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In 2012, a store’s average monthly revenue was $20,000. For the rst 1

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Bunuel
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In 2012, a store’s average monthly revenue was $20,000. For the rst 1 [#permalink] New post   21 Jan 2019, 00:45
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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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Re: In 2012, a store’s average monthly revenue was $20,000. For the rst 1 [#permalink] New post   21 Jan 2019, 05:15
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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 = 20000 * 12 = 240000.

Total of 10 months = 10 * 18000 = 18000

Revenue of November and December = 240000 - 180000 = 60000.

As per the question : revenue in December was 50% greater than the revenue in November

let the revenue in November is x. Then December = 1.5x.

x + 1.5x = 60000

2.5 x = 60000

x = 60000 / 2.5

x = 24000.

A is the correct answer.
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Re: In 2012, a store’s average monthly revenue was $20,000. For the rst 1 [#permalink] New post   21 Jan 2019, 07:28
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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Re: In 2012, a store’s average monthly revenue was $20,000. For the rst 1 [#permalink] New post   21 Jan 2019, 07:51
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
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Re: In 2012, a store’s average monthly revenue was $20,000. For the rst 1 [#permalink]
21 Jan 2019, 07:51
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