Chris cycled from A to C in a certain part of a country. He travelled for 50 km from A, at a constant speed of $x$ km/h until he reached the point B, where his bicycle chain broke.

1. Find the time taken from A to B using distance divided by speed.

Distance from A to B $=50$ km and speed $=x$ km/h.

$\text{Time}=\frac{\text{distance}}{\text{speed}}=\frac{50}{x}$

1. Find the walking time from B to C using distance divided by speed.

Distance from B to C $=6$ km, speed $=(x-16)$ km/h.

$\text{Time}=\frac{\text{distance}}{\text{speed}}=\frac{6}{x-16}$

1. Form the equation for the total journey time.

From parts (a) and (b),
$\text{total time}=\frac{50}{x}+\frac{6}{x-16}.$
Given that the total time is 4 hours,
$\frac{50}{x}+\frac{6}{x-16}=4.$

2. Clear the denominators by multiplying throughout by $x(x-16)$.

$\frac{50}{x}+\frac{6}{x-16}=4$
Multiply throughout by $x(x-16)$:
$50(x-16)+6x=4x(x-16).$

3. Expand and rearrange to show the required quadratic equation.

$50x-800+6x=4x^2-64x$
$56x-800=4x^2-64x$
Bring all terms to one side:
$4x^2-120x+800=0$
Divide throughout by 4:
$x^2-30x+200=0.$

1. Factorise the quadratic expression.

$x^2-30x+200=0$
$(x-10)(x-20)=0$

2. Solve each linear factor equal to zero.

$x-10=0 \quad \text{or} \quad x-20=0$
$x=10 \quad \text{or} \quad x=20$

1. Test the two possible values of $x$ against the condition that the walking speed must be positive.

Since Chris walked at $(x-16)$ km/h, his walking speed must be positive.

If $x=10$, then
$(x-16)=10-16=-6,$
which is not a possible speed.

So reject $x=10$.

1. Use the valid original cycling speed and find the total distance from A to C.

From part (e), the valid value is $x=20$.

Total distance from A to C:
$50+6=56\text{ km}$

2. Calculate the time for the whole journey by bicycle and convert to hours and minutes.

If Chris cycled the whole journey at $20$ km/h,
$\text{time}=\frac{56}{20}=2.8\text{ h}$

$0.8\text{ h}=0.8\times 60=48\text{ min}$

So the time is $2$ hours $48$ minutes.