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

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.
_________________

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

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

(54820)^2 + (54822)^2 = (54821-1)^2 + (54821+1)^2
= 2 (54821)^2 + 2

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

-----------------------
Thanks
Prashant

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)

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)
_________________

Bunuel how do you know the expression is a multiple of 4? Wouldn't I need the tens and unit digit? Not just units?

\(\frac{54820^2 + 54822^2}{2}\)

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

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

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

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

\(= 54821^2 + 1\)

Answer = D

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

Hope it's clear.
_________________

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

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!

Hi,



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..
_________________

Let 54821 = A

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

ANSWER: D

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
_________________

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) \((54,821)^2 + 1\)
_________________

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________