The DAV Class 8 Maths Book Solutions and **DAV Class 8 Maths Chapter 4 Worksheet 3** Solutions of Direct and Inverse Variation offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 4 WS 3 Solutions

Question 1.

Ramit can finish his work in 25 days, working 8 hours in a day. If he wants to finish the same work in 20 days, how many hours should he work in a day?

Solution:

Let the number of hours required be x.

As there is an inverse variation,

x_{1}y_{1} = x_{2}y_{2}

⇒ 25 × 8 = 20 × x

⇒ x = \(\frac{25 \times 8}{20}\)

⇒ x = 10

Hence, the number of hours required in a day = 10.

Question 2.

Udit can complete his work in 10 days. What amount of work will be completed in 8 days?

Solution:

Let the required amount of work be x.

As there is a direct variation,

x_{1}y_{1} = x_{2}y_{2}

⇒ \(\frac{10}{1}=\frac{8}{x}\)

⇒ 10x = 8

⇒ x = \(\frac{8}{10}=\frac{4}{5}\)

Hence, only \(\frac{4}{5}\) th of the work will be completed in 8 days.

Question 3.

If 20 men can build a wall in 9 days, then how long would it take 12 men to build the same wall?

Solution:

Let the time required be x days.

As there is an inverse variation,

x_{1}y_{1} = x_{2}y_{2}

⇒ 20 × 9 = 12 × x

⇒ x = \(\frac{20 \times 9}{12}\)

⇒ x = 15

Hence the time required is 15 days.

Question 4.

Geetika weaves 20 baskets in 30 days. In how many days she will weave 120 baskets?

Solution:

Let the number of days required be x.

As there is a direct variation,

∴ \(\frac{20}{30}=\frac{30}{x}\)

⇒ 20 × x = 30 × 30

⇒ x = \(\frac{30 \times 30}{20}\)

⇒ x = 45

Hence the number of days required = 45.

Question 5.

A train 280 m long is running at a speed of 42 km/hr. How much time will it take to pass a man standing on a platform?

Solution:

Let the time required be x seconds.

Speed = 42 km/hr

= \(\frac{42 \times 1000}{60 \times 60}=\frac{35}{3} \mathrm{~m} / \mathrm{s}\)

Time = \(\frac{\text { Distance }}{\text { Speed }}=\frac{280}{35 / 3}=\frac{280 \times 3}{35}\)

= 24 seconds.

Hence, the time required is 24 seconds.

Question 6.

A train 350 m long crosses an electric pole in 28 seconds. Find the speed of the train in km/hr.

Solution:

Speed =\(\frac{\text { Distance }}{\text { Time }}=\frac{350}{28}=\frac{25}{2}\) m/s

= \(\frac{25}{2} \times \frac{60 \times 60}{1000}\)

= 45 km/hr

Hence, the required speed = 45 km/hr.

Question 7.

A train 150 m long is running at 72 km/hr. It crosses a bridge in 13 sec. Find the length of the bridge.

Solution:

Let the length of the bridge be x m.

∴ Total length to be crossed = (150 + x) m

Speed = 72 km/hr

= \(\frac{72 \times 1000}{60 \times 60}\) = 20 m/sec

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

⇒ 20 = \(\frac{(150+x)}{13}\)

⇒ 150 + x = 20 × 13

⇒ 150 + x = 260

⇒ x = 260 – 150 = 110m

Hence, the length of the bridge = 110 m.

Question 8.

How long will a train 120 m long take to clear a platform 130 m long, if its speed is 50 kmlhr?

Solution:

Total length to be covered = 120 m + 130 m = 250 m

Let the required time taken be x seconds.

Speed = 50 km/hr

= \(\frac{50 \times 1000}{60 \times 60}=\frac{125}{9}\) m/sec

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

⇒ \(\frac{125}{9}=\frac{250}{x}\)

⇒ 125x = 9 × 250

⇒ x = \(\frac{9 \times 250}{125}\)

⇒ x = 18 seonds.

Hence, the required time is 18 sec.

Question 9.

A train 210 m long took 12 seconds to pass a 90 m long tunnel. Find the speed of the train.

Solution:

Total length = 210 + 90 = 300 m

Speed = \(\frac{\text { Distance }}{\text { Time }}=\frac{300}{12}\) = 25 m/s

= \(\frac{25 \times 60 \times 60}{1000}\) = 90 km/hr

Question 10.

A train 270 m long is running at 80 km/hr. How much time will it take to cross a platform 130 m long?

Solution:

Let the required time be x sec.

Speed = 80 km/hr

= \(\frac{80 \times 1000}{60 \times 60}=\frac{200}{9}\) m/sec.

and total length = 270 m + 130 m = 400 m

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

\(\frac{200}{9}=\frac{400}{x}\)

⇒ 200 × x = 9 × 400

⇒ x = \(\frac{9 \times 400}{200}\)

⇒ x = 18 sec.

Hence the required time is 18 sec.

### DAV Class 8 Maths Chapter 4 Value Based Questions

Question 1.

Two children were eating pizza at “Master Bakers” having three slices of pizza each. All of a sudden one of their friends came and they shared pizza slices equally among three.

(i) How many slices of pizza would each get?

(ii) Is it good to eat junk food like pizza? Give reason.

Solution:

(i) Let the number of slices of pizza be x.

Clearly, it is a case of inverse variation.

So, 2 × 3 = 3 × x

∴ x = \(\frac{2 \times 3}{3}\) = 2.

(ii) No, it is not good to eat junk food like pizza as it causes obesity and different diseases.

Question 2.

By walking for 30 minutes in the morning, Namita covers two kilometres,

(i) How much distance will she cover in 20 minutes by walking at the same pace?

(ii) What is the importance of morning walk?

Solution:

Let the distance covered in 20 minutes be x.

Clearly, it is a case of direct variation.

Now, \(\frac{2}{30}=\frac{x}{20}\)

⇒ x = \(\frac{2 \times 20}{30}=\frac{4}{3}\)

⇒ x = 1.33 km

(ii) Morning walk keeps the body fit and sound.

### DAV Class 8 Maths Chapter 4 Worksheet 3 Notes

**TIME AND WORK, TIME AND DISTANC**E

Work done by a person varies directly with the time taken to complete a job but work done by the number of persons to complete the same job is inversely proportional to the time taken.

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

Example 1.

A train 450 m long is running at a speed of 36 km/hr. What time will it take to cross a 150 m long bridge?

Solution:

Total length to be covered by the train = Length of the train + Length of the bridge

= 450 m + 150 m = 600 m

Speed of the train = 36 km/hr

= \(\frac{36 \times 1000}{60 \times 60}\) = 10 m/s

Let the time taken by the train to cross the bridge be x seconds.

Distance (m) | 10 | 600 |

Time(sec) | 1 | x |

As it is a direct variation,

\(\frac{10}{1}=\frac{600}{x}\)

10x = 600

∴ x = \(\frac{600}{10}\)

= 60 seconds = 1 minute

Hence, the time taken by the train to cross the bridge is 60 seconds or 1 minute.

Example 2.

A worker weaves 15 chairs in 24 days. In how many days he will weave 105 chairs?

Solution:

Let the time required be x days.

Number of chairs | 15 | 105 |

Number of days | 24 | x |

As there is a direct variation,

\(\frac{15}{24}=\frac{105}{x}\)

⇒ 15x = 24 × 105

⇒ x = \(\frac{24 \times 105}{15}\)

⇒ x = 24 × 7 = 168

Hence, number of days required is 168 days.