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(n+12)^2+(n-16)^2
[#permalink]
09 Sep 2026, 08:05
09 Sep 2026, 08:05
Expert Reply
00:00
Question Stats:
60% (01:23) correct
40% (01:45) wrong based on 10 sessions
Hide Show timer Statistics
Quantity A |
Quantity B |
\((n+12)^2+(n-16)^2\) |
\(52n\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
GRE Prep Club Team Member
Joined: 20 Feb 2017
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Re: (n+12)^2+(n-16)^2
[#permalink]
13 Sep 2026, 11:24
13 Sep 2026, 11:24
Expert Reply
Solution
We compare: \(\text{Quantity A}=(n+12)^2+(n-16)^2\)
\((n+12)^2=n^2+24n+144\)
\((n-16)^2=n^2-32n+256\)
Therefore, \(A=2n^2-8n+400\)
Quantity B: \(B=52n\)
Subtract: \(A-B=2n^2-60n+400\)
Factor: \(A-B=2(n^2-30n+200) =2(n-10)(n-20)\)
So:
If \(n<10, A > B\)
If \(10<n<20, B > A\)
If \(n=10 or 20, A = B\)
If \(n>20, A > B\)
Since no value or restriction for \(n\) is given, the relationship cannot be determined.
Answer: D
We compare: \(\text{Quantity A}=(n+12)^2+(n-16)^2\)
\((n+12)^2=n^2+24n+144\)
\((n-16)^2=n^2-32n+256\)
Therefore, \(A=2n^2-8n+400\)
Quantity B: \(B=52n\)
Subtract: \(A-B=2n^2-60n+400\)
Factor: \(A-B=2(n^2-30n+200) =2(n-10)(n-20)\)
So:
If \(n<10, A > B\)
If \(10<n<20, B > A\)
If \(n=10 or 20, A = B\)
If \(n>20, A > B\)
Since no value or restriction for \(n\) is given, the relationship cannot be determined.
Answer: D
gmatclubot
Re: (n+12)^2+(n-16)^2 [#permalink]
13 Sep 2026, 11:24
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