If a positive fraction numerator and denominator is increased by 2 the fraction increases by $\frac{1}{24}$ find the difference between the numerator and denominator, given the sum of Numerator+Denominator =11?

GIven $$\frac{N+2}{D+2}=\frac{N}{D}+\frac{1}{24}$$

If we solve it using c&d we get $$\frac{N+2}{D+2}=\frac{24N+1}{D24}$$ $$\frac{N+2+D+2}{N+2-D-2}=\frac{24N+1+D24}{24N+1-D24}$$

Which will solve up to

$$\frac{15}{N-D}=\frac{265}{24N-D24-1}$$ $$N-D= 3/19$$

But the solution comes out to be $1$ if we solve without using C&D . any suggestions will be helpful


You have made a mistake (as it's consistent later on, I don't think it's a typo), as your second line should be

$$\frac{N+2}{D+2}=\frac{24N+D}{D24} \tag{1}\label{eq1}$$

Cross-multiplying gives

$$24ND + 48D = 24ND + 48N + D^2 + 2D \; \Rightarrow \; 48\left(D - N\right) = D^2 + 2D \tag{2}\label{eq2}$$

Using $N + D = 11$, use $N = 11 - D$ to get $D - N = 2D - 11$ so \eqref{eq2} becomes

$$96D - 528 = D^2 + 2D \; \Rightarrow \; D^2 - 94D + 528 = 0 \tag{3}\label{eq3}$$

This is a quadratic equation you can solve for $D$ (but make sure that the fraction $\frac{N}{D}$ is positive as the question states). Then substitute the result(s) into the RHS of \eqref{eq2} and divide by $-48$ to get $N - D$, i.e., the difference between the numerator and denominator. I trust you can finish these final calculations.

Using your "c&d" technique (i.e., if $\frac{a}{b} = \frac{c}{d}$, then $\frac{a \; + \; b}{a \; - \; b} = \frac{c \; + \; d}{c \; - \; d}$) instead gives

$$\frac{N+2+D+2}{N+2-D-2}=\frac{24N+D+D24}{24N+D-D24}$$ $$\frac{15}{N-D}=\frac{264+D}{24\left(N-D\right) + D} \tag{4}\label{eq4}$$

Cross-multiplying now gives

$$360\left(N-D\right) + 15D = 264\left(N-D\right) + D\left(N - D\right) \tag{5}\label{eq5}$$

If you substitute $N - D = 11 - 2D$, you get

$$360\left(11 - 2D\right) + 15D = 264\left(11 - 2D\right) + D\left(11 - 2D\right) \tag{6}\label{eq6}$$

Moving everything to the LHS and dividing by $2$ gives the RHS of \eqref{eq3}. As such, the "c&d" technique does work but, in this case at least, it seems to make the solution a bit longer & more complicated.

  • $\begingroup$ Yes I made that mistake and c&d is not fruitful in this question $\endgroup$ – Nebo Alex Mar 1 at 7:20

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