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M13-03

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M13-03  [#permalink]

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New post 16 Sep 2014, 00:48
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Question Stats:

81% (02:05) correct 19% (02:07) wrong based on 64 sessions

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Re M13-03  [#permalink]

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New post 16 Sep 2014, 00:48
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Official Solution:

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. \(2N - 35\)

After \(N\) tests John had \(70N\) correct answers. Denote \(x\) as the number of questions he has to get right in the last test. \(x\) has to satisfy equation \(\frac{(70N + x)}{(100(N + 1))} = 0.72\). The denominator represents the total number of questions in the \(N + 1\) tests. From this equation \(x = 2N + 72\).

Answer: D
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Re M13-03  [#permalink]

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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]

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New post 29 Nov 2016, 07:15
can you show it substituting ?
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Re: M13-03  [#permalink]

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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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Re: M13-03  [#permalink]

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New post 26 Mar 2017, 21:15
Bunuel wrote:
Official Solution:

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. \(2N - 35\)

After \(N\) tests John had \(70N\) correct answers. Denote \(x\) as the number of questions he has to get right in the last test. \(x\) has to satisfy equation \(\frac{(70N + x)}{(100(N + 1))} = 0.72\). The denominator represents the total number of questions in the \(N + 1\) tests. From this equation \(x = 2N + 72\).

Answer: D


Bunuel, how have you come up with 100(N+1) as total questions? Should it not be 100N?
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Re: M13-03  [#permalink]

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New post 26 Mar 2017, 22:52
OreoShake wrote:
Bunuel wrote:
Official Solution:

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. \(2N - 35\)

After \(N\) tests John had \(70N\) correct answers. Denote \(x\) as the number of questions he has to get right in the last test. \(x\) has to satisfy equation \(\frac{(70N + x)}{(100(N + 1))} = 0.72\). The denominator represents the total number of questions in the \(N + 1\) tests. From this equation \(x = 2N + 72\).

Answer: D


Bunuel, how have you come up with 100(N+1) as total questions? Should it not be 100N?


John took N tests and then one more, thus total of N + 1 tests. Each contains 100 questions, thus total of 100(N+1) questions.
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M13-03  [#permalink]

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New post 10 Oct 2018, 07:17
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In my opinion, the equation should be (N-1)2 + 72 (where 72 is scored in the last test, and rest 2, difference of 72 and 70 will be distributed in preceding test).
So in preceding two tests if the person averages 70, then he needs to score a 76 in the third, similarly if the person averages 70 in first three test then he needs to score a 78 in the fourth, which we can get from the equation above or 2N + 70 Please correct me if this is wrong.
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Re M13-03  [#permalink]

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New post 18 Dec 2018, 17:06
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.
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New post 18 Dec 2018, 22:05
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Re: M13-03  [#permalink]

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New post 23 Jul 2019, 06:22
Bunuel wrote:
Official Solution:

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. \(2N - 35\)

After \(N\) tests John had \(70N\) correct answers. Denote \(x\) as the number of questions he has to get right in the last test. \(x\) has to satisfy equation \((70N + x)/(100(N + 1) = 0.72\). The denominator represents the total number of questions in the \(N + 1\) tests. From this equation \(x = 2N + 72\).



Answer: D


Hi Bunuel!

Though the plugging in approach is clear, could you please explain how did you derive the highlighted part? I cannot understand how does the average formula - Sum of all instances/total no. of instances is used here.

Also, why do we have to \((70N+x)/(100(N+1))\) why cant we just divide by N+1 ?

Thanks
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Re: M13-03  [#permalink]

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New post 23 Jul 2019, 06:44
578vishnu wrote:
Bunuel wrote:
Official Solution:

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. \(2N - 35\)

After \(N\) tests John had \(70N\) correct answers. Denote \(x\) as the number of questions he has to get right in the last test. \(x\) has to satisfy equation \((70N + x)/(100(N + 1) = 0.72\). The denominator represents the total number of questions in the \(N + 1\) tests. From this equation \(x = 2N + 72\).



Answer: D


Hi Bunuel!

Though the plugging in approach is clear, could you please explain how did you derive the highlighted part? I cannot understand how does the average formula - Sum of all instances/total no. of instances is used here.

Also, why do we have to \((70N+x)/(100(N+1))\) why cant we just divide by N+1 ?

Thanks


There are \(N\) tests, each containing 100 questions, so there are 100N questions in total.
John had an average of 70% of correct answers, so he answered 0.7*100N = 70N questions correctly.

100(N+1) represents the total number of questions in N + 1 tests (each containing 100 questions). While 70N+x represents the number of correct answers in N + 1 tests.
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Re: M13-03  [#permalink]

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New post 09 Oct 2019, 19:39
We can do it by basic arithmetic within 30 seconds:

lets say he gave 3 exams so N is 3, he scored 70,70,70, in next attempt we want the number as such that it increases past three attempts score by 72, 2*3 + 72, the average for next attempt, so answer should give 78

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Re: M13-03   [#permalink] 09 Oct 2019, 19:39
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