SSC GD 2024 PYQ — Question 52

The correct answer is option 2 i.e. 60km/hr.

Let ( x ) be the usual speed of the car (in km/h).

Time at usual speed = 210 / x hours

If the car reduces its speed by 10 km/h, the new speed is $x - 10$ km/h. 

Time at reduced speed = 210 / x - 10 hour

The time taken at the reduced speed is 0.7 hours less than the time at the usual speed. 

⇒ 210 / x - 10 = 210 / x - 0.7

⇒ $x(x - 10) \cdot \frac{210}{x - 10} = x(x - 10) \cdot \left( \frac{210}{x} - 0.7 \right)$

⇒ 210x = 210(x - 10) - 0.7x(x - 10)

⇒ 210x = 210x - 2100 - 0.7x(x - 10)

⇒ 0 = -2100 - 0.7x(x - 10)

⇒ x(x - 10) = x² - 10x

⇒ 0 = -2100 - 0.7(x² - 10x)

⇒ 0 = -2100 - 0.7x² + 7x

⇒ 0.7x^​​​2 - 7x - 2100 = 0

⇒ x² - 10x - 3000 = 0

⇒ x = $\frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(-3000)}}{2(1)}$

⇒ x = $\frac{10 \pm \sqrt{100 + 12000}}{2}$

⇒x = $\frac{10 \pm \sqrt{12100}}{2}$

⇒ x = $\frac{10 \pm 110}{2}$

⇒ x = $\frac{10 + 110}{2} = \frac{120}{2} = 60$

⇒ x = $\frac{10 - 110}{2} = \frac{-100}{2} = -50$

Since speed cannot be negative, the usual speed of the car is ( 60) km/h.