Question

Two numbers are in the ratio 3:5. If each is increased by 10 , the ratio between the new numbers so formed is 5:7 . Find the original numbers.​

Answer 1

Given :-

• Ratio of two numbers is 3:5.
• If each is increased by 10 , the ratio between the new numbers so formed is 5:7.

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To find :-

Original numbers?

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Solution :-

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☯ Let's consider the two numbers be x and y.

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Now,

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$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

Ratio of two numbers is 3:5.

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$:\implies\sf x : y = 3:5$

$:\implies\sf \dfrac{x}{y} = \dfrac{3}{5}$

$:\implies\sf 5x = 3y$

$:\implies\sf \pink{x = \dfrac{3y}{5}}\qquad\quad\bigg\lgroup\bf eq (1)\bigg\rgroup$

And,

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If each is increased by 10 , the ratio between the new numbers so formed is 5:7.

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Numbers after increasing by 10,

(x + 10) and (y + 10)

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Therefore,

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$:\implies\sf (x + 10) : (y + 10) = 5:7$

$:\implies\sf \dfrac{(x + 10)}{(y + 10)} = \dfrac{5}{7}$

$:\implies\sf 7(x + 10) = 5(y + 10)$

$:\implies\sf 7x + 70 = 5y + 50$

$:\implies\sf 7x - 5y = 50 - 70$

$:\implies\sf 7x - 5y = - 20$

$:\implies\sf \pink{7x + 20 = 5y} \qquad\quad\bigg\lgroup\bf eq (2)\bigg\rgroup$

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Now, Put value of x From eq (1) in eq (2),

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$:\implies\sf 7 \bigg( \dfrac{3y}{5} \bigg) + 20 = 5y$

$:\implies\sf \dfrac{21y}{5} + 20 = 5y$

$:\implies\sf \dfrac{21y}{5} - 5y = - 20$

$:\implies\sf \dfrac{21y - 25y}{5} = -20$

$:\implies\sf \dfrac{- 4y}{5} = - 20$

$:\implies\sf - 4y = -20 \times 5$

$:\implies\sf - 4y = - 100$

$:\implies\sf 4y = 100$

$:\implies\sf y = \cancel{\dfrac{100}{4}}$

$:\implies{\underline{\boxed{\frak{\purple{y = 25}}}}}\;\bigstar$

Now, Putting value of y in eq (1),

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$:\implies\sf x = \dfrac{3 \times 25}{5}$

$:\implies\sf x = \cancel{ \dfrac{75}{5}}$

$:\implies{\underline{\boxed{\frak{\purple{x = 15}}}}}\;\bigstar$

$\therefore$ Hence, The original numbers are 15 and 25.

Answer 2

Answer:

answer-

Step-by-step explanation:

$let \: the \: first \: number \: be \: 3x \: \\ let \: the \: second \: number \: be \: 5x \\ if \: each \: is \: increased \: by \: 10 \\ then \: the \: no.s. \: will \: be \: = \frac{3x + 10}{5x + 10} = \frac{5}{7} \\ \: we \: will \: cross \: multiply \: it \: \\ = 7(3x + 10) \: = 5(5x + 10) \\ = 21x + 70 = 25x + 50 \\ = 21x - 25x = - 70 + 50 \\ = - 4x = - 20 \\ = x = \frac{20}{4} \\ = x = 5 \\ now \: original \: no. = 3 \times 5 = 15\\ = 5\times 5 = 25$