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28 Jul 2017, 09:08
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Question Stats:
86% (01:53) correct
14%
(02:24)
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Set A contains 120 terms with an average of 8.2. Set B contains 240 terms with an average of 10.6. If Set A and B are combined what is the resulting average?
(A) 8.4
(B) 8.8
(C) 9.0
(D) 9.8
(E) 10.1
(A) 8.4
(B) 8.8
(C) 9.0
(D) 9.8
(E) 10.1
28 Jul 2017, 10:17
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Bunuel
Set A contains 120 terms with an average of 8.2. Set B contains 240 terms with an average of 10.6. If Set A and B are combined what is the resulting average?
(A) 8.4
(B) 8.8
(C) 9.0
(D) 9.8
(E) 10.1
(A) 8.4
(B) 8.8
(C) 9.0
(D) 9.8
(E) 10.1
Weighted average = (120*8.2 + 240*10.6) / 120 +240
=984+ 2544 / 360
=3528/360 = 9.8
The above method is calculation intensive.
Alternatively ,
Set A has 120 terms with average 8.2
Set B has 240 terms with average 10.6
Difference between average of set A and B = 2.4
Number of terms of B : Number of terms of A = 2:1
2x+x = 2.4
=> x = .8
So the weighted average will be in opposite ratio of relative weights
8.2+ 2*.8 = 9.8
Answer D
28 Jul 2017, 11:00
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Bunuel
Set A contains 120 terms with an average of 8.2. Set B contains 240 terms with an average of 10.6. If Set A and B are combined what is the resulting average?
(A) 8.4
(B) 8.8
(C) 9.0
(D) 9.8
(E) 10.1
(A) 8.4
(B) 8.8
(C) 9.0
(D) 9.8
(E) 10.1
Avg. A = 8.2
n (A) = 120
Total A = 120 * 8.2 = 12 * 82 = 984
Avg. B = 10.6
n (B) = 240
Total B = 240 * 10.6 = 24 * 106 = 2544
Resulting average = (12* 82 + 24 * 106 ) /(120 +240) = (12* 82 + 24 * 106 ) /(360) = (82 +212)/30 = 294/30 = 98/10 = 9.8
Answer D
