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A store's monthly sales total declined j percent from October to Novem

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A store's monthly sales total declined j percent from October to Novem [#permalink]

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New post 21 Nov 2017, 10:19
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Difficulty:

  65% (hard)

Question Stats:

54% (02:08) correct 46% (01:22) wrong based on 48 sessions

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A store's monthly sales total declined j percent from October to November. If the store's total sales in December were equal to its total sales in October, by what percent, in terms of j, did the store's sales increase from November to December?

A. j

B. \(\frac{100j}{100+j}\)

C. \(\frac{10,000j}{100-j}\)

D. \(\frac{100j}{100-j}\)

E. \(\frac{j}{1-j}\)
[Reveal] Spoiler: OA

Last edited by abansal1805 on 21 Nov 2017, 12:21, edited 1 time in total.

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Re: A store's monthly sales total declined j percent from October to Novem [#permalink]

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New post 21 Nov 2017, 11:29
If we take an arbitrary value for j = 20.
Assuming the store's monthly sales for October to be 1000.
Hence, the sale in November is 800.
If the sales in December was the same as that of October,
we would need an increase of 200, and the increase will be \(\frac{200}{800}*100 = 25\)%

When we substitute the value of j=20, Option D(\(\frac{100j}{100-j}\)) gives the correct answer

abansal1805 IMO the OA must be D, not C. Please check!
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Kudos [?]: 754 [0], given: 20

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Re: A store's monthly sales total declined j percent from October to Novem [#permalink]

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New post 21 Nov 2017, 12:23
Dear pushpitkc,
You are correct. Thank you for pointing out the error.
It has been fixed.
Much appreciated.

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A store's monthly sales total declined j percent from October to Novem [#permalink]

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New post 27 Nov 2017, 09:32
pushpitkc wrote:
If we take an arbitrary value for j = 20.
Assuming the store's monthly sales for October to be 1000.
Hence, the sale in November is 800.
If the sales in December was the same as that of October,
we would need an increase of 200, and the increase will be \(\frac{200}{800}*100 = 25\)%

When we substitute the value of j=20, Option D(\(\frac{100j}{100-j}\)) gives the correct answer.


Thanks for your answer. I had a quick question. I saw that you picked values and got 25% for an answer. I followed your method, but when I read 25%, I wrote it as 1/4. When you plug that into your answer choice, I get a different answer.

j=20, Option D(\(\frac{100(20)}{100-20}\)) does not equal 1/4. I'm confused.

Thanks for reading this.

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Re: A store's monthly sales total declined j percent from October to Novem [#permalink]

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New post 03 Dec 2017, 20:55
Lateralus14 wrote:
pushpitkc wrote:
If we take an arbitrary value for j = 20.
Assuming the store's monthly sales for October to be 1000.
Hence, the sale in November is 800.
If the sales in December was the same as that of October,
we would need an increase of 200, and the increase will be \(\frac{200}{800}*100 = 25\)%

When we substitute the value of j=20, Option D(\(\frac{100j}{100-j}\)) gives the correct answer.


Thanks for your answer. I had a quick question. I saw that you picked values and got 25% for an answer. I followed your method, but when I read 25%, I wrote it as 1/4. When you plug that into your answer choice, I get a different answer.

j=20, Option D(\(\frac{100(20)}{100-20}\)) does not equal 1/4. I'm confused.

Thanks for reading this.


Hi Lateralus14, the question is asking for percentage increase, and hence the 100. If you're solving it with fraction, then 100 has to be omitted from the calculation.

So, \(\frac{100*(20)}{(100-20)}\) will become \(\frac{20}{80}\)*100, or \(\frac{1}{4}\)*100 or 25% increase.

Hope this clears the confusion.

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Re: A store's monthly sales total declined j percent from October to Novem   [#permalink] 03 Dec 2017, 20:55
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