An algebra problem by A Former Brilliant Member

speed = s, distance = d, time = t

Considering the first part of the question,

$s_A$ = $\frac{d_A}{t_A}$ and $s_B$ = $\frac{d_B}{t_B}$

Here, both $t_A$ = $t_B$ = 5 hours

Thus, $s_A$ = $\frac{d_A}{5}$ and $s_B$ = $\frac{d_B}{5}$

Or, $5s_A$ = $d_A$ and $5s_B$ = $d_B$

But, $d_A$ = 100 km + $d_B$

Thus, $5s_A$ = 100 km + $d_B$

Hence, $5s_A$ = 100 km + $5s_B$

Or, $s_A$ = 20 km + $s_B$ -- Equation 1

Considering the second part of the question,

$s_A$ = $\frac{d_A}{t_A}$ and $s_B$ = $\frac{d_B}{t_B}$

Here, both $t_A$ = $t_B$ = 1 hour

$s_A$ = $d_A$ and $s_B$ = $d_B$

But, $d_A$ = 100 km - $d_B$

Thus, $s_A$ = 100 km - $d_B$

Hence, $s_A$ = 100 km - $s_B$ -- Equation 2

Equating both equations, $s_A$ or u = 60 and $s_B$ or v = 40

Hence, $\frac{u-v}{2}$ = $\frac{60-40}{2}$ = $\frac{20}{2}$ = $\boxed{10}$

From the first part of the problem, the cars meet in 5 hours when they move in the same direction. This means that car with speed u km/hr will cover 5u km distance and the car with speed v will cover 5v km distance.

From the problem description, originally they were 100 km apart. Since they met in 5 hours, the initial difference in their distance should be 100 km, therefore:

5u - 5v = 100 since it is given that u>v,

which means 5(u - v) = 100

or u - v = 20

and hence (u - v)/2 = 10

We don't even need the second piece of information in the problem, that is, it took them an hour to meet when they moved towards each other, to solve it!

100 km difference decreases in 5 hours. So, 100/5 =20=u-v=Difference of speed.

So, (u-v)/2 =20/2 =10

in same directions of two cars, speed = u - v = d/t speed =u - v = 100/5 = 20.........(1) but in opposite directions, speed = u + v = d/t speed = u + v =100/1 =100........(2) by solving eq. (1) and eq. (2) we get, u = 60 ,v = 40 and the term , u - v /2 = 10........ans.

5u = 5v + 100 (eq. 1). u + v = 100 (eq. 2). solve for u and v. u = 60, v= 40. (60 - 40)/2 = 10