrote:Official Guide

If a certain toy store's revenue in November was $2/5$ of its revenue in December and its revenue in January was $1/4$ of its revenue in November, then the store's revenue in December was how many times the average (arithmetic mean) of its revenues in November and January?

A. $1/4$
B. $1/2$
C. $2/3$
D. $2$
E. $4$

OA E

We can let n = revenue in November, d = revenue in December, and j = revenue in January. We can create the following equations:

n = (2/5)d

j = (1/4)n

So, j = (1/4)(2/5)d = (1/10)d

We now determine that the average revenue for November and January:

[(2/5)d + (1/10)d]/2

= [(4/10)d + (1/10)d]/2

= (5/10)d/2

= (1/4)d

Since the average revenue for November and January is Â¼ the December revenue, we see that December's revenue is 4 times the average revenue of November and January.

Alternative Solution:

We can let the revenue in November = 40, thus the revenue is December = 40/(2/5) = 40 x 5/2 = 100 and the revenue in January = 1/4 x 40 = 10.

We see that the average of the revenues in November and January is (40 + 10)/2 = 25 and December's revenue, 100, is 4 times the average revenue of November and January.

Answer: E

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