The DAV Class 8 Maths Book Solutions and **DAV Class 8 Maths Chapter 4 Worksheet 1** Solutions of Direct and Inverse Variation offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 4 WS 1 Solutions

Question 1.

Fill in the missing terms in the following tables, if x and y vary directly.

(i)

Solution:

(i) Let us put with p, q and s in the blanks

\(\frac{6}{18}=\frac{8}{p}\)

⇒ 6p = 8 × 18

⇒ p = \(\frac{8 \times 18}{6}\)

⇒ p = 8 × 3 = 24

\(\frac{6}{18}=\frac{12}{q}\)

⇒ 6q = 18 × 12

⇒ q = \(\frac{18 \times 12}{6}\)

⇒ q = 3 × 12 = 36

and \(\frac{6}{18}=\frac{s}{63}\)

⇒ 18s = 6 × 63

⇒ s = \(\frac{6 \times 63}{18}\)

⇒ s = 21.

(ii)

Solution:

Let us put a, b and c in blanks of the given table.

\(\frac{10}{5}=\frac{a}{10}\)

⇒ 5 × a = 10 × 10

⇒ a = \(\frac{10 \times 10}{5}\) = 20

\(\frac{10}{5}=\frac{30}{b}\)

⇒ 10 × b = 5 × 30

⇒ b = \(\frac{5 \times 30}{10}\) = 15

and \(\frac{10}{5}=\frac{46}{c}\)

⇒ 10 × c = 5 × 46

⇒ c = \(\frac{5 \times 46}{10}\)

⇒ c = 23

(iii)

Solution:

Let us fill the blanks with l and m in the given table.

∴ \(\frac{6}{15}=\frac{10}{l}\)

⇒ 6l = 10 × 15

⇒ l = \(\frac{10 \times 15}{6}\) = 25

and \(\frac{6}{15}=\frac{m}{40}\)

⇒ 15 × m = 6 × 40

⇒ m = \(\frac{6 \times 40}{15}\) = 16.

Question 2.

Five bags of rice weigh 150 kg. How many such bags of rice will weigh 900 kg?

Solution:

Let the number of bags be x.

As there is a direct variation,

\(\frac{5}{100}=\frac{x}{900}\)

⇒ 100 x = 5 × 900

⇒ x = \(\frac{5 \times 900}{100}\) = 45 bags

Hence, the required bags are 45.

Question 3.

The cost of 3 kg of sugar is ₹ 54. What will be the cost of 15 kg of sugar?

Solution:

Let the required cost be ₹ x.

As there is a direct variation,

\(\frac{3}{54}=\frac{15}{x}\)

⇒ 3x = 15

⇒ x = \(\frac{15 \times 54}{3}\)

⇒ x = 5 × 54 = 270.

Hence, the required cost is ₹ 270.

Question 4.

A motor boat covers 20 km in 4 hours. What distance will it cover in 7 hours (speed remaining the same)

Solution:

Let the distance covered be x km.

As there is a direct variation,

∴ \(\frac{20}{4}=\frac{x}{7}\)

⇒ 4x = 20 × 7

⇒ x = \(\frac{20 \times 7}{4}\)

⇒ 4x = 5 × 7 = 35 km

Hence, the required distance = 35 km.

Question 5.

A motor bike travels 210 km on 30 litres of petrol. How far would it travel on 7 litres of petrol?

Solution:

Let the distance travelled be x km.

As there is a direct variation,

∴ \(\frac{210}{30}=\frac{x}{7}\)

⇒ 30x = 210 × 7

⇒ x = \(\frac{210 \times 7}{30}\)

⇒ x = 7 × 7 = 49 km

Hence, the required distance is 49 km.

Question 6.

If 12 women can weave 15 m of cloth in a day, how many metres of cloth can be woven by 20 women in a day?

Solution:

Let the required length of the cloth be x m

As there is a direct variation,

∴ \(\frac{12}{15}=\frac{20}{x}\)

⇒ 12 × x = 15 × 20

⇒ x = \(\frac{15 \times 20}{12}\)

⇒ x = 25

Hence the length of the required cloth is 25 m.

Question 7.

If the weight of the 5 sheets of paper is 20 gram, how many sheets of the same paper would weigh 2.5 kg?

Solution:

Let the number of sheets required be x.

As there is a direct variation,

∴ \(\frac{5}{20}=\frac{x}{2500}\)

⇒ 20 × x = 5 × 2500

⇒ x = \(\frac{5 \times 2500}{20}\)

⇒ x = 625

Hence the number of sheets required = 625.

Question 8.

Reena types 600 words in 4 minutes. How much time will she take to type 3150 words?

Solution:

Let the time required be x minutes.

As there is a direct variation,

∴ \(\frac{600}{4}=\frac{3150}{x}\)

⇒ 600 × x = 4 × 3150

⇒ x = \(\frac{4 \times 3150}{600}\)

⇒ x = 21

Hence, the required time is 21 minutes.

Question 9.

Rohan takes 14 steps in covering a distance of 2.8 m. What distance would he cover in 150 steps?

Solution:

Let the required distance be x m.

As there is a direct variation,

∴ \(\frac{14}{2.8}=\frac{150}{x}\)

⇒ 14 × x = 2.8 × 150

⇒ x = \(\frac{2.8 \times 150}{14}\)

⇒ x = 30.

Hence, the required distance is 30 m.

Question 10.

A dealer finds that 48 refined oil cans (of 5 litres each) can be packed in 8 cartons of the same size. How many such cartons will be required to pack 216 cans?

Solution:

Let the number of cartons required be x.

As there is a direct variation,

∴ \(\frac{48}{8}=\frac{216}{x}\)

⇒ 48 × x = 8 × 216

⇒ \(\frac{8 \times 216}{48}\)

⇒ x = 36.

Hence, the number of cartons required is 36.

Question 11.

The total cost of 15 newspapers is ₹ 37.50. Find the cost of 25 newspapers.

Solution:

Let the required cost be ₹ x.

As there is a direct variation,

∴ \(\frac{15}{37.50}=\frac{25}{x}\)

⇒ 15 × x = 25 × 37.50

⇒ \(\frac{25 \times 37.50}{15}\)

⇒ x = 62.50

Hence, the cost of 25 newspapers is ₹ 62.50.

Question 12.

A labourer gets ₹ 675 for 9 days work. How many days should he work to get ₹ 900?

Solution:

Let the number of days required be x.

∴ \(\frac{675}{9}=\frac{900}{x}\)

⇒ 675 × x = 9 × 900

⇒ x = \(\frac{9 \times 900}{675}\)

⇒ x = 12.

Hence, the number of days required is 12 days.

### DAV Class 8 Maths Chapter 4 Worksheet 1 Notes

Direct Variation:

- Let x be the number of members in a particular family and their monthly expenditure is ₹ y.
- If the number of the family members, i.e. x increases, their monthly expenditure, i.e. y also increases.
- On the other hand, if the number of members, i.e. x decreases, their monthly expenditure, i.e. y also decreases.

This is called the Direct Variation.

Inverse Variation:

Suppose, x men finish a work in y days. If the number of men increases, then the work will be finished in less than y days and if the number of men decreases, the days will be increased to finish the same work in time. So, we can say that x and y are inversely proportional to each other which is known as Inverse Variation.

**DIRECT VARIATION**

Example 1.

Find the value of a and b in the following table, if x and y vary directly.

x | 3 | a | 18 |

y | 5 | 35 | b |

Solution:

Given that x and y vary directly.

∴ \(\frac{3}{5}=\frac{a}{35}=\frac{18}{b}\)

⇒ \(\frac{3}{5}=\frac{a}{35}\)

⇒ 5a = 3 × 35

⇒ a = \(\frac{3 \times 35}{5}\)

⇒ a = 3 × 7 = 21

Now taking \(\frac{3}{5}=\frac{18}{b}\)

⇒ 3b = 5 × 18

⇒ b = \(\frac{5 \times 18}{3}\)

⇒ b = 5 × 6

∴ b = 30

Example 2.

In the following table, verify that x and y are directly proportional.

Solution:

Here,

\(\frac{x}{y}=\frac{11}{22}=\frac{1}{2}\)

\(\frac{x}{y}=\frac{8}{16}=\frac{1}{2}\)

\(\frac{x}{y}=\frac{5}{10}=\frac{1}{2}\)

\(\frac{x}{y}=\frac{17}{34}=\frac{1}{2}\)

\(\frac{x}{y}=\frac{1}{2}\) (k constant)

∴ x and y are directly proportional to each other.

Example 3.

The cost of 5 metres of cloth is ₹ 210. Tabulate the cost of 2, 4, 10 and 13 metres of cloth of the same type.

Solution:

Let the length of cloth be x m and its cost be ₹ y.

Forming the table,

As x and y are directly proportional.

∴ \(\frac{2}{y_1}=\frac{5}{210}\)

⇒ 5y_{1} = 2 × 210

⇒ y_{1} = \(\frac{2 \times 210}{5}\)

= 2 × 42 = ₹ 84

Similarly,

\(\frac{4}{y_2}=\frac{5}{210}\)

⇒ 5y_{2} = 4 × 210

⇒ y_{2} = \(\frac{4 \times 210}{5}\)

⇒ y_{2} = 4 × 42 = ₹ 168

\(\frac{10}{y_3}=\frac{5}{210}\)

⇒ 5y_{3} = 10 × 210

⇒ y_{3} = \(\frac{10 \times 210}{5}\)

⇒ y_{3} = ₹ 420

\(\frac{13}{y_4}=\frac{5}{210}\)

⇒ 5y_{4} = 13 × 210

⇒ y_{4} = \(\frac{13 \times 210}{5}\)

⇒ y_{4} = 13 × 42 = ₹ 546.

Example 4.

If the weight of 10 sheets of thick paper is 50 gram, how many sheets of the same paper would weigh 2\(\frac{1}{2}\) kg?

Solution:

Let the number of sheets required be x.

Number of sheets | 10 | x |

Weight. of sheets (in gram) | 50 | 2500 |

As there is a direct variation,

\(\frac{10}{50}=\frac{x}{2500}\)

⇒ 50 × x =10 × 2500

⇒ x = \(\frac{10 \times 2500}{50}\)

⇒ x = 500

Hence, the required number of sheets is 500.