Central Tendency & DispersionMTP Nov 20Question 3239 of 473

If $\displaystyle x$ and $\displaystyle y$ are related as $\displaystyle 3x + 4y = 20$ and the quartile deviation of $\displaystyle x$ is $\displaystyle 16$ , then the QD of $\displaystyle y$ is

\displaystyle 16

\displaystyle 14

\displaystyle 10

\displaystyle 12

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Correct Answer

✅ Option d — $\displaystyle 12$

We are given the linear relationship between variables

\displaystyle x

and

\displaystyle y

3x + 4y = 20 \implies 4y = -3x + 20 \implies y = -0.75x + 5

Since quartile deviation is independent of change of origin but affected by change of scale, the relationship is:

\text{Q.D.}_y = |a| \times \text{Q.D.}_x

Substitute the given values (

\displaystyle a = -0.75

and

\displaystyle \text{Q.D.}_x = 16

\text{Q.D.}_y = |-0.75| \times 16 = 0.75 \times 16 = 12

Hence, **Option D** is the correct answer.

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