A car can cover a particular distance in 45 minutes. Now when it reduc

1.	A car can cover a particular distance in 45 minutes. Now when it reduces its speed by 20% it covers 80% of the distance in 45 minutes. What is the original speed and distance, if the reduced speed is 54 kmph?1. 15 m/s, 90 km2. 18.75 m/s, 50.625 km3. 15 m/s, 56.625 km4. 18.75 m/s, 56.625 km
Answer» Correct Answer - Option 2 : 18.75 m/s, 50.625 km Given: T = 45 minutes Reduced speed = 54 km/hr Percentage reduction in speed = 20% 80% Distance is covered in 45 minutes with reduced speed Formula used: Distance = Speed × Time Calculations: Let the original speed be x km/hr Reduced speed = 54 km/hr x × (100 - 20)/100 = 54 ⇒ x × (4/5) = 54 ⇒ x = 54 × (5/4) = 135/2 km/hr As time is 45 minutes = 45/60 = 3/4 hour Distance = Speed × Time = (135/2) × (3/4) = 405/8 ⇒ Distance = 50.625 km ∴ Required speed and distance is 18.75 m/s and 50.625 km

A car can cover a particular distance in 45 minutes. Now when it reduces its speed by 20% it covers 80% of the distance in 45 minutes. What is the original speed and distance, if the reduced speed is 54 kmph?1. 15 m/s, 90 km2. 18.75 m/s, 50.625 km3. 15 m/s, 56.625 km4. 18.75 m/s, 56.625 km

Answer» Correct Answer - Option 2 : 18.75 m/s, 50.625 km

Given:

T = 45 minutes

Reduced speed = 54 km/hr

Percentage reduction in speed = 20%

80% Distance is covered in 45 minutes with reduced speed

Formula used:

Distance = Speed × Time

Calculations:

Let the original speed be x km/hr

Reduced speed = 54 km/hr

x × (100 - 20)/100 = 54

⇒ x × (4/5) = 54

⇒ x = 54 × (5/4) = 135/2 km/hr

As time is 45 minutes = 45/60 = 3/4 hour

Distance = Speed × Time = (135/2) × (3/4) = 405/8

⇒ Distance = 50.625 km

∴ Required speed and distance is 18.75 m/s and 50.625 km

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