Question

13. 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

Given:

• The sum of three numbers is 98.
• Ratio of First number to Second number = 2:3
• Ratio of second number to the third number = 5:8

To Find:

• All the three numbers.

Solution:-

Let the ,

• First Number be = X
• Second Number be = Y
• Third Number be = Z

According to the question,

X + Y + Z = 98 -- (1)

1 ) Ratio of First number to Second Number = 2:3

That is:-

→ First number/Second number = 2/3

→ x/y = 2/3

→ x = 2/3 × y

→ x = 2y/3 ---(i)

2) Ratio of Second number to third number = 5:8

That is:-

→ Second number/Third number = 5/8

→ y/z = 5/8

→ z = 8/5 × y

→ z = 8y/5 ---(ii)

3) Now putting this value of X and Z obtained above in 1

We have:-

→ x + y + z = 98

→ 2y/3 + y + 8y/5 = 98

→ 5(2y) + 15(y) + 3(8y) /15 = 98

→ 49y/15 = 98

→ y = 98 × 15/49

→ y = 30

Putting value of y = 30 in (i) & (ii),

→ x = 2y/3

→ x = 2(30)/3

→ x = 60/3

→ x = 20

and,

→ z = 8y/5

→ z = 8(30)/5

→ z = 240/5

→ z = 48

Hence,

• Required numbers are 20, 30 and 48.

Answer 2

Given

• There are three numbers.
• The ratio of the first number to the second number is 2:3.
• The ratio of second number to the third number is 5:8.

⠀

To find

• The three numbers.

⠀

Solution

⠀⠀⠀⠀❍ Let the

• First number be x
• Second number be y
• Third number be z

⠀

⠀⠀❍ Sum of three numbers is 98 ❍

$\tt\longrightarrow{x + y + z = 98}$⠀⠀....[1]

⠀

❍ Ratio of first and second number is 2:3 ❍

$\tt\longrightarrow{\dfrac{x}{y} = \dfrac{2}{3}}$

⠀

$\tt\longrightarrow{3x = 2y}$

⠀

$\bf\longrightarrow{x = \dfrac{2y}{3}}$⠀⠀....[2]

⠀

❍ Ratio of second and third number is 5:8 ❍

$\tt\longrightarrow{\dfrac{y}{z} = \dfrac{5}{8}}$

⠀

$\tt\longrightarrow{8y = 5z}$

⠀

$\tt\longrightarrow{z = \dfrac{8y}{5}}$⠀⠀.....[3]

⠀

★ From [1]

$\large{\tt{\longmapsto{x + y + z = 98}}}$

⠀

★ Putting the values of x and z from [2] and [3]

$\tt\longmapsto{\dfrac{2y}{3} + y + \dfrac{8y}{5} = 98}$

⠀

★ Taking LCM

$\tt\longmapsto{\dfrac{10y + 15y + 24y}{15} = 98}$

⠀

$\tt\longmapsto{10y + 15y + 24y = 1470}$

⠀

$\tt\longmapsto{49y = 1470}$

⠀

$\tt\longmapsto{y = \dfrac{1470}{49}}$

⠀

$\bf\longmapsto{y = 30}$

⠀

★ Putting the value of y in [2]

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

⠀

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

⠀

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

⠀

★ Putting the value of y in [3]

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

⠀

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

⠀

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

⠀

Hence,

⠀⠀⠀⠀⠀⠀❍ The values are

• First number (x) = 20
• Second number (y) = 30
• Third number (z) = 48

━━━━━━━━━━━━━━━━━━━━━━