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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]  19 Jun 2018, 09:24
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40% (01:45) correct 59% (02:30) wrong based on 22 sessions
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

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Re: A student's score on the final test in a certain course was [#permalink]  19 Jun 2018, 22:14
Carcass 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. $$160%$$

Director
Joined: 07 Jan 2018
Posts: 597
Followers: 7

Kudos [?]: 525 [2] , given: 88

Re: A student's score on the final test in a certain course was [#permalink]  21 Jun 2018, 22:11
2
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We can assume two values for score in the first and second test.
To make calculation simple Let us say the student's score in the first test was 50 and in the second test was also 50.
The average of the two test would be 50.
The question mentions that the student's score on the final test was 60% more than the average of the two tests.
Hence the final test score of the student was 50 * 1.6 = 80

Now we need to find the average of the three tests because we have to compare this with the final test score.
Avg of the three tests = $$\frac{50 + 50 + 80}{3}$$ = $$\frac{180}{3}$$ = 60
We have all the information needed to answer the question.
Therefore final score/avg of three scores * 100 = the final score as a % of avg of the three scores.
$$\frac{80}{60}$$ * $$100$$ = $$133.33%$$
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This is my response to the question and may be incorrect. Feel free to rectify any mistakes

Manager
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Kudos [?]: 7 [0], given: 60

Re: A student's score on the final test in a certain course was [#permalink]  17 Dec 2018, 10:10

Last edited by QuantumWonder on 17 Dec 2018, 10:13, edited 1 time in total.
Manager
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Kudos [?]: 7 [0], given: 60

Re: A student's score on the final test in a certain course was [#permalink]  17 Dec 2018, 10:12
Carcass 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. $$160%$$

student's final test score: x1
avg of 2 other tests: (x2+x3)/2

The student's score on the final test was what percent of the student's average test score for the entire course?

x1/(avg of 3 tests)
.8(x2 + x3)
----------
.6(x2 + x3)

= 1.33 = 133%

x1 = 1.6 (x2+x3)/2
x1 = .8(x2+ x3)

the average of all three tests: (x1 + x2 + x3)/3
.8(x2 + x3) + x2 + x3
----------------------
3

1.8(x2 + x3)
-------------
3

= .6(x2 + x3)

The student's score on the final test was what percent of the student's average test score for the entire course?

Well you get that by dividing student's final score to the total average

.8(x2+x3)
----------
.6(x2 + x3)

= 130%
Manager
Joined: 02 Dec 2018
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Kudos [?]: 7 [0], given: 60

Re: A student's score on the final test in a certain course was [#permalink]  17 Dec 2018, 10:14
amorphous wrote:
We can assume two values for score in the first and second test.
To make calculation simple Let us say the student's score in the first test was 50 and in the second test was also 50.
The average of the two test would be 50.
The question mentions that the student's score on the final test was 60% more than the average of the two tests.
Hence the final test score of the student was 50 * 1.6 = 80

Now we need to find the average of the three tests because we have to compare this with the final test score.
Avg of the three tests = $$\frac{50 + 50 + 80}{3}$$ = $$\frac{180}{3}$$ = 60
We have all the information needed to answer the question.
Therefore final score/avg of three scores * 100 = the final score as a % of avg of the three scores.
$$\frac{80}{60}$$ * $$100$$ = $$133.33%$$

My question is, when do you know when it is safe to plug in numbers? It wasn't obvious to me that it would have been safe to plug in numbers for this problem without changing the answer.
Director
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Re: A student's score on the final test in a certain course was [#permalink]  19 Dec 2018, 00:43
1
KUDOS
QuantumWonder wrote:
amorphous wrote:
We can assume two values for score in the first and second test.
To make calculation simple Let us say the student's score in the first test was 50 and in the second test was also 50.
The average of the two test would be 50.
The question mentions that the student's score on the final test was 60% more than the average of the two tests.
Hence the final test score of the student was 50 * 1.6 = 80

Now we need to find the average of the three tests because we have to compare this with the final test score.
Avg of the three tests = $$\frac{50 + 50 + 80}{3}$$ = $$\frac{180}{3}$$ = 60
We have all the information needed to answer the question.
Therefore final score/avg of three scores * 100 = the final score as a % of avg of the three scores.
$$\frac{80}{60}$$ * $$100$$ = $$133.33%$$

My question is, when do you know when it is safe to plug in numbers? It wasn't obvious to me that it would have been safe to plug in numbers for this problem without changing the answer.

When the ans required is in percent then try to work with numbers because we would be comparing in relative basis rather than absolute value.
_________________

This is my response to the question and may be incorrect. Feel free to rectify any mistakes

Re: A student's score on the final test in a certain course was   [#permalink] 19 Dec 2018, 00:43
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