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Concentration: International Business, General Management

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On a scale that measures the intensity of a certain phenomen [#permalink]

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14 Jan 2012, 03:06

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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?

Hey Guys i didn't understand the solution to the problem in OG 12 PS#98, can anyone please explain me the solution???

kotela it's better to post the question itself to get prompt replies. I guess you are talking about the following question:

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

Basically we are given that each number on the scale is 10 times the previous number. For example: If reading of n=3 is 5 then the reading of n=4 would be 5*10 --> the reading of n=5 would be 5*10^2 --> the reading of n=6 would be 5*10^3. Therefore the reading of 8 will be 10^5 times as great as the reading of 3 (the power of 10 goes up by 1 for each step in reading and as there are 5 steps from 3 to 8 then the reading of 8 will be 10^5 times as great as the reading of 3).

Or think about it this way: we have functional relationship: when we increase the reading by 1 the intensity increases 10 times the previous one: \(f(n+1)=10*f(n)\).

So if \(f(3)=x\), then \(f(4)=10*f(3)=10x\) and so on. Therefore \(f(8)=10^5*f(3)\).

Hey Guys i didn't understand the solution to the problem in OG 12 PS#98, can anyone please explain me the solution???

kotela it's better to post the question itself to get prompt replies. I guess you are talking about the following question:

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

Basically we are given that each number on the scale is 10 times the previous number. For example: If reading of n=3 is 5 then the reading of n=4 would be 5*10 --> the reading of n=5 would be 5*10^2 --> the reading of n=6 would be 3*10^3. Therefore the reading of 8 will be 10^5 times as great as the reading of 3 (the power of 10 goes up by 1 for each step in reading and as there are 5 steps from 3 to 8 then the reading of 8 will be 10^5 times as great as the reading of 3).

Or think about it this way: we have functional relationship: when we increase the reading by 1 the intensity increases 10 times the previous one: \(f(n+1)=10*f(n)\).

So if \(f(3)=x\), then \(f(4)=10*f(3)=10x\) and so on. Therefore \(f(8)=10^5*f(3)\).

Can anyone give me proper explanation of this question? 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

Dear abusaleh, This is a tricky question. It is based on logarithmic scales --- both the Richter scale for earthquakes and the decibel scale for the volume of sound follow this pattern.

It tell us if we we go from n to n+1 on the scale, the intensity is 10 times greater.

Start at 3. If we go from 3 to 4 on the scale, the intensity increases 10 times. Then from 4 to 5, it increases 10 times. Same, from 5 to 6, from 6 to 7, and from 7 to 8. There are five "steps" from 3 to 8, and each one of these steps increases the intensity by a factor of 10. That means, when we go from 3 to 8 on the scale, we multiply the intensity by 10 five times ---- 10*10*10*10*10 = 10^5

Re: On a scale that measures the intensity of a certain phenomen [#permalink]

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19 Jul 2014, 02:47

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A soln. easy to understand-

For n=3, consider the intensity to be x.

For n+1 = 4 , we have [intensity*10 ]= 10x For n+2 = 5 , we have [10x*10 ]= 100x For n+3 = 6 , we have [100x*10 ]= 1000x For n+4 = 7 , we have [1000x*10 ]= 10000x For n+5 = 8 , we have [10000x*10 ]= 100000x

Now , for n = 8 -->100000x = 10^5 x which is 10^5 times x, the original intensity at n=3

Re: On a scale that measures the intensity of a certain phenomen [#permalink]

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13 Aug 2015, 16:12

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Re: On a scale that measures the intensity of a certain phenomen [#permalink]

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14 Aug 2015, 03:23

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mydreammba wrote:

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

n = 10^(n-1)*k , where n=1,2,3,....... 8 = 10^7*k 3 = 10^2*k It implies 8 is 10^5 times of 3

Re: On a scale that measures the intensity of a certain phenomen [#permalink]

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28 Feb 2016, 11:27

mydreammba wrote:

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

According to Manhattan-Gmat, the GMAT is a test of foreign language. Any reading is 10 times the previous reading, if the reading is increasing by 1.

Re: On a scale that measures the intensity of a certain phenomen [#permalink]

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30 Apr 2016, 23:04

Hello all, I was wondering why we can't solve for n. If n +1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n wouldn't that mean n + 1 = 10 * n?

Hello all, I was wondering why we can't solve for n. If n +1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n wouldn't that mean n + 1 = 10 * n?

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

To solve this problem we need to examine the information in the first sentence. We are told that “a reading of n + 1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n.”

Let’s practice this idea with some real numbers. Let’s say n is 2. This means that n + 1 = 3. With the information we were given we can say that a reading of 3 is ten times as great as the intensity of a reading of 2.

Furthermore, we can say that a reading of 4 is actually 10 x 10 = 10^2 times as great as the intensity of a reading of 2.

Increasing one more unit, we can say that a reading of 5 is 10 x 10 x 10 = 10^3 times as great as the intensity of a reading of 2.

We have found a pattern, which can be applied to the problem presented in the stem:

3 is “one” unit away from 2, and thus a reading of 3 is 10^1 times as great as the intensity of a reading of 2.

4 is “two” units away from 2, and thus a reading of 4 is 10^2 times as great as the intensity of a reading of 2.

5 is “three” units away from 2, and thus a reading of 5 is 10^3 times as great as the intensity of a measure of 2.

We can use this pattern to easily answer the question. Here we are being asked for the number of times the intensity corresponding to a reading of 8 is as great as the intensity corresponding to a reading of 3. Because 8 is 5 units greater than 3, a reading of 8 is 10^5 times as great as the intensity corresponding to a reading of 3.

Answer C.
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Re: On a scale that measures the intensity of a certain phenomen [#permalink]

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21 Mar 2017, 04:12

mydreammba wrote:

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

we can say that f(n) = 10^n f(8)/ f(3) = 10^8/10^3 = 10^5

Re: On a scale that measures the intensity of a certain phenomen [#permalink]

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04 Aug 2017, 10:25

Bunuel wrote:

kotela wrote:

Hey Guys i didn't understand the solution to the problem in OG 12 PS#98, can anyone please explain me the solution???

kotela it's better to post the question itself to get prompt replies. I guess you are talking about the following question:

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

Basically we are given that each number on the scale is 10 times the previous number. For example: If reading of n=3 is 5 then the reading of n=4 would be 5*10 --> the reading of n=5 would be 5*10^2 --> the reading of n=6 would be 3*10^3. Therefore the reading of 8 will be 10^5 times as great as the reading of 3 (the power of 10 goes up by 1 for each step in reading and as there are 5 steps from 3 to 8 then the reading of 8 will be 10^5 times as great as the reading of 3).

Or think about it this way: we have functional relationship: when we increase the reading by 1 the intensity increases 10 times the previous one: \(f(n+1)=10*f(n)\).

So if \(f(3)=x\), then \(f(4)=10*f(3)=10x\) and so on. Therefore \(f(8)=10^5*f(3)\).

Answer: C.

Hope it's clear.

I believe the highlighted part should be 10 * (5 *10^2) = 5 * 10^3

Hey Guys i didn't understand the solution to the problem in OG 12 PS#98, can anyone please explain me the solution???

kotela it's better to post the question itself to get prompt replies. I guess you are talking about the following question:

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

Basically we are given that each number on the scale is 10 times the previous number. For example: If reading of n=3 is 5 then the reading of n=4 would be 5*10 --> the reading of n=5 would be 5*10^2 --> the reading of n=6 would be 3*10^3. Therefore the reading of 8 will be 10^5 times as great as the reading of 3 (the power of 10 goes up by 1 for each step in reading and as there are 5 steps from 3 to 8 then the reading of 8 will be 10^5 times as great as the reading of 3).

Or think about it this way: we have functional relationship: when we increase the reading by 1 the intensity increases 10 times the previous one: \(f(n+1)=10*f(n)\).

So if \(f(3)=x\), then \(f(4)=10*f(3)=10x\) and so on. Therefore \(f(8)=10^5*f(3)\).

Answer: C.

Hope it's clear.

I believe the highlighted part should be 10 * (5 *10^2) = 5 * 10^3

_______________________ Edited the typo. Thank you.
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