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M21-28

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Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132595 [0], given: 12326

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16 Sep 2014, 01:15
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Difficulty:

35% (medium)

Question Stats:

81% (01:34) correct 19% (01:44) wrong based on 16 sessions

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After $$M$$ students took a test, there was a total of 64% of correct answers. If the test contains 50 questions, what is the least number of questions that the next student have to get right to bring the total of correct answers to 70% ?

A. 3M + 20
B. 3M + 35
C. 4M + 15
D. 4M + 20
E. 4M + 45
[Reveal] Spoiler: OA

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Kudos [?]: 132595 [0], given: 12326

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132595 [0], given: 12326

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16 Sep 2014, 01:15
Expert's post
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Official Solution:

After $$M$$ students took a test, there was a total of 64% of correct answers. If the test contains 50 questions, what is the least number of questions that the next student have to get right to bring the total of correct answers to 70% ?

A. 3M + 20
B. 3M + 35
C. 4M + 15
D. 4M + 20
E. 4M + 45

Denote $$x$$ as the required number of correct answers. $$x$$ must satisfy the equation $$\frac{0.64*M*50 + x}{50M + 50} = \frac{7}{10}$$ or $$350M + 350 = 320M + 10x$$ or $$x = 3M + 35$$.

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Intern
Joined: 08 Jan 2015
Posts: 7

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Location: United States (CA)
Schools: Wharton '20 (II)
GMAT 1: 730 Q47 V42
GPA: 3.63

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19 Feb 2015, 15:24
Bunuel wrote:
Official Solution:

After $$M$$ students took a test, there was a total of 64% of correct answers. If the test contains 50 questions, what is the least number of questions that the next student have to get right to bring the total of correct answers to 70% ?

A. 3M + 20
B. 3M + 35
C. 4M + 15
D. 4M + 20
E. 4M + 45

Denote $$x$$ as the required number of correct answers. $$x$$ must satisfy the equation $$\frac{0.64*M*50 + x}{50M + 50} = \frac{7}{10}$$ or $$350M + 350 = 320M + 10x$$ or $$x = 3M + 35$$.

Can you split out that formula and explain why each term is in there? I don't understand how you got to $$\frac{0.64*M*50 + x}{50M + 50} = \frac{7}{10}$$

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Manager
Joined: 14 Jul 2014
Posts: 97

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26 Feb 2015, 06:34
rizzy wrote:
Bunuel wrote:
Official Solution:

After $$M$$ students took a test, there was a total of 64% of correct answers. If the test contains 50 questions, what is the least number of questions that the next student have to get right to bring the total of correct answers to 70% ?

A. 3M + 20
B. 3M + 35
C. 4M + 15
D. 4M + 20
E. 4M + 45

Denote $$x$$ as the required number of correct answers. $$x$$ must satisfy the equation $$\frac{0.64*M*50 + x}{50M + 50} = \frac{7}{10}$$ or $$350M + 350 = 320M + 10x$$ or $$x = 3M + 35$$.

Can you split out that formula and explain why each term is in there? I don't understand how you got to $$\frac{0.64*M*50 + x}{50M + 50} = \frac{7}{10}$$

Consider the required equation

0.64 * M * 50 + x = 7/10 * 50 ( M + 1)

0.64 * M * 50 = 50 M Questions were answered off which 64% were accurate
x = the least number of questions that the next student have to get right to bring the total of correct answers to 70%

7/10 * 50 ( M + 1) = total correct answers must be 70% . Here in (M+1) = +1 is for the next student

Hope it helps

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Intern
Joined: 21 Dec 2016
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05 Sep 2017, 12:24
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Hi Bunuel,

I took the shortest way possible I guess but I don´t know if I´m correct at all....

Assuming M=0, the only person in doing the exam needs a 70% of correct answers.

70% of 50 = 0,7 * 50 = 35; With M=0; Answer B

Is that possible?

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Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132595 [1], given: 12326

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14 Oct 2017, 04:17
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Framarper wrote:
Hi Bunuel,

I took the shortest way possible I guess but I don´t know if I´m correct at all....

Assuming M=0, the only person in doing the exam needs a 70% of correct answers.

70% of 50 = 0,7 * 50 = 35; With M=0; Answer B

Is that possible?

In this case it worked but what would you do if one of the choices were 4M + 35?
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