Chapter 7 - Variation
Practice set 7.1
1. Write the following statements using the symbol of variation.
(1) Circumference (c) of a circle is directly proportional to its radius (r).
Solution:
c ∝ r
(2) Consumption of petrol (l) in a car and distance travelled by that car (d) are in direct variation.
Solution:
l ∝ d
2. Complete the following table considering that the cost of apples and their number are in direct variation.
Solution:
x ∝ y …[k is constant of variation]
∴ x = ky
When x = 1 then y = 8 …[Given]
∴ 1 = k × 8
∴ k = \(\large \frac {1}{8}\)
The equation of variation is x = \(\large \frac {1}{8}\) y
(i) When y = 56 then x = ?
Substituting y = 56 in equation of variation
x = \(\large \frac {1}{8}\) × 56
∴ x = 7
(ii) When x = 12 then y = ?
Substituting x = 12 in equation of variation
∴ 12 = \(\large \frac {1}{8}\) × y
∴ 12 × 8 = y
∴ y = 96
(iii) When y = 160 then x = ?
Substituting y = 160 in equation of variation.
∴ x = \(\large \frac {1}{8}\) × 160
∴ x = 20
3. If m ∝ n and when m = 154, n = 7. Find the value of m, when n = 14
Solution:
m ∝ n …[Given]
∴ m = kn …[k is constant of variation]
Substituting m = 154 and n = 7 we get,
154 = k × 7
∴ k = \(\large \frac {154}{7}\)
∴ k = 22
The equation of variation is m = 22 × n
Substituting n = 14 in equation of variation, we get,
m = 22 × 14
∴ m = 308
Ans: The value of m is 308
4. If n varies directly as m, complete the following table.
Solution:
n ∝ m …[Given]
∴ n = km …[k is constant of variation]
Substituting m = 3 and n = 12 we get,
12 = k × 3
\(\large \frac {12}{3}\) = k
∴ k = 4
∴ The equation of variation is n = 4 × m
(i) If m = 6.5 then n = ?
Substituting m = 6.5 in equation of variation
n = 4 × 6.5
∴ n = 26
(ii) If n = 28 then m = ?
Substituting n = 28 in equation of variation.
28 = 4 m
∴ \(\large \frac {28}{4}\) = m
∴ m = 7
(iii) If m = 1.25 then n = ?
Substituting m = 1.25 in equation of variation
n = 4 × 1.25
∴ n = 5
5. y varies directly as square root of x. When x = 16, y = 24. Find the constant of variation and equation of variation.
Solution:
y varies directly as square root of x …[Given]
y ∝ \(\sqrt{x}\)
y = k \(\sqrt{x}\) …[k is constant of variation]
If x = 16 then y = 24 …[Given]
∴ 24 = k × \(\sqrt{16}\)
∴ 24 = k × 4
∴ \(\large \frac {24}{4}\) = k
∴ k = 6
Constant of variation is 6
Equation of variation is y = 6 \(\sqrt{x}\)
Ans: The constant of variation is 6 and the equation of variation is y = 6 \(\sqrt{x}\)
6. The total remuneration paid to labourers employed to harvest soyabeen is in direct variation with the number of labourers. If the remuneration of 4 labourers is ₹ 1000, find the remuneration of 17 labourers.
Solution:
Let the total remuneration paid be m and the number of labourers be n.
m ∝ n …[Given]
∴ m = kn …[k is constant of variation]
If n = 4 then m = 1000 …[Given]
∴ 1000 = k × 4
∴ \(\large \frac {1000}{4}\) = k
∴ k = 250
∴ The equation of variation is m = 250 × n
If n = 17 then m = ?
Substituting the value of n in equation of variation.
m = 250 × 17
∴ m = 4250
Ans: The remuneration paid to 17 labourers is ₹4250.
Practice set 7.2
1. The information about numbers of workers and number of days to complete a work is given in the following table. Complete the table.
Solution:
More the number of workers then less days are required to complete the work.
So, the number of workers and days vary in inverse proportion.
Let the number of workers be m and the number of days be n.
m ∝ \(\large \frac {1}{n}\)
∴ m × n = k …[k is constant of variation]
When m = 30 then n = 6 …[Given]
30 × 6 = k
∴ k = 180
The equation of variation is m × n = 180
(i) If n = 12 then m = ?
Substituting n = 12 in equation of variation.
m × 12 = 180
∴ m = \(\large \frac {180}{12}\)
∴ m = 15
(ii) If m = 10 then n = ?
Substituting m = 10 in equation of variation.
10n = 180
∴ n = \(\large \frac {180}{10}\)
∴ n = 18
(iii) If n = 36 then m = ?
Substituting n = 36 in equation of variation
m × 36 = 180
∴ m = \(\large \frac {180}{36}\)
∴ m = 5
2. Find constant of variation and write equation of variation for every example given below.
(1) p ∝ \(\large \frac {1}{q}\) ; if p = 15 then q = 4
Solution:
p ∝ \(\large \frac {1}{q}\)
∴ p × q = k …[k is constant of variation]
Substituting p = 15 and q = 4
15 × 4 = k
k = 60
Constant of variation = 60
Equation of variation = p × q = 60
Ans: The constant of variation is 60 and the equation of variation is p × q = 60.
(2) z ∝ \(\large \frac {1}{w}\) ; when z = 2.5 then w = 24
Solution:
z ∝ \(\large \frac {1}{w}\)
∴ z × w = k …[k is constant of variation]
Substituting z = 2.5 and w = 24, we get,
2.5 × 24 = k
∴ k = 60
Constant of variation = 60
The equation of variation is z × w = 60
Ans: The constant of variation is 60 and the equation of variation is z × w = 60.
(3) s ∝ \(\large \frac {1}{t²}\) ; if s = 4 then t = 5
Solution:
s ∝ \(\large \frac {1}{t²}\)
s × t² = k …[k is constant of variation]
Substituting s = 4 and t = 5
4 × 5² = k
∴ k = 4 × 25
∴ k = 100
Constant of variation = 100
Equation of variation is s × t² = 100
Ans: The constant of variation is 100 and the equation of variation is s × t² = 100.
(4) x ∝ \(\large \frac {1}{\sqrt{2}}\) ; if x = 15 then y = 9
Solution:
x ∝ \(\large \frac {1}{\sqrt{2}}\)
∴ x × \(\sqrt{y}\) = k …[k is constant of variation]
Substituting x = 15 and y = 9
15 × \(\sqrt{9}\) = k
∴ k = 15 × 3
∴ k = 45
Constant of variation = 45
Equation of variation is x × \(\sqrt{y}\) = 45
Ans: The constant of variation is 45 and the equation of variation is x × \(\sqrt{y}\) = 45.
3. The boxes are to be filled with apples in a heap. If 24 apples are put in a box then 27 boxes are needed. If 36 apples are filled in a box how many boxes will be needed ?
Solution:
Let the number of apples filled in each box be x and number of boxes be y.
More the apples filled in each box, less number of boxes will be required.
So, this is an example of inverse variation.
x ∝ \(\large \frac {1}{y}\)
∴ xy = k …[k is constant of variation]
Substituting x = 24 and y = 27
24 × 27 = k
k = 648
The equation of variation is x × y = 648
Substituting x = 36 in the equation of variation.
36 × y = 648
∴ y = \(\large \frac {648}{36}\)
∴ y = 18
Ans: 18 boxes will be needed.
4. Write the following statements using symbol of variation .
(1) The wavelength of sound (l) and its frequency (f) are in inverse variation.
Ans:
l ∝ \(\large \frac {1}{f}\)
(2) The intensity (I) of light varies inversely with the square of the distance (d) of a screen from the lamp.
Ans:
I ∝ \(\large \frac {1}{d²}\)
5. x ∝ \(\large \frac {1}{y}\) and when x = 40 then y = 16. If x = 10, find y.
Solution:
x ∝ \(\large \frac {1}{y}\)
∴ x × \(\sqrt{y}\) = k …[k is constant of variation]
Substituting x = 40 and y = 16, we get,
40 \(\sqrt{16}\) = k
∴ k = 40 * 4
∴ k = 160
∴ The equation of variation is x × \(\sqrt{y}\) = 160
Substituting x = 10 in equation of variation, we get,
10 × \(\sqrt{y}\) = 160
∴ \(\sqrt{y}\) = \(\large \frac {160}{10}\)
∴ \(\sqrt{y}\) = 16
∴ \(\sqrt{y²}\) = 16²
∴ y = 256
Ans: The value of y is 256.
6. x varies inversely as y, when x = 15 then y = 10, if x = 20 then y = ?
Solution:
x varies inversely as y …[Given]
x ∝ \(\large \frac {1}{y}\)
∴ x × \(\sqrt{y}\) = k …[k is constant of variation]
Substituting x = 15 and y = 10, we get,
15 × 10 = k
∴ k = 150
The equation of variation is x × y = 150
Substituting x = 20 in equation of variation, we get,
20 × y = 150
y = \(\large \frac {150}{20}\)
∴ y = 7.5
Ans: The value of y is 7.5
Practice set 7.3
1. Which of the following statements are of inverse variation?
(1) Number of workers on a job and time taken by them to complete the job.
Solution:
This is an example of inverse variation.
(2) Number of pipes of same size to fill a tank and the time taken by them to fill the tank.
Solution:
This is an example of inverse variation.
(3) Petrol filled in the tank of a vehicle and its cost
Solution:
This is an example of direct variation.
(4) Area of circle and its radius.
Solution:
This is an example of direct variation.
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 hours be t and number of workers be n.
Number of workers = n = 15
Time = t = 48 hours
The number of hours to build a wall is inversely proportional to the number of workers.
∴ t ∝ \(\large \frac {1}{n}\)
∴ tn = k …[k is constant of variation]
Substituting t = 48 and n = 15, we get
48 × 15 = k
∴ k = 720
∴ The equation of variation is t × n = 720
Substituting t = 30 in equation of variation, we get,
30n = 720
∴ n = \(\large \frac {720}{30}\)
∴ n = 24
Ans: 24 workers can build the wall in 30 hours.
3. 120 bags of half litre milk can be filled by a machine within 3 minutes. Find the time to fill such 1800 bags ?
Solution:
Let the number of bags be n and time required to fill them be t.
Number of bags = n = 120
Time = t = 3 hours
More the number of bags to be filled by machine, more time is required.
So, this is an example of direct variation.
∴ n ∝ t
∴ n = kt …[k is constant of variation]
Substituting n = 120 and t = 3, we get,
120 = k × 3
∴ k = \(\large \frac {120}{3}\)
∴ k = 40
The equation of variation is n = 40 × t
Substituting n = 1800 in equation of variation, we get,
1800 = 40t
∴ t = \(\large \frac {1800}{40}\)
∴ t = 45
Ans: 1800 bags of milk can be filled by the machine in 45 minutes.
4. A car with speed 60 km/hr takes 8 hours to travel some distance. What should be the increase in the speed if the same distance is to be covered in 7 \(\large \frac {1}{2}\) hours?
Solution:
Let the speed of the car be s and time required to cover some distance be t.
There is an inverse variation between speed and time.
∴ s ∝ \(\large \frac {1}{t}\)
∴ st = k …[k is constant of variation]
Substituting s = 60 and t = 480 …[1 hour = 60 minutes]
∴ 60 × 480 = k
∴ k = 28800
t = 7 \(\large \frac {1}{2}\) hours
∴ t = 7 × 60 + 30
∴ t = 450 minutes
The equation of variation is s × t = 28800
Substituting t = 450 minutes in equation of variation, we get,
s × 450 = 28800
∴ s = \(\large \frac {28800}{450}\)
∴ s = 64
Hence, if speed of car is 64 km/hr then the same distance can be covered in 7 \(\large \frac {1}{2}\) hours
∴ Increase in speed = 64 – 60
∴ Increase in speed = 4 km/hrs.
Ans: Speed of the car should be increased by 4 km/hr.