(a) $\frac{1}{x}+\frac{1}{y}=5.....(1)$
$\frac{1}{y}-\frac{1}{x}=1........(2)$
Substitute p and q for $\frac{1}{x}$ and $\frac{1}{y}$ respectively.
$p+q=5.........(1a)$
$q-p=1.......(2a)$
From (2a), q = 1 + p.
Putting into (1a), we have:
$p+(1+p)=5$
$2p+1=5⟹2p=5-1=4$
$p=2$
$q=1+p=1+2=3$
$p=\frac{1}{x}=2$
$⟹x=\frac{1}{2}$
$q=\frac{1}{y}=3$
$⟹y=\frac{1}{3}$
Hence, $x=\frac{1}{2};y=\frac{1}{3}$.
(b) Total distance travelled = 48 km
Total time taken = 45 minutes = $\frac{45}{60}=0.75hr$
Let the man travel a distance of x km on a good surface. Then he travel a distance of (48 - x) km on bad surface.
Let the time taken to travel on good surface be t hours, then the time taken to travel on bad surface = (0.75 - t) hrs.
Using $speed=\frac{distance}{time}$ in each case
On good surface : $72=\frac{x}{t}$
$⟹x=72t...(1)$
On bad surface : $48=\frac{48-x}{0.75-t}$
$48(0.75-t)=48-x$
$⟹36-48t=48-x......(2)$
Substitute 72t = x in (2) :
$36-48t=48-72t⟹72t-48t=48-36$
$24t=12⟹t=\frac{12}{24}=0.5hours$
From (1), x = 72t
$x=72\times 0.5=36km$
Hence, the man travels a distance of 36 km on good road surface.