M21-28

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

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?

+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.

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.

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,
Brent

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

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

Posted from my mobile device

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.