Class 8 RD Sharma Solutions - Chapter 10 Direct And Inverse Variations - Exercise 10.1 | Set 1

Chapter 10 of RD Sharma's Class 8 Mathematics textbook focuses on Direct and Inverse Variations. Exercise 10.1 specifically deals with direct variation (also known as direct proportion). In this exercise, students learn to identify, represent, and solve problems involving quantities that vary directly with each other.

Question 1: Explain the concept of direct variation.

Solution:

If the values of two quantities depend on each other in such a way that if we increase the value of one quantity the value of the other quantity also increase, similarly if we decrease the value of one quantity the value of other quantity also decreases, therefore if the ratio between the two variables remains constant, it is said to be in direct variation.

Question 2: Which of the following quantities vary directly with each other?

(i) Number of articles (x) and their price (y).

(ii) Weight of articles (x) and their cost (y).

(iii) Distance x and time y, speed remaining the same.

(iv) Wages (y) and number of hours (x) of work.

(v) Speed (x) and time (y) distance covered remaining the same).

(vi) Area of a land (x) and its cost (y).

Solution:

(i) Number of articles (x) and their price (y)

The number of articles is directly proportional to their price, therefore, when the number of articles increase, then the cost of article will also increase. So it is a case of direct proportion.

(ii) Weight of articles (x) and their cost (y).

The weight (x) of the articles is directly proportional to their cost (y), therefore, If weight of the article is increasing then the cost of article will also increase. So it is a case of direct proportion.

(iii) Distance x and time y, speed remaining the same.

On increasing the distance between objects, the time required to cover them will also increase, therefore, on constant speed, time increases when distance increases. So it is a case of direct proportion

(iv) Wages (y) and number of hours (x) of work.

If the workers work for more hours, they will be paid more wages, therefore, wages increases if the number of working hours increase. So it is a case of direct proportion

(v) Speed (x) and time (y) distance covered remaining the same.

Time is inversely proportional to distance , that is, keeping the same distance, the time taken will reduce if speed is increased, here one quantity is decreasing when we are increasing the other. So it is not a case of direct proportion.

(vi) Area of a land (x) and its cost (y).

On increasing the area of the land available, its cost will also increase and multiply. So it is a case of direct proportion

Question 3: In which of the following tables x and y vary directly?

(i)

(ii)

(iii)

(iv)

Solution:

(i) Directly proportional.

If we clearly notice the values in the table, then the value in column b is thrice the value of column a. Therefore, the rows ‘a’ and ‘b’ are directly proportional, in this case.

(ii) Directly proportional.

If we clearly notice the values in the table, then the value in column b is half the value of column a. Therefore, ‘a’ and ‘b’ are directly proportional, in this case.

(iii) Not directly proportional.

If we clearly notice the values in the table, then the value in column b is thrice the value of column a, only in first three columns and not in others. Therefore, ‘a’ and ‘b’ are not directly proportional, in this case.

(iv) Not directly proportional.

If we clearly notice the values in the table, then the value in column b differ by different constant amounts with respect to values in column a. Therefore, ‘a’ and ‘b’ are not directly proportional, in this case.

Question 4: Fill in the blanks in each of the following so as to make the statement true:

(i) Two quantities are said to vary…. with each other, if they increase (decrease) together in such a way that the ratio of the corresponding values remains same.

(ii) x and y are said to vary directly with each if for some positive number k, ………= k.

(iii) if u = 3v, then u and v vary…. with each other.

Solution:

(i) directly 

(ii) k = x/y where k is a positive number.

(iii) directly

Question 5. Complete the following tables given that x varies directly as y.

(i)

x 2.5 … … 15

y 5 8 12 …

(ii)

x 5 … 10 35 25 …

y 8 12 … … … 32

(iii)

x 6 8 10 … 20

y 15 20 … 40 …

(iv)

x 4 9 … … 3 …

y 16 … 48 36 … 4

(v)

x 3 5 7 9

y … 20 28 …

Solution:

(i)

We know k = x/y

2.5/5 = x₁/8

By cross-multiplying

8(2.5) = 5×1

20 = 5x₁

x₁ = 20/5

= 4

We know k = x/y

4/8 = x₂/12

By cross-multiplying

12(4) = 8x₂

48 = 8x₂

x₂ = 48/8

= 6

We know k = x/y

6/12 = 15/y₁

By cross-multiplying

6y₁ = 15(12)

6y₁ = 180

y₁ = 180/6

= 30

x 2.5 4 6 15

y 5 8 12 30

(ii)

We know k = x/y

5/8 = x₁/12

By cross-multiplying

12(5) = 8x₁

60 = 8x₁

x₁ = 60/8

= 7.5

We know k = x/y

7.5/12 = 10/y₁

By cross-multiplying

7.5y₁ = 10(12)

7.5y₁ = 120

y₁ = 120/7.5

= 16

We know k = x/y

10/16 = 35/y₂

By cross-multiplying

10y₂ = 35(16)

10y₂ = 560

y₂ = 560/10

= 56

We know k = x/y

35/56 = 25/y₃

By cross-multiplying

35y₃ = 56(25)

35y₃ = 1400

y₃ = 1400/35

= 40

We know k = x/y

25/40 = x₂/32

By cross-multiplying

25(32) = 40x₂

800 = 40x₂

x₂ = 800/40

= 20

x 5 7.5 10 35 25 20

y 8 12 16 56 40 32

(iii)

We know k = x/y

8/20 = 10/y₁

By cross-multiplying

8y₁ = 10(20)

8y₁ = 200

y₁ = 200/8

= 25

We know k = x/y

10/25 = x₁/40

By cross-multiplying

10(40) = 25x₁

400 = 25x₁

x₁ = 400/25

= 16

We know k = x/y

16/40 = 20/y₂

By cross-multiplying

16y₂ = 20(40)

16y₂ = 800

y₂ = 800/16

= 50

x 6 8 10 16 20

y 15 20 25 40 50

(iv)

We know k = x/y

4/16 = 9/y₁

By cross-multiplying

4y₁ = 9(16)

= 144

y₁ = 144/4

= 36

We know k = x/y

9/36 = x₁/48

By cross-multiplying

9(48) = 36x₁

432 = 36x₁

x₁ = 432/36

= 12

We know k = x/y

12/48 = x₂/36

By cross-multiplying

12(36) = 48x₂

432 = 48x₂

x₂ = 432/48

= 9

We know k = x/y

9/36 = 3/y₂

By cross-multiplying

9y₂ = 3(36)

= 108

y₂ = 108/9

= 12

We know k = x/y

3/12 = x₃/4

By cross-multiplying

3(4) = 12x₃

12 = 12x₃

x₃ = 12/12

= 1

x 4 9 12 9 3 1

y 16 36 48 36 12 4

(v)

We know k = x/y

3/y₁ = 5/20

By cross-multiplying

3(20) = 5y₁

60 = 5y₁

y₁ = 60/5

= 12

We know k = x/y

7/28 = 9/y₂

By cross-multiplying

7y₂ = 9(28)

= 252

y₂ = 252/7

= 36

x 3 5 7 9

y 12 20 28 36

Question 6. Find the constant of variation from the table given below:

Set up table and solve the following problems. Use unitary method to verify the answer.

Solution:

Dividing the second column values by first column values, y/x we get the corresponding ratios 

C1	C2	C3	C4
y/x	12/3 = 4	20/5 = 4	28/7 = 4	36/9 = 4

Therefore, for all the columns y is four times x.

∴ The constant of variation in the given table is x/y = 1/4.

Question 7. Rohit bought 12 registers for Rs 156, find the cost of 7 such registers.

Solution:

Cost of 12 registers = Rs 156

Cost of 1 register = Rs 156/12 

                              = Rs 13 (Taking 12 to RHS)

Cost of 7 registers = Cost of 1 register * number of req. registers 

=> Rs 13 × 7

=> Rs 91

Therefore, 7 registers cost Rs 91. 

Question 8. Anupama takes 125 minutes in walking a distance of 100 meter. What distance would she cover in 315 minutes?

Solution:

Distance covered in 125 minutes = 100 meter

Distance covered in 1 minute = 100 m / 125 min 

Now distance covered in 315 minutes = Distance covered in 1 minute x 315 minutes 

=> distance covered in 315 minutes = (100/125) x 315 

Solving the equation, 

=> distance covered in 315 minutes = (31500/125) =252

∴ The distance covered in 315min is 252 meters.

Question 9. If the cost of 93 m of a certain kind of plastic sheet Rs 1395, then what would it cost to buy 105 m of such plastic sheet?

Solution:

Cost of 93m plastic sheet = Rs 1395 

Cost of 1m plastic sheet = Rs 1395/93

Cost of 105m plastic sheet = Cost of 1m plastic sheet x required length of plastic sheet

=> Cost of 105m plastic sheet = Rs (1395/93) x 105

=> Cost of 105m plastic sheet = 146475/93

= Rs 1575

∴ The cost of 105m plastic sheet is Rs 1575

Question 10. Suneeta types 1080 words in one hour. What is her GWAM (gross words a minute rate)?

Solution:

No of words typed in one hour (60 minutes) = 1080 (because 1 hour = 60 min)

No of words typed in 1 minute = 1080 /60 

Solving, we get

=>No of words typed in 1 minute =18

∴ Number of words typed in one minute is 18

Summary

These practice questions cover various applications of direct variation. They include problems related to cost and quantity, speed and distance, time and work, and more. Solving these problems requires students to recognize the direct proportional relationship between quantities and apply the concept to find unknown values.