word problem

[gmatnmein2010]

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

[komal]

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

this is like linear equation problem
where r =sa +b --(1) or y = mx + c
put r=6 then s=30
and r= 24 then s =60
thus 2 equations
6 =30a+b-------------(A)
24=60a+b------------(B)
solve (A) and (B)
thus we get
a =18/30 and c =-12
thus if s =100 then r = (18/30)*100-12 =48

Answer is C

[shashank.ism]

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

R-scale and S-scale are related linearly,

Let value on R-scale = x and value on S-scale = y to give the points (6, 30) and (24, 60).

slope of the line connecting them = m=(60 - 30) / (24 - 6)= 5/3

Now that you have the slope you can use the point-slope form of a linear equation: y - y1 = m(x - x1) where m = slope. So,
y - 60 = 5/3(x - 24)
y = (5/3)x - 40 + 60
y = (5/3)x + 20

Now that you have the equation, plug in 100 for y since 100 is the S-scale value you want to find:

(5/3)x + 20 = 100 -->x = 48

So, the R-scale value is 48, or answer C.

[vijay_venky]

Linear value means let us say r=ms+c

6=30m+c
24=60m+c

means 30m=18 and m=0.6

c=-12

and so s=0.6r-12 and r=100 so s=48

[thephoenix]

Let R = mS + c. Then 6 = m*30 + c and 24 = m*60+c; substituting for c, c = 6-30*m we get 24 = 60*m + 6-30*m, so m = 18/30 = 3/5. Solving for c, c = -12. So for S = 100, R = 3/5*100 -12 = 48.
C