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M21-28

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M21-28 [#permalink]

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New post 16 Sep 2014, 01:15
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A
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Difficulty:

  15% (low)

Question Stats:

87% (01:57) correct 13% (01:47) wrong based on 97 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

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Re M21-28 [#permalink]

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New post 16 Sep 2014, 01:15
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 \(x = 3M + 35\).

Answer: B
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M21-28 [#permalink]

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New post 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 \(x = 3M + 35\).

Answer: B


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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Re: M21-28 [#permalink]

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New post 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 \(x = 3M + 35\).

Answer: B


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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Re: M21-28 [#permalink]

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New post 05 Sep 2017, 12:24
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?

Thank you in advance.
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Re: M21-28 [#permalink]

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New post 14 Oct 2017, 04:17
1
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?

Thank you in advance.


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

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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Re: M21-28 [#permalink]

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New post 11 Dec 2017, 03:18
+1 for option B. Given that there are 50 students and they answer 50 questions each. We have 50M questions. 64% of these are correct. we need to find the number of questions the 51st student needs to answer correctly to get the tally to 70%. (0.64*50M+x)/(50(M+1))=0.7. Solve and get x as 3M+35. Hence option B.
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Re: M21-28   [#permalink] 11 Dec 2017, 03:18
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