The DAV Class 8 Maths Book Solutions and **DAV Class 8 Maths Chapter 4 Worksheet 2** Solutions of Direct and Inverse Variation offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 4 WS 2 Solutions

Question 1.

In the following tables, a and b vary inversely. Fill in the missing values.

(i)

Solution:

Let us put m and n in the blanks.

As there is an inverse variation,

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

⇒ 7 × 8 = m × 4

⇒ m = \(\frac{7 \times 8}{4}\)

⇒ m = 14

similarly, x_{1}y_{1} = x_{2}y_{2}

⇒ 7 × 8 = 28 × n

⇒ n = \(\frac{7 \times 8}{28}\)

⇒ n = 2.

Hence, the missing values are 14 and 2.

(ii)

Solution:

Let us put p and q in the blanks.

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

⇒ 2.5 × 8 = 4 × p

⇒ p = \(\frac{2.5 \times 8}{4}\)

⇒ p = 5

Similarly, x_{1}y_{1} = x_{2}y_{2}

⇒ 2.5 × 8 = 0.5 × q

⇒ q = \(\frac{2.5 \times 8}{0.5}\)

⇒ q = 40.

Hence, the missing values are 5 and 40.

(iii)

Solution:

Let us put s and t in the blanks.

As there is an inverse variation,

∴ x_{1}y_{2} = x_{2}y_{1}

⇒ 10 × 6 = s × 15

⇒ s = \(\frac{10 \times 6}{15}\)

⇒ s = 4.

Similarly, x_{1}y_{1} = x_{2}y_{2}

⇒ 10 × 6 = 12 × t

⇒ t = \(\frac{10 \times 6}{12}\)

⇒ t = 5

Hence, the missing values are 4 and 5.

Question 2.

The science teacher asked the students of class-VIII to make a project report on pollution. When 10 students work on it, the work gets finished in 3 days. How many students are required so that the work finishes in 2 days?

Solution:

Let the number of students required be x.

As there is an inverse variation,

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

⇒ 10 × 3 = x × 2

⇒ x = \(\frac{10 \times 3}{2}\)

⇒ x = 15

Hence, the number of students required is 15.

Question 3.

Running at an average speed of 40 km/hr, a bus completes a journey in 4\(\frac{1}{2}\) hours. How much time will the return journey take if the speed is increased to 45 km/hr?

Solution:

Let the time required be x hours.

As there is an inverse variation,

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

⇒ 40 × \(\frac{9}{2}\) = 45 × x

⇒ x = \(\frac{40 \times \frac{9}{2}}{45}\)

⇒ x = 4

Hence, the time required is 4 hours.

Question 4.

Disha cycles to her school at an average speed of 12 km/hr. It takes her 20 minutes to reach the school. If she wants to reach her school in 15 minutes, what should be her average speed?

Solution:

Let the required speed be x km/hr.

As there is an inverse variation,

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

⇒ 12 × 20 = x × 15

⇒ x = \(\frac{12 \times 20}{15}\)

⇒ x = 16

Hence, the required speed is 16 km/hr.

Question 5.

If 15 men can repair a road in 24 days, then how long will it take 9 men to repair the same road?

Solution:

Let the required time be x days.

As there is an inverse variation,

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

⇒ 15 × 24 = 9 × x

⇒ x = \(\frac{15 \times 24}{9}\)

⇒ x = 40

Hence, the number of days required is 40.

Question 6.

If 30 goats can graze a field in 15 days, then how many goats will graze the same field in 10 days?

Solution:

Let the number of goats be x

As there is an inverse variation,

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

⇒ 30 × 15 = x × 10

⇒ x = \(\frac{30 \times 15}{10}\)

⇒ x = 45

Hence the required number of goats is 45.

Question 7.

A contractor with a work force of 420 men can complete a work of construction of a building in 9 months. Due to request by the owners, he was asked to complete the job in 7 months. How many extra men, he must employ to complete the job?

Solution:

Let number of extra men required be x.

As there is an inverse variation,

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

⇒ 420 × 9 = x × 7

⇒ x = \(\frac{420 \times 9}{7}\)

⇒ x = 540

Extra men required = 540 – 420 = 120

Hence the extra men required = 120.

Question 8.

Uday can finish a book in 25 days, if he reads 18 pages every day. How many days will he take to finish it, if he reads 15 pages every day?

Solution:

Let the number of days required be x

As there is an inverse variation,

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

⇒ 18 × 25 = 15 × x

⇒ x = \(\frac{40 \times 125}{100}\)

⇒ x = 30

Hence, the number of days required = 30 days.

Question 9.

A {shopkeeper has enough money to buy 40 books, each costing ₹ 125. How many books he can buy if he gets a discount of ₹ 25 on each book?

Solution:

Let the number of books required be x.

Cost after discount = 125 – 25 = ₹ 100

As there is an inverse variation,

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

⇒ 40 × 125 = x × 100

⇒ x = \(\frac{40 \times 125}{100}\)

⇒ x = 50

Hence, the number of books = 50.

Question 10.

Six pumps working together empty a tank in 28 minutes. How long will it take to empty the tank if 4 such pumps are working together?

Solution:

Let the required time be x minutes.

As there is an inverse variation.

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

⇒ 6 × 28 = 4 × x

⇒ x = \(\frac{6 \times 28}{4}\)

⇒ x = 42

Hence, the time required is 42 minutes.

Question 11.

A train moving at a speed of 75 km/hr covers a certain distance in 4.8 hours. What should be the speed of the train to cover the same distance in 3 hours?

Solution:

Let the required speed be x km/hr.

As there is an inverse variation,

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

⇒ 75 × 4.8 = x × 3

⇒ x = \(\frac{75 \times 4.8}{3}\)

⇒ x = 120

Hence, the speed need is 120 km/hr.

Question 12.

A garrison of 120 men has provision for 30 days. At the end of 5 days, 5 more men joined them. How many days can they sustain on the remaining provision?

Solution:

Let the number of days required be x.

Now the number of men becomes 120 + 5 = 125

Number of days left = 30 – 5 = 25

As there is an inverse variation,

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

⇒ 120 × 25 = 125 × x

⇒ x = \(\frac{120 \times 25}{125}\)

⇒ x = 24

Hence, the provision will last for 24 days.

### DAV Class 8 Maths Chapter 4 Worksheet 2 Notes

**Inverse Variation**

Example 1.

6 Pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5 pipes of the sanie type are used to fill the same tank?

Solution:

Let the required time be x hrs.

Number. of pipes | 6 | 5 |

Time (in hours) | 4/3 | x |

[∵ 1 hour 20 minutes = 1 + \(\frac{1}{3}\) = \(\frac{4}{3}\) hours]

As there is inverse variation,

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

⇒ 6 × \(\frac{4}{3}\) = 5 × x

⇒ x = \(\frac{6 \times \frac{4}{3}}{5}\)

⇒ x = \(\frac{8}{5}\) = 1 hour 36 minutes.

Hence, the required time is 1 hour 36 minutes.

Example 2.

If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?

Solution:

Let the number of workers required be x.

Number of Workers | 15 | x |

Time (in hrs.) | 48 | 30 |

As there is inverse variation,

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

⇒ 15 × 48 = x × 30

⇒ x = \(\frac{15 \times 48}{30}\)

⇒ x = 24

Hence the number of workers required is 24.

Example 3.

A car takes 2 hours to reach a destination by travelling at the speed of 60 km/hr. How long will it take when the car travels at the speed of 80 km/hr?

Solution:

Let the time required be x hrs.

Time (in hours) | 2 | x |

Speed (km/hr) | 60 | 80 |

As there is an inverse variation,

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

⇒ 2 × 60 = x × 80

⇒ x = \(\frac{2 \times 60}{80}\)

⇒ x = \(\frac{3}{2}\) hrs

= 1 hour 30 minutes.

Hence, the required time is 1 hour 30 minutes.

Example 4.

If 52 men can do a piece of work in 35 days. In how many days will 28 men complete the same work?

Solution:

Let the number of days required be x.

Number of men | 52 | 28 |

Number of days | 35 | x |

As there is an inverse variation,

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

⇒ 52 × 35 = 28 × x

⇒ x = \(\frac{52 \times 35}{28}\)

⇒ x = 65

Hence, the number of days required is 65.