In a fraction, if the numerator is decreased by 1 and the denominator is increased by 1,

Correct option is (B) 7

Let the required fraction be \(\frac xy.\)

According to first condition, we have

\(\frac{x-1}{y+1}=\frac12\)

\(\Rightarrow\) 2x - 2 = y + 1    (By cross multiplication)

\(\Rightarrow\) 2x - y - 3 = 0   ___________(1)

According to second condition, we have

\(\frac{x+1}{y-1}=\frac45\)

\(\Rightarrow\) 5 (x+1) = 4 (y - 1)  (By cross multiplication)

\(\Rightarrow\) 5x + 5 = 4y - 4

\(\Rightarrow\) 5x - 4y + 9 = 0   ___________(2)

Multiply equation (1) by 4, we get

8x - 4y - 12 = 0   ___________(3)

Subtract equation (2) from (3), we get

(8x - 4y - 12) - (5x - 4y + 9) = 0 - 0

\(\Rightarrow\) 3x - 21 = 0

\(\Rightarrow\) x = \(\frac{21}3\) = 7

Hence, the numerator of fraction \(\frac xy\) is x = 7.

Correct option is B) 7