# A boy was going home for 12km. We went 2km/h faster than last time and now spent 1 hour less on the road. How fast was the boy going?

Mar 19, 2016

$4$ km/h last time (taking $3$ hours) and $6$ km/h this time (taking $2$ hours).

#### Explanation:

If the boy travelled at $x$ km/h last time then:

$\frac{12}{x} - \frac{12}{x + 2} = 1$

Multiplying through by $x \left(x + 2\right)$ we find:

$12 \left(x + 2\right) - 12 x = x \left(x + 2\right)$

That is:

$24 = {x}^{2} + 2 x$

Adding $1$ to both sides we get:

$25 = {x}^{2} + 2 x + 1 = {\left(x + 1\right)}^{2}$

Hence $x + 1 = \pm \sqrt{25} = \pm 5$

So $x = - 1 \pm 5$

That is: $x = 4$ or $x = - 6$.

Since the time must be non-negative, the applicable solution is $x = 4$, meaning that the boy travelled at $4$ km/h last time and $6$ km/h this time.

Mar 19, 2016

The later traveling speed was 6 km/h

#### Explanation:

Let initial velocity be $\text{ "v_o" }$ in Km/h
Let new velocity be $\text{ "color(white)(.)v_n" }$ in Km/h

Let initial time be ${t}_{o}$ in hours
Let new time be ${t}_{n}$ in hours

Known that velocity x time = distances

$\implies {v}_{o} \times {t}_{o} = 12$

Given that ${v}_{n} = {v}_{o} + 2$
Given that ${t}_{n} = {t}_{o} - 1$

$\textcolor{b l u e}{\text{Putting it all together}}$

$\textcolor{b r o w n}{{v}_{o} \times {t}_{o} = 12}$................................................................(1)

$\textcolor{b r o w n}{{v}_{n} \times {t}_{n} = 12} \text{ } \textcolor{b l u e}{\to \left({v}_{o} + 2\right) \left({t}_{o} - 1\right) = 12}$.......(2)

Using equation(1) substitute for ${t}_{o}$ in equation (2)

Set ${t}_{o} = \frac{12}{v} _ o$

$\left({v}_{o} + 2\right) \left(\frac{12}{v} _ o - 1\right) = 12$

$12 - {v}_{o} + \frac{24}{v} _ o - 2 = 12$

$\implies \frac{24}{v} _ o - {v}_{o} - 2 = 0$

Multiply by ${v}_{o}$

$24 - {\left({v}_{o}\right)}^{2} - 2 {v}_{o} = 0$

$- {\left({v}_{o}\right)}^{2} - 2 {v}_{o} + 24 = 0$

Multiply by -1

${\left({v}_{o}\right)}^{2} + 2 {v}_{o} - 24 = 0$

$\left({v}_{o} - 4\right) \left({v}_{o} + 6\right) = 0$

${v}_{o} = - 6 \text{ }$is not logical

${v}_{o} = + 4$ is a more sensible walking speed

We need ${v}_{n} = {v}_{o} + 2 = 6 \text{ }$km/h