On a scale that measures the intensity of a certain phenomen

As the intensity increases 10 times every time

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

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"

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

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

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.

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

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

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

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

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.