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After M students took the 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% ?

We know that M students took the test and there is a total of 64% of the answers correct. This means that 64% of 50 questions were right, on average. We know it's on average because they tell us the total of correct answers. So if you add up all correct answers for M test takers and divde that by the number of test takers * 50, it gives you 64%.

64% of 50 is 32 questions correct on average. So now, we need to figure out how many questions would be correct on 1 test to get 70% So 70% of 50 questions is 35 correct answers. In order for the next test taker to bring the total correct to 70%, the test taker must get 35 questions correct AND make up the difference of the other test takers. By difference, I mean these other test takers got 32 questions right, and to get 70% they should have gotten 35 correct. Each prior test taker is 3 short. So this next one must 35 correct for their own 70% score plus 3 questions to make up the difference.

If you had 1 prior test taker, out of 50 questions the prior taker got 32 right, for a score of 64%. The next one needs to get 35 questions right for their own 70%, but if you added both together, you'd have 35 + 32 correct for 67 out of 100, and that's just 67%. So, the second test taker must get 3 extra questiosn correct so that the total is 70 out of 100. so since there was 1 test taker in this example, M = 1 here. So the answer is 3M + 35.

3 questions per prior test taker to make up the average + 35 questions correct to make sure this test taker gets 70% too!

Hope this explanation is helpful. You might also see a variation of this question like this:

18 students took a test with 50 questions. The 18 students averaged 64% correct answers. If 3 more students take the exam, is it possible for these 3 students to raise the average to 70% correct answers?

You'd figure this out by realizing how many correct answers the first 18 students were short to get 70%. Because we're using the same numbers, we know they average 32 out of 50. So they were each on average 3 questions short. If 18 students were 3 questions short, that's a total of 48 questions. The question tells us that we have 3 more students to take the exam. Each of these new students needs to get 35 (for 70%) correct + some to make up the difference. Since there are 15 questions that are not required to get a 70%, these are questions that can be answered correctly to increase the average of all students. If there are 15 questions that can help to raise the average and there are 3 students remaining to take the exam, then there are 45 questions left that can be used to raise the average of the others. We already figured out that we need 48 questions correct in addition to the questions answered to get the 70% average. There are not enough coming test takers to raise the average to 70% from 64%.

arjtryarjtry wrote:

i could not understand this .. pls guide, by plugging in some values....

After M students took the 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% ?

Club - m21#28

* 3M + 20 * 3M + 35 * 4M + 15 * 4M + 20 * 4M + 45

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

i could not understand this .. pls guide, by plugging in some values....

After M students took the 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% ?

Club - m21#28

* 3M + 20 * 3M + 35 * 4M + 15 * 4M + 20 * 4M + 45

my quick approach: 64% of 50 by M students = 32M 70% of 50 by (M+1) students = 35M + 35

so the the least number of questions that the next student has to answer correctly: 35M + 35 - 32M = 3M + 35
_________________

Thanks for noticing. I know that when i started studying for the GMAT, the concepts weren't overly difficult, but sometimes I would read a post that used 3 sentences to explain a difficult concept and I still had no idea how they did it. I decided that I wanted to make sure I clearly covered even the basic stuff because many people need an explanation that explains the simple parts of the question. This community is as much for someone that wants to go from 490 to 590 as it is someone that wants to go from 650 to 720!

AND....I'm seriously thinking of getting back into study-mode so I can take the GMAT again (yeah, kind of crazy, but it's not to improve on my 720 for appliations.). Helping out on GMATClub has shown me that I really enjoy helping others with this stuff and would LOVE to be a GMAT teacher / tutor. MGMAT pays the best, but I'd have to move to a lot bigger city than Oklahoma City. I think I'd be good at it, but I will need a 760+, more like a 780 to really be able to do it.

dzyubam wrote:

Sometimes I can't help wondering how thorough Allen can be. Thank you very much for your valuable contribution to our community.

Thanks everyone!

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

i could not understand this .. pls guide, by plugging in some values....

After M students took the 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% ?

Club - m21#28

* 3M + 20 * 3M + 35 * 4M + 15 * 4M + 20 * 4M + 45

64% correct answers = 0.64*50*M= 32M to make 70% correct answers = 0.7*50*M= 35M

When we add additional person = it should be 35M+35 (total correct answers to make 70%)

additon person has to score atleast to make 70% = 35M+35 -32M =3M+35
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

I didn't feel like dealing with abstract concepts so I used the following method: M=100 students; Q=50 Questions Total Number of Q's correct = .64Q x M = 3,200 Prompt answer: M+1 x .70Q = 3,535 3,535 - 3,200 = 335 or 3M + 35

I realize my process looks a lot harder on paper, but a lot of this was done in my brain...

If C is the total correct answers provided by M students then C/((50)*M) = 0.64 or C = 32M; If C' is the correct answer provided by M+1 students then C'/((50)*(M+1)) = 0.7 or C' = 35M+35 The extra correct answers by (M+1)th student will be C' - C = 3M + 35 ... Answer B ..

Hope it helps .. .
_________________

Only those who will risk going too far can possibly find out how far one can go. ... T.S. Eliot

I guess my question really is this---if 10 students had taken the test, isn't this implying that the next student has to get a 65/50 (3M+35)? Which wouldn't that be impossible?

I'm not sure that I"m reading this question correctly, but it seems as if it is saying that 1 student has to make up the difference?

If 10 students had taken the test , then total q = 50*10=500

current % is 64 which is 32 q per test , so total correct qs are 32*10 =320 questions

Now one more stud is added and the required % is 70( this 70% is for the entire set of students) i.e. 35 per test so total questions would be 35*11= 385 q

So that 11 student need to take 385-320 =65 which is his 35 questions + 3*10 i.e. 3M+35

Also the question is indeed asking as if, that 11th guy has to do everything .. I hope this clarifies you ...

I am too stupid for abstract solutions, so facing such questions, I always use real-life example. In that case I set M=2, and calculated the number of correct solutions, and then in A-E I found a variant that will match my number.

After M students took the 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% ?

I will give my two cents ( my first two cents for this forum which is awesome!!) so i considered it as a weighted average (64%)*M (students) + x (our value to find out)*(1) (only one student) / # of students (M+1) = 70%

multiplying for 100 and simplyfing the denominator 64m + x = 70m +70 x=6m+70 (diving for 2) x= 3m+35 (time around 40 secs to one min)

So even if there was only one student who got 64%...the other student will need to get 76% (38 correct) to bring up the average to 70% (35 right answers).

If we look at the answer choices -- we can right away eliminate the wrong answers -- by just substituting (M= 1)...

My aproach: 64% correct of 50q = 32q 70% correct of 50q = 35q It´s a mean question (how much is the next studend grade for the mean be 70% or 35q): [(PM*PMs)+(NM*NMs)/(PM+NM)]=35 [(M*32)+(1*x)/(M+1)]=35 32M+x=35(M+1) 32M+x=35M+35 x=35M-32M+35 x=3M+35 PM = number of previous studends = M PMs = score of previous studends = 32 NM = number of next studends = 1 NMs = score of next studends = x =)
_________________

Well, its very simple.. 64% of 50 questions = 32 70% of 50 questions = 35.

to make the total avg 70% the next student must score 70% for himself + cover up the lackings of right answers to make the 'M' previous students' avg 70% also. as each previous M# of students lack by 3 questions to avg. 70%, the new student must score 3*M extra right answers besides his 70%(35 right answers).

total no of questions = M(no. of students) * No. of total questions answered by each test taker = M*50 64% correct answers = 64% of 50M = 32M Now there comes comes an additional student therefore no. of students = M+1 and total no. of questions = 50(M+1) 70% of 50(M+1) = 35M+35

therefore that one kid needs : 35M+35 - 32M= 3M+35

If M students get an average score of 64% out of 50, that is 32 marks. Then how much does the next student need to score to increase the average marks to 70% of 50, that is 35.

so the equation can be put as

32*M + x = 35*(M+1),

where 32*m is the total score of m students and x is the score needed by the next student. 35*(m+1) is the total score by M+1 students.