An algebra problem by Javed Siddiqui

Let the numbers be $a,b,c$ . We have

$a+b+c=98$ $\color{#D61F06}(1)$

$\dfrac{a}{b}=\dfrac{2}{3}$ $\implies$ $a=\dfrac{2}{3}b$ $\color{#D61F06}(2)$

$\dfrac{b}{c}=\dfrac{5}{8}$ $\implies$ $c=\dfrac{8}{5}b$ $\color{#D61F06}(3)$

Substitute $\color{#D61F06}(2)$ and $\color{#D61F06}(3)$ in $\color{#D61F06}(1)$ , we have

$\dfrac{2}{3}b+b+\dfrac{8}{5}b=98$

$\dfrac{49}{15}b=98$

$\boxed{b=30}$

Solution:

Let the three numbers be x, y and z.

Sum of the numbers is 98.

x + y + z = 98………………(i)

The ratio of the first to the second is 2/3.

x/y = 2/3.

x = 2/3 × y.

x = 2y/3.

The ratio of the second to the third is 5/8.

y/z = 5/8.

z/y = 8/5.

z = 8/5 × y.

z = 8y/5.

Put the value of x = 2y/3 and z = 8y/5 in (i).

2y/3 + y + 8y/5 = 98

49y/15 = 98.

49y = 98 × 15.

49y = 1470.

y = 1470/49.

y = 30 .

Therefore, the second number is 30. so the answer is c 30