Question

Two numbers are in the ratio 5:8 and if their sum is 169. Find the numbers. ​

Answer 1

Answer :

• The numbers are 65 and 104

Given that :

• Two numbers are in the ratio 5 : 8
• If their sum is 169

To find :

• The numbers

Solution :

Given,

Two numbers are in the ratio 5 : 8 so,

• Let the two numbers be 5x and 8x

Given, sum is 169

➞ 5x + 8x = 169

➞ 13x = 169

➞ x = 169/13

➞x = 13

Finding the numbers :

➞ 5x

➞ 5(13)

➞ 65

➞ 8x

➞ 8(13)

➞ 104

Hence, The numbers are 65 and 104

Verification :

➞ 5x + 8x = 169

➞ 5(13) + 8(13) = 169

➞ 65 + 104 = 169

➞ 169 = 169

Hence, Verified

Answer 2

Given :

• Ratio of two numbers is 5 : 8

• Sum of the two numbers is 169

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To find :

• The two numbers.

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Assumption :

Let one of the number be x.

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Solution :

Given -

Sum of the two numbers is 169

$\tt\:x~+~Other~number~=~169$

$\tt\color{purple}{\implies~Other~number~=~(169-x)}$

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Now by the question -

Ratio of two numbers = 5 : 8

$\tt\therefore\:\frac{One~number}{Other~number}=\frac{5}{8}$

Plugging the values respectively

$\tt\implies\:\frac{x}{169-x}=\frac{5}{8}$

Cross multiplying

$\tt\implies\:8x=5(169-x)$

$\tt\implies\:8x=845-5x$

Transposing -5x to LHS it becomes +5x

$\tt\implies\:8x+5x=845$

$\tt\implies\:13x=845$

Transposing 13 to RHS it becomes to the denominator

$\tt\implies\:x=\frac{845}{13}$

Reducing the fraction to the lower terms

$\tt\implies\:x=\cancel{\frac{845}{13}}$

$\small{\underline{\boxed{\mathrm{\implies\:x\:=\:65}}}}$ ★

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Therefore :

$\sf\:One~number~=~x~=$ $\small\fbox{65}$

$\sf\:Other~number~=~(169-x)~=169-65~=$ $\small\fbox{104}$

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Verification :

Plugging the values in the equation framed earlier -

$\tt\therefore\:\frac{One~number}{Other~number}=\frac{5}{8}$

$\tt\twoheadrightarrow\:\frac{65}{104}=\frac{5}{8}$‎

Cross multiplying

$\tt\twoheadrightarrow\:104 \times 5 = 65 \times 8$

$\tt\twoheadrightarrow\:520= 520$

$\bf\twoheadrightarrow\:LHS= RHS$

Hence Verified !~

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Henceforth :

⠀‎ ⠀⠀ ⠀⠀ ⠀❒ One number is $\large{\mathfrak\pink{65}}$

⠀‎ ⠀⠀ ⠀⠀ ⠀❒ Other number is $\large{\mathfrak\pink{104}}$

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