Two measuring scales R and S. When R is 6, S is 30 ; when R

See, it is difficult to define a relationship with two sets of value. I assumed it to be a linear relationship.Made 2 =ns

6a+b=30
24a+b=60 solve for a and b; a= 5/3; b=20 Therefore, for S=100; R=48

R = 60.

We're looking for rate in change of R with respect to S, and vice versa depending on the value given.

Rate = (R2 - R1) / (S2 - S1)
= (24 - 6) / (60 - 30)
= 18 / 30
= 0.6

Prediction:
R3 = S3 * Rate
R3 = 100 * 0.6
R3 = 60

I disagree with you here.

When We are saying that rate of change is constant, then we can't multiply by values. Here the approach could have been

Since we know that rate of change of R / Rate of change of S = 0.6.

(R3-6) /(S3-6) = 0.6 should also be true. and also

(R3-24)/(S3-60) =0.6 be true.

Here R3,S3 are unknown and 100 respectively.

If we solve any of the two above R3 comes as 48. This is based on assumption that Rate of change remains constant and R & S has a linear relation. If you assume that R ans S vary as their squares. We'll have foll. 2 eqns.
6^2a + b = 30
and 24^2a + b = 60.

Then for S=100, R = 36.

Hence, I think we are missing something in this question.

We have as many solutions as posts..But this is how the question is posted..

AKAMAI , can you please help?

Thanks
Praetorian

This is an absurd question. First of all, there is no reason or justification to assume, based on the wording of the problem, that any relationship exists whatsoever between R and S. For all we know, these could be random numbers.

Even if we assume that there is, in fact, some relationship between R and S, there are an INFINITE number of equations that can relate R and S based on a sample set of TWO and no reason or justification to assume that the relationship is linear.