Question 11:
George departed for Kuantan from Kuala Lumpur at 9.30 a.m. He stopped at a rest station for 45 minutes. Then, he continued his journey and arrived at his destination at 3.45 p.m. How long did he travel on the road.
Solution:
Time taken on the road
= (1545 – 0930) – 0045 ← [3.45 p.m. change to 24-hour system = 1545 (1200 + 0345)]
= 0615 – 0045
= 0530
= 5 hours 30 minutes
George departed for Kuantan from Kuala Lumpur at 9.30 a.m. He stopped at a rest station for 45 minutes. Then, he continued his journey and arrived at his destination at 3.45 p.m. How long did he travel on the road.
Solution:
Time taken on the road
= (1545 – 0930) – 0045 ← [3.45 p.m. change to 24-hour system = 1545 (1200 + 0345)]
= 0615 – 0045
= 0530
= 5 hours 30 minutes
Question 12:
Rita began doing her homework at 8.25 in the morning. She took a break for 45 minutes to have some exercises and ½ hour to have lunch. If she finished her homework at 4.45 in the evening on the same day, how long did she take to do her homework?
Solution:
Time taken for Rita to finish her homework
= (1645 – 0825) – (0030 + 0045) ← [4.45 p.m. = 1645 (1200 + 0445)]
= 0820 – 0115
= 0705
= 7 hours 5 minutes
Rita began doing her homework at 8.25 in the morning. She took a break for 45 minutes to have some exercises and ½ hour to have lunch. If she finished her homework at 4.45 in the evening on the same day, how long did she take to do her homework?
Solution:
Time taken for Rita to finish her homework
= (1645 – 0825) – (0030 + 0045) ← [4.45 p.m. = 1645 (1200 + 0445)]
= 0820 – 0115
= 0705
= 7 hours 5 minutes
Question 13:
Roslan attended a tuition class for 2½ hours. The class started at 7.15 p.m. The class finished 35 minutes earlier due to power failure.
At what time did the class finish?
Solution:
Class finish at
= (1915 + 0230) – 0035 ← [7.15 p.m. = 1915 (1200 + 0715)]
= 2145 – 0035
= 2110 → (9.10 p.m.)
Roslan attended a tuition class for 2½ hours. The class started at 7.15 p.m. The class finished 35 minutes earlier due to power failure.
At what time did the class finish?
Solution:
Class finish at
= (1915 + 0230) – 0035 ← [7.15 p.m. = 1915 (1200 + 0715)]
= 2145 – 0035
= 2110 → (9.10 p.m.)
Question 14:
An aeroplane took off from Kuala Lumpur to Singapore at 2.45 p.m. The flight normally takes 55 minutes. The aeroplane was delayed and landed at Singapore at 4.50 p.m.
How long, in minutes, was the aeroplane delayed?
Solution:
Expected arrival time
= 2.45 p.m. + 55 minutes
= 3.40 p.m.
Time delayed
= 4.50 p.m. – 3.40 p.m.
= 1 hour 10 minutes
= 70 minutes
An aeroplane took off from Kuala Lumpur to Singapore at 2.45 p.m. The flight normally takes 55 minutes. The aeroplane was delayed and landed at Singapore at 4.50 p.m.
How long, in minutes, was the aeroplane delayed?
Solution:
Expected arrival time
= 2.45 p.m. + 55 minutes
= 3.40 p.m.
Time delayed
= 4.50 p.m. – 3.40 p.m.
= 1 hour 10 minutes
= 70 minutes
Question 15:
The flight from Kuala Lumpur to Kota Kinabalu takes 1 hour 45 minutes.
Jacky’s flight should leave at 2.15 p.m. but is delayed by 50 minutes.
At what time, in the 24-hour system, he will reach Kota Kinabalu?
Solution:
2.15 p.m. = 1415 hours
Actual departure time
= 1415 hours + 50 minutes
= 1505 hours
Arrival time at Kota Kinabalu
= 1505 + 0145
= 1650
The flight from Kuala Lumpur to Kota Kinabalu takes 1 hour 45 minutes.
Jacky’s flight should leave at 2.15 p.m. but is delayed by 50 minutes.
At what time, in the 24-hour system, he will reach Kota Kinabalu?
Solution:
2.15 p.m. = 1415 hours
Actual departure time
= 1415 hours + 50 minutes
= 1505 hours
Arrival time at Kota Kinabalu
= 1505 + 0145
= 1650