BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

OG 11th edition, Question 99

Expert replies
by Catfish75 » Mon Jun 08, 2009 8:30 pm
This is Question 99 on p. 165 of the OG, 11th ed. Maybe it's deceptively simple, but for some reason, I'm not grasping the logic behind this. Can someone please explain it to me?

"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 a 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 a reading of 3?"

(A) 5
(B) 50
(C) 10^5
(D) 5^10
(E) 8^10 - 3^10

OA: The answer is C


Here's the explanation given in the OG: "Since each increase of 1 in the scale creates an intensity increase of a factor of 10, the intensity reading of 8 is 10^8/10^3" ... which yields 10^5.
Join the discussion
Source: — Problem Solving |

by yogami » Tue Jun 09, 2009 1:36 am
Lets say the initial intensity for the scale at reading 1 is n. For 2 it is 10n for 3 it is 100n i.e. 10^2 n
4 - 10^3 n
5 - 10^4 n
..
.
.
.
.8 10^7 n

if you compare 3 and 8 you will see that 10^7 is 10^5 times 10^2 hence the answer is C.
200 or 800. It don't matter no more.
Join the discussion

by Catfish75 » Tue Jun 09, 2009 12:13 pm
Thanks for the reply and the explanation, this makes sense to me now.
Join the discussion