**Collins Maths Solutions Class 7 Chapter 10 Speed Distance and Time**

Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Collins Maths Class 7 Mathematics, Chapter 10, Speed, Distance and Time. Here students can easily find chapter 7 solutions with exercise wise explanation. Students will find proper solutions for Exercise 10.1 and 10.2 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 10.1 Solution :**

**Question no – (1) **

**Solution : **

Distance = 4.32 km

= 4.32 × 1000

= 4320 m

Time 4 minute 48 second.

= (4 × 60) 48

= (240 + 48)

= 280 Second

**∴** Speed = 4320/288 m/s

= 15 m/s

Therefore, The speed of the boy will be 15 m/s.

**Question no – (2) **

**Solution : **

Time = 25 minute

= 25/60

= 5/12 h

Speed = 48 km/h

**∴ **Distance,

= 48 × 5/12

= 20 km

Hence, Jaya will cover 20 km in 25 minute.

**Question no – (3) **

**Solution : **

Speed = 41 km/h

Time = 2 hours 30 minute.

= 2 1/2

= 5/2 hours

Distance,

= (4 × 5/2)

= 10 ;km

**Now,** Speed = 12 km

Distance = 10 km

Time = 10/12

= 5/6 × 60

= 50 min

Therefore, Rashmi will take 50 minute.

**Question no – (4) **

**Solution : **

Speed 1 = 90 km/h

Speed 2 = 120 km/h

**∴** Distance,

= (90 + 120/2)

= (90 + 60) km

= 150 km

So, to complete the race 150 km will cover.

**Question no – (5) **

**Solution : **

Average speed = D/T

= 250 + 128 + 210/5 + 2 + 3

= 594/10

= 59.4 km/hr.

Therefore, the average speed for the whole journey will be 59.4 km/hr.

**Question no – (6) **

**Solution : **

Average speed = D/T

= 2 × 35 × 25/(35 + 25)

= 2 × 35 × 25/60

= 175/6

= 29.2 km/h

Therefore, the average speed for the whole journey will be 29.2 km/h

**Question no – (7) **

**Solution : **

Let distance x km

= x/5 – x/6 = 15 + 10/60

= 25/60

= 5/12

= 6x – 5x/30 = 5/12

= x = 30/12 × 5

= x = 150/12

= x = 12.5 km

Therefore, the distance of his office from his house will be 12.5 km.

**Question no – (8) **

**Solution : **

Let, distance = x km

= x/40 – x/50 = (11 +5/60)

= 5x – 4x/200 = 40/60

= 200/10

= 20 km

**Question no – (9) **

**Solution : **

Speed = 54 km/h

= (54 × 5/18

= 15 m/s

Now (a) Time = 180/15

= 12s

**(b) Distance = (180 + 540)**

= 720 m

**∴** Time = 720/15

= 48 second.

**Question no – (10) **

**Solution : **

Length of bridge = 125 m

Let, length of train = x

Speed = x/15 m/s

**∴** x/15 = x + 125/30

= 30x = 15x + (15 × 125)

= 30x – 15x = 15 × 125

= x = 15 × 125/15

= x = 125 km

Now Speed = 125/15

= 8 5/15

= 8 1/3 m/s

Therefore, length of the train will be 125 m and its speed will be 8 1/3 m/s.

**Question no – (11) **

**Solution : **

Speed = 90 km/h

= (90 × 5/18) m/s

= 25 m/s

Distance = (150 + 200)

= 350 m

**∴** Time = 350/25

= 14 second.

Therefore, the required time will be 14 second.

**Question no – (12) **

**Solution : **

Speed = 160/18 m/s

= 80/9 m/s

Time = 100 / 80/9

= 100 × 9/80

= 90/8

= 11.25 second.

Hence, it will take 11.25 second to pass another platform.

**Speed, Distance and Time Exercise 10.2 Solution :**

**Question no – (1) **

**Solution : **

Time = 12 min

= 12/60

= 1/5 h

Relative Speed = (8 – 6) km/h

= 2 km/h

**∴** Distance = 2 × 1/5

= 2/5 km/h

= 2000/5

= 400 m

Therefore, the distance between them will be 400 m.

**Question no – (2) **

**Solution : **

**1st** = (90 km/h × 5/18)

= 25 m/s

**2nd** = 12.5 m/s

**∴ **Relative speed = (25 – 12.5)m/s

= 12.5 m/s

Total distance = (195 + 180) m

= 370 m

= Time = 370/12.5

= 29.6 second

= 30 second.

**(b) opposite direction,**

Relative speed = (25 + 12.5) m/s

= 37.5 m/s

Time = 370/37.5

= 9.866

= 10 seconds

**Question no – (3) **

**Solution : **

**Total distance, **

= (137 + 163) m

= 300 m

**1st train speed,**

= (48 × 5/18)m/s

= 8 × 5/3

= 40/ m/s

**2nd train, **

= 3 minute

= (3 × 60)

= 180 second

Speed = 300/180

= 10/6

= 5/3 m/s

Relative speed,

= (40/3 + 5/3)

= 45/3 m/s

**∴ Time**

= 300 × 3/45

= 20 second.

**Question no – (4) **

**Solution : **

**1st speed,**

= (45 × 5/18) m/s

= 75/6 m/s

**2nd speed,**

= (36 × 5/18)

= 10 m/s

Relative speed,

= (75/6 – 10)

= 75 – 60/6

= 15/6 m/s

Distance,

= (15/6 × 80)

= 200 m

**∴ **2nd train length,

= (200 – 120)

= 80 m

Therefore, the length of the second train will be 80 m.

**Question no – (5) **

**Solution : **

Upstream speed,

= (9 – 2) km/h

= 7 km/h

Time = 35/7 h

= 5 hour.

Therefore, the boat will take 5 hours to go 35 km upstream.

**Question no – (6) **

**Solution : **

Let, the speed of boat is still water is = x km/h

S_{p} = ( x + 2) km/h

S_{a }= ((x / 2) km/h

= x + 2 = 34.5/3

= 3x + 6 = 34.5

= x = 34.5 – 6/3

= 9.5 km/h

Therefore, the speed of the boat in the still water is 9.5km/h

**Question no – (7) **

**Solution : **

Upstream speed,

= (10 – 3) km/h

= 7 km/h

**∴** In 13 hours distance cover,

= (7 × 13)

= 91 km.

Downstream speed,

= (10 + 3)

= 13 km/h

Time = 91/3

= 7 h

Therefore, the distance rows is 91 km and time taken 7 hours.

**Next Chapter Solution : **

👉 Chapter 11 👈