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A student's score on the final test in a certain course was

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A student's score on the final test in a certain course was [#permalink]

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05 Feb 2014, 12:58
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A student's score on the final test in a certain course was 60% greater than the average (arithmetic mean) score of the 2 other tests the student took in the course. The student's score on the final test was what percent of the student's average test score for the entire course?

A. $$33\frac{1}{3}%$$

B. 40%

C. 75%

D. $$133\frac{1}{3}%$$

E. 160%
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Apr 2014, 06:33, edited 2 times in total.
Edited the question.

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Re: A student's score on the final test in a certain course was [#permalink]

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05 Feb 2014, 17:55
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Given :
Let scores be s1,s2,s3
$$s3 = 1,6 (\frac{s1}{2}+ \frac{s2}{2})$$

(1) rearranging this :
$$s1+s2 = \frac{5}{4}*s3$$

(2) Average of 3 score is =
$$\frac{1}{3}*(s1+s2+s3)$$

We can use the information at (1) to replace $$s1+s2$$ into equation (2)

$$Average =\frac{1}{3} * ( \frac{5}{4}*s3 +s3)$$

$$Average = \frac{1}{3} * ( \frac{9}{4}*s3)$$

$$Average = \frac{3}{4}*s3$$

$$s3 = \frac{4}{3}*Average$$

$$s3 = 1.33333 * Average$$

making s3 133% of the Average

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Re: A student's score on the final test in a certain course was [#permalink]

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06 Mar 2014, 01:07
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Say Test1 = 100

Test2 = 100

Avg of Test1 & Test2 = 100

Final test is 60% more than above average = 160

Avg of all 3 tests combined $$= \frac{360}{2} = 120$$

% increase = $$\frac{160}{120}* 100 = 133 \frac{1}{3}$$

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Re: A student's score on the final test in a certain course was [#permalink]

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06 Mar 2014, 01:51
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Expert's post
goodyear2013 wrote:
A student's score on the final test in a certain course was 60% greater than the average (arithmetic mean) score of the 2 other tests the student took in the course. The student's score on the final test was what percent of the student's average test score for the entire course?

A. 33(1/3)%
B. 40%
C. 75%
D. 133(1/3)%
E. 160%

Hi, I want to request the solution for this question, please.

You can use the concept of averages here: The score in the third test was 60% higher (say, it was 160) which means you have 60% extra which you should give equally to each of the three tests to get the average up to 120.
We need Final test score as a % of updated average score = (160/120)*100 = 133.33%

To read up on the concept of averages, check: http://www.veritasprep.com/blog/2012/04 ... etic-mean/
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Re: A student's score on the final test in a certain course was [#permalink]

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24 Oct 2017, 06:12
goodyear2013 wrote:
A student's score on the final test in a certain course was 60% greater than the average (arithmetic mean) score of the 2 other tests the student took in the course. The student's score on the final test was what percent of the student's average test score for the entire course?

A. $$33\frac{1}{3}%$$

B. 40%

C. 75%

D. $$133\frac{1}{3}%$$

E. 160%

We can let a be the score on the final test and b and c be the scores on the other two tests. Thus, a = 1.6(b + c)/2 = 0.8(b + c), and the average score of all three tests is:

(a + b + c)/3 = (0.8(b + c) + b + c)/3 = 1.8(b + c)/3 = 0.6(b + c)

Thus, the final test score is 0.8(b + c)/[0.6(b + c)] x 100 = 0.8/0.6 x 100 = 8/6 x 100 = 133 ⅓ percent of the average score.

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Re: A student's score on the final test in a certain course was   [#permalink] 24 Oct 2017, 06:12
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