Question

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

Answer 1

Answer:

THE THIRD NUMBER IS 30.

Answer 2

Answer :

$\boxed{\boxed{\mathsf{The \: Numbers \: are \: = 20 , 30\: and \: 48}}}$

Step-by-step Explanation :

Given

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

To Find

• All the three numbers.

Solution

Let the ,

$\mathsf{\longrightarrow \: First \: Number \: be = X}$

$\mathsf{\longrightarrow \: Second \: Number \: be = Y}$

$\mathsf{\longrightarrow \: Third \: Number \: be = Z}$

Now According To Question

$\large \mathsf{X + Y + Z = 98}$ ----- (1)

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

That is --

$\mathsf{\longrightarrow \: \dfrac{First\: Number}{Second\: Number} = \dfrac{2}{3}}$

$\mathsf{\longrightarrow \: \dfrac{x}{y} = \dfrac{2}{3}}$

$\mathsf{\longrightarrow \: x = \dfrac{2}{3} \times y}$

$\mathsf{\longrightarrow \: x = \dfrac{2y}{3}}$ -- (i)

2 ) Ratio of Second Number to the third Number = 5:8

That is ,

$\mathsf{\longrightarrow \: \dfrac{Second\: Number}{Third\: Number} = \dfrac{5}{8}}$

$\mathsf{\longrightarrow \: \dfrac{y}{z} = \dfrac{5}{8}}$

$\mathsf{\longrightarrow \: z = \dfrac{8}{5} \times y }$

$\mathsf{\longrightarrow \: z = \dfrac{8y}{5} }$ ---- (ii)

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

We have

$\mathsf{\Longrightarrow \: x + y + z = 98}$

$\mathsf{\longrightarrow \: \dfrac{2y}{3} + y + \dfrac{8y}{5} = 98}$

$\mathsf{\longrightarrow \: \dfrac{2y}{3} + y + \dfrac{8y}{5} = 98}$

$\mathsf{\longrightarrow \: \dfrac{5(2y) + 15(y) + 3(8y)}{15} = 98}$

$\mathsf{\longrightarrow \: \dfrac{49y}{15} = 98}$

$\mathsf{\longrightarrow \: y = 98 \times \dfrac{15}{49}}$

$\mathsf{\longrightarrow \: y = \cancel{98} \times \left(\dfrac{ 15}{ \cancel{49}} \right)}$

$\mathsf{\longrightarrow \: \boxed{y = 30}}$

$\underline{\underline{\mathsf{Putting\: value \: of \: y = 30 \: in \: (i) (iii)}}}$

$\mathsf{\longrightarrow \: x = \dfrac{2y}{3}}$

$\mathsf{\longrightarrow \: x = \dfrac{2(30)}{3}}$

$\mathsf{\longrightarrow \: x = \dfrac{60}{3}}$

$\mathsf{\longrightarrow \: \boxed{x = 20}}$

and

$\mathsf{\longrightarrow \: z = \dfrac{8y}{5}}$

$\mathsf{\longrightarrow \: z = \dfrac{8(30)}{5}}$

$\mathsf{\longrightarrow \: z = \dfrac{240}{5}}$

$\mathsf{\longrightarrow \: \boxed{z = 48}}$

Therefore Required answer :

$\underline{\boxed{\mathsf{The \: Numbers \: are \: = 20 , 30\: and \: 48}}}$