1.

The sum of numerator and denominator of fraction is 3 less than twice denominator. If each numerator and denominator is decreased by 1 the fraction become 1/2. Find the fraction

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Solution:

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Let \ the \ Numerator \ be x \\ and \ denominator \ be \ y \\\\ Also \ given - \\ x + y = 2y - 3\\ x + y - 2y = -3 \\ x - y = -3 \\ \ x= -3 +y------ (i)\\\\ Also , \ ACQ \\ \frac{x-1}{y-1} = \frac{1}{2} \\\\ 2x - 2 = y - 1\\ 2x - y = -1 + 2 \\ 2x-y = 1 \ \ ------- (ii)\\\\ Putting \ the \ value \ of x \ in \ eq (i) \ in \ eq. \ (ii)\\ 2x - y = 1\\ 2\times (-3+y) -y = 1\\ -6 + 2y -y = 1\\ 2y-y=1+6\\y = 7 \\\\ Putting \ the \ value \ of \ y \in \ equ. \ (i)\\x = -3 + y \\ x= -3 + 7\\ x = 4 \\\\


So, the REQUIRED fraction is 4/7


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