Problem Solving

taneja.niks wrote:A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related
linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of
30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100
on the S-scale?

A. 20
B. 36
C. 48
D. 60
E. 84

How to proceed this one???

An easy way to proceed is to consider the R and S-scales as X and Y-axes respectively.
Then the coordinates are (6, 30), (24, 60), and (x, 100) all lie in the same line.
Then (60 - 30)/(24 - 6) = (100 - 30)/(x - 6)
3/18 = 7/(x - 6)
3(x - 6) = 126
x - 6 = 42 or x = 48

The correct answer is C.

taneja.niks wrote:A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related
linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of
30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100
on the S-scale?

A. 20
B. 36
C. 48
D. 60
E. 84

How to proceed this one???

I'd like to make a brief comment regarding the phrase "related linearly."

Please know that you are not expected to know this specific term (and its implications). While you will find the phrase "linear equation" throughout the OG, you will find no reference to "linear relations" or "related linearly."

If this were an official GMAT question, you would be given additional information explaining that the relationship between the R-scale and S-scale could be represented as a line (or something to that extent).

That's good to know - thanks.
I guess the word "line" hiding in "linear" is enough of a tipoff

Cheers,
Brent

taneja.niks wrote:A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of
30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100
on the S-scale?

A. 20
B. 36
C. 48
D. 60
E. 84

Since R and S are related linearly, it must be true that R = kS + l for some constants k and l.

We are given that when R = 6, S = 30; therefore, 6 = 30k + l.

We are also given that when R = 24, S = 60; therefore, 24 = 60k + l.

Let's subtract the first equation from the second:

18 = 30k

k = 3/5

Thus, R = (3/5)S + l. To find the value of l, we can substitute either R = 6 and S = 30 or R = 24 and S = 60 in the equation. Let's use the first pair:

6 = (3/5)(30) + l

6 = 18 + l

l = -12

Thus, R and S are related via R = (3/5)S - 12. When S = 100, we find that:

R = (3/5)(100) - 12 = 60 - 12 = 48

Answer: C