Question

$\large\underbrace{\textsf{\textbf{\gray{Q}u\gray{e}s\gray{t}i\gray{o}n\;:}}}$

The ratio of two numbers is 3:5. If both the numbers are increased by 5 their ratio becomes 2:3. Find the numbers.​

Answer 1

★ How to do :-

Here, we are given with the ratio of two numbers. We are said that those two numbers that are in ratio are increased by 5 each. We are also given with the new ratio of those numbers when they are increased we are asked to find those numbers. Here, we use that concept of the cross multiplication of the numbers. By this method we find the value of the value of the variable x and then, we multiply thr value of the variable by both the part of the ratio. So, let's solve!!

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➤ Solution :-

${\tt \leadsto 3 : 5 = 3x : 5x}$

Write the ratio as the fractional form.

${\tt \leadsto \dfrac{3x}{5x}}$

Now, write the other ratio in fractional form.

${\tt \leadsto \dfrac{3x + 5}{5x + 5} = \dfrac{2}{3}}$

Now, cross multiply the numbers.

${\tt \leadsto 3 \: (3x + 5) = 2 \: (5x + 5)}$

Multiply the numbers with both numbers in bracket.

${\tt \leadsto 9x + 15 = 10x + 10}$

Shift the variables on one side and constants on other side.

${\tt \leadsto 9x - 10x = 10 - 15}$

Subtract the values on LHS and RHS.

${\tt \leadsto - 1x = - 5}$

Shift the number on LHS to RHS, changing it's sign.

${\tt \leadsto x = \dfrac{-5}{-1}}$

Simplify the fraction to get the value of x.

${\tt \leadsto x = 5}$

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Now, find the present numbers by multiplying the value of x.

First present number :-

${\tt \leadsto 3x = 3 \times 5}$

${\tt \leadsto 15}$

Second present number :-

${\tt \leadsto 5x = 5 \times 5}$

${\tt \leadsto 25}$

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Now, let's find the value of the numbers after it increased.

First number after incresion :-

${\tt \leadsto 15 + 5 = 20}$

Second number after incresion :-

${\tt \leadsto 25 + 5 = 30}$

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${\red{\underline{\boxed{\bf So, \: the \: numbers \: before \: were \: 15 \: and \: 25}}}}$

${\red{\underline{\boxed{\bf The \: numbers \: after \: incresion \: were \: 20 \: and \: 30}}}}$

Answer 2

$\bold\green{✓\:verified \: answer}$

➤ Solution :-

${\tt \leadsto 3 : 5 = 3x : 5x}$

Write the ratio as the fractional form.

${\tt \leadsto \dfrac{3x}{5x}}$

• Now, write the other ratio in fractional form.

${\tt \leadsto \dfrac{3x + 5}{5x + 5} = \dfrac{2}{3}}$

• Now, cross multiply the numbers.

${\tt \leadsto 3 \: (3x + 5) = 2 \: (5x + 5)}$

• Multiply the numbers with both numbers in bracket.

${\tt \leadsto 9x + 15 = 10x + 10}$

• Shift the variables on one side and constants on other side.

${\tt \leadsto 9x - 10x = 10 - 15}$

• Subtract the values on LHS and RHS.

${\tt \leadsto - 1x = - 5}$

• Shift the number on LHS to RHS, changing it's sign.

${\tt \leadsto x = \dfrac{-5}{-1}}$

• Simplify the fraction to get the value of x.

${\tt \leadsto x = 5}$

Now, find the present numbers by multiplying the value of x.

• First present number :-

${\tt \leadsto 3x = 3 \times 5}$

${\tt \leadsto 15}$

• Second present number :-

${\tt \leadsto 5x = 5 \times 5}$

${\tt \leadsto 25}$

Now, let's find the value of the numbers after it increased.

• First number after incresion :-

${\tt \leadsto 15 + 5 = 20}$

• Second number after incresion :-

${\tt \leadsto 25 + 5 = 30}$

${\blue{\underline{\boxed{\bf So, \: the \: numbers \: before \: were \: 15 \: and \: 25}}}}$

${\green{\underline{\boxed{\bf The \: numbers \: after \: incresion \: were \: 20 \: and \: 30}}}}$