Question

If 1 is added to both numerator and denominator of a fraction , it becomes 7/8 and if 1 is subtracted from both numerator and denominator of the same fraction , it becomes 6/7.Find the fraction.

Answer 1

GIVEN :–

• If 1 is added to both numerator and denominator of a fraction , it becomes 7/8.

• If 1 is subtracted from both numerator and denominator of the same fraction , it becomes 6/7.

TO FIND :–

• The fraction = ?

SOLUTION :–

• Let the fraction is $\: \sf \: \dfrac{x}{y} \:$

• According to the first condition –

$\\ \implies \sf \dfrac{x + 1}{y + 1} = \dfrac{7}{8}$

$\\ \implies \sf 8(x + 1) = 7(y +1 )$

$\\ \implies \sf 8x + 8 = 7y +7$

$\\ \implies \sf 8x = 7y - 1$

$\\ \implies \sf x = \dfrac{1}{8} (7y - 1) \: \: \: - - - eq.(1)$

• According to the second condition –

$\\ \implies \sf \dfrac{x - 1}{y - 1} = \dfrac{6}{7}$

$\\ \implies \sf 7(x - 1) = 6(y - 1 )$

$\\ \implies \sf 7x - 7 = 6y - 6$

$\\ \implies \sf 7x = 6y + 1$

• Now using eq.(1) –

$\\ \implies \sf 7 \left[\dfrac{1}{8} (7y - 1) \right] = 6y + 1$

$\\ \implies \sf 7 (7y - 1) = 8(6y + 1 )$

$\\ \implies \sf \: 49y - 7 = 48y + 8$

$\\ \implies \sf \: 49y - 48y = 7 + 8$

$\\ \implies \sf \large y = 15$

• Put the value of 'y' in eq.(1) –

$\\ \implies \sf x = \dfrac{1}{8} (7 \times 15 - 1)$

$\\ \implies \sf x = \dfrac{1}{8} ( 105 - 1)$

$\\ \implies \sf x = \dfrac{1}{8} ( 104)$

$\\ \implies \sf \large x = 13$

• Hence , The fraction is $\: \sf \: \dfrac{x}{y} = \dfrac{13}{15} . \:$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

• If 1 is added to both numerator and denominator of fraction,it becomes 7/8.
• If 1 is subtracted from both numerator and denominator of the same fraction,it becomes 6/7.

${\bf{\blue{\underline{To\:Find}}}}$

• Fraction =?

${\bf{\blue{\underline{Now:}}}}$

• Let the fraction =r/m

$\star \: \: \huge\underline{\mathfrak{ case \: 1:-}}$

If 1 is added to both numerator and denominator of fraction then it becomes 7/8.

$: \implies{\sf{ \frac{r + 1}{m + 1} = \frac{7}{8} }}$

$: \implies{\sf{ 8(r + 1) = 7(m + 1) }}$

$: \implies{\sf{ 8r + 8 = 7m + 7 }}$

$: \implies{\sf{ 8r - 7m = 7 - 8 }}$

$: \implies{\sf{ 8r - 7m = -1.......(1) }}$

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$\star \: \: \huge \underline{\mathfrak{ case \: 2:-}}$

If 2 is subtracted to both numerator and denominator of fraction then it becomes 6/7.

$: \implies{\sf{ \frac{r - 1}{m - 1} = \frac{6}{7} }}$

$: \implies{\sf{ 7(r - 1) = 6(m - 1) }}$

$: \implies{\sf{ 7r - 7 = 6m -6 }}$

$: \implies{\sf{ 7r - 6m = 7-6}}$

$: \implies{\sf{ 7r - 6m = 1.......(2) }}$

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From (1) and (2),

Take (1),

$: \implies{\sf{ 8r = -1 + 7m }}$

$: \implies{\sf{ r= \frac{7m - 1}{8}....(3) }}$

Put the Value of r in equation (2),

$: \implies{\sf{ 7 \bigg(\frac{7m - 1}{8} \bigg) - 6m = 1 }}$

$: \implies{\sf{ \bigg(\frac{49m - 7}{8} \bigg) - 6m = 1 }}$

$: \implies{\sf{ 49m - 7 - 48m =8 }}$

$: \implies{\sf{ m - 7 =8 }}$

$: \implies{\sf{ m =8 +7}}$

$: \implies\boxed{\sf{ m =15}}$

Now put the Value of m in equation (3),

$: \implies{\sf{ r= \frac{7(15) - 1}{8} }}$

$: \implies{\sf{ r= \frac{105 - 1}{8} }}$

$: \implies{\sf{ r= \frac{104}{8} }}$

$: \implies{\sf\boxed{ r= 13 }}$

$: \dagger \: \: {\sf{ hence \: the \: fraction \: = \frac{13}{15} }}$