# A car travelled 75km at a uniform speed.If its speed is 3km/hr slower,it would have taken 75 min. more.Find the speed of car taking usual speed as x

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The total distance = 75 km

The car's speed = X

The car's speed if it is 3 km/h slower = x-3

The time needed = t

The time needed if (x-3)= t + 75min = t + 1.25 hours

We need to find the speed x.

We know that :

Speed = distance / time

==> x= d/t= 75/ t

==> x= 75/t

==> t= 75/x ......(1)

If the car was 3 km/h slower:

==> speed = distance/time

==> x-3 = 75/ t+1.25

==> t = 75/(x-3) -1.25

From (1) and (2) we find that:

75/x = 75/(x-3) - 1.25

75/x = 75 -1.25x + 3.75 / x-3

Cross multiply:

==> 75(x-3) = x(75 -1.25x +3.75)

==> 75x -225 = 75x -1.25x^2 +3.75x

Reduce similar:

==> 1.25x^2 -3.75x -225=0

x1= 3.75 + sqrt(14.0625- 4(1.25)(-225) / 2.5

= 3.75 + 33.75 / 2.5

= 37.5/2.5 = 15

**The the car's speed = 15 km/h**

We assume the speed of the car at x kmph.

The distance travelled by the car =75kms.

Therefore the time taken by the car to travel 75kms with x kmh speed= 75/x hr .

If the speed is 3 kms slower the time taken would be = 75/(x-3) which is 75 min (= 1.25hr) more than 75/x. So the required equation is:

75/(x-3) - 75/x = 1.25=5/4 . Multiply by x(x-3)*4 both sides.

4*75x-4*75(x-3) = 5x(x-3)

300x - 300x+ 900 =5x^2-15x

900 = 5x^2 -15

5x^2-15x-900 = 0. Divide by 5.

x^2-3x-180 = 0

(x-15)(x+12) = 0.

x=15 is the speed of the car.

Check: At 15 kmph the time taken by the car to travel 75 kms = 75/15 = 5 hr.

At (15-3)kms =12kms the car takes to travel the distance of 75 kms = 75/12 =6.25hr.

So (6.25-5)hr =1.25hr or 72 min