If $p:q$ is the sub-duplicate ratio of $p-x^2:q-x^2$, then $x^2$ is

The correct answer is Option D: None of these. • The problem states that $p:q$ is the sub-duplicate ratio of $p-x^2:q-x^2$. • The sub-duplicate ratio of $a:b$ is $\sqrt{a}:\sqrt{b}$. • Therefore, we can write the given condition as: $\frac{p}{q} = \frac{\sqrt{p-x^2}}{\sqrt{q-x^2}}$ • To eliminate the square roots, we square both sides of the equation: $\left(\frac{p}{q}\right)^2 = \left(\frac{\sqrt{p-x^2}}{\sqrt{q-x^2}}\right)^2$ $\frac{p^2}{q^2} = \frac{p-x^2}{q-x^2}$ • Now, we cross-multiply to solve for $x^2$: $p^2(q-x^2) = q^2(p-x^2)$ • Distribute the terms: $p^2q - p^2x^2 = q^2p - q^2x^2$ • Group the terms containing $x^2$ on one side and the other terms on the other side: $q^2x^2 - p^2x^2 = q^2p - p^2q$ • Factor out $x^2$ from the left side: $x^2(q^2 - p^2) = q^2p - p^2q$ • Factor out $pq$ from the right side: $x^2(q^2 - p^2) = pq(q - p)$ • We know that $q^2 - p^2 = (q-p)(q+p)$. Substitute this into the equation: $x^2(q-p)(q+p) = pq(q-p)$ • Assuming $q \neq p$, we can divide both sides by $(q-p)$: $x^2(q+p) = pq$ • Finally, solve for $x^2$: $x^2 = \frac{pq}{q+p}$ or $x^2 = \frac{pq}{p+q}$ • Comparing this result with the given options: (A) $\frac{p}{p+q}$ - Incorrect. (B) $\frac{q}{p+q}$ - Incorrect. (C) $\frac{qp}{p-q}$ - Incorrect, as our denominator is $p+q$. • Since our derived value for $x^2$ is $\frac{pq}{p+q}$, and this does not match options (A), (B), or (C), the correct answer is (D) None of these. • Option (C) is incorrect because it has $p-q$ in the denominator, whereas our calculation yields $p+q$. The sign difference is crucial. Options (A) and (B) are incorrect because their numerators are $p$ and $q$ respectively, not $pq$.

Ratio, Proportion, Indices, LogarithmPYQ May 18Question 787 of 211

All Questions

If $\displaystyle p:q$ is the sub-duplicate ratio of $\displaystyle p-x^2:q-x^2$ , then $\displaystyle x^2$ is

Options

\displaystyle \frac{p}{p+q}

\displaystyle \frac{q}{p+q}

\displaystyle \frac{qp}{p-q}

DNone of these

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Correct Answer

✅ Option D — None of these

All Options:

A $\displaystyle \frac{p}{p+q}$
B $\displaystyle \frac{q}{p+q}$
C $\displaystyle \frac{qp}{p-q}$
DNone of these ✓

Detailed Solution & Explanation

• The problem states that

\displaystyle p:q

is the sub-duplicate ratio of

\displaystyle p-x^2:q-x^2

. • The sub-duplicate ratio of

\displaystyle a:b

\displaystyle \sqrt{a}:\sqrt{b}

. • Therefore, we can write the given condition as:

\displaystyle \frac{p}{q} = \frac{\sqrt{p-x^2}}{\sqrt{q-x^2}}

• To eliminate the square roots, we square both sides of the equation:

\displaystyle \left(\frac{p}{q}\right)^2 = \left(\frac{\sqrt{p-x^2}}{\sqrt{q-x^2}}\right)^2

\displaystyle \frac{p^2}{q^2} = \frac{p-x^2}{q-x^2}

• Now, we cross-multiply to solve for

\displaystyle x^2

\displaystyle p^2(q-x^2) = q^2(p-x^2)

• Distribute the terms:

\displaystyle p^2q - p^2x^2 = q^2p - q^2x^2

• Group the terms containing

\displaystyle x^2

on one side and the other terms on the other side:

\displaystyle q^2x^2 - p^2x^2 = q^2p - p^2q

• Factor out

\displaystyle x^2

from the left side:

\displaystyle x^2(q^2 - p^2) = q^2p - p^2q

• Factor out

\displaystyle pq

from the right side:

\displaystyle x^2(q^2 - p^2) = pq(q - p)

• We know that

\displaystyle q^2 - p^2 = (q-p)(q+p)

. Substitute this into the equation:

\displaystyle x^2(q-p)(q+p) = pq(q-p)

• Assuming

\displaystyle q \neq p

, we can divide both sides by

\displaystyle (q-p)

\displaystyle x^2(q+p) = pq

• Finally, solve for

\displaystyle x^2

\displaystyle x^2 = \frac{pq}{q+p}

\displaystyle x^2 = \frac{pq}{p+q}

• Comparing this result with the given options: (A)

\displaystyle \frac{p}{p+q}

- Incorrect. (B)

\displaystyle \frac{q}{p+q}

- Incorrect. (C)

\displaystyle \frac{qp}{p-q}

- Incorrect, as our denominator is

\displaystyle p+q

. • Since our derived value for

\displaystyle x^2

\displaystyle \frac{pq}{p+q}

, and this does not match options (A), (B), or (C), the correct answer is (D) None of these. • Option (C) is incorrect because it has

\displaystyle p-q

in the denominator, whereas our calculation yields

\displaystyle p+q

. The sign difference is crucial. Options (A) and (B) are incorrect because their numerators are

\displaystyle p

and

\displaystyle q

respectively, not

\displaystyle pq

About This Chapter: Ratio, Proportion, Indices, Logarithm

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Ratio, Proportion, Indices, Logarithms

This chapter covers Ratio, Proportion, Indices, Logarithms and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 5-7 Marks weightage. Focus on understanding core concepts rather than memorizing.

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If p:q\displaystyle p:qp:q is the sub-duplicate ratio of p−x2:q−x2\displaystyle p-x^2:q-x^2p−x2:q−x2, then x2\displaystyle x^2x2 is