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# M13-03

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Math Expert
Joined: 02 Sep 2009
Posts: 58445

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16 Sep 2014, 00:48
00:00

Difficulty:

25% (medium)

Question Stats:

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

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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. $$2N - 35$$

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16 Sep 2014, 00:48
2
2
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 $$x = 2N + 72$$.

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05 Aug 2015, 21:52
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 04 Apr 2014
Posts: 2

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29 Nov 2016, 07:15
can you show it substituting ?
Math Expert
Joined: 02 Sep 2009
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29 Nov 2016, 07:38
2
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.

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Location: India
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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 $$x = 2N + 72$$.

Bunuel, how have you come up with 100(N+1) as total questions? Should it not be 100N?
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Joined: 02 Sep 2009
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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 $$x = 2N + 72$$.

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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10 Oct 2018, 07:17
1
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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18 Dec 2018, 17:06
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.
Math Expert
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18 Dec 2018, 22:05
ridhi24 wrote:
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.

The question is fine. There are TWO different solutions above. Please ask more specific questions. Thank you.
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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 $$x = 2N + 72$$.

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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Please give Kudos if you agree with my approach :)
Math Expert
Joined: 02 Sep 2009
Posts: 58445

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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 $$x = 2N + 72$$.

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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Concentration: General Management, Accounting
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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

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

Moderators: chetan2u, Bunuel