WBBSE Solutions For Class 7 Maths Arithmetic Chapter 6 Time And Distance

Arithmetic Chapter 6 Time And Distance

Question 1. Choose the correct answer:-

1. The time taken to travel 48kmat 36 km/h is.

1. 1hr somin.
2. 1hr 20min
3. 1hr 40 min
4. None of these.

Solution:

Distance = 48Km

speed = 36 km/h

Time =?

Speed = \(\frac{\text { Distance }}{\text { Time }}\)

Time = \(\frac{\text { Distance }}{\text { speed }}\)

⇒ \(\frac{48}{36}\)

= 1.33

Time = 1-33hr ⇒ 1hr.30 min

∴ The option (1) Thy 30 min is the correct answer.

Class 7 Maths Arithmetic Chapter 6 Solutions

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2. The distance covered in 4.5hɣmpat 8.4km/hr is.

1. 37.8km
2. 36 km
3. 36.8 km
4. 37Km.

Time (T) = 4.5hrs

Distance =?

Speed(s) = 8.4 km/hr.

Speed\(=\frac{\text { Distance covered }}{\text { Time taken. }}\)

Distance covered = Speed x Time taken.

= 8.4×4.5

Distance covered = 37.8 Km

∴ The option (1) 37.8 km is the correct answer.

3. If the bus covers a distance of 50 km in 3hrs 45 mins. then its speed is.

1. 45km/hr.
2. 42.5km/hr
3. 40km/hr
4. None of these.

Distance = 150 KM.

Time = 3hrs 45mins ⇒ 1hr = 60 min

= 45 ⇒ \(\frac{45}{60} \times 1\) = 0.75

Speed =?

Speed = \(\frac{\text { Distance covered }}{\text { time taken }}\)

⇒ \(\frac{150}{3.75} \mathrm{~km} / \mathrm{hr}\)

= 40Km/hr

∴ The option (3) 40km/hr is the correct answer.

Question 2. Write true or False:

1. Speed remaining fixed the distance traveled and time taken are inversely proportional.

→ False

Speed = \(\frac{\text { Distance }}{\text { time }}\)

Speed ∞ Distance

⇒ \(\text { speed } \infty \frac{1}{\text { time. }}\)

2. Speed = Distance travelled × Time taken.

→ False

Speed = \(\frac{\text { Distance traveled. }}{\text { Time taken }}\)

3. when two objects move in the same direction their relative speed will be different of their actual speed.

→ True

Class 7 Arithmetic Chapter 6 Questions

Question 3. Fill in the blanks.

1. If two objects move in opposite directions, then relative speed is equal to the sum of their speed

2. \(4 \frac{1}{2}\) km/hr = 1.25 m/sec

⇒ \(4 \frac{1}{2} \Rightarrow \frac{9}{2}\) = 4.5km/hy

1Km = 1000meters

1hr = 60min = 3600sec

1 minute = 60 sec

60minute = ? 3600sec

4KM → 4000 meters

⇒ \(\frac{1}{2}\) km → 500 meters.

Total distance = 4500 meters

1hr = 3600 sec

m/sec= \(\frac{4500}{3600}\)

= 1.25

3. The time taken to travel 60km at 15 km/hr is 4 hr.

Distance = 60Km

Speed = 15km/hr

Time =?

Speed \(=\frac{\text { Distance covered }}{\text { Time taken }}\)

Time = \(\frac{\text { Distance covered }}{\text { speed }}\)

⇒ \(\frac{60}{15}\)

Time = 4hr

The time taken to travel 60 km at 15km/hr is 4hr

Question 4. A train of length 150m, moving with a speed of 75km/hr passes a tree calculate how long will it take to do so.
Solution:

Given:

Length of train = 150m

Speed of train = 75km/hr

⇒ \(75 \times \frac{5}{18}\) = 20.83 m/sec.

Time \(=\frac{\text { Distance covered }}{\text { Time taken }}\)

⇒ \(\frac{150}{20.83}\)

Time = 7.2 seconds.

So it will take approximately 7.2 seconds for the train to pass the tree.

Question 5. A train with a length of 100m takes 20 sec to pass a light post calculate the speed of the train.
Solution:

Length of the train = 100m

Time (t) = 20sec.

speed= ?

Speed \(=\frac{\text { Distance covered }}{\text { time taken }}\)

⇒ \(\frac{100 \mathrm{~m}}{20 \mathrm{sec}}\)

Speed = 5 m/sec

⇒ \(5 \times \frac{18}{5}\)

Speed = 18km/hr

∴ The speed of the train is 18km/hr

Class 7 Arithmetic Textbook Solutions

Question 6. A 120m long train with a speed of 45km/hr passes a platform in 30 sec. calculate the length of the Pat platform.
Solution:

Length of train = 120m

speed of train = 45 km/hr ⇒ \(45 \times \frac{5}{18}\) = 12.5m/s

Time required to pass the platform = 30 sec.

Let ‘x’ be the length of the platform.

The total length to pass the platform is (x+120)m

Speed \(=\frac{\text { Distance covered. }}{\text { Time taken to cover Distance }}\)

12.5 = \(\frac{x+120}{30}\)

375 = X+120

x = 375-120

x = 255m

∴ The length of the platform is 255m

Question 7. A train takes 6sec to pass a lamp post and 30 sec to pass a 280m long platform. Find the length of the train and also its speed.
Solution:

Let L be the length of the train

S be the speed of the train

For passing a lamp post

L = S × 6

For passing the platform

L+280 = S×30

6S+280 = S×30

280 = 30S-6S

280 = 24S

S = \(\frac{280}{24}\)

S = 11.67 m/s

S = \(11.67 \times \frac{18}{5}\)

S = 42.01Km/h

L = 6×S ⇒ 6 × 11.67

L≈70m

∴ The Speed of the train is 11.67mls of 42 km/h

The length of the train is 70m

Class 7 Maths Arithmetic Problems

Question 8. Two trains 120m and 105m long respectively come running at the rate of 60 km/hr and 45km/hr. respectively. How long will they take to cross each other if the two trains are running in the same direction?
Solution:

After the two trains meet, they would pass each other ie, the two trains would simultaneously pass a distance equal to the sum of their own length

The distance two trains will cover (120+105) mo8 225m

The relative speed. in the same direction

If two objects are moving in the same direction then the

relative speed = difference of their speeds.

= 60-45

relative speed = 15 km/hr

The time to cross each other.

T \(=\frac{\text { Total length }}{\text { Relative speed }}\)

⇒ \(\frac{225}{15}\)

= 15 sec

The time to cross each other is 15 sec

Question 9. A train passes two platforms of length. 275m and 140m In 28sec and 16 sec respectively calculate the length and Speed of the train
Solution:

Let the length of the train be xm

when the train passes a platform of length 275m Then the train has to cover a distance of length (275+2) m

ly

To pass the second platform of length 140m

Then the train has to cover a distance of length (140+x) m

Mathematically represent

Speed remains constant, and time and distance are in direct proportion.

28:16:: (x+275): (X +140).

⇒ \(\frac{28}{16}=\frac{x+275}{x+140}\)

⇒ 28(x+140) = 16(2+275)

⇒  28(x+140) = 16(x+275)

⇒ 28x+ 28×140 = 16x+16×275

⇒ 28x+3920 = 16x +4400

⇒ 28x-16X = 4400-3920

⇒ 12x = 480

x = \(\frac{480}{12}\)

x = 40 meters.

∴ The length of the train is 40 meters

In 28 sec the train travels (40+275) =315m

In 1 Sec the train travels \(\frac{315}{28} \mathrm{~m}\)

In 1hr or 3600 sec the train travels = \(\frac{315 \times 3600}{28}\)

⇒ \(\frac{1134,000}{28}\)

⇒ 40,500meters

⇒ 40.5km/hr.

∴ The length of the train is 40 meters.

The Speed of the train is 40.5 km/hr.

Class 7 Arithmetic Formulas

Question 10. Two trains of equal length running in opposite directions, pass a man standing by a side of railway line in 18 secs and 12 secs respectively. At what time will the two trains cross each other?
Solution:

Let’s denote the length of each train by L meters.

Let the speed of the first train be V₁ meters per second

The speed of the second train be V₂ meters per second

First train

L = V₁ x 18 ⇒ V₁ = \(\frac{L}{18}\)

second train

L = V₂ x 12 ⇒ V₂ = \(\frac{L}{12}\)

Relative speed is the sum of their speeds.

The relative speed V_r is given by

V_r = V₁+V₂

V_r = \(\frac{L}{18}+\frac{4}{12}\)

⇒ \(\frac{2 L+3 L}{36}\)

⇒ \(\frac{5 L}{36}\)

The total length to be covered when the two trains cross each other is the sum of their lengths

Total length = L + L = 2L

The time taken to cross each other is the total length divided by the relative speed:

T \(=\frac{\text { Total length }}{\text { Relative speed }}\)

⇒ \(\frac{2 L}{\frac{5 L}{36}}\)

⇒ \(2 L \times \frac{36}{5 L}\)

⇒ \(\frac{24}{\frac{5 L}{36}}\)

⇒ \(2 L \times \frac{36}{5L}\)

⇒ \(\frac{36 \times 2}{5}\)

= 14.4 Sec

∴ The two trains cross each other at 14.4 Sec