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Set A contains 120 terms with an average of 8.2. Set B contains 240

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Set A contains 120 terms with an average of 8.2. Set B contains 240  [#permalink]

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New post 28 Jul 2017, 09:08
1
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

89% (01:57) correct 11% (02:11) wrong based on 72 sessions

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Re: Set A contains 120 terms with an average of 8.2. Set B contains 240  [#permalink]

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New post 28 Jul 2017, 10:17
2
1
Bunuel wrote:
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


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
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Re: Set A contains 120 terms with an average of 8.2. Set B contains 240  [#permalink]

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New post 28 Jul 2017, 11:00
Bunuel wrote:
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


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
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Re: Set A contains 120 terms with an average of 8.2. Set B contains 240  [#permalink]

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New post 30 Jul 2017, 00:17
2
Bunuel wrote:
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


Total terms in set B is 240 and set A is 120
Now Average will be more towards 10.6

8.2................10.6
Difference is 2.4 divide it into 3 parts (2:1) = 0.8
10.6-0.8 = 9.8

D
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Set A contains 120 terms with an average of 8.2. Set B contains 240  [#permalink]

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New post 31 Jul 2017, 07:47
1
Bunuel wrote:
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

I used neither alligation nor difference between averages, but I simplified the arithmetic.

Number of terms determines the weight of each value.

120 terms and 240 terms have the same weight as 1 term and 2 terms, respectively. \(\frac{120}{120}\) = 1, \(\frac{240}{120}\) = 2

1 term * average 8.2 = \(Sum_1\) of 8.2
2 terms * ave. 10.6 = \(Sum_2\) of 21.2

Total sum: (8.2 + 21.2) = 29.4
Total terms: (1 + 2) = 3

Average = total sum/ total # of terms

\(\frac{29.4}{3}\) = 9.8, Answer D
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Set A contains 120 terms with an average of 8.2. Set B contains 240   [#permalink] 31 Jul 2017, 07:47
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Set A contains 120 terms with an average of 8.2. Set B contains 240

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