Question 1000284
.
Two ports A and B are 300 km apart. Two {{{highlight(ships)}}} leave A for B such that the 2nd leaves 8 hours after the first. The ships arrive at  B simultaneously. 

Find the time the {{{highlight(slower)}}} ship spent on the trip if the speed of one of them is 10 km/h higher than that of the other.
---------------------------------------------------------------------
<pre>
Let u = the speed of the slower ship.

Then the speed of the faster ship is u + 10.

The slower ship spent {{{300/x}}} hours for the trip.

The faster ship spent {{{300/(u+1)}}} hours.

According to the condition, the difference of these times is 8 hours. It gives an equation


(((300/u}}} - {{{300/(u+1)}}} = 8.

To solve it, multiply both sides by u*(u+1) and then simplify.

You will get

300*(u+10) - 300u = 8u*(u+10),

3000 = {{{8u^2}}} + {{{80u}}},

{{{8u^2}}} + {{{80u}}} - {{{3000}}} = {{{0}}}.

{{{u^2}}} + {{{10u}}} - {{{375}}} = {{{0}}}.

Factorize:

(u-15)*(u+25) = 0.

The roots are u = 15 and u = -25.

The solution is 15 {{{km/h}}} for the slower ship.

The time is {{{300/15}}} = 20 hours for the slower ship.
</pre>