Question

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?

Answer 1

`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:

`12/x - 12/(x+2) = 1`

Multiplying through by `x(x+2)` we find:

`12(x+2)-12x = x(x+2)`

That is:

`24 = x^2+2x`

Adding `1` to both sides we get:

`25 = x^2+2x+1 = (x+1)^2`

Hence `x+1 = +-sqrt(25) = +-5`

So `x = -1+-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.

Answer 2

The later traveling speed was 6 km/h

Explanation:

Let initial velocity be `" "v_o" "` in Km/h
Let new velocity be `" "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

`=>v_o xxt_o=12`

Given that `v_n=v_o +2`
Given that `t_n =t_o -1`

`color(blue)("Putting it all together")`

`color(brown)(v_o xxt_o=12)`................................................................(1)

`color(brown)(v_n xxt_n=12)" "color(blue)(->(v_o +2)(t_o -1)=12)`.......(2)

Using equation(1) substitute for `t_o` in equation (2)

Set `t_o=12/v_o`

`(v_o +2)(12/v_o-1)=12`

`12-v_o +24/v_o -2=12`

`=>24/v_o -v_o -2=0`

Multiply by `v_o`

`24-(v_o)^2 -2v_o=0`

`-(v_o)^2-2v_o + 24=0`

Multiply by -1

`(v_o)^2+ 2v_o-24=0`

`(v_o -4)(v_o +6)=0`

`v_o=-6" "`is not logical

`v_o = +4` is a more sensible walking speed

We need `v_n = v_o +2 = 6 " "`km/h