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M21 #28

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Re: M21 #28 [#permalink] New post 27 Jun 2012, 21:39
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% ?

(A) 3M + 20
(B) 3M + 35
(C) 4M + 15
(D) 4M + 20
(E) 4M + 45

i could not understand this .. pls guide, by plugging in some values....[/quote]


the ans is B..
I solved it by taking M as 1 . M scored 64% that is he scored 32 out of 50 to get a total of 70% after another student appear for exam he/she should score 38 ( (32 + x)/100 * 100 = 70 ; solving for x, we get 38 ). Now eliminating other options we get B
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Re: M21 #28 [#permalink] New post 28 Jun 2012, 12:03
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 .. .
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Re: M21 #28 [#permalink] New post 02 Jul 2012, 11:36
im not sure why every body is making such an easy problem sound so complicated
this can be made simpler by substitution method

the test contains 50 questions and currently M students keep it at 64% so i dont want to multiply some no withh 64 so i make the no of students=2
so total questions 100
no of correct answers in them =64%*100=64
one more student comes in so no of questions=150 in total
70% of 150=105
=> 105-64=41 questions have to be answered correctly by other student

we assumed m=2 i.e no if students

so by substitution in answer choices 3M+35 (option B ) only holds true this might seem long but a lot easier to do

hope it helps afterall we want to get answer the fastest way not the ideal way :)
Re: M21 #28   [#permalink] 02 Jul 2012, 11:36
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