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

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The average of (54,820)^2and (54,822)^2 =  [#permalink]  14 Jun 2008, 22:27
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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
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Re: PS: Manhattan Gmat Math [#permalink]  14 Jun 2008, 22:39
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

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Re: PS: Manhattan Gmat Math [#permalink]  14 Jun 2008, 22:43
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

nice approach. anymore approaches????????????
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Re: PS: Manhattan Gmat Math [#permalink]  14 Jun 2008, 22:46
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

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]  15 Jun 2008, 06:24
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]  15 Jun 2008, 09:41
Thanks everybody. both approaches are nice ones.
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Re: PS: Manhattan Gmat Math [#permalink]  07 May 2011, 09:34
1
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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

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Re: PS: Manhattan Gmat Math [#permalink]  07 May 2011, 13:16
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]  07 May 2011, 14:01

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

= 54820^2+2*54820+2

= (54821)^2 +1
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Re: PS: Manhattan Gmat Math [#permalink]  07 May 2011, 19:55
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
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Re: PS: Manhattan Gmat Math [#permalink]  08 May 2011, 06:48
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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]  09 May 2011, 04:17
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: PS: Manhattan Gmat Math   [#permalink] 09 May 2011, 04:17
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