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M13-03

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Bunuel
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M13-03 [#permalink] New post   16 Sep 2014, 00:48
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25% (medium)

Question Stats:

82% (01:58) correct 18% (02:20) wrong based on 88 sessions
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After completing \(N\) tests, each with 100 questions, John achieved an average (arithmetic mean) of 70% correct answers. In terms of \(N\), how many questions must John answer correctly on the next test to increase his average to 72%?

A. \(N - 35\)
B. \(N + 72\)
C. \(2N + 70\)
D. \(2N + 72\)
E. \(2N - 35\)
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D
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Re M13-03 [#permalink] New post   16 Sep 2014, 00:48
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Official Solution:

After completing \(N\) tests, each with 100 questions, John achieved an average (arithmetic mean) of 70% correct answers. In terms of \(N\), how many questions must John answer correctly on the next test to increase his average to 72%?

A. \(N - 35\)
B. \(N + 72\)
C. \(2N + 70\)
D. \(2N + 72\)
E. \(2N - 35\)


Let \(x\) be the number of questions John needs to answer correctly in the next test.

We know that John answered 70% of \(100N\) questions correctly, which is \(0.70*100N=70N\). He needs his average after \((N+1)\) tests to be 72%, so:

\(\frac{70N + x}{100(N+1)} = 0.72\)

Solving for \(x\) gives us:

\(x = 2N + 72\)


Answer: D
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amirzohrevand
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Re: M13-03 [#permalink] New post   05 Aug 2015, 21:52
I think this is a high-quality question and I agree with explanation.
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Re: M13-03 [#permalink] New post   29 Nov 2016, 07:15
can you show it substituting ?
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Bunuel
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Re: M13-03 [#permalink] New post   29 Nov 2016, 07:38
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Pakku wrote:
can you show it substituting ?


After taking N tests, each containing 100 questions, John had an average of 70% of correct answers. How much does John need to score on the next test to make his average equal 72%?

A. N−35
B. N+72
C. 2N+70
D. 2N+72
E. 2N−35

Say N=1.

So, after 1 test John has 70 correct answers.
In 2 tests, so in 200 questions he needs to have 0.72*200=144 correct answers, so in the second test he must get 144-70=74 questions correctly.

Now, plug N=1 into the answer choices to see which one yields 74. Only option D fits.

Answer: D.
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AntonioGalindo
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Re: M13-03 [#permalink] New post   13 Jan 2022, 09:48
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Average = Sum / N . Where N is the number of terms

70 = Sum / N "Tests", hence Sum = 70N

72= Sum / N + 1 "Tests"

x= score required on the next test.

Substituting

72= (70N + x) / (N+1)

Solve for X
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Re: M13-03 [#permalink] New post   28 Jul 2023, 07:30
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M13-03 [#permalink]
28 Jul 2023, 07:30
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