Question

3. In a fraction the denominator of a rational number is greater than its numerator by 7. If the
numerator is increased by 17 and the denominator decreased by 6, the new number becomes 2.
Find the original number.​

Answer 1

$\underline{\huge{\gray{\tt{Answer :-}}}}$

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$\sf\large\bold{\orange{\underline{\blue{ Given :-}}}}$

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• The denominator of a rational number is greater than its numerator by 7

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• When the numerator is increased by 17 and the denominator decreased by 6, the new number becomes 2

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$\sf\large\bold{\orange{\underline{\blue{ Solution :-}}}}$

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• Let the numerator be "n"

• Let the denominator be "d"

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$\underline{\bold{\texttt{Original fraction :}}}$

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➠ $\sf \dfrac { n } { d }$ ⚊⚊⚊⚊ ⓵

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$\purple{\underline {According \: to \: the \ question :}}$

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The denominator of a rational number is greater than its numerator by 7

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➠ d = n + 7 ⚊⚊⚊⚊ ⓶

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$\underline{\bold{\texttt{New numerator :}}}$

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Numerator is increased by 17

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➠ n + 17

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$\underline{\bold{\texttt{New denominator :}}}$

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Denominator is decreased by 6

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➠ d - 6

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Or,

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➜ n + 7 - 6

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➜ n + 1

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Also given that , when the numerator is increased by 17 and the denominator decreased by 6, the new number becomes 2

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So,

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➜ $\sf \dfrac { n + 17 } { n + 1 } = 2$

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➜ 2n + 2 = n + 17

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➜ 2n - n = 17 - 2

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➨ n = 15 ⚊⚊⚊⚊ ⓷

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• Hence the numerator is 15

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⟮ Putting n = 15 from ⓷ to ⓶ ⟯

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➜ d = n + 7

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➜ d = 15 + 7

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➨ d = 22 ⚊⚊⚊⚊ ⓸

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• Hence the denominator is 22

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⟮ Putting the values of equation ⓷ & ⓸ to ⓵ ⟯

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➨ $\sf \dfrac { 15} { 22}$

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• Hence the original fraction is $\sf \dfrac { 15 } { 22 }$

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Answer 2

Given:

The denominator of a rational number is greater than its numerator by

When the numerator is increased by 17 and the denominator decreased by 6, the new number becomes 2

Solution:−

Let the numerator be "n"

Let the denominator be "d"

Original fraction :

d

n

⚊⚊⚊⚊ ⓵

ATQ

The denominator of a rational number is greater than its numerator by 7

➠ d = n + 7 ⚊⚊⚊⚊ ⓶

New numerator :

Numerator is increased by 17

➠ n + 17

New denominator

Denominator is decreased by 6

➠ d - 6

Or,

➜ n + 7 - 6

➜ n + 1

Also given that , when the numerator is increased by 17 and the denominator decreased by 6, the new number becomes 2

So,

➜{ n + 17 } { n + 1 } = 2

n+1

n+17

=2

➜ 2n + 2 = n + 17

➜ 2n - n = 17 - 2

➨ n = 15 ⚊⚊⚊⚊ ⓷

Hence the numerator is 15

⟮ Putting n = 15 from ⓷ to ⓶ ⟯

➜ d = n + 7

➜ d = 15 + 7

➨ d = 22 ⚊⚊⚊⚊ ⓸

Hence the denominator is 22

⟮ Putting the values of equation ⓷ & ⓸ to ⓵ ⟯

➨ { 15} { 22}

22

15

Hence the original fraction is \sf \dfrac { 15 } { 22 }

22

15

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