Question

The sum of three numbers is 98. If the ratio of the first to the second is 2:3 and that of the second to the third is 5:8, find the three numbers.​

Answer 1

From the Question,

Let,

• The three numbers be ${x,\:y}$ and ${z}$

Given data,

• $\sf{x\:+\:y\:+\:z\:=\:98}$
• $\sf{x\:{:}\:y\:=\:2\:{:}\:3,\:y\:{:}\:z\:=\:5\:{:}\:8}$
• $\sf{{\dfrac{x}{y}}\:=\:{\dfrac{2}{3}},\:{\dfrac{y}{z}}\:=\:{\dfrac{5}{8}}}$

From this, we get,

• $\sf{x}\:=\:{\dfrac{2y}{3}}\:----(1)$
• $\sf{z}\:=\:{\dfrac{8y}{5}}\:----(2)$

We know that, $\sf{x\:+\:y\:+\:z\:=\:98}$

Substituting the known values, we get,

$\implies\sf{{\dfrac{2y}{3}}\:+\:y\:+\:{\dfrac{8y}{3}}\:=\:98}$

$\implies\sf{{\dfrac{10y\:+\:15y\:+\:24y}{15}}\:=\:98}$

$\implies\sf{{\dfrac{49y}{15}}\:=\:98}$

$\implies\sf{49y\:=\:98\:\times\:15}$

$\implies\sf{49y\:=\:1470}$

$\implies\sf{y\:=\:{\dfrac{1470}{49}}}$

$\implies\sf{y\:=\:{\cancel{{\dfrac{1470}{49}}}}}$

$\implies\sf{y\:=\:30}$

Substituting 'y' value in (1), we get,

$\implies\sf{x}\:=\:{\dfrac{2y}{3}}$

$\implies\sf{x}\:=\:{\dfrac{2(30)}{3}}$

$\implies\sf{x}\:=\:{\dfrac{60}{3}}$

$\implies\sf{x}\:=\:{\cancel{{\dfrac{60}{3}}}}$

$\implies\sf{x\:=\:20}$

Substituting 'y' value in (2), we get,

$\implies\sf{z}\:=\:{\dfrac{8y}{5}}$

$\implies\sf{z}\:=\:{\dfrac{8(30)}{5}}$

$\implies\sf{z}\:=\:{\dfrac{240}{5}}$

$\implies\sf{z}\:=\:{\cancel{{\dfrac{240}{5}}}}$

$\implies\sf{z\:=\:48}$

Hence, x = 20, y = 30 and z = 48.