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Question

 A car and a bike left the village for a city simultaneously. The distance between the city and village is 90 km. Find the speeds of both car and bike if the car reached the city 1 hour earlier than the bike, and it's speed was 15 km/hour greater.

Answer 1

The solution to this problem is based on 2 math concepts:

Distance traveled equals rate * time

And solving simultaneous equations by substitution=

Let R = Rate of the car

R-15 = Rate of the bike

Let T= Time of the car

T+1 = Time of the bike

RT = 90 Km (Distance of the car)

90 = (R-15)(T+1) = RT + R-15T-15

RT =RT +R - 15T-15 =>0 = R = 15 + 15T

(15+15T)T =90 (substituting the value of R in RT = 90

Solving the quadratic equation 15T^2 +15T+90 gives

T = 2 hours and R = 45km/hour Rate of the car

T+1 = 3 hours and R-15 = 30 Km/hr Rate of the bike

Answer 2

There may be an easier way to solve this, but this method works.

T_b=time for bike

T_c=time for car=T_b-1

V_b=velocity bike

V_c=velocity of car=V_b+15

T_c*V_c=90=(T_b-1)(V_b+15)

T_bV_b=90

V_bT_b+15T_b-V_b-15=90 => 15T_b-V_b-15 = 0

T_b=90/V_b

(90/V_b)-V_b-15=0

Solve this quadratic for V_b and ignore the extraneous root. Then you will be able to find the other 3 parameters.