A certain quantity is measured on two different scales, the R-scale

Please don't paraphrase questions - by not using the actual wording of the question, a LOT gets lost in translation. As written, the question is completely unanswerable.

Also, please provide answer choices, so we can discuss not only pure algebra, but the key alternative strategies that will get you a great score on test day.

Using my magical powers, I'll post the actual question:



First, we have to understand what "linearly" means. It's not a straight ratio (since 6:30 does NOT equal 24:60). We need to look at the increases in each measurement to see what the scalar actually is.

From 6 to 24 we have an increase of 18. From 30 to 60 we have an increase of 30. Therefore, the increase ratio is 18:30 or 3:5. In other words, for every 3 that R increases, S increases by 5.

We know that S is 100. To get from 60 to 100, we went up by 40, or 8 "jumps" of 5; therefore, R will go up by 8 "jumps" of 3.

24 + 8(3) = 24 + 24 = 48: choose (c).

Note that (a) makes no sense, since if S=60 corresponds to R=24, how could S=100 correspond to a lower value for R?

As R increases by 18 (from 6 to 24), S increases by 30 (from 30 to 60). Thus increase of 18 in R corresponds to increase of 30 in S.

Therefore change of 70 in S from 30 (when R is 6) to 100 must correspond to change of 70/30*18=42 in R. Hence, 100 in S corresponds to 6+42=48 in R.

Answer: C.
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This not a good question. In order to calculate R for S=100 we must know how are the scales of R ans S related.

If they are related linearly, then we would have: \(42=24m+b\) and \(60=30m+b\). The question: if \(100=Rm+b\), then \(R=?\)

Solving system of equations for \(m\) and \(b\) --> \(m=3\) and \(b=-30\) --> substituting these values in \(100=Rm+b\) --> \(R=\frac{130}{3}\).


But if they are not related linearly and for example are related like: \(S=\frac{m}{R}+b\), then we would have \(42=\frac{m}{24}+b\) and \(60=\frac{m}{30}+b\). The question: if \(100=\frac{m}{R}+b\), then \(R=?\)

Solving system of equations for \(m\) and \(b\) --> \(m=-2160\) and \(b=132\) --> substituting these values in \(100=\frac{m}{R}+b\) --> \(R=\frac{135}{2}\).

Hope it's clear.
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Consider a line built on R scale.

--------------------------|-----------------------------------|-------------------------------------------
6 24

The increase in value in R scale -=24-6=18 points
Now consider a line built on scale S. For the same increase in value, the increase in S scale is from 30-60 i.e30 points(60-30)

So, we know that an increase of 18 points on R scale corresponds to 30 points on S scale. We need to find our what measurement on R scale corresponds to a value of 100 on S scale.

an increase from 30 to 100 value on S scale= 70 points

R scale S scale
18 point increase= 30 point increase
? = 70 point increase

cross multiply

?=70*18/30=42

Thus, an increase of 70 points on S scale equals an increase of 42 points on R scale. So, the actual value=6+42=48

Since there is a linear relationship between R and S....we can assume R =aS + b (or any other linear equation in terms of R and S)
Now, when R= 6, S= 30....so the equation becomes 6 = 30a + b---(i)
When R = 24, S = 60...so the equation becomes 24 = 60a + b----(ii)
Solving the above two equations we get a = 3/5, b= -12...so the equation becomes R = (3/5)S - 12....now put the value of S =100 in the equation...u'll get the value of R as 48...

Linear equations so:
30= 6x+b (x variable, b contant)
60= 24x+b

solving these two-- b= 20, x= 5/3
100= 5/3 (x)+ 20
80= 5/3 (x)
x= 48

Think of the graphical approach here.

Related linearly means that their relation represents a line. e.g. x and y co-ordinates are related linearly on a line segment. We know how to deal with lines.

So think of it as two points (6, 30) and (24, 60) lying on a line. So what will be (r, 100) on the same line?
We see that an increase of 18 in x co-ordinate causes an increase of 30 in y co-ordinate. So y increases by 30/18 = 5/3 for every 1 point increase in x co-ordinate.
From 60 to 100, the increase in y co-ordinate is 40. So x co-ordinate will increase from 24 to 24 + 40*(3/5) = 48.

Answer (C)
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R scale and S are linearly related i.e, y = mx + c, and y is the S scale and x is the R scale.
Given,
30 = m*6 + C
60 = m*24 + C
Solving these 2 equation for m and C gives, m =\(\frac{5}{3}\) and C = 20

so solve for S scales of 100

100 = \(\frac{5}{3}\) * x + 20
x = 48

Answer is C)


+1 for kudos

6 > 24 > x
30 > 60 > 100

Through concept of gradient we can solve in less than 30 seconds.
(24-6)/(60-30)=(x-24)/(100-60)
==>x=48

Hi All,

The answer choices to this question are 'spread out' enough that you can use a bit of logic to estimate the correct answer.

We're told that the relationship between the values on the R-Scale and S-Scale are LINEAR, which means that as one value increases, the other value will increase by a fixed amount. We're then told the relationship between the R-Scale and S-Scale for two sets of values (6 and 30; 24 and 60). Notice how that when the R-scale value increases from 6 to 24 (an increase of 18), the S-scale value increases from 30 to 60 (an increase of 30). The question asks for the relative R-scale value when the S-scale value is 100.

Since an increase of 30 on the S-scale = an increase of 18 on the R-scale, when we go from 60 to 100 on the S-scale, we're increasing by 40 (a little more than 30)....so the increase on the R-scale should be a little more than 18....

24 + (a bit more than 18)..... = a bit more than 42....

There's only one answer that matches:

Final Answer:

GMAT assassins aren't born, they're made,
Rich

This question is very simple but the GMAT has cleverly disguised it to make this a harder question. The explanation is long, but the solution is intuitive.

The question relates to the concept of lines in geometry. We know that we are dealing with geometrical lines because the question uses the words "related linearly". If we can recognize these words and properly translate into algebraic forms, then the question is easy to solve.

The simplest algebraic representation of a line is of the following form:

Y = mX + B

where Y = y-intercept, X = x-intercept and B = constant that represents the value of the y-intercept when X = 0.

From the above equation, we can also derive the slope of the line. The slope will be the value of m when the constant B equals zero. Hence:

Slope = m = Y/X

Coming to the problem, we should recognize that we are given two points: (6, 30) and (24, 60), if we represent R as X and S as Y. Sunstituting these values of the two points in the equation of the line will provide us with two equations in the two unknowns, m and B.

Once we obtain the equation of the line, we can calculate the X-intercept when S. i.e., Y equals 100.

In summary, there are multiple ways of looking at this problem:

1. Since we are given two points that lie on a line, we can calculate the slope of the line. From this, we can calculate the slope when we are given the third point on the line, i.e., (X, 100).

2. We can set up algebraic equations, calculate the slope and the intercept and finally, use the equation of the line to find out the x-intercept when the y-intercept equals 100.

Thus:

Y = mX + B

=> 30 = 6X + B, and
60 = 24X + B

Solving:

m = 30/18 and B = 20.

So, the equation of the line is:

Y = (30/18)X + 20

The question asks us to calculate the value of X when Y = 100.

100 = (30/18)X + 20 or X = 48

The key to the question is to recognize that we are dealing with a geometrical line and that we are provided with two equations in two unknowns.

VeritasKarishma how should i know 100 is x or y ?

It depends on what you assume. Note that in my solution above, I have assumed that y is the S scale and x is the R scale.
That is how I made these equations:
30 = m*6 + C
60 = m*24 + C

Now I am given that 100 is the value of S scale so y = 100. Now, knowing m and C, I will calculate the value of x.

You can very well flip them and still the answer will be the same. You can assume x to be the S scale and y to be the R scale and then make equations:
6 = 30m + C
24 = 60m + C
Now you will get m = 3/5 and C = -12

Now your 100 is x so
y = 100*(3/5) - 12 = 48

Value on R scale is still = 48
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GIVEN:

1. Two scales, R-scale and S-scale, that are related linearly.
2. 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)
3. 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:

• 6 = a(30) + b ----(IV)

Similarly, putting values of R and S from (II) into (III), we get:

• 24 = a(60) + b ----(V)

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 = \(\frac{3}{5}(100) - 12\)
  
  • R = 60 – 12
  • R = 48

Correct Answer: Option C


Best,
Aditi Gupta
Quant expert, e-GMAT
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A linear relationship between two variables means that every pair of observations (R,S) must lie on the same line in a coordinate system.

The slope of the line is:

∆R/∆S = (24 – 6)/(60 – 30) = 18/30 = 3/5

Since the slope must be the same between any two different points of a line, we have:

(r – 6)/(100 – 30) = 3/5

r – 6 = 42

r = 48

Answer: C
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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
­­­

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in simple terms I discovered that 30-6 = 24 and 24 is the second r term. Since it's linear it means either + or - and nothing else. We don't know what 100 would be but we could assume that 60-24 = 36 would correspond with 90 since 30-60... 90. Then since we're looking for a little bit larger of a number we see 48 is the perfect choice.