Selina Solutions Concise Mathematics Class 6 Chapter 12 provides the basic idea about the concepts, which are vital from the exam point of view. In-depth practice of textbook questions, help students to gain interest in Mathematics, along with improving their analytical thinking. The various examples present in the textbook help students to understand the method of solving the difficult questions effortlessly. Download the free PDF of Selina Solutions Concise Mathematics Class 6 Chapter 12 Proportion, from the links given below
Class 12 has problems on the main concept Proportions. If the two ratios are equal, then they are said to be in Proportion. Solutions provided in simple language help students to gain a better hold on the fundamental concepts. This also helps them to perform well in the annual examination.
Selina Solutions Concise Mathematics Class 6 Chapter 12: Proportion Download PDF
Exercises of Selina Solutions Concise Mathematics Class 6 Chapter 12: Proportion
Access Selina Solutions Concise Mathematics Class 6 Chapter 12: Proportion
Exercise 12(A)
1. In each of the following, check whether or not the given ratios form a proportion:
(i) 8: 16 and 12: 15
(ii) 16: 28 and 24: 42
(iii) 12 ÷ 3 and 8 ÷ 2
(iv) 25: 40 and 20: 32
(v) 15 / 18 and 10 / 12
Solution:
(i) 8: 16 and 12: 15
The above expression can be written as follows:
8: 16 = 8 / 16
= 1 / 2 and
12: 15 = 12 / 15
= 4 / 5
Since 8: 16 ≠12: 15
Therefore they are not in proportion
(ii) 16: 28 and 24: 42
The above expression can be written as follows:
16: 28 = 16 / 28
= 4 / 7 and
24: 42 = 24 / 42
= 4 / 7
Since 16: 28 = 24: 42
Therefore they form a proportion
(iii) 12 ÷ 3 and 8 ÷ 2
The above expression can be written as follows:
12 ÷ 3 = 12 / 3
= 4 and
8 ÷ 2 = 8 / 2
= 4
Since 12: 3 = 8: 2
Therefore they form a proportion
(iv) 25: 40 and 20: 32
The above expression can be written as follows:
25: 40 = 25 / 40
= 5 / 8 and
20: 32 = 20 / 32
= 5 / 8
Since 25: 40 = 20: 32
Therefore they form a proportion
(v) 15 / 18 and 10 / 12
15 / 18 = 5 / 6 and
10 / 12 = 5 / 6
Since 15 / 18 = 10 / 12
Therefore they form a proportion
2. Find the value of x in each of the following proportions:
(i) x: 4 = 6: 8
(ii) 14: x = 7: 9
(iii) 4: 6 = x: 18
(iv) 8: 10 = x: 25
(v) 5: 15 = 4: x
Solution:
(i) x: 4 = 6: 8
The given proportion can be calculated for the value of x as follows:
x: 4 = 6: 8
x / 4 = 6 / 8
⇒ x × 8 = 4 × 6
⇒ x = (4 × 6) / 8
⇒ x = 3
Therefore the value of x is 3
(ii) 14: x = 7: 9
The given proportion can be calculated for the value of x as follows:
14: x = 7: 9
14 / x = 7 / 9
⇒ x × 7 = 14 × 9
⇒ x = (14 × 9) / 7
⇒ x = 18
Therefore the value of x is 18
(iii) 4: 6 = x: 18
The given proportion can be calculated for the value of x as follows:
4 / 6 = x / 18
⇒ x × 6 = 4 × 18
⇒ x = (4 × 18) / 6
⇒ x = 12
Therefore the value of x is 12
(iv) 8: 10 = x: 25
The given proportion can be calculated for the value of x as follows:
8 / 10 = x / 25
⇒ 10 × x = 25 × 8
⇒ x = (25 × 8) / 10
⇒ x = 20
Therefore the value of x is 20
(v) 5: 15 = 4: x
The given proportion can be calculated for the value of x as follows:
5 / 15 = 4 / x
⇒ 5 × x = 15 × 4
⇒ x = (15 × 4) / 5
⇒ x = 12
Therefore the value of x is 12
3. Find the value of x so that the given four numbers are in proportion:
(i) x, 6, 10 and 15
(ii) x, 4, 15 and 30
(iii) 2, x, 10 and 25
(iv) 4, x, 6 and 18
(v) 9, 12, x and 8
Solution:
(i) x, 6, 10 and 15
The given proportion can be calculated for the value of x as follows:
x: 6: 10: 15
⇒ x ×15 = 6 × 10
⇒ x = (6 × 10) / 15
⇒ x = 60 / 15
⇒ x = 4
Therefore the value of x is 4
(ii) x, 4, 15 and 30
The given proportion can be calculated for the value of x as follows:
x: 4: 15: 30
⇒ x × 30 = 4 × 15
⇒ x = (4 × 15) / 30
⇒ x = 60 / 30
⇒ x = 2
Therefore the value of x is 2
(iii) 2, x, 10 and 25
The given proportion can be calculated for the value of x as follows:
2: x: 10: 25
⇒ x × 10 = 2 × 25
⇒ x = (2 × 25) / 10
⇒ x = 50 / 10
⇒ x = 5
Therefore the value of x is 5
(iv) 4, x, 6 and 18
The given proportion can be calculated for the value of x as follows:
4: x: 6: 18
⇒ x × 6 = 18 × 4
⇒ x = (18 × 4) / 6
⇒ x= 72 / 6
⇒ x = 12
Therefore the value of x is 12
(v) 9, 12, x and 8
The given proportion can be calculated for the value of x as follows:
9: 12: x: 8
⇒ 12 × x = 9 × 8
⇒ x = (9 × 8) / 12
⇒ x = 72 / 12
⇒ x = 6
Therefore the value of x is 6
4. The first, second and the fourth terms of a proportion are 6, 18 and 75, respectively. Find its third term
Solution:
Given
First term = 6
Second term = 18
Fourth term = 75
Third term =?
Let the third term be x
6: 18: x: 75
⇒ x × 18 = 6 × 75
⇒ x = (6 × 75) / 18
⇒ x = 25
Therefore the value of third term is 25
5. Find the second term of the proportion whose first, third and fourth terms are 9, 8 and 24 respectively.
Solution:
Given
First term = 9
Third term = 8
Fourth term = 24
Second term =?
Let the second term be x
9: x: 8: 24
⇒ x × 8 = 9 × 24
⇒ x = (9 × 24) / 8
⇒ x = 216 / 8
⇒ x = 27
Therefore the value of x is 27
6. Find the fourth term of the proportion whose first, second and third terms are 18, 27 and 32 respectively.
Solution:
Given
First term = 18
Second term = 27
Third term = 32
Fourth term =?
Let the fourth term be x
18: 27: 32: x
⇒ x × 18 = 32 × 27
⇒ x = (32 × 27) / 18
⇒ x = 48
Therefore the value of x is 48
7. The ratio of the length and the width of a school ground is 5: 2. Find the length, if the width is 40 metres.
Solution:
Given
The ratio of the length and the width of a school ground = 5: 2
The width of the school ground = 40 metre
Let the length of the school ground be x metre
Hence the length of the ground can be calculated as follows:
Ratio of length to width of a school ground = x: 40
According to the given statement
5: 2 = x: 40
⇒ 2 × x = 40 × 5
⇒ x = (40 × 5) / 2
⇒ x = 200 / 2
⇒ x = 100 m
Therefore the length of the school ground is 100 m
8. The ratio of the sale of eggs on a Sunday and that of the whole week at a grocery shop was 2: 9. If the total value of the sale of eggs in the same week was Rs 360, find the value of the sale of eggs that Sunday.
Solution:
Given
Ratio of sale of eggs on a Sunday and whole week at a grocery shop = 2: 9
Total sale of eggs in the same week = Rs 360
Let the sale of eggs on Sunday be x
Hence the sale of eggs on Sunday can be calculated as follows:
2: 9 = x: 360
⇒ 9 × x = 360 × 2
⇒ x = (360 × 2) / 9
⇒ x = 720 / 9
⇒ x = 80
Therefore the value of the sale of eggs on Sunday is of Rs 80
9. The ratio of copper and zinc in an alloy is 9: 8. If the weight of zinc, in the alloy, is 9.6 kg, find the weight of copper in the alloy.
Solution:
Given
Ratio of copper and zinc in an alloy = 9: 8
Weight of zinc in an alloy = 9.6 kg
Let x kg be the weight of copper in the alloy
Hence the weight of copper can be calculated as below
9: 8 = x: 9.6
⇒ 8 × x = 9 × 9.6
⇒ x = (9 × 9. 6) / 8
⇒ x = 86.4 / 8
⇒ x = 10.8
Therefore the weight of copper in the alloy is 10.8 kg
10. The ratio of the number of girls to the number of boys in a school is 2: 5. If the number of boys is 225; find:
(i) the number of girls in the school
(ii) the number of students in the school.
Solution:
Given
Ratio of girls to the boys in a school = 2.5
Number of boys in a school = 225
(i) Let x be the number of girls in a school
Hence number of girls in a school can be calculated as follows:
2: 5 = x: 225
⇒ 5 × x = 2 × 225
⇒ x = (2 × 225) / 5
⇒ x = 450 / 5
⇒ x = 90
Therefore the number of girls in the school is 90
(ii) Total number of students in a school becomes
Total student = Total boys + Total girls
= 225 + 90
= 315
Therefore total number of students in the school is 315
11. In a class 1 out of every 5 students pass. If there are 225 students in all the sections of a class, find how many pass?
Solution:
Given
Total number of students in all the sections of a class = 225
And 1 out of every 5 students pass
So, total number of pass students can be calculated as follows:
Total student pass = 225 × 1 / 5
= 45
Therefore total pass students are 45
12.Make set of all possible proportions from the numbers 15, 18, 35 and 42
Solution:
Given
Numbers are 15, 18, 35 and 42
Hence the possible proportions are as follows:
(i) 15: 18:: 35: 42
(ii) 15: 35:: 18: 42
(iii) 42: 18:: 35: 15
(iv) 42: 35:: 18: 15
Exercise 12(B)
1. If x, y and z are in continued proportion, then which of the following is true:
(i) x: y = x: z
(ii) x: x = z: y
(iii) x: y = y: z
(iv) y: x = y: z
Solution:
Here option (iii) is in continued proportion i.e. x: y = y: z
2. Which of the following numbers are in continued proportion:
(i) 3, 6 and 15
(ii) 15, 45 and 48
(iii) 6, 12 and 24
(iv) 12, 18 and 27
Solution:
Here options (iii) and (iv) are in continued proportion
(iii) 6 / 12 = 12 / 24
1 / 2 = 1 / 2
(iv) 12 / 18 = 18 / 27
2 / 3 = 2 / 3
Therefore (iii) and (iv) are in continued proportion
3. Find the mean proportion between
(i) 3 and 27
(ii) 0.06 and 0.96
Solution:
(i) 3 and 27
The mean proportion between 3 and 27 can be calculated as below
Mean proportion = √ 3 × 27
= √ 81
= 9
Therefore mean proportion between 3 and 27 is 9
(ii) 0.06 and 0.96
The mean proportion between 0.06 and 0.96 can be calculated as below
Mean proportion = √ 0.06 × 0.96
= √.0576
= 0.24
Therefore the mean proportion between 0.06 and 0.96 is 0.24
4. Find the third proportional to:
(i) 36, 18
(ii) 5.25, 7
(iii) Rs 1. 60, Rs 0.40
Solution:
(i) 36, 18
Let the required third proportional be x
Hence, 36, 18, x are in continued proportion
⇒ 36: 18 = 18: x
⇒ 36x = 18 × 18
⇒ x = (18 × 18) / 36
⇒ x = 324 / 36
⇒ x = 9
Therefore the required third proportional is 9
(ii) 5.25, 7
Let the required third proportional be x
Hence 5.25, 7, x are in continued proportion
⇒ 5.25: 7 = 7: x
⇒ 5.25x = 7 × 7
⇒ x = (7 × 7) / 5.25
⇒ x = 49 / 5.25
⇒ x = 28 / 3
⇒ x =
Therefore the required third proportional is
(iii) Rs 1.60, Rs 0.40
Let the required third proportional be x
Hence Rs 1.60, 0.40, x are in continued proportion
⇒ 1.60: 0.40 = 0.40: x
⇒ 1.60 × x = 0.40 × 0.40
⇒ x = (0.40 × 0.40) / 1.60
⇒ x = 0.1
Therefore required third proportional is 0.1
5. The ratio between 7 and 5 is same as the ratio between Rs x and Rs 20.50; find the value of x
Solution:
Given
Ratio between 7 and 5 is same as the ratio between Rs x and Rs 20.50
Hence, the value of x can be calculated as below
7: 5 = x: 20.50
5x = 7 × 20.50
x = (7 × 20.50) / 5
x = (143.5) / 5
x = 28.7
Therefore the value of x is 28.7
6. If (4x + 3y): (3x + 5y) = 6: 7, find:
(i) x: y
(ii) x, if y = 10
(iii) y, if x = 27
Solution:
(i) x: y
Given
(4x + 3y): (3x + 5y) = 6: 7
We can calculate x: y as below
7(4x + 3y) = 6(3x + 5y)
28x + 21y = 18x + 30y
28x – 18x = 30y – 21y
10x = 9y
x / y = 9 / 10
Therefore x: y is 9: 10
(ii) x, if y = 10
Given
(4x + 3y): (3x + 5y) = 6: 7
And y = 10
Hence we can calculate x as below
7(4x + 3 × 10) = 6(3x + 5 × 10)
7(4x + 30) = 6(3x + 50)
28x + 210 = 18x + 300
28x – 18x = 300 – 210
10x = 90
x = 90 / 10
x = 9
Therefore the value of x is 9
(iii) y, if x = 27
Given
(4x + 3y): (3x + 5y) = 6: 7
And x = 27
We can calculate x as below
7(4 × 27 + 3y) = 6(3 × 27 + 5y)
7(108 + 21y) = 6(81 + 5y)
756 + 21y = 486 + 30y
9y = 756 – 486
9y = 270
y = 270 / 9
y = 30
Therefore the value of y is 30
7. If (2y + 5x) / (3y – 5x) = 
, find:
(i) x: y
(ii) x, if y = 70
(iii) y, if x = 33
Solution:
(i) x: y
Given
(2y + 5x) / (3y – 5x) =
Hence x: y can be calculated as below
(2y + 5x) / (3y – 5x) =
We get
(2y + 5x) / (3y – 5x) = 5 / 2
2 [(2y + 5x)] = 5[(3y – 5x)]
4y + 10x = 15y – 25x
25x + 10x = 15y – 4y
35x = 11y
x / y = 11 / 35
x: y = 11: 35
Therefore x: y is 11: 35
(ii) x, if y = 70
Given
(2y + 5x) / (3y – 5x) =
And y = 70
Hence x can be calculated as below
(2 × 70 + 5x) / (3 × 70 – 5x) =
On calculating further, we get
(2 × 70 + 5x) / (3 × 70 – 5x) = 5 / 2
2 [140 + 5x] = 5 [210 – 5x]
280 + 10x = 1050 – 25x
25x + 10x = 1050 – 280
35x = 770
x = 770 / 35
x = 22
Therefore x is 22
(iii) y, if x = 33
Given
(2y + 5x) / (3y – 5x) =
And x = 33
Hence y can be calculated as below
(2y + 5 ×33) / (3y – 5 × 33) =
(2y + 5 ×33) / (3y – 5 × 33) = 5 / 2
2 [(2y + 165)] = 5 [(3y – 165)]
4y + 330 = 15y – 825
11y = 1155
y = 1155 / 11
y = 105
Therefore y is 105
Exercise 12(C)
1. Are the following numbers in proportion:
(i) 32, 40, 48 and 60?
(ii) 12, 15, 18 and 20?
Solution:
(i) 32, 40, 48 and 60
Given
Numbers are 32, 40, 48 and 60
If 32: 40 = 48: 60 then the ratios are in continued proportion
Hence we can find out as shown below
32: 40 = 48: 60
32 × 60 = 40 × 48
1920 = 1920
Since they are equal
Therefore they are in continued proportion
(ii) 12, 15, 18 and 20
Given
Numbers are 12, 15, 18 and 20
If 12: 15 = 18: 20 then the ratios are in continued proportion
Hence we can find out as shown below
12: 15 = 18: 20
12 × 20 = 15 × 18
240 = 270
240 ≠270
Since they are not equal
Therefore they are not in a continued proportion
2. Find the value of x in each of the following such that the given numbers are in proportion
(i) 14, 42, x and 75
(ii) 45, 135, 90 and x
Solution:
(i) 14, 42, x and 75
Given
Numbers are 14, 42, x and 75
If 14: 42 = x: 75 then they are in continued proportion
Hence we can find out as shown below
14: 42 = x: 75
14 × 75 = x × 42
x = (14 × 75) / 42
x = 1050 / 42
x = 25
Therefore the value of x is 25
(ii) 45, 135, 90 and x
Given
Numbers are 45, 135, 90 and x
If 45: 135 = 90: x then they are in continued proportion
Hence we can find out value of x as shown below
45: 135 = 90: x
45 × x = 90 × 135
x = (90 × 135) / 45
x = 12150 / 45
x = 270
Therefore the value of x is 270
3. The costs of two articles are in the ratio 7: 4. If the cost of the first article is Rs 2800; find the cost of the second article
Solution:
Given
Costs of two articles are in the ratio = 7: 4
Cost of first article = Rs 2800
Let us assume the cost of second article be x
Hence the value of second article can be calculated as shown below
7: 4 = 2800: x
7 / 4 = 2800 / x
7 × x = 2800 × 4
x = (2800 × 4) / 7
x = 11200 / 7
x = 1600
Therefore the cost of second article is Rs 1600
4. The ratio of the length and the width of a rectangular sheet of paper is 8: 5. If the width of the sheet is 17.5 cm; find the length
Solution:
Given
Ratio of the length and the width of a rectangular sheet of paper = 8: 5
Width of the sheet = 17.5 cm
Hence the value of length can be calculated as shown below
8: 5 = x: 17.5
8 / 5 = x / 17.5
5 × x = 8 × 17.5
x = (8 × 17.5) / 5
x = 140 / 5
x = 28
Therefore the length of sheet is 28 cm
5. The ages of A and B are in the ratio 6: 5. If A’s age is 18 years, find the age of B.
Solution:
Given
The ages of A and B are in the ratio = 6: 5
Age of A = 18 years
Let the age of B be x years
Hence the value of B’s age can be calculated as shown below
6: 5 = 18: x
6 / 5 = 18 / x
6 × x = 18 ×5
x = (18 × 5) / 6
x = 90 / 6
x = 15
Therefore the age of B is 15 years
6. A sum of Rs 10, 500 is divided among A, B and C in the ratio 5: 6: 4. Find the share of each.
Solution:
Given
Amount divided among A, B and C in the ratio = 5: 6: 4
Total amount divided = Rs 10500
Hence the sum of ratios can be calculated as below
Sum of ratios = 5 + 6 + 4
= 15
Hence the share’s of A, B and C can be calculated as below
A’s share = Rs (10500) / 15 × 5
= 700 × 5
= Rs 3500
B’s share = Rs (10500) / 15 × 6
= 700 × 6
=4200
C’s share = Rs (10500) / 15 × 4
= 700 × 4
= Rs 2800
Therefore the share of A, B and C are Rs 3500, Rs 4200 and Rs 2800
7. Do the ratios 15 cm to 2 m and 10 sec to 3 minutes form a proportion?
Solution:
Given
15 cm: 2 m:: 10 sec: 3 min
We know that,
1 m = 100 cm
1min = 60 sec
Hence 2 m = 200 cm
3 min = 180 sec
We can check that they form the proportion or not as shown below:
15: 200:: 10: 180
3: 40:: 1: 180
Therefore they do not form the proportion
8. Do the ratios 2 kg: 80 kg and 25 g: 625 g form a proportion?
Solution:
Given
2 kg: 80:: 25 g: 625 g
We can check that they form the proportion or not as shown below
2: 80:: 25: 625
1: 40:: 1: 25
Therefore they do not form the proportion
9. 10 kg sugar cost Rs 350. If x kg sugar of the same kind costs Rs 175, find the value of x
Solution:
Given
Cost of 10 kg sugar = Rs 350
Cost of x kg sugar = Rs 175
The value of x can be calculated as shown below
10 kg: x kg:: 350: 175
10 × 175 = 350 × x
350 x = 1750
x = 1750 / 350
x = 5
Therefore, 5 kg of sugar costs Rs 175
10. The length of two ropes are in the ratio 7: 5. Find the length of:
(i) shorter rope, if the longer one is 22.5 m
(ii) longer rope, if the shorter is 9.8 m
Solution:
(i) shorter rope, if the longer one is 22.5 m
Given
Ratio of length of two rope = 7: 5
And longer rope = 22.5 m
Let the shorter rope length be x
Hence, length of shorter rope can be calculated as below
7: 5 = 22.5: x
7x = 22.5 × 5
x = (22.5 × 5) / 7
x = 112.5 / 7
x = 16.07 m
Therefore the length of shorter rope is 16.07 m
(ii) longer rope, if the shorter is 9.8 m
Given
Ratio of length of two rope = 7: 5
And shorter rope = 9.8 m
Let the length of longer rope be x
Hence, the length of longer rope can be calculated as below
7: 5 = x: 9.8
5x = 9.8 × 7
x = (9.8 × 7) / 5
x = 68.6 / 5
x = 13.72 m
Therefore the length of longer rope is 13.72 m
11. If 4, x and 9 are in continued proportion, find the value of x
Solution:
Given
Numbers are 4, x and 9 are in continued proportion
Hence we can find out value of x as below
4: x = x: 9
x2 = 9 × 4
x = √36
x = 6
Therefore the value of x is 6
12. If 25, 35 and x are in continued proportion, find the value of x
Solution:
Given
Numbers are 25, 35 and x are in continued proportion
Hence we can find out value of x as below
25: 35 = 35: x
25 × x = 35 × 35
x = (35 × 35) / 25
x = 1225 / 25
x = 49
Therefore the value of x is 49
Comments