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At a certain school, all 118 juniors have an average (arithm
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30 Aug 2018, 07:28
30 Aug 2018, 07:28
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D
Question Stats:
81% (01:03) correct
18% (01:12) wrong based on 144 sessions
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At a certain school, all 118 juniors have an average (arithmetic mean) final exam score of 88 and all 100 seniors have an average final exam score of 92.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The average (arithmetic mean) final exam score for all of the juniors and seniors combined |
90 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: At a certain school, all 118 juniors have an average (arithm
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25 Nov 2018, 11:19
25 Nov 2018, 11:19
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Isn't this a lot of calculation for a GRE question?
Re: At a certain school, all 118 juniors have an average (arithm
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25 Nov 2018, 14:02
25 Nov 2018, 14:02
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Answer: B
__________________________________________________
WHAT WE HAVE:
118 juniors
Average(mean) score : 88
100 seniors
Average(mean) score : 92
__________________________________________________
FIRST APPROACH (no need for calculations)
If the number of juniors was equal to seniors then mean would be mean of 92 and 88, so it would be 90, but now juniors are more than seniors and as their mean is less than 90, the total mean would be something less than 90. So B is bigger than A.
__________________________________________________
SECOND APPROACH (with calculations)
A: The average (arithmetic mean) final exam score for all of the juniors and seniors combined
A = (total seniors score + total juniors score)/ (seniors population + juniors population)
A = (118*88 + 100*92) / (118+100) = (118*44 + 50*92) / ( 109) = 89.83
B = 90
So B is bigger than A.
__________________________________________________
WHAT WE HAVE:
118 juniors
Average(mean) score : 88
100 seniors
Average(mean) score : 92
__________________________________________________
FIRST APPROACH (no need for calculations)
If the number of juniors was equal to seniors then mean would be mean of 92 and 88, so it would be 90, but now juniors are more than seniors and as their mean is less than 90, the total mean would be something less than 90. So B is bigger than A.
__________________________________________________
SECOND APPROACH (with calculations)
A: The average (arithmetic mean) final exam score for all of the juniors and seniors combined
A = (total seniors score + total juniors score)/ (seniors population + juniors population)
A = (118*88 + 100*92) / (118+100) = (118*44 + 50*92) / ( 109) = 89.83
B = 90
So B is bigger than A.
