appy001 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
GIVEN: - Two scales, R-scale and S-scale, that are related linearly.
- A measurement of 6 on the R-scale corresponds to a measurement of 30 on S-scale.
- Simply put, if R = 6, S = 30 (assuming R represents the R-scale measurement and S represents the S-scale measurement) ----(I)
- Measurement of 24 on R-scale corresponds to measurement of 60 on S-scale.
- Simply put, if R = 24, S = 60 ----(II)
TO FIND: - Measurement on R-scale for which corresponding S-scale measurement is 100.
- That is, find R when S = 100.
CONCEPT RECALL: “Two quantities are related linearly” implies that the relationship between these two quantities can be defined using a
linear equation.
So, if x and y are two variables related linearly, then their relation can be expressed as y = ax + b, for constants a and b.
WORKING OUT: We are given that the R-scale and S-scale measurements are related linearly, we can say:
- R = aS + b for some constants ‘a’ and ‘b’ ----(III)
Now, from (I), R = 6 when S = 30. Putting (I) into (III), we get:
Similarly, putting values of R and S from (II) into (III), we get:
Now, (IV) and (V) are two linear equations with two unknowns, a and b. Solving this system of equations can get us the values of ‘a’ and ‘b’ and hence, the complete relationship between R and S. This will then help us answer the final question.
Let’s dive right into solving the system!
Solving (IV) and (V) to find ‘a’ and ‘b’:
6 = a(30) + b --- (IV)
24 = a(60) + b --- (V)Subtracting (IV) from (V), we get 18 = 30a. Thus,
a = 3/5.
Using this value of ‘a’, we can find ‘b’ from any of the equations.
If we use a = 3/5 in (IV), we get 6 = \(\frac{3}{5}(30) + b\)
⇒ 6 – 18 = b ⇒
b = -12 Writing final relationship between the two scales: Putting values of ‘a’ and ‘b’ in (III), the final relation between R and S can be written as
- R = \(\frac{3}{5}S – 12\) ----(VI)
Finding ‘R’ for given corresponding ‘S’: We need to find ‘R’ for S = 100. Using S = 100 in (VI), we get:
- R = \(\frac{3}{5}(100) - 12\)
Correct Answer: Option CBest,
Aditi Gupta
Quant expert,
e-GMAT