Aman goes to his office by car at the speed of 80 km/hr and reaches 10 minutes earlier. If he goes at the speed 30 km/hr, then he reaches 10 minutes late. What will be the speed of the car to reach on time?

a) 480/11 km/hr
b) 240/6 km/hr
c) 450/7 km/hr
d) 480/13 km/hr

Ans : a) 480/11 km/hr

SSC GD 24 Feb 2024 Paper – Mathematics

Solution :

Let Distance = D, and Time taken = T

Distance = Speed x Time,
10 Minutes = 10/60 = 1/6 hrs

Case -1 : Aman goes to his office by car at the speed of 80 km/hr and reaches 10 minutes earlier.
D = $ 80 \times (T- \frac 16) $

Case 2 : If he goes at the speed 30 km/hr, then he reaches 10 minutes late.
D = $ 30 \times (T+ \frac 16) $

From both cases D is same, therefore,
$ 80 \times (T- \frac 16) $ = $ 30 \times (T+ \frac 16) $
$ 80 \times ( \frac {6T – 1}{6}) $ = $ 30 \times ( \frac {6T + 1}{6}) $
48T – 8 = 18T +3
->30T = 11
T = 11/30

Now find the distance from Case 1
D = $ 80 \times (\frac {11}{30}- \frac 16) $
D = $ 80 \times (\frac {11-5}{30}) $
D = $ 80 \times (\frac {6}{30}) $ = 16 Km

Find the required Speed
Speed = D/T = $ \frac{ 16 \times 30} {11} $
Speed = 480/11 km/hr

Short Trick –

When both time are equal

Use the formula for the required speed to reach on time:
Required Speed= $ \frac{2 \times \text{Speed}_1 \times \text{Speed}_2}{\text{Speed}_1 + \text{Speed}_2}​​ $

Substitute the values:Required Speed= $ \frac{2 \times 80 \times 30}{80 + 30} = \frac{2 \times 80 \times 30}{110} $

Simplify: Required Speed = $ \frac{480}{11} $