Question

A man made a trip of 480 km in 9 hrs.Some part of the trip was covered at 45 km/h and the remaining at 60km/h .Find the part of the trip covered by him at 60km/h

Answer 1

Answer :-

The part of the distance covered by man when speed is 60 Km/h is 300 Km.

Solution :-

Total distance covered by man = 480 km

Total tme taken = 9 hours

When speed is 60 Km/h

Speed of the man in remaining part of the trip = 60 Km/h

Let the distance covered when speed is 60 Km/h be 'x' Km/h

Time taken = Distance/Speed

= x/60

When speed is 45 Km/h

Speed of the in some part of the trip = 45 Km/h

Distance covered by the when speed is 45 Km/h = (Total distance - x) = (480 - x) Km

Time taken = Distance/Speed

= (480 - x)/45

We know that

Time taken when speed is 60 Km/h + Time taken when speed is 45 Km/h = Tatal time taken

$\sf \implies \dfrac{x}{60} + \dfrac{480 - x}{45} = 9 \\ \\ \\ \sf \implies \dfrac{3x+ 4(480 - x)}{180} = 9 \\ \\ \\ \sf \implies \dfrac{3x + 1920 - 4x}{180} = 9 \\ \\ \\ \sf \implies - x + 1920 = 9(180) \\ \\ \\ \sf \implies - x + 1920 = 1620 \\ \\ \\ \sf \implies 1920 - 1620 = x \\ \\ \\ \sf \implies x = 300$

Therefore the part of the distance covered by man when speed is 60 Km/h is 300 Km.

Answer 2

SOLUTION:-

Given:

Let the part of the trip covered at 45km/hr be x km.

Then the part covered at 60 km/hr will be:

=) (480 - x)km

Now, we know that time,

$= > time = \frac{Distance}{Speed}$

$= > \frac{x}{45} + \frac{480 - x}{60} = 9 \: hrs \\ \\ = > \frac{4x + 1440 - 3x}{180} = 9 \\ \\ = > \frac{x + 1440}{180} = 9 \\ \\ = > x + 1440 = 1620 \\ \\ = > x = 1620 - 1440 \\ \\ = > x = 180 \: km/hrs$

Hence,

The part covered at 60km/hrs;

=) (480- 180)km/hrs

=) 300km

Hope it helps ☺️