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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]

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14 Jun 2008, 22:27
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68% (01:15) correct 32% (01:17) wrong based on 272 sessions

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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

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, 11:29, 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]

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14 Jun 2008, 22:39
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GMAT TIGER wrote:
The average 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

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]

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14 Jun 2008, 22:43
mandy12 wrote:
GMAT TIGER wrote:
The average 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

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]

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14 Jun 2008, 22:46
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GMAT TIGER wrote:
The average 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 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]

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15 Jun 2008, 06:24
GMAT TIGER wrote:
The average 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 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]

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15 Jun 2008, 09:41
Thanks everybody. both approaches are nice ones.
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Re: PS: Manhattan Gmat Math [#permalink]

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07 May 2011, 09:34
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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 $$54822^2$$= $$54821^2 +1$$

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Re: PS: Manhattan Gmat Math [#permalink]

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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]

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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]

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07 May 2011, 19:55
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GMAT TIGER wrote:
The average 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 $$(a+1)^2$$ = $$\frac{[(a-1)^2 + (a+1)^2]}{2} = a^2 + 1$$
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Re: PS: Manhattan Gmat Math [#permalink]

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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]

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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: The average of (54,820)^2and (54,822)^2 =  [#permalink]

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14 Sep 2013, 11:18
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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14 Sep 2013, 11:30
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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.

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, 11:30
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