Class 8 RD Sharma – Chapter 10 Direct And Inverse Variations – Exercise 10.2 | Set 1
Question 1. In which of the following tables x and y vary inversely:
i)
x | 4 | 3 | 12 | 1 |
y | 6 | 8 | 2 | 24 |
Solution:
In order to show x and y vary inversely,
x and y should be equal to Constant i.e x × y = Constant
So, 4 × 6 = 24
3 × 8 = 24
12 × 2 = 24
1 × 24 = 24
Since the product of xy in all cases is 24 (i.e Constant)
Hence, in this question, x and y vary inversely
ii)
x | 5 | 20 | 10 | 4 |
y | 20 | 5 | 10 | 25 |
Solution:
In order to show x and y vary inversely,
x and y should be equal to Constant i.e x × y = Constant
So, 5 × 20 = 100
20 × 5 = 100
10 × 10 = 100
4 × 25 = 100
Since the product of xy in all cases is 100 (i.e Constant)
Hence, in this question, x and y vary inversely
iii)
x | 4 | 3 | 6 | 1 |
y | 9 | 12 | 8 | 36 |
Solution:
In order to show x and y vary inversely,
x and y should be equal to Constant i.e x × y = Constant
So, 4 × 9 = 36
3 × 12 = 36
6 × 8 = 48
1 × 36 = 36
Since the product of xy in all cases is not Constant.
Hence, in this question, x and y don’t vary inversely
iv)
x | 9 | 24 | 15 | 3 |
y | 8 | 3 | 4 | 25 |
Solution:
In order to show x and y vary inversely,
x and y should be equal to Constant i.e x × y = Constant
So, 9 × 8 = 72
24 × 3 = 72
15 × 4 = 60
3 × 25 = 75
Since the product of xy in all cases is not Constant.
Hence, in this question, x and y don’t vary inversely
Question 2. It x and y vary inversely, fill in the following blanks:
i)
x | 12 | 16 | ….. | 8 | ….. |
y | ….. | 6 | 4 | …… | 0.25 |
Solution:
Since it is given that x and y vary inversely,
So, x × y = Constant
x | 12 | 16 | b | 8 | d |
y | a | 6 | 4 | c | 0.25 |
Finding the value of ‘a’
12 × a = 16 × 6
=> a = 8
Finding the value of ‘b’
16 × 6 = b × 4
=> b = 24
Finding the value of ‘c’
24 × 4= 8 × c
=> c = 12
Finding the value of ‘d’
8 × 12 = d × 0.25
=> d = 384
Hence, the values of a, b, c, and d are 8, 24, 12, and 384.
ii)
x | 16 | 32 | 8 | 128 |
y | 4 | ….. | ….. | 0.25 |
Solution:
Since it is given that x and y vary inversely,
So, x × y = Constant
x | 16 | 32 | 8 | 128 |
y | 4 | a | b | 0.25 |
Finding the value of ‘a’
16 × 4 = 32 × a
=> a = 2
Finding the value of ‘b’
32 × 2 = 8 × b
=> b = 8
Hence, the values of a and b are 2 and 8.
iii)
x | 9 | ….. | 81 | 243 |
y | 27 | 9 | ….. | 1 |
Solution:
Since it is given that x and y vary inversely,
So, x × y = Constant
x | 9 | a | 81 | 243 |
y | 27 | 9 | b | 1 |
Finding the value of ‘a’
9 × 27 = a × 9
=> a = 27
Finding the value of ‘b’
27 × 9 = 81 × b
=> b = 3
Hence, the values of a and b are 27 and 3.
Question 3. Which of the following quantities vary inversely to each other?
(i) The number of x men hired to construct a wall and the time y took to finish the job.
(ii) The length x of a journey by bus and price y of the ticket.
(iii) Journey (x km) undertaken by a car and the petrol (y literes) consumed by it.
Solution:
i) If more men will be hired then the time taken will be less and if fewer men will be hired, the time taken to finish work will be more. So, here x and y vary inversely.
ii) If the length of the journey is more than the price of the ticket will also be more and vice-versa. So, here x and y vary directly.
iii) If the length of the journey is more than the consumption of petrol will also be more and vice-versa. So, here x and y vary directly.
Question 4. It is known that for a given mass of gas, the volume v varies inversely as the pressure p. Fill in the missing entries in the following table:
x | ….. | 48 | 60 | ….. | 100 | …… | 200 |
y | 2 | ….. | 3/2 | 1 | …… | 1/2 | …… |
Solution:
Since it is given that volume ‘v’ and pressure ‘p’ vary inversely,
So, v × p = Constant
x | a | 48 | 60 | c | 100 | e | 200 |
y | 2 | b | 3/2 | 1 | d | 1/2 | f |
Finding the value of ‘a’
a × 2 = 60 × 3/2
=> a = 45
Finding the value of ‘b’
48 × b = 60 × 3/2
=> b = 15/8
Finding the value of ‘c’
60 × 3/2 = c × 1
=> c = 90
Finding the value of ‘d’
90 × 1 = 100 × d
=> d = 0.9
Finding the value of ‘e’
100 × 0.9 = e × 1/2
=> e = 180
Finding the value of ‘f’
180 × 1/2 = 200 × f
=> f = 9/20
Hence, the values of a, b, c, d, e and f are 45, 15/8, 90, 0.9, 180 and 9/20.
Question 5. If 36 men can do a piece of work in 25 days, in how many days will 15 men do it?
Solution:
As we know if the number of men will increase, they will take fewer days. So, the number of men and days vary inversely. Let 15 men can finish the work in a days
Number of Men | 36 | 15 |
Days to finish work | 25 | a |
So, 36 × 25 = 15 × a
=> a = 60 days
Hence, 15 men can finish the work in 60 days
Question 6. A workforce of 50 men with a contractor can finish a piece of work in 5 months. In how many months the same work can be completed by 125 men?
Solution:
As we know if the number of men will increase, they can finish the work in fewer months. So, the number of men and months vary inversely. Let 125 men can finish the work in a months
Number of Men | 50 | 125 |
Months to finish work | 5 | a |
So, 50 × 5 = 125 × a
=> a = 2 months
Hence, 125 men can finish the work in 2 months
Question 7. A work-force of 420 men with a contractor can finish a certain piece of work in 9 months. How many extra men must he employ to complete the job in 7 months?
Solution:
As we know if the number of men will increase, they can finish the work in fewer months. So, the number of men and months vary inversely. Let a man can finish the work in 7 months
Number of Men | 420 | a |
Months to finish work | 9 | 7 |
So, 420 × 9 = a × 7
=> a = 540 men
Total men = 540
Number of men already working = 420
So, number of extra men = 540 – 420 = 120 men
Hence, 120 extra men can finish work in 7 months.
Question 8. 1200 men can finish a stock of food in 35 days. How many more men should join them so that the same stock may last for 25 days?
Solution:
As we know if the number of men will increase, they can finish the stock of food in fewer days. So, the number of men and the stock of food vary inversely. Let a man can finish the stock of food in 25 days
Number of Men | 1200 | a |
Days to finish stock of food | 35 | 25 |
So, 1200 × 35 = a × 25
=> a = 1680 men
Total men = 1680
Number of men already present = 1200
So, number of more men required = 1680 – 1200 = 480 men
Hence, 480 more men should join to finish the stock of food within 25 days.
Question 9. In a hostel of 50 girls, there are food provisions for 40 days. If 30 more girls join the hostel, how long will these provisions last?
Solution:
As we know if the number of girls will increase, they can finish the food provisions in fewer days. So, the number of girls and food provisions vary inversely. Let the food provisions will end in a days
Number of Girls | 50 | 80 |
Days to finish food provision | 40 | a |
So, 50 × 40 = 80 × a
=> a = 25 days
Hence, if 30 more girls join the hostel the food provision will end in 25 days
Chapter 10 Direct And Inverse Variations – Exercise 10.2 | Set 2
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