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Last year, the price of a vacation package was P. At the beginning of

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Last year, the price of a vacation package was P. At the beginning of [#permalink]

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New post 12 Sep 2016, 06:16
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

65% (02:14) correct 35% (01:12) wrong based on 34 sessions

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Last year, the price per transaction of certain company increases P percent from the year before, and the number of transactions increase N percent from the year before. Total revenue is simply the number of transactions times the price per transaction. Assume this one transaction is this company's only source of revenue. Which of the following is the percent increase in revenue last year, from the year before?

A) \(P*N\) %

B) \((P+N)\)

C) \((P+N+\)\(\frac{P*N}{100})\) %

D) \((P+N-\)\(\frac{P*N}{100})\) %

E) \((P*N+\)\(\frac{(P*N)^2}{10000})\) %
[Reveal] Spoiler: OA

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Re: Last year, the price of a vacation package was P. At the beginning of [#permalink]

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New post 12 Sep 2016, 07:50
Assume
x = price per transaction for previous year
y = price per transaction for this year

Relationship between x and y
y =x (1+P/100)

Similarly, if
n1= no. of transactions for last year
n2 = no. of transactions for this year

n2 = n1(1+N/100)

Now, the question is [(n2y- n1x) /n1x] * 100

n2y = n1x(1+P/100)(1+N/100)

So, [(n2y - n1x)/n1x] * 100 = (n1x[(1+P/100) (1+N/100) - 1]/ n1x) * 100

D is the answer.
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Re: Last year, the price of a vacation package was P. At the beginning of [#permalink]

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New post 19 Sep 2017, 10:14
I believe C is the answer
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Re: Last year, the price of a vacation package was P. At the beginning of [#permalink]

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New post 19 Sep 2017, 11:26
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MyTurnNow wrote:
Last year, the price per transaction of certain company increases P percent from the year before, and the number of transactions increase N percent from the year before. Total revenue is simply the number of transactions times the price per transaction. Assume this one transaction is this company's only source of revenue. Which of the following is the percent increase in revenue last year, from the year before?

A) \(P*N\) %

B) \((P+N)\)

C) \((P+N+\)\(\frac{P*N}{100})\) %

D) \((P+N-\)\(\frac{P*N}{100})\) %

E) \((P*N+\)\(\frac{(P*N)^2}{10000})\) %



Revenue from the year before last year = xy , where x is the transaction and y is the price

Revenue last year = xy ( 1 + P%) (1+N%)

Percentage change =[(1+P%)(1+N%) -1] * 100 = [1 + P% + N% + P%*N% - 1]*100 = P + N + P*N/100


Correct answer is Option C
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Re: Last year, the price of a vacation package was P. At the beginning of [#permalink]

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New post 19 Sep 2017, 11:31
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Squib17 wrote:
MyTurnNow wrote:
Last year, the price per transaction of certain company increases P percent from the year before, and the number of transactions increase N percent from the year before. Total revenue is simply the number of transactions times the price per transaction. Assume this one transaction is this company's only source of revenue. Which of the following is the percent increase in revenue last year, from the year before?

A) \(P*N\) %

B) \((P+N)\)

C) \((P+N+\)\(\frac{P*N}{100})\) %

D) \((P+N-\)\(\frac{P*N}{100})\) %

E) \((P*N+\)\(\frac{(P*N)^2}{10000})\) %




Revenue from the year before last year = xy , where x is the transaction and y is the price

Revenue last year = xy ( 1 + P%) (1+N%)

Percentage change =[(1+P%)(1+N%) -1] * 100 = [1 + P% + N% + P%*N% - 1]*100 = P + N + P*N/100


Correct answer is Option C



Well done. That's my approach as well. Instead of keeping it as P% (or N%), I made it ugly by denoting P/100 (or N/100). But, I eventually arrived at the same expression.
Re: Last year, the price of a vacation package was P. At the beginning of   [#permalink] 19 Sep 2017, 11:31
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