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

Question

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

Bunuel · Accepted Answer

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.

mbaiseasy · Answer

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

Sudhanshuacharya · Answer

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

=>  /2 (using a^2 + b^2 = (a-b)^2 + 2ab)
=>  /2
=>  /2
=>  
=>   (using (a+b)(a-b) = a^2-b^2)
=>  
=> 54,821^2 +1

and D

Prashant1984 · Answer

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

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

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

priyamne · Answer

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)

KarishmaB · Answer

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{ }{2} = a^2 + 1 
Hence answer (D)

bankerboy30 · Answer

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

PareshGmat · Answer

\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

Bunuel · Answer

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

Hope it's clear.

sagarjain90 · Answer

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

truongynhi · Answer

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!

chetan2u · Answer

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..

dabaobao · Answer

Let 54821 = A

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

ANSWER: D

MathRevolution · Answer

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

Basshead · Answer

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

pdfff · Answer

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

avigutman · Answer

I don't understand,  pdfff . The two terms should have equal weights here.

Abhishek009 · Answer

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)

GMATinsight · Answer

Use some example to see th epattern such as

\frac{2^2 + 4^2 }{2} = 10 =3^2 +1 
 \frac{3^2 + 5^2 }{2} = 17 =4^2+1 
 \frac{4^2 + 6^2 }{2} = 26 =5^2+1

i.e.  \frac{54820^2 + 54822^2 }{2} = 54821^2+1

Hence Option D

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

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

Prep Toolkit

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