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Founder  Joined: 18 Apr 2015
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The total weight of m peanuts at a weight of n + 3 mg eac [#permalink]
Expert's post 00:00

Question Stats: 91% (00:47) correct 8% (00:27) wrong based on 24 sessions

 Quantity A Quantity B The total weight of m peanuts at a weight of n + 3 mg each The total weight of n almonds at a weight of m + 3 mg each

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

_________________
GRE Instructor Joined: 10 Apr 2015
Posts: 1962
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Kudos [?]: 1792 , given: 9

Re: The total weight of m peanuts at a weight of n + 3 mg eac [#permalink]
Expert's post
Carcass wrote:

 Quantity A Quantity B The total weight of m peanuts at a weight of n + 3 mg each The total weight of n almonds at a weight of m + 3 mg each

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

The total weight of m peanuts at a weight of (n + 3) milligrams each
Total weight = (m)(n + 3)

The total weight of n almonds at a weight of (m + 3) milligrams each
Total weight = (n)(m + 3)

So, we have:
Quantity A: (m)(n + 3)
Quantity B: (n)(m + 3)

Expand both quantities:
Quantity A: mn + 3m
Quantity B: nm + 3n

Subtract mn (aka nm) from both quantities:
Quantity A: 3m
Quantity B: 3n

Divide both quantities by 3 to get:
Quantity A: m
Quantity B: n

Since we are not given any information about the value of m and n, the correct answer must be D

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com  Intern Joined: 14 Jan 2019
Posts: 31
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Kudos [?]: 8  , given: 5

Re: The total weight of m peanuts at a weight of n + 3 mg eac [#permalink]
1
KUDOS
A= m*(n+3) = mn+3m
B= n*(m+3) = mn+3m

Since we don't know the value of m and n we can't determine which is greater. Re: The total weight of m peanuts at a weight of n + 3 mg eac   [#permalink] 10 Feb 2019, 08:56
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