Here we have to assume that the relation is linear.

From the given condition, we can frame a linear eqn as -
S=Rx+y......................(1)

42=24x+y
60=30x+y

solving the above eqns, we get
x=3, y=-30

Substituting x,y in the eqn(1)
S=3R-30......................(2)

Now, we have to find R for S=100.

Substituting S in Eqn(2), we get

100=3R-30

Therefore, R=130/3

Plz confirm the OA

Hi.
Seems there is one more solution apart from the one provided by you.
Below it is .

Let the scales R and S are not linearly related. and let the relation be in the form S = (a/R) + b
From the question stem
42 = (a/24) + b => 1008 = a + 24b
60 = (a/30) + b => 1800 = a + 30b
Subtract the first equation from the second.
1800 - 1008 = a + 30b - (a + 24b)
792 = 6b
a = -2160 and b = 132
So if S = 100 R = -2160/(100-132) = 67.5

Link for the solution.
https://www.beatthegmat.com/tough-math-p ... 22454.html

Thanks for the alternate solution.

As there are different answers by following the linear and non-linear approach, what is the original answer ?

I guess, if such a question comes in the exam then there would've been clearly mentioned on the same whether the relation is linear or non-linear.