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In a particular course, Arif appeared in 10 quizzes.The average of his

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Joined: 31 Oct 2013
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Concentration: Accounting, Finance
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In a particular course, Arif appeared in 10 quizzes.The average of his  [#permalink]

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New post 27 Jan 2019, 02:39
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76% (02:39) correct 24% (01:54) wrong based on 21 sessions

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In a particular course, Arif appeared in 10 quizzes.The average of his best 9 quizzes is 10% more than the average of all the quizzes he attended. The total marks obtained in best 9 quizzes is what % of the total marks obtained in 10 quizzes?


A) 80%
B) 88%
C) 90%
D) 95%
E) 99%
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Re: In a particular course, Arif appeared in 10 quizzes.The average of his  [#permalink]

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New post 27 Jan 2019, 04:09
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selim wrote:
In a particular course, Arif appeared in 10 quizzes.The average of his best 9 quizzes is 10% more than the average of all the quizzes he attended. The total marks obtained in best 9 quizzes is what % of the total marks obtained in 10 quizzes?


A) 80%
B) 88%
C) 90%
D) 95%
E) 99%


Assuming the average of 10 quizzes as 100 and the average of 9 quizzes as 110(10% more than avg of 10 quizzes).
Total marks of 9 quizzes is 990(110*9) and total marks of 10 quizzes is 1000(100*10).
So answer is 990/1000 = 99% - D

Algebraically, let x be the sum of 9 quizzes and y be the sum of 10 quizzes.

Given x/9 - y/10 = (1/10) * (y/10)

On solving, we get x/y=99/100 - D

Cheers!
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Re: In a particular course, Arif appeared in 10 quizzes.The average of his  [#permalink]

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New post 27 Jan 2019, 04:29
selim wrote:
In a particular course, Arif appeared in 10 quizzes.The average of his best 9 quizzes is 10% more than the average of all the quizzes he attended. The total marks obtained in best 9 quizzes is what % of the total marks obtained in 10 quizzes?

A) 80%
B) 88%
C) 90%
D) 95%
E) 99%


IMO E

sum of x1 to x9 / 9 = (1.1 * x1 to x10)/ 10

= sum of x1 to x9 = 1.1 * 9 * sum of x1 to x10 ) * 100 / 10

= 99 %
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Re: In a particular course, Arif appeared in 10 quizzes.The average of his   [#permalink] 27 Jan 2019, 04:29
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