Question

To travel 100 miles, it takes Sue, riding a moped, 3 hours less time than it takes Doreen to travel 42 miles riding a bicycle. Sue travels 19 miles per hour faster than Doreen. Find the times and rates of both girls.

Answer 1

Let us assume Doreen rate = x miles per hour.

Sue travels 19 miles per hour faster than Doreen.

Therefore,

Rate of Sue = (x+19) miles per hour.

We know time, rate and distance relation as

Time = Distance / rate.

Therefore, time taken by Doreen to travel  42 miles at the rate x miles per hour =

$\frac{42}{x}$.

And time taken by Sue to travel  100 miles at the rate (x+19) miles per hour =

$\frac{100}{(x+19)}$.

Sue takes 3 hours less time than it takes Doreen.

Therefore,

$\frac{42}{x}-\frac{100}{(x+19)}=3$

We need to solve the equatuion for x now.

$\frac{42}{x}-\frac{100}{\left(x+19\right)}=3$

$\mathrm{Find\:Least\:Common\:Multiplier\:of\:}x,\:x+19:\quad x\left(x+19\right)$

$\mathrm{Multiply\:by\:LCM=}x\left(x+19\right)$

$\frac{42}{x}x\left(x+19\right)-\frac{100}{x+19}x\left(x+19\right)=3x\left(x+19\right)$

$42\left(x+19\right)-100x=3x\left(x+19\right)$

$-58x+798=3x^2+57x$

$3x^2+57x=-58x+798$

$3x^2+57x-798=-58x+798-798$

$3x^2+57x-798=-58x$

$\mathrm{Add\:}58x\mathrm{\:to\:both\:sides}$

$3x^2+57x-798+58x=-58x+58x$

$3x^2+115x-798=0$

$\mathrm{Solve\:with\:the\:quadratic\:formula}$

$\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-115+\sqrt{115^2-4\cdot \:3\left(-798\right)}}{2\cdot \:3}:\quad 6$

$x=\frac{-115-\sqrt{115^2-4\cdot \:3\left(-798\right)}}{2\cdot \:3}:\quad -\frac{133}{3}$

$x=6,\:x=-\frac{133}{3}$

We can't take rates as negative numbers.

So, the rate of Doreen (x) = 6 miles per hour.

Rate of Sue = x+19 = 6+19 = 25 miles per hour.

Time taken by Doreen @ 6 miles per hour to cover 42 miles = 42/6 = 7 hours.

Time taken by Sue @ the rate 25 miles per hour to cover 100 miles = 100/25 = 4 hours.