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

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Senior Manager
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Set A contains 120 terms with an average of 8.2. Set B conta [#permalink]  15 Oct 2017, 22:56
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99% (01:05) correct 0% (00:00) wrong based on 7 sessions
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

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[Reveal] Spoiler: OA
Director
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Re: Set A contains 120 terms with an average of 8.2. Set B conta [#permalink]  15 Oct 2017, 23:41
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We can solve it using weighted average as $$\frac{120*8.2+240*10.4}{360}=9.8$$

Answer D
Manager
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Re: Set A contains 120 terms with an average of 8.2. Set B conta [#permalink]  23 Jan 2019, 23:45
Expert's post
If you wanna avoid a calculator ( which I recommend for easy computations like this)

lets use some basic arithmetic.

(8.2*120 + 10.6*240)/360 can be simplified since the numbers are multiples of 120, if you take that number out of the expression we have.

(8.2+ 10.6*2)/3 = (8.2+21.2)/3= 29.4/3= 9.8
Re: Set A contains 120 terms with an average of 8.2. Set B conta   [#permalink] 23 Jan 2019, 23:45
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# Set A contains 120 terms with an average of 8.2. Set B conta

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