Question

if 1 is added to both numerator and denominator of a fraction it becomes equal to 7 by 8 if one subtractor from both numerator and denominator it becomes 6 by 7 find the fraction​

Answer 1

Step-by-step explanation:

• 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} \:

y

x

• According to the first condition –

\begin{gathered} \\ \implies \sf \dfrac{x + 1}{y + 1} = \dfrac{7}{8} \\ \end{gathered}

⟹

y+1

x+1

=

8

7

\begin{gathered} \\ \implies \sf 8(x + 1) = 7(y +1 ) \\ \end{gathered}

⟹8(x+1)=7(y+1)

\begin{gathered} \\ \implies \sf 8x + 8 = 7y +7 \\ \end{gathered}

⟹8x+8=7y+7

\begin{gathered} \\ \implies \sf 8x = 7y - 1 \\ \end{gathered}

⟹8x=7y−1

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

⟹x=

8

1

(7y−1)−−−eq.(1)

• According to the second condition –

\begin{gathered} \\ \implies \sf \dfrac{x - 1}{y - 1} = \dfrac{6}{7} \\ \end{gathered}

⟹

y−1

x−1

=

7

6

\begin{gathered} \\ \implies \sf 7(x - 1) = 6(y - 1 ) \\ \end{gathered}

⟹7(x−1)=6(y−1)

\begin{gathered} \\ \implies \sf 7x - 7 = 6y - 6 \\ \end{gathered}

⟹7x−7=6y−6

\begin{gathered} \\ \implies \sf 7x = 6y + 1 \\ \end{gathered}

⟹7x=6y+1

• Now using eq.(1) –

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

⟹7[

8

1

(7y−1)]=6y+1

\begin{gathered} \\ \implies \sf 7 (7y - 1) = 8(6y + 1 ) \\ \end{gathered}

⟹7(7y−1)=8(6y+1)

\begin{gathered} \\ \implies \sf \: 49y - 7 = 48y + 8 \\ \end{gathered}

⟹49y−7=48y+8

\begin{gathered} \\ \implies \sf \: 49y - 48y = 7 + 8 \\ \end{gathered}

⟹49y−48y=7+8

\begin{gathered} \\ \implies \sf \large y = 15 \\ \end{gathered}

⟹y=15

• Put the value of 'y' in eq.(1) –

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

⟹x=

8

1

(7×15−1)

\begin{gathered} \\ \implies \sf x = \dfrac{1}{8} ( 105 - 1) \\ \end{gathered}

⟹x=

8

1

(105−1)

\begin{gathered} \\ \implies \sf x = \dfrac{1}{8} ( 104) \\ \end{gathered}

⟹x=

8

1

(104)

\begin{gathered} \\ \implies \sf \large x = 13 \\ \end{gathered}

⟹x=13

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

y

x

=

15

13

.