Question 1185921
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The distance between two cities A and B is 140 km. A car driving from A to B left at the
same time as a car driving from B to A. The cars met after one hour, then the first car reached
city B 35 minutes later than the second car reached city A. Find the speed of the car from city A to B.
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<pre>
Let  "a"  be the speed of the car A, in km/h;

let  "b"  be the speed of the car B, in km/h.


Then you have these two equation


    a + b = 140  kilometers           (1)     (The cars met after one hour)

    {{{140/b}}} - {{{140/a}}} = {{{35/60}}}  of an hour    (2)    (first car reached city B 35 minutes later than the second car reached city A)


From the first equation, express b = 140-a  and substitute it into the second equation.  You will get then


    {{{140/(140-a)}}} - {{{140/a}}} = {{{7/12}}}

    {{{20/(140-a)}}}  - {{{20/a}}} = {{{1/12}}}


At this point, I just can to guess the solution  MENTALLY :  a = 60 km/h.


An alternative way is to reduce the last equation to the standard quadratic equation form and solve it formally.


<U>ANSWER</U>.  The speed of the car A is 60 km/h.


<U>CHECK</U>.   a) the speed of the car B is then 140-60 = 80 km/h.

         b) the time equation (2) is then  {{{140/80}}} - {{{140/60}}} = {{{7/4}}} - {{{7/3}}} = {{{(7*4)/12 - (7*3)/12}}} = {{{7/12}}} = {{{35/60}}} = 35 minutes.

            ! Correct !
</pre>

Solved.