the sum of numerator and the denominator of a fraction is 4 more than twice the numerator if 3 is added to each of the numerator and denominator the ratio becomes 2 : 3 find the fraction?
Answers
Step-by-step explanation:
The given fraction will be 5/9.
Solution:
Let n be the numerator and d be the denominator.
Given,
The sum of the given n and d is equal to twice the numerator plus 4, i.e.
\begin{lgathered}\begin{array} { c } { n + d = 2 n + 4 } \\\\ { n + d - 2 n - 4 = 0 } \\\\ { - n + d = 4 } \\\\ { n - d = - 4 \ldots ( i ) } \end{array}\end{lgathered}
n+d=2n+4
n+d−2n−4=0
−n+d=4
n−d=−4…(i)
From given, the numerator n and the denominator d must be increased by 3 to get the ratio 2:3
\begin{lgathered}\begin{array} { c } { \frac { n + 3 } { d + 3 } = \frac { 2 } { 3 } } \\\\ { 3 n + 9 = 2 d + 6 } \\\\ { 3 n - 2 d = 6 - 9 } \\\\ { 3 n - 2 d = - 3 \quad \ldots . ( i i ) } \end{array}\end{lgathered}
d+3
n+3
=
3
2
3n+9=2d+6
3n−2d=6−9
3n−2d=−3….(ii)
From equations (i) and (ii), we get,
\begin{lgathered}\begin{aligned} ( i ) \times 3 \Rightarrow & 3 n - 3 d = - 12 \\\\ (ii) & \Rightarrow 3 n - 2 d = - 3 \end{aligned}\end{lgathered}
(i)×3⇒
(ii)
3n−3d=−12
⇒3n−2d=−3
Solving (i) and (ii) we get,
\begin{lgathered}-d=-9\\\\d=9\end{lgathered}
−d=−9
d=9
Substituting d=9 in equation (i), we get,
\begin{lgathered}n-d=-4\\\\n-9=-4\\\\n=-4+9\\\\n=5\end{lgathered}
n−d=−4
n−9=−4
n=−4+9
n=5
∴ The given fraction is
\frac {n}{d}=\frac{5}{9}
d
n
=
9
5