Variation And Proportion
Class 8th Mathematics (old) MHB Solution
Exercise 32
Question 1.In which of the following sets are the numbers in proportion?
(1) 28, 16, 21, 12
(2) 12, 32, 15, 35
(3) 39, 26, 57, 38
Answer:
Numbers a, b, c, d are said to be in proportion if
(1) Here,
a = 28, b = 16, c = 21, d = 12
On dividing numerator and denominator by 4 gives,
Now,
On dividing numerator and denominator by 3 gives,
From (i) and (ii)
∴ 28, 16, 21, 12 are in proportion
2) Here,
a = 12, b = 32, c = 15, d = 35
On dividing numerator and denominator by 4 gives,
Now,
On dividing numerator and denominator by 5 gives,
From (i) and (ii)
∴ 12, 32, 15, 35 are not in proportion
(3) Here,
a = 39, b = 26, c = 57, d = 38
On dividing numerator and denominator by 13 gives,
Now,
On dividing numerator and denominator by 19 gives,
From (i) and (ii)
∴ 39, 26, 57, 38 are in proportion
Question 2.
If the numbers in each of the following groups are in proportion, what is the value of the variable in each group?
(1) 6, x, 10, 15
(2) p, 9, 8, 6
(3) 25, 35, 35,m
Answer:
(1) 6, x, 10, 15 in proportion which means 
cross multiplying, we get
⇒ 6 × 15 = x × 10
⇒ 90 = 10x
∴ x = 9
(2) p, 9, 8, 6 in proportion which means 
cross multiplying, we get
⇒ p × 6 = 9 × 8
⇒ 6p = 72
∴ p = 12
(3) 25, 35, 35, m in proportion which means 
cross multiplying, we get
⇒ 25 × m = 35 × 35
⇒ m = 
= 
= 7 × 7
∴ m = 49
Question 3.
There is direct variation between the sale of books and the commission received for the sale. The table below gives the values of some sales and some commissions. Write the remaining values and complete the table.
Answer:
Let the missing values be x, y, z as shown
Let ‘a’ be a value in Sale and ‘b’ be a value in Commission
As there is direct variation between sale and communication
⇒ a ∝ b
⇒ a = kb
⇒ 
= k(constant)
When sale is 50 commission is 10 i.e. a = 50 and b = 10
⇒ = 
= 5 = k
∴ k = 5
For all value of sale and commission the constant k does not change since it is direct variation therefore
= 
= 
= k = 5
Consider = 5
∴ x = 50
Consider = 5
⇒ y = 125 × 5
∴ y = 625
Consider = 5
∴ z = 165
The table becomes
Exercise 33
Question 1.In each example, find the constant of proportionality and write the equation of variation.
(1) The quantity y varies directly as x. When the value of y is 20, the value of x is 4.
(2) p α q. When p is 12, the value of q is 18.
(3) c α d. When c = 28, d = 21.
(4) m α n. When m = 7.5, n = 10.
Answer:
(1) given y ∝ x
Constant of proportionality = 
When y is 20 x is 4 therefore
= 
= 5 …(i)
Therefore, constant of proportionality = 5
Equation of variation is given by taking the denominator ‘x’ in equation (i) from L.H.S to R.H.S i.e. y = 5x
(2) given p ∝ q
Constant of proportionality = 
When p is 12 q is 18 therefore
= 
= 
…(i)
Therefore, constant of proportionality =
Equation of variation is given by taking the denominator ‘q’ in equation (i) from L.H.S to R.H.S
i.e. p = q
(3) given c ∝ d
Constant of proportionality = 
When c is 28 d is 21 therefore
= 
= 
…(i)
Therefore, constant of proportionality =
Equation of variation is given by taking the denominator ‘d’ in equation (i) from L.H.S to R.H.S
i.e. c = d
(4) given m ∝ n
Constant of proportionality = 
When m is 7.5 n is 10 therefore
= 
= 0.75 …(i)
Therefore, constant of proportionality = 0.75
Equation of variation is given by taking the denominator ‘n’ in equation (i) from L.H.S to R.H.S i.e. m = 0.75n
Exercise 34
Question 1.Complete the following table showing an instance of inverse proportion. The table shows the number of workers carrying out a task and the number of days they take to do it.
Answer:
Let the missing values be x, y, z as shown
Let ‘a’ be some value in number of workers and ‘b’ be some value in days
Given a ∝ 
⇒ a = k × ⇒ ab = k
When number of workers = 30 i.e. a = 30
days require = 4 i.e. b = 4
using k = ab
⇒ k = 30 × 4
∴ k = 120
As by given it is inverse proportion
⇒ 8x = 10y = 24z = 120
Consider 8x = 120
⇒ x = 
= 15
∴ x = 15
Consider 10y = 120
⇒ y = 
= 12
∴ y = 12
Consider 24z = 120
⇒ z = 
= 5
∴ z = 5
The table becomes
Exercise 35
Question 1.In each example, find the constant of proportionality and write the equation of variation.
(1) 
When y = 15, x = 14
(2) 
When w = 24, z = 2.5
(3) 
When t = 5.5, s = 4.4
Answer:
(1) y ∝ 
⇒ y = k × ⇒ xy = k(constant of proportionality)
When y = 15, x = 14
⇒ k = xy = 15 × 14
∴ k = 210
Constant of proportionality = 210
Equation of variation is xy = 210
(2) z ∝ 
⇒ z = k × ⇒ zw = k(constant of proportionality)
When z = 2.5, w = 24
⇒ k = zw = 2.5 × 24
∴ k = 60
Constant of proportionality = 60
Equation of variation is zw = 60
(3) s ∝ 
⇒ s = k × ⇒ st = k(constant of proportionality)
When s = 4.4, t = 5.5
⇒ k = st = 4.4 × 5.5
∴ k = 24.2
Constant of proportionality = 24.2
Equation of variation is st = 24.2
Question 2.
Sixty crates, each of which can hold 36 mangoes, are required to pack a certain number of mangoes. How many crates, each of which can hold 48 mangoes, will be required for the same number of mangoes? What kind of proportion is there between the number of crates and the number of mangoes in each crate?
Answer:
Let number of mangoes be ‘n’
1 crate of hold 36 mangoes
Therefore, 60 crates can hold 60×36 = 2160 mangoes
n = 2160
now 1 crate holds 48 mangoes
Let number of crates required be ‘c’
number of mangoes are same which is n = 2160
Therefore, c × 48 = 2160
⇒ c = 
= 
= 45
45 crates are required
From table 60×36 = 45×48 = 2160
If we consider x as a value in number of crates and y as a value in number of mangoes in each crate we can write xy = 2160
⇒ xy = k
⇒ y = k × ⇒ y ∝
Hence there is inverse proportion between number of crates and number of mangoes in each crate
In this question we can directly say that it is inverse proportion because it is obvious that when the holding capacity increases the number of containers reduces.