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A certain quantity is measured on two different scales, the R-scale 18 Jan 2025, 10:29
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Scale R: Jump from 6 to 24 is an increment of 18, next would be 42 (24+18) (since it is given linear)
Scale S: Jump from 30 to 60 is an increment of 30, next would be 90 (60+30) (since it is given linear)

Now 100 on Scale S is 90 + 10 i.e. 1/3 of the increment (30/3)
So 1/3 of increment of Scale R is 18/3 = 6

Hence equivalent of 100 on Scale R would be 42 + 6 = 48 (C)
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A certain quantity is measured on two different scales, the R-scale 14 Apr 2025, 22:06
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QS says 6 and 24 correspond to 30 and 60 .. That means that 18 difference on R scale would give 30 difference on S scale . 1 unit on R scale is 5/3 on S scale . Now 40 difference(100 on S Scale - 60 on S Scale) on S scale would give 40/5/3 units of difference on R scale => 24 units of difference on R scale . So S scale value for 60 i.e 24 + difference of 24 =>48 units.
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File comment: QS says 6 and 24 correspond to 30 and 60 .. That means that 18 difference on R scale would give 30 difference on S scale . 1 unit on R scale is 5/3 on S scale . Now 40 difference(100 on S Scale - 60 on S Scale) on S scale would give 40/5/3 units of difference on R scale => 24 units of difference on R scale . So S scale value for 60 i.e 24 + difference of 24 =>48 units.
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A certain quantity is measured on two different scales, the R-scale 30 May 2025, 02:39
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If you know one of the straight lines equations:
y-y1=(y2-y1/x2-x1) x-x1,

treat the given values as x and y co-ordinate since they are related linearly,
x1,y1=6,30
x2,y2=24,60,

find the equation of line and insert the values you want to find out such as keep 100 in place of y and get the x value which will be 48.


appy001
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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A certain quantity is measured on two different scales, the R-scale 06 Oct 2025, 06:36
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I can see why this problem might feel tricky at first - when you see "related linearly," it's easy to overthink what that means. Let me walk you through this in a way that'll make it click.

Here's how to think about this:

Step 1: Understand what "linearly related" means

When two scales are linearly related, they follow a straight-line relationship. Think of it like the connection between Celsius and Fahrenheit - there's a predictable pattern. You're given two points:
  • When \(R = 6\), then \(S = 30\)
  • When \(R = 24\), then \(S = 60\)

Your goal: Find \(R\) when \(S = 100\).

Step 2: Find the rate of change (slope)

Let's see how both scales change together:
  • R increases from 6 to 24 → that's an increase of 18
  • S increases from 30 to 60 → that's an increase of 30

So the rate of change is \(\frac{30}{18} = \frac{5}{3}\)

This means: for every 3 units R increases, S increases by 5 units.

Step 3: Build the complete equation

Here's the key insight - a linear relationship isn't just about the slope. It follows the pattern: \(S = \frac{5}{3} \times R + \text{constant}\)

To find that constant, plug in one of your known points. Using \(R = 6\) and \(S = 30\):

\(30 = \frac{5}{3} \times 6 + \text{constant}\)
\(30 = 10 + \text{constant}\)
\(\text{constant} = 20\)

So your complete equation is: \(S = \frac{5}{3}R + 20\)

Quick verification with the second point \((R = 24, S = 60)\):
\(S = \frac{5}{3} \times 24 + 20 = 40 + 20 = 60\) ✓

Step 4: Solve for R when S = 100

Now substitute \(S = 100\):

\(100 = \frac{5}{3}R + 20\)
\(80 = \frac{5}{3}R\)
\(R = 80 \times \frac{3}{5} = \frac{240}{5} = 48\)

Answer: C (48)

The biggest trap here? Many students assume it's just a simple ratio problem and try something like \(R \times 5 = S\), forgetting that linear relationships include that constant term (+20 in this case). That's what makes the difference!

Want to master this systematically?

The approach I showed you works, but there's actually a more sophisticated framework for tackling all linear relationship problems efficiently. The complete solution on Neuron shows you the strategic approach that works across different variations of this question type, plus the time-saving patterns that help you spot the right setup instantly. You can check out the detailed explanation on Neuron to see the full framework and understand how this pattern applies to other problems. You can also explore comprehensive solutions for similar official questions on Neuron with practice quizzes that help you build consistent accuracy on these concepts.

Hope this helps!
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