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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 14 Jun 2008, 21:27
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65% (02:03) correct 35% (01:02) wrong based on 52 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

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-mean-of-54-820-2-and-123303.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 14 Sep 2013, 10:29, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: PS: Manhattan Gmat Math [#permalink] New post 14 Jun 2008, 21:39
GMAT TIGER wrote:
The average of (54,820)^2and (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


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

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


Taking the average of above 2 , we get (54821)^2 +1

hence the answer is D
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Re: PS: Manhattan Gmat Math [#permalink] New post 14 Jun 2008, 21:43
mandy12 wrote:
GMAT TIGER wrote:
The average of (54,820)^2and (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


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

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


Taking the average of above 2 , we get (54821)^2 +1

hence the answer is D



nice approach. anymore approaches????????????
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Re: PS: Manhattan Gmat Math [#permalink] New post 14 Jun 2008, 21:46
GMAT TIGER wrote:
The average of (54,820)^2and (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


Answer D.
Let x=54,820 and y = 54,822 = x + 2.
Then average is (x^2 + y^2) / 2 = [(x^2) + (x + 2)^2] / 2 = (x^2 + x^2 + 4x + 4) /2 = x^2 +2x + 2 = (x + 1)^2 + 1
Now, sub into original numbers the average is (54,820 + 1)^2 + 1 = 54,821^2 = 1.
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Re: PS: Manhattan Gmat Math [#permalink] New post 15 Jun 2008, 05:24
GMAT TIGER wrote:
The average of (54,820)^2and (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 think the best way to deal with these type of problems

try to make em look simplified..suppose you have (a)^2+b^2.. you may want to make x-y=a, x+y=b..

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

2*54821^2+2/2=54821^2 + 1

D it is..
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Re: PS: Manhattan Gmat Math [#permalink] New post 15 Jun 2008, 08:41
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Re: PS: Manhattan Gmat Math [#permalink] New post 07 May 2011, 08:34
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One can try plugging numbers and see the pattern:

Take 2 and 4: average= 3
Now, for 2^2 and 4^2, average = 10 =3^2 +1

Take 4 and 6: average= 5
Now, for 4^2 and 6^2: average= 26= 5^2+1

We can quickly realize that average of 54820 and 54822= 54821
So, as per the pattern derived above, average of 54820^2 and 54822^2= 54821^2 +1

Answer: D
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Re: PS: Manhattan Gmat Math [#permalink] New post 07 May 2011, 12:16
54820 = x

54822 = x+2

[x^2 + (x+2) ^2]/2 = average = x^2 + 2x + 2
means the last digit has to be 2.

only option d gives that.
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Re: PS: Manhattan Gmat Math [#permalink] New post 07 May 2011, 13:01
Answer is D.


(54820^2 + (54820+2)^2)/2

= 54820^2+2*54820+2

= (54821)^2 +1
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Re: PS: Manhattan Gmat Math [#permalink] New post 07 May 2011, 18:55
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GMAT TIGER wrote:
The average of (54,820)^2and (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: PS: Manhattan Gmat Math [#permalink] New post 08 May 2011, 05:48
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I tried the following approach :

54820^2 - unit's digit = 0

54822^2 - unit's digit = 4

So average will have unit's digit as (4+0)/2 = 2

Only choice D has that, so answer - D
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Re: PS: Manhattan Gmat Math [#permalink] New post 09 May 2011, 03:17
subhashghosh wrote:
I tried the following approach :

54820^2 - unit's digit = 0

54822^2 - unit's digit = 4

So average will have unit's digit as (4+0)/2 = 2

Only choice D has that, so answer - D


Nicely thought! +1
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Re: The average of (54,820)^2and (54,822)^2 =  [#permalink] New post 14 Sep 2013, 10:18
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink] New post 14 Sep 2013, 10:30
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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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-mean-of-54-820-2-and-123303.html
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Re: The mean of (54,820)^2 and (54,822)^2 =   [#permalink] 14 Sep 2013, 10:30
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