Question

19. The numerator of a fraction is 1 less than the denominator.

Answer 1

Let the original fraction be

$\dfrac{n-1}{n}$

Now let's increase both numerator and denominator by 2:

$\dfrac{n+1}{n+2}$

This produces a 1/12 increase:

$\dfrac{n+1}{n+2}-\dfrac{n-1}{n}=\dfrac{1}{12}$

The left hand side can be rearranged as

$\dfrac{2}{n^2+2n}=\dfrac{1}{12}$

Invert both sides:

$\dfrac{n^2+2n}{2}=12 \iff n^2+2n=24$

Solve the quadratic equation:

$n^2+2n-24=0 \iff n=-6 \lor n=4$

So, in the first case, the original fraction is

$\dfrac{n-1}{n}=\dfrac{-6-1}{-6}=\dfrac{7}{6}$

In the second case, we have

$\dfrac{n-1}{n}=\dfrac{4-1}{4}=\dfrac{3}{4}$