On a scale that measures the intensity of a certain phenomen

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

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

mikemcgarry · Accepted Answer

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 = C Does this make sense? Mike :-)

Bunuel · Answer

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

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.

gnan · Answer

As the intensity increases 10 times every time

from 3 to 8, it increases 5 times. So,answer is 10^5

Zarrolou · Answer

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"

gallactico12 · Answer

A soln. easy to understand-

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

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

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

mathivanan · Answer

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

JeffTargetTestPrep · Answer

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.

EMPOWERgmatRichC · Answer

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:  C

GMAT assassins aren't born, they're made, 
Rich

Arpitkumar · Answer

Yeah , but whats the logic behind multiplying , why not add ?

EMPOWERgmatRichC · Answer

Hi Arpitkumar,

A few posts in this thread explain how the 'math' works (including the one directly above your post) - we're told that each increase of "+1" in a 'reading' translates to a "10 TIMES" increase in intensity. Since the question asks for how many 'TIMES GREATER' one INTENSITY is than another, then we are clearly going to be using multiplication. If the question had asked for the difference one READING over another, then we would use addition.

GMAT assassins aren't born, they're made,
Rich

BrentGMATPrepNow · Answer

STRATEGY: We   could   try to derive a formula that will help us calculate the intensity corresponding to a reading of 8. 
However, it would be simple (and relatively fast) to use the "table method" to list each successive intensity reading. 
This way, it will be easy to spot any mistakes we might make .

Let   x   the intensity corresponding to a reading of 3

So, our table, in the form  reading | intensity , will look like this: 
 3 |  x  
 4 | ( x )(10) 
 5 | ( x )(10)(10) 
 6 | ( x )(10)(10)(10) 
 7 | ( x )(10)(10)(10)(10) 
 8 | ( x )(10)(10)(10)(10)(10)

The intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3?  
Answer =  ( x )(10)(10)(10)(10)(10) /  x    
 = (10)(10)(10)(10)(10) 
 = 10⁵

Answer: C

ArnauG · Answer

Let's analyze the given information. We know that a reading of n+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n.

Therefore, for every increase of 1 on the scale, the intensity increases by a factor of 10.

To find the intensity corresponding to a reading of 8, we can start from a reading of 3 and calculate the factor by which the intensity increases for each increase of 1 on the scale:

From 3 to 4: The intensity increases by a factor of 10.
From 4 to 5: The intensity increases by a factor of 10.
From 5 to 6: The intensity increases by a factor of 10.
From 6 to 7: The intensity increases by a factor of 10.
From 7 to 8: The intensity increases by a factor of 10.

Since the intensity increases by a factor of 10 for each increase of 1, the intensity corresponding to a reading of 8 is 10×10×10×10×10=10^5 times as great as the intensity corresponding to a reading of 3.

Therefore, the correct answer is (C) 10^5.

Jyotsnakalia · Answer

Can we say that this works on the concept of a Geometric progression?
The term is always being multiplied with the same r
Lets assume that the first term is a and r is 10
a1 = a
a2 = ar
a3 = ar^2
a4 = ar^ 3
.
.
a8 = ar ^7

a8 / a3 = (ar^2)/(ar^7)
= r^5
= 10^5

This approach can work, if the ratio between any 2 terms was asked

EnriqueDandolo · Answer

Hello experts  KarishmaB   Bunuel  I hope you are doing fine. I was wondering if you could please help point out why my reasoning to solve this question is flawed. The question tells me the following:

"...a reading of n+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n..."

What I understand from this is that: n+1=10n.

Therefore if I were to plug in n=2 and n=7 I would get the following

10*7/10*2=3.5

Which isnt on the answer choices.

Thanks in advance for your help.

KarishmaB · Answer

n+1=10n is not correct.

"...a reading of n+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n..."
They are talking about a logarithmic scale. Think about Richter scale to measure the intensity of earthquakes. A reading of 4 on Richter scale is 10 times stronger as compared to a reading of 3. A reading of 5 on Richter scale is 10 times stronger as compared to a reading of 4. 
So reading of 8 is 10*10*10*10*10 times stronger compared with a reading of 3.

Answer (C)

totaltestprepNick · Answer

C https://www.youtube.com/watch?v=BG9263LZez8 Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience tutoring@totaltestprep.net

On a scale that measures the intensity of a certain phenomen

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