Question

Two numbers are in the ratio 7: 9. If each number is increased by 18, the ratio becomes 9 : 11 . Find the numbers.

Answer 1

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• Ratio of 2 no's = 7 : 9
• If 18 is added ratio = 9 : 11

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

• Find the numbers = ??

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

• let the ratio be 7x and 9x

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ATQ :-

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$\sf: \implies\: {\bold{\dfrac{7x + 18}{9x +18} = \dfrac{9}{11}}}$

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$\sf: \implies\: {\bold{ 11( 7x + 18) = 9( 9x + 18)}}$

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$\sf: \implies\: {\bold{77x + 198 = 81x + 162}}$

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$\sf: \implies\: {\bold{ 81x - 77x = 198 - 162}}$

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$\sf: \implies\: {\bold{ 4x = 36 }}$

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$\sf: \implies\: {\bold{ x = \dfrac{36}{4}}}$

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$\sf: \implies\: {\bold{ x = 9 }}$

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• Putting value of x in numbers

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7x = 7 × 9 = 63

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9x = 9 × 9 = 81

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

• Original numbers are 63 and 81

Answer 2

Given :-

Ratio of the numbers = 7 : 9

Each number is increased by 18.

Ratio after increased = 9 : 11

To Find :-

The first number.

The second number.

Analysis :-

Consider the common ratio as a variable. Multiply the variable to each of the numbers.

Make an equation when the numbers.

Get the value of the value and substitute it to the numbers.

Solution :-

Let the common ratio be 'x'. Then the numbers would be '7x' and '9x'.

When the numbers are increased by 18,

$\sf 7x+\dfrac{18}{9x} +18=\dfrac{9}{11}$

Next,

$\sf 77x + 198 = 81x + 162$

$\sf 77x - 81x = 162 - 198$

$\sf -4x = -36$

Finding the value of x,

$\sf x=\dfrac{-36}{-4}$

$\sf x=9$

Finding the numbers,

First numbers = 7x

= 7 × 9 = 63

Second number = 9x

= 9 × 9 = 81

Therefore, the two numbers are 63 and 81.