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C
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Difficulty:
55%
(hard)
Question Stats:
69% (02:38) correct
31%
(02:21)
wrong
based on 35
sessions
History
Date
Time
Result
Not Attempted Yet
The arithmetic mean of the scores of a group of students in a test was 52. The top 20% of the students obtained a mean score of 80, while the bottom 25% obtained a mean score of 31. What was the mean score of the remaining 55% of the students?
(A) 45%
(B) 50%
(C) 51.4%
(D) 54.6%
(E) 58.2%
(A) 45%
(B) 50%
(C) 51.4%
(D) 54.6%
(E) 58.2%
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12 Sep 2019, 14:40
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Average score= \(52-[\frac{(20*28)-(25*21)}{55}]\)
Average score=\(52-{\frac{35}{55}}\)= 52-0.64=51.36
D
Average score=\(52-{\frac{35}{55}}\)= 52-0.64=51.36
D
Hovkial
12 Sep 2019, 15:03
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Quote:
Answer is D.
The question tries to throw you off by mentioning "top" and "bottom" scorers. Distribution does not matter in the mean, it is simply an average. The information that the question provides is that the OVERALL (100%) mean of 52 is made up of 3 smaller components, 25% of the 100% has a mean of 31 and 20% of the 100% has a mean of 80. The remaining 55% has a mean of "x", which is what we are trying to find.
Write this out in a simple equation:
52 = 0.2(80) + 0.25(31) + 0.55(x)
Multiply using mental math:
52 = 16 + 7.75 + 0.55x
52 = 23.75 + 0.55x
28.25 = 0.55x
Multiply the entire equation by 100 (shift decimals to the right by two places) and...
2825 = 55x
Therefore x = 2825/55 = 565/11
To approximate, lets say 565/10. We know 565/11 is smaller than 565/10 because the denominator is larger. Therefore the answer must be slightly smaller than but fairly close to 56.5%... The answer is D.
