Question

(i) If (x – 9): (3x + 6) is the duplicate ratio of 4: 9, find the value of x.
(ii) If (3x + 1): (5x + 3) is the triplicate ratio of 3: 4, find the value of x.

(iii) If (x + 2y): (2x – y) is equal to the duplicate ratio of 3: 2, find x: y.

Answer 1

Duplicate of ratio means square of ratio and triplicate of ratio means cube of ratio.

So,

$\left(i\right)$ $\frac{(x–9)}{(3x+6)}={\left(\frac{4}{9}\right)}^2$

So we get

$\frac{(x–9)}{(3x+6)}=\frac{16}{81}$

By cross multiplication

$81x–729=48x+96$

$81x–48x=96+729$

So we get

$33x=825$

$\Rightarrow x=\frac{825}{33}=25$

$\left(ii\right)$ $\frac{(3x+1)}{(5x+3)}={\left(\frac{3}{4}\right)}^3$

So we get

$\frac{(3x+1)}{(5x+3)}=\frac{27}{64}$

By cross multiplication

$64(3x+1)=27(5x+3)$

$192x+64=135x+81$

$192x–135x=81–64$

$57x=17$

So we get

$x=\frac{17}{57}$

$\left(iii\right)$ $\frac{(x+2y)}{(2x–y)}={\left(\frac{3}{2}\right)}^2$

So we get

$\frac{(x+2y)}{(2x–y)}=\frac{9}{4}$

By cross multiplication

$9(2x–y)=4(x+2y)$

$18x–9y=4x+8y$

$18x=4x=8y+9y$

So we get

$14x=17y$

$\frac{x}{y}=\frac{17}{14}$

$x:y=17:14$