MSBSHSE Solutions For Class 9 Maths Part 1 Chapter 4 Ratio and Proportion

MSBSHSE Solutions For Class 9 Maths Part 1 Chapter 4 Ratio and Proportion are useful for students as it helps them to score well in the class exams. BYJU’S provides the students with precise solutions, which are prepared by our dedicated specialists. All solutions from MSBSHSE Solutions For Class 9 Maths Part 1 Chapter 4 are provided here for free. These solutions are designed by subject matter experts who have assembled model questions, covering all the exercise questions from the textbook. These solutions help students to familiarize themselves with the ratio and proportion of a given number. This chapter also deals with properties of ratios, operations on equal ratios, theorem of equal ratios and Continued proportion. Solutions are available in PDF format on our website.

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Access answers to Maths MSBSHSE Solutions For Class 9 Part 1 Chapter 4 – Ratio and Proportion

Practice set 4.1 Page no: 61

1. From the following pairs of numbers, find the reduced form of ratio of first number to second number.
i. 72, 60
ii. 38, 57
iii. 52, 78

Solution:

(i) Given 72,60

Reduced form of ratio of first number to second number:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 1

(ii) Given 38, 57

Reduced form of ratio of first number to second number:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 2

(iii) Given 52, 78

Reduced form of ratio of first number to second number:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 3

2. Find the reduced form of the ratio of the first quantity to second quantity.
i. 700 Rs, 308Rs.
ii. 14Rs, 12 Rs.40 paise.
iii. 5 litre, 2500 ml
iv. 3 years 4 months, 5 years 8 months
v. 3.8 kg, 1900 gm
vi. 7 minutes 20 seconds, 5minutes 6 seconds.

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 4

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 5

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 6

(iv) Given 3 years 4 months, 5 years 8 months

3 years = 3 × 12 = 36 months

∴ 3 years 4 months = 40 months

5 years = 5 × 12 = 60

∴ 5 years 8 months = 68 months

∴ Reduced form of the ratio of 40 months and 68 months is:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 7

(v) Given 3.8 kg, 1900 gm

3.8 kg = 3.8 × 1000 = 3800 gm

∴ Reduced form of the ratio of 3800 gm and 1900 gm is:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 8

(vi) Given 7 minutes 20 seconds, 5minutes 6 seconds.

7 minutes = 7 × 60 = 420 seconds

∴ 7 minutes 20 seconds = 440 seconds

5minutes = 5 × 60 = 300 seconds

∴ 5 minutes 6 seconds = 306 seconds

∴ Reduced form of the ratio of 440 seconds and 306 seconds is:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 9

3. Express the following percentages as ratios in the reduced form.
(i) 75: 100
(ii) 44: 100
(iii) 6.25%
(iv) 52: 100
(v) 0.64%

Solution:

(i) Given 75: 100

Reduced form of the ratio of 75:100 is:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 10

(ii) Given 44:100

Reduced form of the ratio of 44:100 is:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 11

(iii) Given 6.25%

Reduced form of 6.25% is:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 12

(iv) Given 52:100

Reduced form of the ratio of 52:100 is:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 13

(v) Given 0.64%

Reduced form of 0.64% is:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 14

4. Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required?

Solution:

Let the persons required to build a house in 6 days be x.

Days required to build a house and number of persons are in inverse proportion.

∴ 6 × x = 8 × 3

∴ 6 x = 24

∴ x = 4

∴ 4 persons are required to build the house in 6 days.

5. Convert the following ratios into percentage.
(i) 15: 25
(ii) 47: 50
(iii) 7/10
(iv) 546/600
(v) 7/16

Solution:

(i) Given 15: 25

= 15/25

= ((15/25) × 100) %

= (15 × 4) %

= 60 %

(ii) Given 47: 50

= 47/50

= ((47/50) × 100) %

= (47 × 2) %

= 94 %

(iii) Given 7/10

= ((7/10) × 100) %

= (7 × 10) %

= 70 %

(iv) Given 546/600

= ((546/600) × 100) %

= (546/6) %

= 91 %

(v) Given 7/16

= ((7/16) × 100) %

= (7 × 6.25) %

= 43.75 %

6. The ratio of ages of Abha and her mother is 2: 5. At the time of Abha’s birth her mother’s age was 27 year. Find the present ages of Abha and her mother.

Solution:

Given the ratio of ages of Abha and her mother is 2 : 5.

Let the common multiple be x.

∴ Present age of Abha = 2x years and

Present age of Abha’s mother = 5x years

According to the given condition, the age of Abha’s mother at the time of Abha’s birth = 27 years

∴ 5x – 2x = 27

∴ 3x = 27

∴ x = 9

∴ Present age of Abha = 2x = 2 x 9 = 18 years

∴ Present age of Abha’s mother = 5x =5 x 9 = 45 years

The present ages of Abha and her mother are 18 years and 45 years respectively.

7. Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5: 4?

Solution:

Given present age of Vatsala = 14 years

Present age of Sara = 10 years

Let after x years, the ratio of their ages will be 5:4.

∴ Age of Vatsala after x years = (14 + x) years

Age of Sara after x years = (10 + x) years

Ratio of their ages = 5:4

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 15

On cross multiplying, we get:

56 + 4x = 50 + 5x

⇒ 5x – 4x = 56 – 50

⇒ x = 6

∴ After 6 years, their ages will be 20 years and 16 years and ratio of their ages will be 5:4.

8. The ratio of present ages of Rehana and her mother is 2: 7. After 2 years, the ratio of their ages will be 1: 3. What is Rehana’s present age?

Solution:

The ratio of present ages of Rehana and her mother is 2 : 7

Let the common multiple be x.

∴ Present age of Rehana = 2x years and Present age of Rehana’s mother = 7x years

After 2 years,

Rehana’s age = (2x + 2) years

Age of Rehana’s mother = (7x + 2) years

According to the given condition,

After 2 years, the ratio of their ages will be 1 : 3

∴ 2x+27x+2 = 13

∴ 3(2x + 2) = 1(7x + 2)

∴ 6x + 6 = 7x + 2

∴ 6 – 2 = 7x – 6x

∴ 4 = x

∴ x = 4

∴ Rehana’s present age = 2x = 2 x 4 = 8 years

∴ Rehana’s present age is 8 years.

Practice set 4.2 Page no: 63

1. Using the property Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 16, fill in the blanks substituting proper numbers in the following.

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 17

Solution:

(i) Let
Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 18

∴ on comparing first two equalities, we get:

5/7 = x/28

Cross multiply and get:

7x = 28 × 5

⇒ x = 4 × 5 = 20

Now, compare the first and third equalities and get:

5/7 = 35/y

Cross multiply and get:

5y = 7 × 35

⇒ y = 7 × 7 = 49

Now, compare the first and fourth equalities and get:

5/7 = z/3.5

Cross multiply and get:

7z = 5 × 3.5

⇒ 7z = 5 × (35/10)

⇒ z = 5 × (5/10)

⇒ z = 25/10 = 2.5


Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 19

(ii) Let
Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 20

∴ On comparing first two equalities, we get:

9/14 = 4.5/x

Cross multiply and get:

9x = 14 × 4.5

⇒ x = 14 × 0.5 = 7

Now, compare the first and third equalities and get:

9/14 = y/42

Cross multiply and get:

14y = 9 × 42

⇒ y = 9 × 3 = 27

Now, compare the first and fourth equalities and get:

9/14 = z/3.5

Cross multiply and get:

14z = 9 × 3.5

⇒ z = 9 × (3.5/14)

⇒ z = 9 × (0.25)

⇒ z = 2.25


Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 21

2. Find the following ratios.
(i) The ratio of radius to circumference of the circle.
(ii) The ratio of circumference of circle with radius r to its area.
(iii) The ratio of diagonal of a square to its side, if the length of side is 7 cm.
(iv) The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of its perimeter to area.

Solution:

(i) Let r be the radius of a circle.

Circumference of circle = 2πr

Ratio of radius to circumference of circle = r/2πr

= 1/2π

= 1: 2π

(ii) Let r be the radius of a circle.

Circumference of circle = 2πr

Area of the circle =2πr2

Ratio of radius to circumference of circle = 2πr/πr2

= 2/r

= 2: r

(iii) Side of square = 7 cm

Diagonal of square = √2 × side = 7√2 cm

Ratio of diagonal of a square to its side = 7/7√2

= 1/√2

= 1: √2

(iv) Length of rectangle = 5 cm

Breadth of rectangle = 3.5 cm

Perimeter of rectangle = 2(Length + Breadth)

= 2(5+3.5)

= 2(8.5)

= 17 cm

Area of rectangle = Length × Breadth

= 5 × 3.5

= 16.5 cm2

Ratio of Perimeter to area of rectangle

= 17/16.5

= 170/165

= 34/33

3. Compare the following pairs of ratios.

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 22

v. Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 23

Solution:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 24

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 25

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 26

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 27

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 28

Therefore, both the ratios are equal, according to the ratio comparison rules.


Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 29

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 30

Practice set 4.3 Page no: 70

1. If a/b = 7/3 then find the values of the following ratios.

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 31

Solution:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 32

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 33

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 34

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 35

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 36

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 37

2. If Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 38then find the values of the following ratios.

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 39

Solution:

(i) Given:
Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 40

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 41

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 42

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 43

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 44

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 45

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 46

Practice set 4.4 Page no: 73

1. Fill in the blanks of the following

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 47

Solution:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 48

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 49

2. 5m-n=3m+4n, then find the values of the following expressions.

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 50

Solution:

(i) Given: 5m – n = 3m + 4m

⇒ 5m – 3m = 4n + n

On simplifying we get

⇒ 2m = 5n

On rearranging

⇒ m/n = 5/2

⇒ m2/n2 = 25/4

Apply componendo and dividendo:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 51

= 29:21

(ii) Given: 5m – n = 3m + 4m

⇒ 5m – 3m = 4n + n

On simplifying we get

⇒ 2m = 5n

On rearranging

⇒ m/n = 5/2

⇒ 3m/4n = 15/8

Apply componendo and dividendo:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 52

= 23: 7

Practice set 4.5 Page no: 77

1. Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?

Solution:

Let x be the number that should be subtracted from 12, 16, 21 so that the numbers remain in continued proportion.

Numbers a, b and c are said to be continued proportion if b2 = ac.

∴ From the definition of continued proportion, we get:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 53

⇒ (16 – x)2 = (12 – x) (21 – x)

On expanding the above equation, we get

⇒ 256 + x2 – 32x = 252 – 33x + x2

⇒-32x + 33x = 252 – 256

⇒ x= -4

∴ -4 should be subtracted from 12, 16, 21 so that the numbers remain in continued proportion.

2. If (28 – x) is the mean proportional of (23 – x) and (19 – x) then find the value of x.

Solution:

A number b is said to be mean proportional of two numbers a and c if b2 = ac.

∴ From the definition of mean proportion, we get:

(28 – x)2 = (23 – x) (19 – x)

On expanding the above equation, we get

⇒ 784 + x2 – 56x = 437 – 42x + x2

⇒-56x + 42x = 437 – 784

⇒ -14x= -347

⇒ x = 347/14

3. Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.

Solution:

Let the numbers be x, y, z.

As the numbers are in continued proportion, therefore

y2 = xz ……………… (1)

Also, the mean proportion = 12

∴ y = √xz = 12

⇒ xz = 144 …………… (2)

It is given that the sum of remaining two numbers = 26

∴ x + z = 26

⇒ x = 26 – z

Put the value of x in equation (2):

(26 – z) z = 144

Expanding and simplifying we get

⇒ 26z – z2 = 144

⇒ z2 – 26z + 144 = 0

⇒ z2 – 8z – 18z + 144 = 0

⇒ z (z – 8) – 18 (z – 8) = 0

⇒ (z – 8) (z – 18) = 0

⇒ z = 8 or z = 18

∴ x = 26 – 8 or x = 26 – 18

⇒ x = 18 or x = 8

y = 12

∴ The numbers in proportion be 8, 12, 18 or 18, 12, 8.

4. If (a + b+ c) (a –b + c) = a2 + b2 + c2, show that a, b, c are in continued proportion.

Solution:

Given: (a + b+ c) (a –b + c) = a2 + b2 + c2

On multiplying we get

⇒ a2 –ab + ac + ab – b2 + bc + ca – bc + c2 = a2 + b2 + c2

⇒ a2 –ab + ac + ab – b2 + bc + ca – bc + c2 – a2 – c2= b2 + b2

⇒ 2ac = 2b2

⇒ b2 = ac

∴ a, b, c are in continued proportion.

5. If a/b = b/c and a, b, c >0, then show that,
i. (a + b+ c) (b – c) = ab – c2
ii. (a2 + b2) (b2 + c2) = (ab + bc)2
iii. (a2 + b2)/ab = (a + c)/b

Solution:

(i) Given: a/b = b/c

⇒ b2 = ac

Consider (a + b+ c) (b – c) = ab – ac + b2 – bc + cb – c2

= ab – ac + ac – c2 (∵ b2 = ac)

= ab – c2

(ii) Given a/b = b/c

⇒ b2 = ac

Consider (a2 + b2) (b2 + c2) = a2b2 + a2 c2 + b2 b2 + b2c2

= a2b2 + ac(ac)+ b2(ac)+ b2c2 (∵ b2 = ac)

= a2b2 + b2(ac)+ b2(ac)+ b2c2 (∵ b2 = ac)

= a2b2 + 2b2(ac)+ b2c2

= a2b2 + 2ab2c+ b2c2

= (ab + bc)2

(iii) Given: a/b = b/c

⇒ b2 = ac

Consider (a2 + b2)/ab = (a2 + ac)/ab (∵ b2 = ac)

= (a + c)/b

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 54

6. Find mean proportional of 

Solution:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 55

On simplifying we get

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 56

Problem set 4 Page no: 77

1. Select the appropriate alternative answer for the following questions.
i. If 6: 5 = y: 20 then what will be the value of y?
A. 15
B. 24
C. 18
D. 22.5

Solution:

B. 24

Explanation:

Given 6:5 = y:20

⇒ 6/5 = y/20

Cross multiply and get:

5y = 6 × 20

⇒ y = 6 × 4 = 24

∴ Option B is correct.

ii. What is the ratio of 1 mm to 1 cm?
A. 1: 100
B. 10: 1
C. 1: 10
D. 100: 1

Solution:

A. 1: 100

Explanation:

We know that 1 cm = 100 mm

∴ 1mm: 1cm

⇒ 1mm: 100mm

= 1: 100

∴ Option A is correct.

iii. The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin’s age to Mohasin’s age?
A. 3: 2
B. 2: 3
C. 4: 3
D. 3: 4

Solution:

B. 2: 3

Explanation:

Given Nitin’s age = 24 years

Mohasin’s age = 36 years

∴ Ration of Nitin’s age to Mohasin’s age = 24:36

= 24/36

= 2/3

= 2:3

∴ Option B is correct.

iv. 24 Bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get?
A. 8
B. 15
C. 12
D. 9

Solution:

D. 9

Explanation:

Total bananas = 24

Ratio in which the bananas are divided = 3:5

Let number of bananas Shubham got = 3x

∴ Number of bananas Anil got = 5x

∴ 3x+5x = 24

⇒ 8x = 24

⇒ x = 3

∴ Shubham got (3 × 3) = 9 bananas.

Thus, option D is correct.

v. What is the mean proportional of 4 and 25?
A. 6
B. 8
C. 10
D. 12

Solution:

C. 10

Explanation:

Mean proportional of two numbers a and b = √(ab)

∴ Mean proportional of 4 and 25 = √ (4 × 25)

= √100 = 10

2. For the following numbers write the ratio of first number to second number in the reduced form.
i. 21, 48

ii. 36, 90
iii. 65, 117

iv. 138, 161
v. 114, 133

Solution:

(i) Ratio of 21 and 48 in the reduced form:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 57

(To simplify, break the numbers in simpler form)

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 58

∴ Ratio of 21 and 48 in reduced form is 1:4.

(ii) Ratio of 36 and 90 in the reduced form:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 59

(To simplify, break the numbers in simpler form)

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 60

∴ Ratio of 36 and 90 in reduced form is 2:5.

(iii) Ratio of 65 and 117 in the reduced form:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 61

(To simplify, break the numbers in simpler form)

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 62

∴ Ratio of 65 and 117 in reduced form is 5:9.

(iv) Ratio of 138 and 161 in the reduced form:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 63

(To simplify, break the numbers in simpler form)

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 64

∴ Ratio of 138 and 161 in reduced form is 6:7.

(v) Ratio of 114 and 133 in the reduced form:

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 65 (To simplify, break the numbers in simpler form)

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 66

∴ Ratio of 114 and 133 in reduced form is 6:7.

3. Write the following ratios in the reduced form.
i. Radius to the diameter of a circle.
ii. The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.
iii. The ratio of perimeter to area of a square, having side 4 cm.

Solution:

(i) Let r be the radius of the circle.

Let d be the diameter of the circle.

Diameter = 2 × Radius

∴ Ratio of radius to diameter in the reduced form = Radius: Diameter

Maharashtra Board Solutions for Class 9 Maths Part 1 Chapter 4 - Image 67

∴ Ratio of radius to diameter in the reduced form = 1:2

(ii) Given: Length of rectangle = l = 4 cm

Breadth of rectangle = b = 3 cm

Diagonal of rectangle = √ (l2 + b2)

= √ (16 + 9)

= √25 = 5

∴ Diagonal of rectangle = 5 cm

Ratio of diagonal to the length of a rectangle = 4:5

(iii) Given: Side of square = 4 cm

Perimeter of square = 4 × Side = 4 × 4 = 16 cm2

Area of the square = (Side)2 = (4)2 = 14 cm2

The ratio of perimeter to area of a square = 16:14 = 8:7

4. Check whether the following numbers are in continued proportion.
i. 2, 4, 8
ii. 1, 2, 3
iii. 9, 12, 16
iv. 3, 5, 8

Solution:

(i) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a: b∷ b:c

or b2 = ac

Here, a = 2, b = 4 and c = 8

∴ (4)2 = 2 × 8

⇒ 16 = 16, which holds true.

∴ 2, 4, 8 are in continued proportion.

(ii) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a: b∷ b:c

or b2 = ac

Here, a = 1, b = 2 and c = 3

∴ (2)2 = 1 × 3

⇒ 4 = 3, which does not hold true.

∴ 1, 2, 3 are not in continued proportion.

(iii) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a: b∷ b: c

or b2 = ac

Here, a = 9, b = 12 and c = 16

∴ (12)2 = 9 × 16

⇒ 144 = 144, which holds true.

∴ 9, 12, 16 are in continued proportion.

(iv) Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion,

i.e. a: b∷ b: c

or b2 = ac

Here, a = 3, b = 5 and c = 8

∴ (5)2 = 3 × 8

⇒ 25 = 24, which does not hold true.

∴ 3, 5, 8 are not in continued proportion.

5. a, b, c are in continued proportion. If a = 3 and c = 27 then find b.

Solution:

Given: a, b, c are in continued proportion.

Three numbers ‘a’, ‘b’ and ‘c’ are said to be continued proportion if a, b and c are in proportion, i.e. a: b∷ b:c

or b2 = ac

Here, a = 3, c = 27

∴ (b)2 = 3 × 27

⇒ b2 = 81

⇒ b = ±√81 = ±9

∴ b = -9 or 9

6. Convert the following ratios into percentages.
i. 37: 500

ii. 5/8
iii. 22/30

iv. 5/16
v. 144/1200

Solution:

(i) Given 37: 500 = 37/500

= ((37/500) × 100) %

= (37/5) %

= 7.4 %

(ii) Given 5/8 = ((5/8) × 100) %

= (5 × 12.5) %

= 62.5 %

(iii) Given 22/30 = ((22/30) × 100) %

= (220/3) %

= 73.33 %

(iv) Given 144/1200 = ((144/1200) × 100) %

= (144/12) %

= 12 %

A ratio is a mathematical expression of comparing two similar or different quantities by division. This expression can be expressed from ratio to percentage form by conversion method. It is denoted by ‘:’ symbol. For example, 2:3, 5:6, 3:7, etc. are ratios. It is also represented by ‘/’ symbol. Hence, the ratio is basically a type of division method. The concept related to Ratio and Proportion is explained in BYJU’S to help the students be thorough with the topics.

 

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