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The arithmetic average of 20 numbers is A. The arithmetic average of

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The arithmetic average of 20 numbers is A. The arithmetic average of  [#permalink]

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20 Mar 2018, 01:20
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The arithmetic average of 20 numbers is A. The arithmetic average of 10 of the number is 16. In terms of A, what is the arithmetic average of the remaining 10 numbers?

(A) A - 8
(B) 16 - A
(C) 16 - 2A
(D) 2A - 16
(E) 20A - 6

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The arithmetic average of 20 numbers is A. The arithmetic average of  [#permalink]

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20 Mar 2018, 01:35
Bunuel wrote:
The arithmetic average of 20 numbers is A. The arithmetic average of 10 of the number is 16. In terms of A, what is the arithmetic average of the remaining 10 numbers?

(A) A - 8
(B) 16 - A
(C) 16 - 2A
(D) 2A - 16
(E) 20A - 6

Questions involving average and standard deviation can often be solved without explicit calculation, by using the underlying properties.
We'll look for such a solution, a Logical approach.

Since the average of all 20 numbers is A and we divided the numbers into two equal sets, we know they must 'balance out' to reach A.
That is, if the first 10 numbers have an average above A, the second must have an average below A, by the same amount.
So our remaining 10 numbers have an average of A + (A - 16) if A>16 or A - (16 - A) if 16>A
Either way, this comes to 2A - 16.

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Re: The arithmetic average of 20 numbers is A. The arithmetic average of  [#permalink]

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20 Mar 2018, 01:59
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Bunuel wrote:
The arithmetic average of 20 numbers is A. The arithmetic average of 10 of the number is 16. In terms of A, what is the arithmetic average of the remaining 10 numbers?

(A) A - 8
(B) 16 - A
(C) 16 - 2A
(D) 2A - 16
(E) 20A - 6

I like to solve average related questions by using the formula -

Average * no. of elements = Total.

By this method -> Sum of 20 numbers = 20*A.
Sum of first 10 numbers -> 10*16 = 160.
Sum of remaining 10 numbers -> 20*A - 160

Hence average = sum / # of elements

$$Average = (20*A - 160)/10$$
$$Average = 2*A - 16$$

Hence D.

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Re: The arithmetic average of 20 numbers is A. The arithmetic average of  [#permalink]

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20 Mar 2018, 09:17
Bunuel wrote:
The arithmetic average of 20 numbers is A. The arithmetic average of 10 of the number is 16. In terms of A, what is the arithmetic average of the remaining 10 numbers?

(A) A - 8
(B) 16 - A
(C) 16 - 2A
(D) 2A - 16
(E) 20A - 6

Total of the 20 numbers is 20A & The Total of the 10 numbers is 160

So, Total of the remaining 10 numbers is 20A - 160

Thus, the average of the remaining 10 numbers is 2A - 16, Answer must be (D)
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Re: The arithmetic average of 20 numbers is A. The arithmetic average of  [#permalink]

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20 Mar 2018, 10:04
Bunuel wrote:
The arithmetic average of 20 numbers is A. The arithmetic average of 10 of the number is 16. In terms of A, what is the arithmetic average of the remaining 10 numbers?

(A) A - 8
(B) 16 - A
(C) 16 - 2A
(D) 2A - 16
(E) 20A - 6

Arithmetic average of the remaining 10 numbers = $$\frac{(20*A - 10*16)}{10}$$ = 2*A - 16 => Answer (D)
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Re: The arithmetic average of 20 numbers is A. The arithmetic average of  [#permalink]

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20 Mar 2018, 11:03
Bunuel wrote:
The arithmetic average of 20 numbers is A. The arithmetic average of 10 of the number is 16. In terms of A, what is the arithmetic average of the remaining 10 numbers?

(A) A - 8
(B) 16 - A
(C) 16 - 2A
(D) 2A - 16
(E) 20A - 6

let x=average of remaining 10 numbers
(16+x)/2=A
x=2A-16
D
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Re: The arithmetic average of 20 numbers is A. The arithmetic average of  [#permalink]

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20 Mar 2018, 11:34
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Bunuel wrote:
The arithmetic average of 20 numbers is A. The arithmetic average of 10 of the numbers is 16. In terms of A, what is the arithmetic average of the remaining 10 numbers?

(A) A - 8
(B) 16 - A
(C) 16 - 2A
(D) 2A - 16
(E) 20A - 6

Here's a different approach.
Let's create a nice set of values that satisfy the given conditions.

Let's say the set of 20 numbers is {16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16}

The arithmetic average of 20 numbers is A
So, A = 16

The arithmetic average of 10 of the numbers is 16
So, the average of {16, 16, 16, 16, 16, 16, 16, 16, 16, 16} is clearly 16

What is the arithmetic average of the remaining 10 numbers?
In other words, what is the arithmetic average of {16, 16, 16, 16, 16, 16, 16, 16, 16, 16}
Well, the average is 16

So, we want an answer choice that yields an OUTPUT of 16 when A = 16

(A) 16 - 8 = 8. No good. We want an output if 16
(B) 16 - 16 = 0. No good. We want an output if 16
(C) 16 - 2(16) = -16. No good. We want an output if 16
(D) 2(16) - 16 = 16. PERFECT!
(E) 20(16) - 6 = 314. No good. We want an output if 16

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Re: The arithmetic average of 20 numbers is A. The arithmetic average of  [#permalink]

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21 Mar 2018, 16:16
Bunuel wrote:
The arithmetic average of 20 numbers is A. The arithmetic average of 10 of the number is 16. In terms of A, what is the arithmetic average of the remaining 10 numbers?

(A) A - 8
(B) 16 - A
(C) 16 - 2A
(D) 2A - 16
(E) 20A - 6

We know that average = sum/number, or average x number = sum. Thus, the sum of the 20 numbers is 20A. The sum of 10 of the numbers is 160.

So the sum of the remaining 10 numbers is 20A - 160 and the average is (20A - 160)/10 = 2A - 16.

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Re: The arithmetic average of 20 numbers is A. The arithmetic average of  [#permalink]

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22 Mar 2018, 10:54
totla sum of 20 numbers will be 20*A
sum of 10 Numbers will be 10*16=160

so sum of remaining 10 numbers will be 20*A- 160

So average will be 20*A-160/10= 2A-16

Re: The arithmetic average of 20 numbers is A. The arithmetic average of   [#permalink] 22 Mar 2018, 10:54
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