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20 Jan 2019, 23:45
1
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Question Stats:

80% (01:32) correct 20% (00:00) wrong based on 9 sessions

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In 2012, a store’s average monthly revenue was $20,000. For the first 10 months of the year, the store’s average monthly revenue was$18,000. If the revenue in December was 50% greater than the revenue in November, what was the store’s revenue in November?

A 24,000
B 30,000
C 36,000
D 48,000
E 56,000

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Joined: 31 Oct 2013
Posts: 1157
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)

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21 Jan 2019, 06:28
Bunuel wrote:
In 2012, a store’s average monthly revenue was $20,000. For the first 10 months of the year, the store’s average monthly revenue was$18,000. If the revenue in December was 50% greater than the revenue in November, what was the store’s revenue in November?

A 24,000
B 30,000
C 36,000
D 48,000
E 56,000

Total Revenue for the Year 2012 is $$2,40,000$$
Total revenue for the first 10 years ( Jan 2012 - Oct 2012 ) is $$1,80,000$$

So, The Revenue for November 2012 & December 2012 is $$60,000$$

Let revenue for Nov be $$x$$ , so revenue in Dec is $$\frac{3x}{2}$$

Thus, $$\frac{3x}{2} + x = \frac{5x}{2} = 60,000$$

So, $$x = 24,000$$ , Answer must be (A) 24,000
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Abhishek....

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Re: In 2012, a store’s average monthly revenue was $20,000. For the rst 1 [#permalink] ### Show Tags 21 Jan 2019, 06:51 Bunuel wrote: In 2012, a store’s average monthly revenue was$20,000. For the first 10 months of the year, the store’s average monthly revenue was $18,000. If the revenue in December was 50% greater than the revenue in November, what was the store’s revenue in November? A 24,000 B 30,000 C 36,000 D 48,000 E 56,000 Total sale of store = 20,000 * 12 = 2,40,000 10 months sale = 18,000 * 10 = 1,80,000 nov=x and dec = 1.5x 18*10^4+2.5x = 24*10^4 solve for x = 6* 10^4/2.5 x = 24,000 IMO A _________________ If you liked my solution then please give Kudos. Kudos encourage active discussions. Re: In 2012, a store’s average monthly revenue was$20,000. For the rst 1   [#permalink] 21 Jan 2019, 06:51
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