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After M students took a test, there was a total of 64% of

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After M students took a test, there was a total of 64% of [#permalink] New post   05 Jun 2012, 22:57
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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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B
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Re: After M students took a test, there was a total of 64% of [#permalink] New post   06 Jun 2012, 00:08
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As 64% of 50 questions are correct, Total correct answers = 0.64 x 50 x M = 32.M

Let the next student get N correct answers. And after he gives the test, the %ge of correct answers is 70%

So, Total correct answers after the next student writes the test = 0.7 x 50 x (M+1)

Equating the Total correct answers, we get

0.7 x 50 x (M+1) = 32.M + N

35.M + 35 = 32.M + N

N = 3.M + 35
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Re: After M students took a test, there was a total of 64% of [#permalink] New post   06 Jun 2012, 01:03
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rakp wrote:
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

Please explain your method.


It is a weighted average question!
Out of 50M questions, 64% are correct. Out of 50 questions, you need x% to be correct so that the total % of correct answers is 70%.

w1/w2 = 50M/50 = (x -70)/(70 - 64)
6M = x - 70
x% = (6M + 70)%

Therefore, out of 50 questions, (6M + 70)% i.e. (6M + 70)/100 * 50 = (3M + 35) questions must be correct.

Answer (B)
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Re: After M students took a test, there was a total of 64% of [#permalink] New post   06 Jun 2012, 08:05
Thank you for both the replies.

Gnan, I have a silly question. When M is the number of students who took the test initially then in third step why are we calculating 0.7 x 50 x (M+1)? Shouldn't M+1 be the number of students?

Thank you for clarifying.
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Re: After M students took a test, there was a total of 64% of [#permalink] New post   06 Jun 2012, 08:59
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rakp wrote:
Thank you for both the replies.

Gnan, I have a silly question. When M is the number of students who took the test initially then in third step why are we calculating 0.7 x 50 x (M+1)? Shouldn't M+1 be the number of students?

Thank you for clarifying.


I think that equation is equal to the number of questions that the new student and the old students got correct.

70% = (old correct answers + new correct answers)/(old number of questions + new questions)

Old Correct Answers = 64% * (50M) where 50M is the total number of old questions

New Correct Answers = x, the variable we are trying to find

Old Number of Questions = 50M

New Number of Questions = 50

So with that:

70% = (32M + x)/(50M + 50)

Then you can solve for x.
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Re: After M students took a test, there was a total of 64% of [#permalink] New post   06 Jun 2012, 11:42
Thank you that made it very simple :)
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Re: After M students took a test, there was a total of 64% of [#permalink] New post   04 Jan 2017, 07:36
rakp wrote:
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


let m = 100 thus answer choices becomes a) 320 b) 335 c) 415 d)20 in terms of numbers

Now 100 students answered right question = 64/100* (5000) = 50*64 = 3,200 questions out of 5000

now an extra student is to be added to bring the % of correct questions to 70% of all answered. all answered including the new student = 5, 050

thus

x/5050 = 7/10 thus x = 3,535 question right so we need to increase number of right questions to 3,535 from 3200 that is by answering 335 right questions ... option B
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Re: After M students took a test, there was a total of 64% of [#permalink] New post   27 Feb 2018, 11:17
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Hi All,

I'm a big fan of TESTing VALUES (the approach used in the other explanations). This question can also be solved with Algebra:

Since there are M students who averaged 64% correct on a 50 question test, we have....

M(.64)(50) = 32M points

We're going to add 1 more student; that student has to score enough points to raise the average to 70%.

X = points scored by that 1 student

Using the Average Formula, we have....

32M + X points
M + 1 students

(32M + X)/(M + 1) = .7(50)
(32M + X)/(M + 1) = 35
32M + X = 35M + 35
X = 3M + 35

Final Answer:
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B


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Re: After M students took a test, there was a total of 64% of [#permalink] New post   01 Feb 2023, 03:13
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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

I picked easy numbers.
Let No of students M = 1
No of questions = 50, 64% correct implies 32 questions are correct
Plugging in
A) No of questions right = 3(1) + 20 = 23 questions to bring up the average to 70? No
B) No of questions right = 3(1) + 35 = 38 questions minimum from 32 questions to bring up the average to 70% Yes \((32 + 38)/100\)= 70
B
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Re: After M students took a test, there was a total of 64% of [#permalink] New post   22 Jul 2023, 04:44
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After \(M\) students took a test, there were a total of 64% of correct answers. If the test contains 50 questions, what is the least number of questions that the next student has 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


Let's denote \(x\) as the required number of correct answers for the next student to raise the total percentage of correct answers to 70%. The equation that \(x\) must satisfy is \(\frac{0.64*M*50 + x}{50M + 50} = \frac{7}{10}\), which can be simplified to \(350M + 350 = 320M + 10x\). Solving for \(x\), we get \(x = 3M + 35\).


Answer: B
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Re: After M students took a test, there was a total of 64% of [#permalink] New post   26 Mar 2024, 09:25
It’s important to recognise this as an ‘average based ‘ question. Initial average 64% and new average is 70%. Total of 50 questions. In other words 32 is current average and 35 is the new average. The new student’s score should lift the current students average score by 3 I.e 3M and also the new student should score 35 to keep the average 35. 3M+35.

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Re: After M students took a test, there was a total of 64% of [#permalink] New post   03 Apr 2024, 10:30
Bunuel wrote:
After \(M\) students took a test, there were a total of 64% of correct answers. If the test contains 50 questions, what is the least number of questions that the next student has 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


Let's denote \(x\) as the required number of correct answers for the next student to raise the total percentage of correct answers to 70%. The equation that \(x\) must satisfy is \(\frac{0.64*M*50 + x}{50M + 50} = \frac{7}{10}\), which can be simplified to \(350M + 350 = 320M + 10x\). Solving for \(x\), we get \(x = 3M + 35\).


Answer: B

­Hello Bunuel,
Could you please explain where the second 50 in the denominator came from? I'm a little confused, thank you. 
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Re: After M students took a test, there was a total of 64% of [#permalink] New post   03 Apr 2024, 10:48
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TylerFerreira1 wrote:
Bunuel wrote:
After \(M\) students took a test, there were a total of 64% of correct answers. If the test contains 50 questions, what is the least number of questions that the next student has 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


Let's denote \(x\) as the required number of correct answers for the next student to raise the total percentage of correct answers to 70%. The equation that \(x\) must satisfy is \(\frac{0.64*M*50 + x}{50M + 50} = \frac{7}{10}\), which can be simplified to \(350M + 350 = 320M + 10x\). Solving for \(x\), we get \(x = 3M + 35\).


Answer: B

­Hello Bunuel,
Could you please explain where the second 50 in the denominator came from? I'm a little confused, thank you. 

­M students attempted 50M questions, the next students attempts additional 50 questions, so total questions attempted 50M + 50.
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Re: After M students took a test, there was a total of 64% of [#permalink]
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