1. ## Two quadratic problems that need explaining

Hi everyone, I'm just stumped, I've tried but now I'm giving up.

An explanation of how to solve these would be much appreciated.
A tank can be filled in 6 hours using two pipes. The larger pipe alone would fill it 5 hours sooner than the smaller pipe alone. How long would each pipe alone take?

A cyclist travelled at a certain speed to a town 30km away, then reduced speed by 4km/h and came back. If the cyclist had travelled at a steady 8km/h for the whole trip, the total time would have been half an hour less. What was the initial speed of the cyclist?

2. Originally Posted by lucius
Hi everyone, I'm just stumped, I've tried but now I'm giving up.

An explanation of how to solve these would be much appreciated.
A tank can be filled in 6 hours using two pipes. The larger pipe alone would fill it 5 hours sooner than the smaller pipe alone. How long would each pipe alone take?
...
Hello,

let the volume of the tanke be V
let the time which the smaller pipe needs to fill the tank be t

then the time which the larger pipe needs is t-5

The "speed of filling"(?) of the smaller pipe is: $\frac{V}{t}$
and for the larger pipe $\frac{V}{t-5}
$

Both pipes are busy for 6 hours and then the tank is filled:

$6 \cdot \frac{V}{t} + 6 \cdot \frac{V}{t-5} = V$

Divide both sides of the equation by V, multiply both sides by the product of the denominators and you'll get:

$6(t-5) + 6t = t(t-5)$ Rearrange the equation :

$t^2-17t+30=0$ and solve for t.

You'll get 2 solutions: t = 15 or t = 2. Obviously the second solution doesn't satisfy the conditions of your problem.

Therefore: The smaller pipe needs 15 hours and the larger one 10 hours to fill the tank.

3. Originally Posted by lucius
Hi everyone, I'm just stumped, I've tried but now I'm giving up.

...

A cyclist travelled at a certain speed to a town 30km away, then reduced speed by 4km/h and came back. If the cyclist had travelled at a steady 8km/h for the whole trip, the total time would have been half an hour less. What was the initial speed of the cyclist?
Hello,

as you may know the speed is calculated by $speed = \frac{distance}{time}$. Therefore the traveled time is: $t = \frac{d}{s}$

The cyclist traveled 60 km at an average speed of 8 km/h. The total time would be: 60 km / 8 km/h = 7.5 h. According to the problem the cyclist neede half an hour more than the 7.5 h. So the total time was 8 h:

$\frac{30}{s} + \frac{30}{s-4}=8$ Rearrange this equation:

$30(s-4) + 30s = 8s(s-4)$

$8s^2 - 92s +120=0$

Solve for v and you'll get: s = 1.5 km/h or s = 10 km/h. The first solution doesn't satisfy the conditions of your problem.

4. Hello, lucius!

I have a different approach to the first one . . .

A tank can be filled in 6 hours using two pipes.
The larger pipe alone would fill it 5 hours sooner than the smaller pipe alone.
How long would each pipe alone take?

Let $x$ = number of hours for the large pipe to fill the tank (alone).
Then $x+5$ = number of hours for the small pipe to fill the tank.

In one hour, the large pipe can fill: . $\frac{1}{x}$ .of the tank.
In one hour, the small pipe can fill: . $\frac{1}{x+5}$ .of the tank.

Together, in one hour, they can fill: . $\frac{1}{x} + \frac{1}{x+5} \:=\:\frac{2x+5}{x(x+5)}$ .of the tank.

In 6 hours, they fill the entire tank (1 tank).

. . There is our equation! . $6\left(\frac{1}{x} + \frac{1}{x+5}\right) \:=\:1$

Multiply by $x(x+5)\!:\;\;6(x+5) + 6x \:=\:x(x+5)$

We have a quadratic: . $x^2 - 7x - 30 \:=\:0$

. . which factors: . $(x - 10)(x + 3) \:=\:0$

. . and has roots: . $x \:=\:10,\:(-3)$

$\text{Therefore:} \;\begin{array}{c}\text{The large pipe takes 10 hours.} \\ \text{The small pipe takes 15 hours.}\end{array}$

5. Oooh I like Soroban's explanation. Very easy to follow.

6. Great! Thanks a lot to Soroban and earboth, both answers were very helpful.