Question

If $\left(p-q\right):\left(q-x\right)$ be the duplicate ratio of $p:q$, then : $\frac{1}{p}+\frac{1}{q}=\frac{1}{x}$

• A. True
• B. False

Answer 1

The correct option is A True

$(p-x):(q-x)$ is the duplicate ratio of $p:q$

we know,

if $a^2:b^2$ is the duplicate ratio of $a:b$

now a/c to question,

$(p-x):(q-x)$ is the duplicate ratio of $p:q$

so, from above rule,

$(p-x):(q-x)=p^2:q^2$

So,$\frac{p^2}{q^2}=\frac{p-x}{q-x}$

$\Rightarrow \frac{q-x}{q^2}=\frac{p-x}{p^2}$

$\Rightarrow \frac{q}{q^2}-\frac{x}{q^2}=\frac{p}{p^2}-\frac{x}{p^2}$

$\Rightarrow \frac{1}{q}-\frac{x}{q^2}=\frac{1}{p}-\frac{x}{p^2}$

$\Rightarrow \frac{1}{q}-\frac{1}{p}=\frac{x}{q^2}-\frac{x}{p^2}$

$\Rightarrow \frac{p-q}{pq}=\frac{x(p^2-q^2)}{p^2{q}^2}$

$\Rightarrow p-q=\frac{x(p-q)(p+q)}{pq}$

$\Rightarrow 1=\frac{x(p+q)}{pq}$

$\Rightarrow \frac{1}{x}=\frac{(p+q)}{pq}$

$\Rightarrow \frac{1}{x}=\frac{1}{q}+\frac{1}{p}$

$\therefore \frac{1}{p}+\frac{1}{q}=\frac{1}{x}$

Hence the given statement is true.