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Time and Distance Practice Test 4 with solutions
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© examsnet.com
Question : 30
Total: 38
Two points A and B are 150 km apart. A man completes his onward journey from A to B in 3 hours 20 minutes and return journey from B to A in 4 hours 10 minutes. His average speed during the entire journey will be less than his average speed during the journey from A to B by :
5 km/h
7.5 km/h
9 km/h
3 km/h
Validate
Solution:
During onward journey from A to B:
Average speed =
150
10
3
=
150
×
3
10
= 45 km/hr.
During entire jouney :
Average speed =
300
10
3
+
25
6
=
300
×
6
45
= 40 km/hr.
Hence, difference of average speed = 45 – 40 = 5 km/hr
© examsnet.com
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