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by shashank.ism » Thu Feb 11, 2010 7:13 am
If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is:

a.) A+1
b.) A+5
c.) A+25
d.) 2A
e.) 5A
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by sars72 » Thu Feb 11, 2010 8:55 am
1+2+3+4+5 = 15 = A
6+7+8+9+10 = 40

--> a + 25 --> answer choice C

solving algebraically -> x+(x+1)+(x+2)+(x+3)+(x+4) = A
--> 5x+10 = A

sum next 5 integers = (x+5)+(x+6)+(x+7)+(x+8)+(x+9) = 5x+35

(5x+35) - (5x+10) = 25

--> Difference is 25 --> a+25 --> answer choice C
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by sparky_paris » Thu Feb 11, 2010 5:29 pm
Does it have to be positive integers? The answer holds good for any integer
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by money9111 » Thu Feb 11, 2010 8:45 pm
yes +25... all you need to know is that the 1st and 6th number have a difference of 5. the 2nd and 7th number also have a difference of 5 and so on... therefore since there are 5 sets of numbers with a difference of 5... 5*5=25
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by harsh.champ » Thu Feb 18, 2010 10:16 am
shashank.ism wrote:If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is:

a.) A+1
b.) A+5
c.) A+25
d.) 2A
e.) 5A
I solved by this approach:-
Let the 1st term be t.
2nd term = t+1
SO we get that 5t + 10 = A.
The sum of next 5 consecutive integers is (t+5) + (t+6) + .....
=5t +(5+6+7+8+9)
=(5t + 10) + 25
= [spoiler]A + 25
Hence,C should[/spoiler] be the answer.

Hey money9111,
yes +25... all you need to know is that the 1st and 6th number have a difference of 5. the 2nd and 7th number also have a difference of 5 and so on... therefore since there are 5 sets of numbers with a difference of 5... 5*5=25
I didn't get how you took the above bold-faced statement:- 5*5 = 25
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by money9111 » Thu Feb 18, 2010 1:02 pm
ok so we have 10 numbers correct?

The 1st 5 numbers are: x, x+1, x+2, x+3, x+4, x+5 - let's call this Group A
The 2nd 5 numbers are: x+6, x+7, x+8, x+9, x+10 - let's call this Group B

We know that these are the numbers because they're consecutive...

so looking at that.. we know that the difference between the 1st number in Group A and the 1st number in the Group B is 5. ((x+6) - x)=5.

The same goes for the 2nd number in Group A and the 2nd number in Group B. ((x+7)-(x+1))=5.

that happens 5 times.. so 5*5=25...
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by harsh.champ » Thu Feb 18, 2010 2:39 pm
money9111 wrote:ok so we have 10 numbers correct?

The 1st 5 numbers are: x, x+1, x+2, x+3, x+4, x+5 - let's call this Group A
The 2nd 5 numbers are: x+6, x+7, x+8, x+9, x+10 - let's call this Group B

We know that these are the numbers because they're consecutive...

so looking at that.. we know that the difference between the 1st number in Group A and the 1st number in the Group B is 5. ((x+6) - x)=5.

The same goes for the 2nd number in Group A and the 2nd number in Group B. ((x+7)-(x+1))=5.

that happens 5 times.. so 5*5=25...
Okay,I get it now.
Thanks for the shortcut approach.It can really come in very handy.
I guess my problem approach was a bit time taking.
Did you make this approach while solving the question or is this one of a common trick ??
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by money9111 » Thu Feb 18, 2010 2:59 pm
haha no problem! actually i would never have thought of this approach in a million years...i solved it in about double the time that this approach took... my mgmat instructor told us that 700 level test takers think in that manner... so im like whoa... i better start thinking that way...it totally makes sense though
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by shashank.ism » Fri Feb 19, 2010 1:17 am
money9111 wrote:yes +25... all you need to know is that the 1st and 6th number have a difference of 5. the 2nd and 7th number also have a difference of 5 and so on... therefore since there are 5 sets of numbers with a difference of 5... 5*5=25
yeah money1119 that is indeed a very good solution. I would have never thought of in this way. It really shows how we can solve a problem in seconds with some juggling...and score much higher...
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