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Last year, the price per transaction of certain company increases P pe

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Bunuel
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Last year, the price per transaction of certain company increases P pe [#permalink] New post  14 Jun 2019, 00:42
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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. \((P*N+\)\(\frac{(P*N)^2}{10000})\) %
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Re: Last year, the price per transaction of certain company increases P pe [#permalink] New post  14 Jun 2019, 00:53
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Bunuel 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. \((P*N+\)\(\frac{(P*N)^2}{10000})\) %



Let the number of transactions before last year = 100 and price per transaction = 100
Initial Revenue = 100*100 = 10000
Last Year,
Number of transactions increase N percent from the year before = 100 + N
and price per transaction of certain company increases P percent from the year before = 100 + P
Final revenue = (100 + N)*(100 + P) = 10000 + 100(N + P) + NP

% Increase = \(\frac{(Final - Initial)}{Initial}\)*100
= \(\frac{(10000 + 100(N + P) + NP - 10000)}{10000}\)*100
= (100(N + P) + NP)/100
= N + P + NP/100

IMO Option C

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Re: Last year, the price per transaction of certain company increases P pe [#permalink] New post  14 Jun 2019, 00:57
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Bunuel 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. \((P*N+\)\(\frac{(P*N)^2}{10000})\) %


Let's assume the price of a transaction to be x and the number of transactions to be y

So increased price = x (1 + P/100) and increased number of transactions = y (1 + N/100)

Revenue = x (1 + P/100) * y (1 + N/100), and revenue in the year before that is simply x*y

Now, we want to find the percentage increase, so (((x (1 + P/100) * y (1 + N/100)) - x*y) / x*y) * 100

Solving the above equation, x*y will get cancelled from both numerator and denominator, and upon simplification, we get our answer (C)
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Re: Last year, the price per transaction of certain company increases P pe [#permalink] New post  14 Jun 2019, 01:13
Bunuel 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. \((P*N+\)\(\frac{(P*N)^2}{10000})\) %


Lets plug-in numbers and try:
Price of prev. year =100 and P=10 so Current year price = 110
Trans. of prev. year =50 and N=10 so Current year trans. = 55
Prev year revenue = 5000, current year revenue = 110*55=6050; Increase in revenue is 1050/5000*1000=21%
Only C in option provides 21% So IMO C.
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Re: Last year, the price per transaction of certain company increases P pe [#permalink] New post  14 Jun 2019, 02:43
Bunuel 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. \((P*N+\)\(\frac{(P*N)^2}{10000})\) %


let initial price = 10 now p 50% increase new p = 15
and initial txn = 2 N = 50% , new = 3 ;
revenue earlier ; 10*2 ; 20 and new 15* 3; 45
so
total revenue % increase ; 45-20/20 = 125%
seeing options
in C \((P+N+\)\(\frac{P*N}{100})\) %
substitute P = 50 and N = 50 we get 125%
sufficient
IMO C
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Re: Last year, the price per transaction of certain company increases P pe [#permalink] New post  14 Jun 2019, 21:11
Let previous year revenue be R1 and the new revenue be R2

Hence,
R2= R1(1+P/100)(1+N/100)
= R1(1+ P/100+N/100+PN/10000)

Increase = P/100+N/100+ PN/10000
Percentage Increase= Increase*100%
= (P+N + PN/100)
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Last year, the price per transaction of certain company increases P pe [#permalink] New post  20 Sep 2020, 01:06
Let the price per transaction for last month be 100
And the number of transactions be 10
So total revenue for last year = 100*10 = 1000

Let P be 50%, so Price per transaction this year = 150
Let N be 100% so Number of transaction this year = 20
So total revenue for this year = 3000

Percent increase = (3000 - 1000)*100/1000 = 200%

Now plugging in the value of P and N in answer choices.

Let's start with C

(P+N+ (P∗N)/100)%
(50+100 + (50*100)/100)%
(150 + 50)%
200%

This matches the answer we got. Hence correct!!
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Re: Last year, the price per transaction of certain company increases P pe [#permalink] New post  12 Aug 2023, 13:56
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Re: Last year, the price per transaction of certain company increases P pe [#permalink]
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