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What is the value of N?

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by ska7945 » Thu Aug 14, 2008 3:35 pm
------l------l------l------l------l------

4^7 is the first segment.
4^9 is the second segment.
And N is the forth segment.
what is the value of N? (All the segments are same distance)

answer [spoiler]3(4^9)-2(4^7)[/spoiler]
explanation needed.
let's beat GMAT.
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by kshin78 » Thu Aug 14, 2008 5:25 pm
I'm confused. I can see how OA can be correct, but can the answer also be 4^9 + 4^9?
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by iplraf » Thu Aug 14, 2008 5:53 pm
Which exactly is the first segment, the second, the third and the fourth?
Thanks.
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n

by ska7945 » Thu Aug 14, 2008 5:58 pm
------l 4^7

------l------l 4^9

------l------l------l------l N

i am also confused with answer.
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by parallel_chase » Thu Aug 14, 2008 8:07 pm
Here is the answer

We know that all the segments are equal.

therefore each segment will be = 4^9 - 4^7

first = 4^7

second = 4^9

third = 4^9 + (4^9 - 4^7)

fourth = 4^9 + (4^9 - 4^7) + (4^9 - 4^7) = 3(4^9) - 2(4^7)

Hope this helps.
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by cornell » Fri Aug 15, 2008 6:10 am
parallel_chase wrote:Here is the answer

We know that all the segments are equal.

therefore each segment will be = 4^9 - 4^7

first = 4^7

second = 4^9

third = 4^9 + (4^9 - 4^7)

fourth = 4^9 + (4^9 - 4^7) + (4^9 - 4^7) = 3(4^9) - 2(4^7)

Hope this helps.


AGREE...I've the same answer by the same calculation with "parallel_chase"
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by augusto » Fri Aug 15, 2008 8:49 am
I think the problem is not stated correctly

4^7 is the first segment
this is AT MOST one point in a line / segment, and a point is NOT a segment.

4^9 is the second segment.
same as above
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by kshin78 » Fri Aug 15, 2008 9:09 am
parallel_chase wrote:Here is the answer

We know that all the segments are equal.

therefore each segment will be = 4^9 - 4^7

first = 4^7

second = 4^9

third = 4^9 + (4^9 - 4^7)

fourth = 4^9 + (4^9 - 4^7) + (4^9 - 4^7) = 3(4^9) - 2(4^7)

Hope this helps.
is 2(4^7) = 4^9 ?

I think it would be much easier to state it as 4^9 + 4^9 IMO....

but it's kind of hard to tell w/o any multiple choices...
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by anksbhandari » Sun Aug 17, 2008 3:02 am
thanks parallel chase. difficult problem simple solution
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