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If John brought n comic books at $2 each, then brought m comic books

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If John brought n comic books at $2 each, then brought m comic books  [#permalink]

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New post 28 Dec 2017, 00:10
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Question Stats:

77% (01:01) correct 23% (01:06) wrong based on 25 sessions

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If John brought n comic books at $2 each, then brought m comic books at $7 each, and m- n comic books at $13 each, then the average (arithmetic mean) cost, in dollars per comic book, is equal to

A. 22(m + n)/m

B. (20m - 11n)/(2m)

C. 11m

D. (6m - 15n)/(2m - 2n)

E. 22mn/(m + n)

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Re: If John brought n comic books at $2 each, then brought m comic books  [#permalink]

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New post 28 Dec 2017, 03:56
1
Bunuel wrote:
If John brought n comic books at $2 each, then brought m comic books at $7 each, and m- n comic books at $13 each, then the average (arithmetic mean) cost, in dollars per comic book, is equal to

A. 22(m + n)/m

B. (22m - 11n)/(2m)

C. 11m

D. (6m - 15n)/(2m - 2n)

E. 22mn/(m + n)


As all our values are small, we'll go for a straightforward calculation.
This is a Precise approach.

Total money spent is n*2 + m*7 + (m-n)*13 = m*20 - n*11
Total comic books purchased = n + m + (m-n) = 2m
Average price per book is (m*20 - n*11)/2m.

Bunuel are you sure the question is correct? If I'm not mistaken the coefficient of m should be 20 and not 22 in all relevant answers. (Or the question stem should be different)
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Re: If John brought n comic books at $2 each, then brought m comic books  [#permalink]

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New post 28 Dec 2017, 04:06
1
DavidTutorexamPAL wrote:
Bunuel wrote:
If John brought n comic books at $2 each, then brought m comic books at $7 each, and m- n comic books at $13 each, then the average (arithmetic mean) cost, in dollars per comic book, is equal to

A. 22(m + n)/m

B. (22m - 11n)/(2m)

C. 11m

D. (6m - 15n)/(2m - 2n)

E. 22mn/(m + n)


As all our values are small, we'll go for a straightforward calculation.
This is a Precise approach.

Total money spent is n*2 + m*7 + (m-n)*13 = m*20 - n*11
Total comic books purchased = n + m + (m-n) = 2m
Average price per book is (m*20 - n*11)/2m.

Bunuel are you sure the question is correct? If I'm not mistaken the coefficient of m should be 20 and not 22 in all relevant answers. (Or the question stem should be different)


My mistake, you are right. Edited. Thank you.
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If John brought n comic books at $2 each, then brought m comic books  [#permalink]

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New post 28 Dec 2017, 09:30
Bunuel wrote:
If John brought n comic books at $2 each, then brought m comic books at $7 each, and m- n comic books at $13 each, then the average (arithmetic mean) cost, in dollars per comic book, is equal to

A. 22(m + n)/m

B. (20m - 11n)/(2m)

C. 11m

D. (6m - 15n)/(2m - 2n)

E. 22mn/(m + n)


\(\frac{2n + 7m + 13m - 13n}{m + n + m - n}\)

Or, \(\frac{20m - 11n}{2m}\), answer will be (B)
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Re: If John brought n comic books at $2 each, then brought m comic books &nbs [#permalink] 28 Dec 2017, 09:30
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