The denominator of a fraction exceeds its numerator by 3. If one is added to both numerator and denominator, the difference between the new and the original fractions 1 becomes
. Find the original fraction.
Let the denominator be ‘(X+3)’, so the numerator will be ‘X’.
the original fraction is ![]()
the new fraction is ![]()
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On simplifying further,
![]()
24(3) = X2 + 7X + 12
X2 + 7X – 60 = 0 –––––– (i)
On applying Sreedhracharya formula
![]()

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∴ X = –12 or 5.
∴ the numerator is X = 5 and denominator (X+3) will be 8 and the fraction will be
, as taking (–12) will form a fraction i.e.,
(not satisfying the conditions) .
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