 |
| Author |
Message |
|
mww7786
|
Post subject: OG - PS - #99 Posted: Sat Jun 02, 2007 8:43 pm |
|
|
|
|
#99
On a scale that measures the intensity of a certain phenomenon, a reading of n + 1 corresponds to an intensity that is 10 times the intensity corresponding to reading of n. On that scale, the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to 3?
a. 5
b. 50
c. 10^5 correct
d. 5^10
e. (8^10)-(3^10)
|
|
 |
|
 |
|
GMAT 5/18
|
Post subject: Posted: Sun Jun 03, 2007 1:38 pm |
|
|
|
|
mww7786,
I think the best way to solve this would be to use real (smart) numbers. Let's make n = 1. Therefore,
When n = 1, intensity = 1
When n = 2 (n+1), intensity = 10
When n = 3, intensity = 100 (Same as 10^2)
etc, etc
When n = 8, intensity = 10,000,000 (Same as 10^7)
Therefore, the intensity when n = 8 is 10^5 times greater than the intensity when n = 3.
Hope this helps!
|
|
 |
|
 |
|
JadranLee
|
Post subject: Posted: Wed Jun 06, 2007 11:49 am |
|
 |
| ManhattanGMAT Staff |
|
|
Posts: 108 Location: Chicago, IL
|
|
That looks great, GMAT 5/18 . A slight modification: you can start with n=3:
n=3, intensity=x
then n=4, intensity = (10)x
then n=5, intensity = 10(10)x = (10^2) x
then n=6, intensity = 10(10)(10)x = (10^3) x
then n=7, intensity = 10(10)(10)(10)x = (10^4) x
then n=8, intensity = 10(10)(10)(10)(10)x = (10^5) x
As soon as you notice the pattern - that the exponent of the 10 goes up 1 for each step along the way to n=8, you can just count the number of steps to n=8 and realize that the exponent has to be 10^5 times higher at n=8 than at n=3.
-Jad
|
|
 |
|
 |
|
Users browsing this forum: No registered users and 0 guests |
| |
|
|
You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum
|
|
|
 |