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

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enigma123
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The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   17 Nov 2011, 16:45
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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) \((54,821)^2 – 1\)
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D
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Bunuel
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The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   11 Sep 2012, 02:02
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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) \((54,821)^2 – 1\)


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, 03:04
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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!

Answer: D
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   17 Nov 2011, 21:53
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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: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   10 Sep 2012, 23:53
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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
Prashant
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   10 Dec 2012, 04:42
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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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The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   14 Sep 2013, 11:30
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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) \((54,821)^2 – 1\)


If I come across this question in a test, I would just take some small values to convince myself.
Say \(\frac{(2^2 + 4^2)}{2} = 10\)
which can also be represented as \(3^2 + 1\)
A couple more such examples and the pattern would be convincing.
Say \(\frac{(4^2 + 6^2)}{2} =\frac{(16 + 36)}{2} = 26\)
\(5^2 + 1 = 26\)

If you insist of using algebra, average of \((a - 1)^2\) and \((a+1)^2\) = \(\frac{[(a-1)^2 + (a+1)^2]}{2} = a^2 + 1\)
Hence answer (D)
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   24 Jul 2014, 20:13
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, 20:40
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\(\frac{54820^2 + 54822^2}{2}\)

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

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

\(= 54820^2 + 2 * 54820 + 1 + 1\)

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

\(= 54821^2 + 1\)

Answer = D
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   25 Jul 2014, 00:30
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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   29 Sep 2014, 22:33
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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

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

Answer choice D
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   14 Mar 2016, 02:04
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.

Answer: D.


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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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   14 Mar 2016, 02:14
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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.

Answer: D.


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] New post   25 Sep 2019, 05:45
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 54821 = A

((A-1)^2 + (A+1)^2)/2 = A^2 + 1

ANSWER: D
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   01 Oct 2020, 04:31
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Let p = 54,820 . Then , 548,22 = 54,820 + 2 = p + 2

=> \(p^2 + (p + 2)^2 = p^2 + p^2 + 4 + 4p\)

=> \(p^2 + (p + 2)^2 = 2p^2 + 4 + 4p\)

Average: \(\frac{(p^2 + (p + 2)^2) }{ 2} = \frac{(2p^2 + 4 + 4p) }{ 2}\)

=> \(p^2 + 2 + 2p: p^2 + 1 + 4p + 1\)

=> \((p+1)^2\) + 1

=> \((54,820 + 1)^2\) + 1

=> \((54,821)^2\) + 1

Answer D
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   24 Oct 2020, 06:08
My approach:

Plug numbers-

\(2^2 + 4^2 \)= \(\frac{20 }{ 2 }= 10\)

\(10 = 3^2 + 1\)

\(10^2 + 12^2 = \frac{244 }{ 2} = 122\)

\(122 = 11^2 + 1\)

54,821 is between 54,820 and 54,822. Thus, the answer is (D) \((54,821)^2 + 1\)
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   25 Sep 2023, 05:01
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) \((54,821)^2 – 1\)

avigutman sir can i do this question with the help of weighted average approach? i.e.
2^2 + 4^2 = 4(1) + 4(4)
the scale factor of the second one is bigger so the mean will be closer to it
only one answer states it clearly which is d
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   25 Sep 2023, 06:00
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pdfff wrote:
avigutman sir can i do this question with the help of weighted average approach? i.e.
2^2 + 4^2 = 4(1) + 4(4)
the scale factor of the second one is bigger so the mean will be closer to it
only one answer states it clearly which is d

I don't understand, pdfff. The two terms should have equal weights here.
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post   25 Sep 2023, 07:44
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) \((54,821)^2 – 1\)

I will take small no's to solve the big issue....

Mean of \(10^2\) & \(12^2\) = \(\frac{100 + 144}{2}=\frac{244}{2}=122\), Units digit will be \(\frac{0^2 + 2^2}{2}=2\)

Applying the same logic here as well \(\frac{0^2 + 2^2}{2}=2\)

In option (D) \((54,821)^2\) = Units digit as 1 , SO \((54,821)^2 + 1\) = Units digit as 2 (We found out earlier, hence Answer gotta be (D)
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
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