Collins Maths Solutions Class 6 Chapter 9 Speed Distance and Time
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Collins Maths Class 6 Mathematics, Chapter 9, Speed Distance and Time. Here students can easily find Exercise wise solution for chapter 9, Speed Distance and Time. Students will find proper solutions for Exercise 9.1 and 9.2 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.
Speed Distance and Time Exercise 9.1 Solution :
Question no – (1)
Solution :
The speed 18 km/h is the same as 5 m/s. Using this result, answer questions
= 18 km/h = 5 m/s
= 1k m/h = 5/18 m/s
= 1 m/s 18/5 km/h
(a) Express the speeds 16.2 km/h, 45 km/h and 315 km/h in metres per second.
= 16.2 km/h
= 16.2 × 1 km/h
= 45/10 m/s
= 4.5 m/s
= 45 km/h
= 45 × 1 km/h
= 45 × 5/10 m/s
= 25/2 m/s
= 125 m/s
= 315 km/h
= 315 × 1 km/h
= 315 × 5/185 m/s
= 175/2 m/s
= 87.5 m/s
(b) Express the speeds 1.2 m/s. 35 m/s and 625 m/s in kilometres per hour.
= 1.2 m/s
= 1.2 × 1 m/s
= 1.2 × 18/5 km/h
= 1.2/10 × 18/5
= 108/25
= 4.32
= 625 m/s
= 625 × 1 km/
= 625 × 18/5 km/h
= 2250 km/h
Question no – (2)
Solution :
As per the question,
Distance = 10 km
Taken time = 12 min
= 12/60 h
Speed of the train = Distance covered/Time taken
∴ 10 km/12/60h
= 10 × 60/12 km/h
= 50 km/h
Therefore, the speed of the train will be 50 km/h.
Question no – (3)
Solution :
Speed of karan’s bike = 40 km/h
= 40 × 1 km/h
= 40 × 5/18 m/s
= 100/9 m/s
= 100/9 m travelled in 1 s
= 1 m travelled in = 1/100
= 9/100 s
900 m travelled in,
= 900 × 9/100 s
= 81 s
Therefore, Karan takes 81 s to reach his college from home.
Question no – (4)
Solution :
As we know that,
1 min = 60 sec
∴ 1/2 min = 30 s
∴ Distance covered,
= 5 × 30
= 150 m
Therefore, the distance covered by the man will be 150 m.
Question no – (5)
Solution :
According to the question,
Speed of the racer = 90 km/h
= 90 × 1 km/h
= 90 × 5/18 m/s
= 25 m/s
∴ Time 5 min
= 5 × 60 s
= 300 s
∴ Distance covered by the bike racer,
= Speed × Time taken
= 25 × 300 m
= 7500 m
Therefore, the distance cover by the racer will be 7500 m.
Question no – (6)
Solution :
The speed of train was = 14 km/(10/60h)
= 14km/10/60h
= 14 × 69/10 km/h
= 84 km/h
Speed reduced by 4 km/h
∴ New speed will be,
= (84 – 4) km/h
= 80 km/h
= 80 km travelled in 1 h
= 1 km travelled in 1/80 h
= 14 km travelled 14 × 1/80
= 7/40 h
= 7/40 × 3600 s
= 630 s
= 630/60
= 10.5 min
Therefore, the time taken to cover the same distance 10.5 min.
Question no – (7)
Solution :
According to the question,
Dilshad travelled 3.6/2 = 1.8 km
She had to travelled 1.8 km in 5 min
Her speed will be,
= 1.8km/5/60h
= 18/10 60/5
= 216/10
= 21.6 km/h
Therefore, Dilshad must be cycle at a speed of 21.6 km/h to reach the school.
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