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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]

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17 Nov 2011, 15:45
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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]

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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]

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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)

-----------------------
Thanks
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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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.

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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10 Dec 2012, 02:04
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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!

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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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]

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03 Jul 2013, 00:23
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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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]

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24 Jul 2014, 06:56
Hello from the GMAT Club BumpBot!

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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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]

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24 Jul 2014, 19:40
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$$\frac{54820^2 + 54822^2}{2}$$

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

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

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

$$= (54820 + 1)^2 + 1$$

$$= 54821^2 + 1$$

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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24 Jul 2014, 23:30
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]

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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.
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The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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29 Sep 2014, 21:33
Let a = 54820
Then, we need to find
$$\frac{a^2 + (a + 2)^2}{2}$$ => $$\frac{a^2 + a^2 + 4a + 4}{2}$$

$$\frac{2a^2+ 4a + 4}{2}$$ => $$\frac{2(a^2 + 2a + 2)}{2}$$

$$a^2 + 2a + 1 + 1$$ => $$(a + 1)^2 + 1$$

Substitute for a

$$(54820 + 1)^2 + 1$$ => $$(54821)^2 + 1$$

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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29 Sep 2014, 21:52
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

Let a = 54820
Then, we need to find
$$\frac{a^2 + (a + 2)^2}{2}$$ => $$\frac{a^2 + a^2 + 4a + 4}{2}$$

$$\frac{2a^2+ 4a + 4}{2}$$ => $$\frac{2(a^2 + 2a + 2)}{2}$$

$$a^2 + 2a + 1 + 1$$ => $$(a + 1)^2 + 1$$

Substitute for a

$$(54820 + 1)^2 + 1$$ => $$(54821)^2 + 1$$

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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02 Oct 2015, 09:14
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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14 Mar 2016, 01:04
Bunuel wrote:

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.

Hi,

Concerning last digits, I only know that we can add or multiply last digits of various numbers to determine the last digits of their sum or product. Can somebody tell me the rule for division of last digits (as presented by Bunuel)?

Another question: when and only when a multiple of 4 is divided by 2, the last digit of the resulting quotient will always be 2? Is there any other rules to memorize?

Thank you very much!
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The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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14 Mar 2016, 01:14
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truongynhi wrote:
Bunuel wrote:

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.

Hi truongynhi

Concerning last digits, I only know that we can add or multiply last digits of various numbers to determine the last digits of their sum or product. Can somebody tell me the rule for division of last digits (as presented by Bunuel)?

Quote:
Another question: when and only when a multiple of 4 is divided by 2, the last digit of the resulting quotient will always be 2? Is there any other rules to memorize?

Thank you very much!

Hi,

Quote:
Another question: when and only when a multiple of 4 is divided by 2, the last digit of the resulting quotient will always be 2? Is there any other rules to memorize?

This is not CORRECT..
In the given context, last digit 4 and multiple of 4 are interlinked, so just stating one that multiple of 4 when divided by 2 gives 2 will be wrong

reason-

-
8 is a multiple of 4 and will give last digit as 4 and NOT 2..

what is meant here is
we have seen that the last digit of SUM is 4 and this SUM is multiple of 4..
A number Multiple of 4, with last digit 4 is 04,24, 44,64,84 as last two digits, so when div by 2, in each case last digit is 2..
the number cannot be ending in 14,34 etc as Property of 4 is last two digits should be div by 4 if the entire number is to be div by 4..

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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10 Dec 2016, 03:03
Here is my solution to this one -->

Let 54822=t
Hence sum of the two numbers becomes => (t-2)^2+t^2 = 2t^2+4-4t
Hence mean => t^2+2-2t => t^2-2t+1+1 => (t-1)^2+1 = 54821^2 +1

Hence D

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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20 Dec 2016, 05:54
One more cyclicity question.
Fisrt, we should find out the unit digit of the mean of 54820^2 and 54822^2.
To figure it out we need find out last digits of both numbers. Last number of 54820^2 will be 0, last number of 54822^2 will be 4.
Mean will be (0+4)/2 = 2, thus unit digit of the mean number will be 2.
Analyzing all options, it is clear that only one option gives us this value, Option D 54821^2+1
Re: The mean of (54,820)^2 and (54,822)^2 =   [#permalink] 20 Dec 2016, 05:54

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