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=(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}=(X)/(8)`

The total time is;

`(3)/(x)+(x)/(8)=1(1)/(4)`
`\begin{gathered} (3)/(x)+(x)/(8)=(5)/(4) \\ \\ \text{LCM is }8x \\ \end{gathered}`

Multiply all through by the LCM

`\begin{gathered} 8x*(3)/(x)+8x*(x)/(8)=8x*(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.​

The speed X is 4km/h