Question

Jane walked for 3 km at a constant speed of x km/hr. She then jogged for a further x km at a constant speed of 8 km/hr. Her total time was 1/4 hr. Show that x2 - 10x + 24 = 0 Hence find the speed at which Jane walked, if her total distance was less than 8 km.

Answer 1

Given

Part 1

Jane walked for 3 km at a constant speed of x km/hr

speed x km/h and

distance is 3km

$\begin{gathered} \text{Formula} \\ \\ \text{SPEED =}\frac{dis\tan ce}{\text{time}} \\ \\ \end{gathered}$$\begin{gathered} \text{make Time (T) the subject of the formula} \\ \\ \text{Time}=\frac{dis\tan ce}{S\text{ pe}ed} \end{gathered}$$T=\frac{3}{x}$

She then jogged for a further x km at a constant speed of 8 km/hr

Therefore, Distance is X km and

Speed is 8km/hr

$\text{Time}=\frac{X}{8}$

The total time is;

$\frac{3}{x}+\frac{x}{8}=1\frac{1}{4}$$\begin{gathered} \frac{3}{x}+\frac{x}{8}=\frac{5}{4} \\ \\ \text{LCM is }8x \\ \end{gathered}$

Multiply all through by the LCM

$\begin{gathered} 8x\times\frac{3}{x}+8x\times\frac{x}{8}=8x\times\frac{5}{4} \\ simplify \\ 24+x^2=10x \\ \text{rearrange } \\ x^2-10x+24=0 \end{gathered}$

Part 2

We have to solve the quadratic equation

$\begin{gathered} x^2-10x+24=0 \\ By\text{ factorisation} \\ x^2-6x-4x+24=0 \\ x(x-6)-4(x-6)=0 \\ x-6=0 \\ x=6 \\ \text{and} \\ x-4=0 \\ x=4 \end{gathered}$

Recall From the question

if her total distance was less than 8 km.​

$\begin{gathered} 3\operatorname{km}\text{ +6km = 9km} \\ It\text{ is greater than 8km, so it not 6} \\ \\ \text{Then} \\ 4\operatorname{km}\text{ +3km =7km} \\ \text{Then, it is 4} \end{gathered}$

The speed X is 4km/h