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by buoyant » Wed Jan 14, 2015 5:48 pm
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

[spoiler]OA: B[/spoiler]

[spoiler]is it a weighted average question?

i solved as below:
70% of 50 questions= 35 correct answers
chose B
but still don't get why is M in the solution?[/spoiler]
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by MartyMurray » Wed Jan 14, 2015 6:49 pm
buoyant 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

[spoiler]OA: B[/spoiler]

Is it a weighted average question?

I solved as below:
70% of 50 questions= 35 correct answers
chose B
but still don't get why is M in the solution?
Yes, it's basically a weighted average question.

You need M, because the bigger M is the more weight there is to the 64 percent tests.

For instance, if M is 2, there are two tests with 64 percent right answers, and the weighted average will be of three tests, two with 64% right answers and the next one which needs to have enough right answers to get the average to 70%.

If M is 3, the next test will have to have even more right answers in order to make up for three 64% tests and get the average score to 70.

You could do all kinds of math to figure this out, but I just set M = 1. So the one test had 64% of 50 = 32 right answers. To get the average to 70% of 50 = 35 right answers, we need the next person to get 38 right. 38 + 32 = 70 70/2 = 35.

If M is 1, 38 = 35 + 3M. Confirm that none of the other choices somehow work.

Choose B.

Alternatively, you could see that each number of right answers in the first M tests is 32, which is 3 below 35. There are M tests and each of them is 3 below the desired average.

So to create a 35 average, the next test has to have 35 right answers plus enough right answers to make up for the 3 that were not there on each of the other tests. So the score on the next test has to be 35 + 3M. Then the average is [32M + (3M + 35)]/[M + 1] = [35M + 35]/[M + 1] = 35

So that's where the 3M + 35 comes from.
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by GMATGuruNY » Wed Jan 14, 2015 6:57 pm
buoyant 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=2 students.
Since each of these 2 students answers 50 questions, the total number of answered questions = 2*50 = 100.
Since these two students answer 64% of the questions correctly, the total number of correct answers = 64% of 100 = 64.

When the next student takes the 50-question test, the total number of answered questions increases from 100 to 150.
Since the percentage of correct answers increases to 70%, the new total number of correct answers = 70% of 150 = 105.
Since the total number of correct answers increases from 64 to 105, the number of questions answered correctly by the next student = 105-64 = 41.
This is our target.

Now plug M=2 into the answers to see which yields our target of 41.
Only B works:
3M + 35 = 3*2 + 35 = 41.

The correct answer is B.
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by [email protected] » Wed Jan 14, 2015 8:00 pm
Hi buoyant,

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: B

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by Brent@GMATPrepNow » Wed Jan 14, 2015 9:02 pm
buoyant 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
We can solve this question using the WEIGHTED AVERAGES.

Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

We have M students.
The average score in this group is 64 percent correct

Then we have 1 more student.
Let x = the percent of correct responses on the test.

In TOTAL, there are M + 1 students.
So, M/(M+1) = the proportion of students in the group with an average score of 64 percent,
and 1/(M+1) = the proportion of students in the group with an average score of x percent.

We want the COMBINED percent to be 70 percent.
So, applying the formula, we get: 70 = [M/(M+1)][64] + [1/(M+1)][x]
Eliminate the fractions by multiplying each side by M+1 to get: 70(M+1) = 64M + x
Expand to get: 70M + 70 = 64M + x
Solve for x to get: 6M + 70 = x

IMPORTANT: x = the PERCENT of correct responses (our of 50 questions) that the last person got.
So, we now know that 6M + 70 = the PERCENT correct out of 50 questions
In other words, the NUMBER of correct responses = [(6M + 70)/100][50]
= 3M + 35
= B

-------------------------------------

For more information on weighted averages, you can watch this free GMAT Prep Now video: https://www.gmatprepnow.com/module/gmat- ... ics?id=805

Here are some additional practice questions related to weighted averages:
- https://www.beatthegmat.com/weighted-ave ... 17237.html
- https://www.beatthegmat.com/weighted-ave ... 14506.html
- https://www.beatthegmat.com/average-weig ... 57853.html
- https://www.beatthegmat.com/averages-que ... 87118.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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