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The mean of (54,820)^2 and (54,822)^2 =

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The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post 17 Nov 2011, 15:45
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A
B
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D
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Question Stats:

59% (02:02) correct 41% (01:12) wrong based on 140 sessions
The mean of (54,820)^2 and (54,822)^2 =

(A) (54,821)^2
(B) (54,821.5)^2
(C) (54,820.5)^2
(D) (54,821)^2 + 1
(E) (54,821)^2 – 1
[Reveal] Spoiler: OA

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Re: Mean of 2 numbers [#permalink] New post 17 Nov 2011, 20:53
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mean of (54,820)^2 and (54,822)^2

=> [(54,820)^2 +(54,822)^2]/2 (using a^2 + b^2 = (a-b)^2 + 2ab)
=> [(54,82-54,820)^2 + 2*54,820*54,822]/2
=> [2^2 + 2*54,820*54,822]/2
=> [2 + 54,820*54,822]
=> [2 + (54,821-1)*(54,821+1)] (using (a+b)(a-b) = a^2-b^2)
=> [2 + 54,821^2 - 1]
=> 54,821^2 +1

and D
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Re: Mean of 2 numbers [#permalink] New post 10 Sep 2012, 22:53
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(54820)^2 + (54822)^2 = (54821-1)^2 + (54821+1)^2
= 2 (54821)^2 + 2

So mean = (54821)^2 + 1
(D)

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post 11 Sep 2012, 01:02
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enigma123 wrote:
The mean of (54,820)^2 and (54,822)^2 =

(A) (54,821)^2
(B) (54,821.5)^2
(C) (54,820.5)^2
(D) (54,821)^2 + 1
(E) (54,821)^2 – 1


APPROACH #1:

Test some small numbers: \frac{2^2+4^2}{2}=10=3^2+1 or: \frac{4^2+6^2}{2}=26=5^2+1.

APPROACH #2:

Say 54,821=x, then \frac{54,820^2+54,822^2}{2}=\frac{(x-1)^2+(x+1)^2}{2}=x^2+1=54,821^2+1.

APPROACH #3:

The units digit of 54,820^2+54,822^2 is 0+2=4. Now, since 54,820^2+54,822^2 must be a multiple of 4, then \frac{54,820^2+54,822^2}{2} must have the units digit of 2. Only answer choice D fits.

Answer: D.
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post 10 Dec 2012, 02:04
Shrink the monster technique.
The two numbers are two consecutive even numbers raised to 2. Shrink the monster and think of baby numbers such as 0 and 2.

mean=\frac{{0^2 + 2^2}}{2}=2

Substitute you baby numbers to the answer choices.

(A) {0+1}^{2}=1 Eliminate!
(B) {0+1.5}^{2}=2.25 Eliminate!
(C) {0+0.5}^{2}=0.25 Eliminate!
(D) {0+1}^{2}+1=2 Bingo!
(E) {0+1}^{2}-1=0 Eliminate!

Answer: D
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post 10 Dec 2012, 03:42
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Ans: Replace 54820 by x and 54822 by (x+2) and then solve by algebra. Mean of x^2 and (x+2)^2 is x^2+2x+2 which can be written as x^2+2x+1+1. So the answer is (D)
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post 03 Jul 2013, 00:23
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post 04 Jul 2013, 02:58
Let 54820 = x

Equations reduces to

{x^2 + (x+2)^2}/2 = (x^2 + x^2 + 4x + 4)/2 = x^2+2x+2 = x^2 + 2x + 1 + 1 = (X+1)^2 + 1 = 54821^2 + 1
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post 24 Jul 2014, 06:56
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post 24 Jul 2014, 19:13
Bunuel how do you know the expression is a multiple of 4? Wouldn't I need the tens and unit digit? Not just units?
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post 24 Jul 2014, 19:40
\frac{54820^2 + 54822^2}{2}

= \frac{54820^2 + (54820 + 2)^2}{2}

= \frac{54820^2 + 54820^2 + 2 * 54820 * 2 + 2 * 2}{2}

= 54820^2 + 2 * 54820 + 1 + 1

= (54820 + 1)^2 + 1

= 54821^2 + 1

Answer = D
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post 24 Jul 2014, 23:30
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bankerboy30 wrote:
Bunuel how do you know the expression is a multiple of 4? Wouldn't I need the tens and unit digit? Not just units?


54,820^2 + 54,822^2 = even^2 + even^2 = multiple of 4 + multiple of 4 = multiple of 4.

Hope it's clear.
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post 26 Jul 2014, 11:42
enigma123 wrote:
The mean of (54,820)^2 and (54,822)^2 =

(A) (54,821)^2
(B) (54,821.5)^2
(C) (54,820.5)^2
(D) (54,821)^2 + 1
(E) (54,821)^2 – 1



So the question is avg of (x-1)^2 and (x+1)^2. Solve it and you get x^2 + 1. x here is 54,821 so the answer is option D.
Re: The mean of (54,820)^2 and (54,822)^2 =   [#permalink] 26 Jul 2014, 11:42
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