Two numbers are in the ratio $$1\frac{1}{2}$$ : $$2\frac{2}{3}$$ When each of these is increased by 15, they become in the ratio $$1\frac{2}{3}$$ : $$2\frac{1}{2}$$ . The greater of the numbers is = ?

Solution(By Examveda Team)

$$\eqalign{ & {\text{A : B}} \cr & \frac{3}{2}:\frac{8}{3} \cr} $$
take L.C.M. of denominator and multiply)
$$ \Rightarrow \frac{3}{2} \times 6:\frac{8}{3} \times 6 = 9x:16x$$
After adding 15 in each we get
$$\eqalign{ & \therefore \frac{{9x + 15}}{{16x + 15}} = \frac{5}{3} \times \frac{2}{5} \cr & \Rightarrow \frac{{9x + 15}}{{16x + 15}} = \frac{2}{3}\left( {{\text{cross multiply}}} \right) \cr & \Rightarrow 27x + 45 = 32x + 30 \cr & \Rightarrow 5x = 15 \cr & \Rightarrow x = 3 \cr & \therefore {\text{Smaller number}} \cr & = {\text{9}} \times {\text{3 = 27}} \cr & {\text{Greater number}} \cr & {\text{ = 16}} \times {\text{3 = 48}} \cr} $$