Question

A car travels a distance of 840 km at a uniform speed. If the speed of the car is 10 km/hr more then it takes 2 hrs less to cover the same distance. The original speed of car was?

Answer 1

Let the original speed of the car be "S" km/hr.

The car travels 840 km at this speed, so the time it takes at the original speed is 840/S hours.

Now, if the speed of the car is increased by 10 km/hr, its new speed is (S + 10) km/hr.

With this new speed, it takes 2 hours less to cover the same distance of 840 km. So, the time taken at the increased speed is (840/(S + 10)) hours.

We are given that the time taken at the increased speed is 2 hours less than the time taken at the original speed:

840/S - 840/(S + 10) = 2

To solve for S, we can simplify the equation:

840(S + 10) - 840S = 2S(S + 10)

Now, let's solve for S:

840S + 8400 - 840S = 2S^2 + 20S

Combining like terms:

8400 = 2S^2 + 20S

Divide both sides by 2:

4200 = S^2 + 10S

Now, let's rearrange the equation and set it equal to zero:

S^2 + 10S - 4200 = 0

We can solve this quadratic equation using the quadratic formula:

S = (-B ± √(B² - 4AC)) / (2A)

In this case, A = 1, B = 10, and C = -4200. Plug these values into the formula:

S = (-10 ± √(10² - 4(1)(-4200))) / (2(1))

Simplify:

S = (-10 ± √(100 + 16800)) / 2

S = (-10 ± √16900) / 2

S = (-10 ± 130) / 2

Now, calculate both possible values for S:

1. S1 = (-10 + 130) / 2 = 120 / 2 = 60 km/hr
2. S2 = (-10 - 130) / 2 = -140 / 2 = -70 km/hr

Since speed cannot be negative in this context, the original speed of the car was 60 km/hr.

Answer 2

Explanation:

Simplifying this equation, we get: 8400 = 2x (x+10) 8400 = 2x^2 + 20x 2x^2 + 20x - 8400 = 0 x^2 + 10x - 4200 = 0 Using the quadratic formula, we get: x = (-10 ± sqrt (10^2 - 4*1* (-4200)))/ (2*1) x = (-10 ± 110)/2 x = 50 or x = -60 Since speed cannot be negative, we reject the negative solution and conclude that the original speed of the car was 50 km/hr.