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Re: M21-28 [#permalink]
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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]
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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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?
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+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]
Bunuel 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


For LHS denominator why do we add 50 to 50M? Doesn't 50 M already give us the total number of questions attempted?
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Re: M21-28 [#permalink]
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ENEM wrote:
Bunuel 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


For LHS denominator why do we add 50 to 50M? Doesn't 50 M already give us the total number of questions attempted?


M students attempted 50M questions, the next students attempts additional 50 questions, so total questions attempted 50M + 50.
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Re: M21-28 [#permalink]
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Bunuel 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


If M tests were taken, and each test has 50 questions, then the total number of questions answered = 50M

We're told that there was a total of 64% of correct answers.
So, the total number of correctly-answered questions = 64% of 50M = 32M

If a NEW student takes the test, then there will be M+1 tests taken.
Since each test has 50 questions, the NEW total number of questions answered = 50M + 50

Let C = the number of questions the NEW students answer correctly
So, the NEW number of correctly-answered questions = 32M + C

We want to bring the total of correct answers to 70%
That is, we want: (32M + C)/(50M + 50) = 70/100
Simplify right side: (32M + C)/(50M + 50) = 7/10
Multiply both sides by 10 to get: 10(32M + C)/(50M + 50) = 7
Multiply both sides by (50M + 50) to get: 10(32M + C) = 7(50M + 50)
Expand: 320M + 10C =350M + 350
Subtract 320M from both sides: 10C =30M + 350
Divide both sides by 10 to get: C =3M + 35

Answer: B

Cheers,
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Re: M21-28 [#permalink]
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Bunuel wrote:
[b]

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=1 students attempted total questions of 50.

Total correct questions = 32


Let required number of correct answers=\(x\).

\(\frac{32 + x}{100} = \frac{7}{10}\)

\(x=38\)

Apply M in the answer choices, only one fits

3*1 + 35 = 38

Answer: B
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Using the reasoning my behind what the arithmetic mean (average) represents, you can answer the question without having to perform any calculations.

The Mean represents a proxy of each element’s actual value. In other words, if we took all M students and gave them each the Mean/Average Value, the total Sum would not change.

Right now, M students have a correct-per-student average of: 64%(50) = 32

Imagine a list of the M students. We can apportion the total Sum of 32M evenly among each student (since the arithmetic mean is a proxy for each element in a set)

32…….32……32….32…………….32

+1 new student comes in and the average correct-per-student RISES to 70%(50) = 35

This means, each existing student must be “given” an extra +3 points to get their “element” up to the average of 35

32…….32……32……32……………32
+3…….+3……+3…….+3…………..+3
________________________________

therefore, this 1 new student must bring in enough +3 points for each of the M students or:

3(M)

Also, since the 1 new student is the only person adding new points to the sum total, he must bring another +35 points to account for his “average representation”

3(M) + 35

Which is the answer

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Re: M21-28 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M21-28 [#permalink]
Which topic is used here weighted averages or statistics
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Re: M21-28 [#permalink]
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Aryan19852 wrote:
Which topic is used here weighted averages or statistics

­
The question deals wit the concept of wighted average.
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