Question

A takes 3 hours more than B to walk 30 km. But if A doubles his pace, he is ahead of B by 3/2 hours . Find their speed of walking.​

Answer 1

Answer:

Initial speed:

• A: $\displaystyle \rm \frac{10}{3}\; km/ h$.
• B: $\rm 5\; km/ h$.

Step-by-step explanation:

Both equations are concerned about the time differences between A and B. To avoid unknowns in the denominators,

• let $x$ be the initial time in hours for A to walk 30 km, and
• let $y$ be the time in hours for B to walk 30 km.

First equation:

"A takes 3 hours more than B to walk 30 km."

$x = y + 3$.

$x - y = 3$.

When A doubles his pace, he takes only 1/2 the initial time to cover the same distance. In other words, now it takes only $x/2$ hours for A to walk 30 km.

Second equation:

"[A] is ahead of B by 3/2 hours [on their 30-km walk.]"

$\displaystyle \frac{x}{2} + \frac{3}{2} = y$.

$\displaystyle \frac{1}{2}x - y = -\frac{3}{2}$.

Hence the two-by-two linear system:

$\left\{\begin{aligned}&x - y = 3\\&\frac{1}{2}x - y = -\frac{3}{2}\end{aligned}\right.$.

Solve this system for $x$ and $y$:

Subtract the second equation from the first:

$\displaystyle \frac{1}{2}x = \frac{9}{2}$.

$x = 9$.

$y = 6$.

It initially takes 9 hours for A to walk 30 kilometers. The initial speed of A will thus be:

$\displaystyle v = \frac{s}{t} = \rm \frac{30\; km}{9\; h} = \frac{10}{3}\; km/h$.

It takes 6 hours for B to walk 30 kilometers. The speed of B will thus be:

$\displaystyle v = \frac{s}{t} = \rm \frac{30\; km}{6\; h} = 5\; km/h$.