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On a scale that measures the intensity of a certain phenomen [#permalink]
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Updated on: 07 Dec 2012, 06:35
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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? (A) 5 (B) 50 (C) 10^5 (D) 5^10 (E) 8^10  3^10
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Originally posted by mydreammba on 14 Jan 2012, 03:06.
Last edited by Bunuel on 07 Dec 2012, 06:35, edited 2 times in total.
Renamed the topic and edited the question.



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On a scale that measures the intensity of a certain phenomen [#permalink]
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14 Jan 2012, 04:54
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 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 \(f(8)=10^5*f(3)\). Answer: C. Hope it's clear.
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Re: OG12 PS#98 [#permalink]
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14 Jan 2012, 07:28
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. Thanks a lot.....I will
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Re: maths problem [#permalink]
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07 Jun 2012, 01:52
As the intensity increases 10 times every time
from 3 to 8, it increases 5 times. So,answer is 10^5



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Re: maths problem [#permalink]
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07 Jun 2012, 02:09
gnan wrote: As the intensity increases 10 times every time
from 3 to 8, it increases 5 times. So,answer is 10^5 thanks a lot that was very helpful



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Re: Problem solving [#permalink]
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13 Mar 2013, 10:24
abusaleh wrote: 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 Answer = CDoes this make sense? Mike
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Re: Problem solving [#permalink]
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13 Mar 2013, 13:13
We know that: \(f(n+1)=10f(n)\) so \(f(8)=10f(7)\) , \(f(7)=10f(6)\) and so on... \(f(8)=10*10*10*10*10f(3)\) Each 10 represents a "step back"
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Re: On a scale that measures the intensity of a certain phenomen [#permalink]
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19 Jul 2014, 02:47
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



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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
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^(n1)*k , where n=1,2,3,....... 8 = 10^7*k 3 = 10^2*k It implies 8 is 10^5 times of 3



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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 ManhattanGmat, the GMAT is a test of foreign language. Any reading is 10 times the previous reading, if the reading is increasing by 1.



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Re: On a scale that measures the intensity of a certain phenomen [#permalink]
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02 May 2016, 07:18
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 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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20 Mar 2017, 07:17
From question n+1 = N * 10 say the readings are 1,10,100,1000,10^4,10^5 so on... Implies it is 5 times greater
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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



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Re: On a scale that measures the intensity of a certain phenomen [#permalink]
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03 Apr 2017, 00:05
The intensity of 3 is n. 83= 5 Therefore the intensity of 8 will be 5 times the intensity of 3. Therefore the answer is 10^5.
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Re: On a scale that measures the intensity of a certain phenomen [#permalink]
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25 Aug 2017, 00:16
what level question is this?



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Re: On a scale that measures the intensity of a certain phenomen [#permalink]
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25 Aug 2017, 00:18



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Re: On a scale that measures the intensity of a certain phenomen [#permalink]
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02 Dec 2017, 05:42
Honestly this question is very confusing for me. Based on the question stem I cannot recognize that the intensity increases 10 times every time...



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Re: On a scale that measures the intensity of a certain phenomen [#permalink]
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05 Mar 2018, 13:03
Hi All, The logic behind this question is similar to how earthquakes are measured, but you don't need to recognize that to answer the question correctly. Like many questions on the Quant section, this question needs to be "played with" a bit. We're told that with each "+1" increase in the "level of intensity" on a scale, the intensity of a phenomenon increases 10 times. So, for example: Level 1 = Z intensity Level 2 = 10Z intensity Level 3 = 100Z intensity Level 4 = 1,000Z intensity Etc. We're asked how many TIMES greater "level 8" is than "level 3", so I can either continue on with the table or use "exponent math" The table…. Level 3 = 100Z Level 4 = 1,000Z Level 5 = 10,00Z Level 6 = 100,000Z Level 7 = 1,000,000Z Level 8 = 10,000,000Z Level 8/Level 3 = 10,000,000Z/100Z = 10,000 times greater Using "exponent math", we know that every level is "10x" greater than the previous level. Since Level 8 is "5 levels" above Level 3, that would be 10^5 greater. Final Answer: GMAT assassins aren't born, they're made, Rich
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