Problem Solving

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

which ratio is to be used for this.
6:30 | 24:60 |x:100 = 1:5|2:5|x:100
taking 1:5 leads x = 20 which is incorrect
taking 2:5 leads x = 40 not an option

Please suggest how to approach such questions ?

something that's linear will fit in the form

y=mx+b

what you are suggesting is that b=0 so y=mx is a ratio, but that may not be the case. Let's look at the question.

let R scale = x scale
and let S scale = y scale

(x1,y1) = (6,30)
(x2,y2) = (24,60)

to get slope (or m) we can write (60-30)/(24-6) = 30/18 = 5/3

so y= 5/3x + b

you can realize from points given above that y=0 will be at point x=6- (24-6) = 6-18 = -12

so 0=5/3*(-12) + b so b=20

y= 5/3x + 20

now we are given S(or y in our case) = 100

100 = 5/3x + 20
80 = 5/3x
16*3 = x
48 = x or R

so ans is C

m&m wrote:something that's linear will fit in the form

y=mx+b

what you are suggesting is that b=0 so y=mx is a ratio, but that may not be the case. Let's look at the question.

let R scale = x scale
and let S scale = y scale

(x1,y1) = (6,30)
(x2,y2) = (24,60)

to get slope (or m) we can write (60-30)/(24-6) = 30/18 = 5/3

so y= 5/3x + b

you can realize from points given above that y=0 will be at point x=6- (24-6) = 6-18 = -12

so 0=5/3*(-12) + b so b=20

y= 5/3x + 20

now we are given S(or y in our case) = 100

100 = 5/3x + 20
80 = 5/3x
16*3 = x
48 = x or R

so ans is C

Great solution!

Another way we could think of it (although mathematically all you're doing is finding the "slope") is as a sliding ratio.

The R scale measurements went from 6 to 24; the S scale measurements went from 30 to 60.

So, for every increase of 18 in R, we have an increase of 30 in S.

change in R/change in S = 18/30 = 3/5

In our last jump, S goes from 60 to 100 (change of 40). Plugging that into our ratio we get:

change in R/40 = 3/5
change in R = (3/5)40 = 24

R started at 24; 24+24 = 48, choose C.

m&m wrote:
you can realize from points given above that y=0 will be at point x=6- (24-6) = 6-18 = -12

so 0=5/3*(-12) + b so b=20

I don't quite understand how we calculate x-value when y=0. Could anyone explain please? Thank you!

Stuart Kovinsky wrote:

m&m wrote:something that's linear will fit in the form

y=mx+b

what you are suggesting is that b=0 so y=mx is a ratio, but that may not be the case. Let's look at the question.

let R scale = x scale
and let S scale = y scale

(x1,y1) = (6,30)
(x2,y2) = (24,60)

to get slope (or m) we can write (60-30)/(24-6) = 30/18 = 5/3

so y= 5/3x + b

you can realize from points given above that y=0 will be at point x=6- (24-6) = 6-18 = -12

so 0=5/3*(-12) + b so b=20

y= 5/3x + 20

now we are given S(or y in our case) = 100

100 = 5/3x + 20
80 = 5/3x
16*3 = x
48 = x or R

so ans is C

Great solution!

Another way we could think of it (although mathematically all you're doing is finding the "slope") is as a sliding ratio.

The R scale measurements went from 6 to 24; the S scale measurements went from 30 to 60.

So, for every increase of 18 in R, we have an increase of 30 in S.

change in R/change in S = 18/30 = 3/5

In our last jump, S goes from 60 to 100 (change of 40). Plugging that into our ratio we get:

change in R/40 = 3/5
change in R = (3/5)40 = 24

R started at 24; 24+24 = 48, choose C.

Superb method.. stuart.. thanks

something that's linear will fit in the form

y=mx+b

what you are suggesting is that b=0 so y=mx is a ratio, but that may not be the case. Let's look at the question.

let R scale = x scale
and let S scale = y scale

(x1,y1) = (6,30)
(x2,y2) = (24,60)

to get slope (or m) we can write (60-30)/(24-6) = 30/18 = 5/3

so y= 5/3x + b

Nothing more to add after 2 amazing approaches....yet, there is an unanswered Q here

I don't quite understand how we calculate x-value when y=0. Could anyone explain please? Thank you!

once we have the slope-intercept equation, y= 5/3x + b.....simply substitute x and y values of the 2 available points in this equation and figure b

For instance, substitute (6,30) in y= 5/3x + b; we get b = 20 and hence the revised slope-intercept form of the equation is

y = 5/3 x + 20....simply substitute y=100 here and get x = 48...which is option C....done

Stuart Kovinsky wrote:

m&m wrote:something that's linear will fit in the form

y=mx+b

what you are suggesting is that b=0 so y=mx is a ratio, but that may not be the case. Let's look at the question.

let R scale = x scale
and let S scale = y scale

(x1,y1) = (6,30)
(x2,y2) = (24,60)

to get slope (or m) we can write (60-30)/(24-6) = 30/18 = 5/3

so y= 5/3x + b

you can realize from points given above that y=0 will be at point x=6- (24-6) = 6-18 = -12

so 0=5/3*(-12) + b so b=20

y= 5/3x + 20

now we are given S(or y in our case) = 100

100 = 5/3x + 20
80 = 5/3x
16*3 = x
48 = x or R

so ans is C

Great solution!

Another way we could think of it (although mathematically all you're doing is finding the "slope") is as a sliding ratio.

The R scale measurements went from 6 to 24; the S scale measurements went from 30 to 60.

So, for every increase of 18 in R, we have an increase of 30 in S.

change in R/change in S = 18/30 = 3/5

In our last jump, S goes from 60 to 100 (change of 40). Plugging that into our ratio we get:

change in R/40 = 3/5
change in R = (3/5)40 = 24

R started at 24; 24+24 = 48, choose C.

This is stupendous of both of you, Stuart's in particular, his approach could sound peculiar to many as that is some deep anatomy of plain algebra, but presented so beautifully! Nicely put the final nail in the R/S Scale's coffin, vidyasagar. Great thread seen since a long time