Question

2. Mr.Wilson drive 64.8 km from work at a speed of 48 km/hr. Mr.Wilsom drives 81.2km from work ata speed of 58km/hr. They both leave work at the same time.
a. Who arrive home first?
b. How many minutes later is it before the second person gets home?​

Answer 1

Explanation:

time taken by Mr. Wilson to reach home =

64.8 / 48 = 1.35 hour

1 hour + 0.35 hour

1 hour = 60 minutes

0.35 hour = 60 * 0.35 = 21 minutes

Total time taken by mr. wilson = 60 + 21 =

81 minutes.

time taken by Mr Wilsom = 81.2 / 58

time taken by Mr Wilsom= 1.4 hours.

1 hour + 0.4 hour

1 hour = 60 minutes

0.4 hour = 0.4*60 = 24 minutes.

total time taken by Mr. Wilsom = 60+24 = 84 minutes.

Mr. WILSON will reach home earlier by 3 minutes

Answer 2

Explanation:

Distance travel by Mr. Wilson = 64.8 km

Speed = 48 km/h

So, Time = d/s

= 64.8 km/48 km/hr

= 1.35 hr = 81 min.

Now, distance travel by Mr. Wilson = 81.2 km

speed = 58 km/hr

So, Time = d/s

= 81.2 km/58 km/hr

= 1.4 hr = 84 min

Hence, Mr. Wilsom will arrive first by 3 minutes than Mr. Wilson.

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