ML Aggarwal Solutions for Class 10 Maths Chapter 7: Ratio and Proportion

ML Aggarwal Solutions for Class 10 Maths Chapter 7 Ratio and Proportion help students understand all the concepts clearly and develop a strong base in the subject. The solutions improve the academic performance of students and help them pursue their desired courses at higher levels of education. Here, the students can download ML Aggarwal Solutions for Class 10 Maths Chapter 7 Ratio and Proportion free PDF from the links provided below.

This chapter consists of problems on finding the ratio and proportion of various word problems. Students are advised to solve these problems on a daily basis to score well in the annual exam. The ML Aggarwal Solutions provided in PDF format are 100% accurate, based on the latest syllabus.

ML Aggarwal Class 10 Maths Chapter 7:

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Access ML Aggarwal Solutions for Class 10 Maths Chapter 7: Ratio and Proportion

Exercise 7.1

1. An alloy consists of 27 ½ kg of copper and 2 ¾ kg of tin. Find the ratio by weight of tin to the alloy.

Solution:

It is given that

Copper = 27 ½ kg = 55/2 kg

Tin = 2 ¾ kg = 11/4 kg

We know that

Total alloy = 55/2 + 11/4

Taking LCM

= (110 + 11)/ 4

= 121/4 kg

Here

Ratio between tin and alloy = 11/4 kg: 121/4 kg

So we get

= 11: 121

= 1: 11

2. Find the compounded ratio of:

(i) 2: 3 and 4: 9

(ii) 4: 5, 5: 7 and 9: 11

(iii) (a – b): (a + b), (a + b)²: (a² + b²) and (a⁴ – b⁴): (a² – b²)²

Solution:

(i) 2: 3 and 4: 9

We know that

Compound ratio = 2/3 × 4/9

= 8/27

= 8: 27

(ii) 4: 5, 5: 7 and 9: 11

We know that

Compound ratio = 4/5 × 5/7 × 9/11

= 36/77

= 36: 77

(iii) (a – b): (a + b), (a + b)²: (a² + b²) and (a⁴ – b⁴): (a² – b²)²

We know that

Compound ratio = (a – b)/ (a + b) × (a + b)²/ (a² + b²) × (a⁴ – b⁴)/ (a² – b²)²

By further calculation

= (a – b)/ (a + b) × [(a + b) (a + b)]/ (a² + b²) × [(a² + b²) (a + b) (a – b)]/ [(a + b)² (a – b)²]

So we get

= 1/1

= 1: 1

3. Find the duplicate ratio of

(i) 2: 3

(ii) √5: 7

(iii) 5a: 6b

Solution:

(i) 2: 3

We know that

Duplicate ratio of 2: 3 = 2²: 3² = 4: 9

(ii) √5: 7

We know that

Duplicate ratio of √5: 7 = √5²: 7² = 5: 49

(iii) 5a: 6b

We know that

Duplicate ratio of 5a: 6b = (5a)²: (6b)² = 25a²: 36b²

4. Find the triplicate ratio of

(i) 3: 4

(ii) ½: 1/3

(iii) 1³: 2³

Solution:

(i) 3: 4

We know that

Triplicate ratio of 3: 4 = 3³: 4³ = 27: 64

(ii) ½: 1/3

We know that

Triplicate ratio of ½: 1/3 = (1/2)³: (1/3)³ = 1/8: 1/27 = 27: 8

(iii) 1³: 2³

We know that

Triplicate ratio of 1³: 2³ = (1³)³: (2³)³ = 1³: 8³ = 1: 512

5. Find the sub-duplicate ratio of

(i) 9: 16

(ii) ¼: 1/9

(iii) 9a²: 49b²

Solution:

(i) 9: 16

We know that

Sub-duplicate ratio of 9: 16 = √9: √16 = 3: 4

(ii) ¼: 1/9

We know that

Sub-duplicate ratio of ¼: 1/9 = √1/4: √1/9

So we get

= ½: 1/3

= 3: 2

(iii) 9a²: 49b²

We know that

Sub-duplicate ratio of 9a²: 49b² = √9a²: √49b² = 3a: 7b

6. Find the sub-triplicate ratio of

(i) 1: 216

(ii) 1/8: 1/125

(iii) 27a³: 64b³

Solution:

(i) 1: 216

We know that

Sub-triplicate ratio of 1: 216 = ∛1: ∛216

By further calculation

= (1³)^1/3: (6³)^1/3

= 1: 6

(ii) 1/8: 1/125

We know that

Sub-triplicate ratio of 1/8: 1/125 = (1/8)^1/3: (1/125)^1/3

It can be written as

= [(1/2)³]^1/3: [(1/5)³]^1/3

So we get

= ½: 1/5

= 5: 2

(iii) 27a³: 64b³

We know that

Sub-triplicate ratio of 27a³: 64b³ = [(3a)³]^1/3: [(4b)³]^1/3

So we get

= 3a: 4b

7. Find the reciprocal ratio of

(i) 4: 7

(ii) 3²: 4²

(iii) 1/9: 2

Solution:

(i) 4: 7

We know that

Reciprocal ratio of 4: 7 = 7: 4

(ii) 3²: 4²

We know that

Reciprocal ratio of 3²: 4² = 4²: 3² = 16: 9

(iii) 1/9: 2

We know that

Reciprocal ratio of 1/9: 2 = 2: 1/9 = 18: 1

8. Arrange the following ratios in ascending order of magnitude:

2: 3, 17: 21, 11: 14 and 5: 7

Solution:

It is given that

2: 3, 17: 21, 11: 14 and 5: 7

We can write it in fractions as

2/3, 17/21, 11/14, 5/7

Here the LCM of 3, 21, 14 and 7 is 42

By converting the ratio as equivalent

2/3 = (2 × 14)/ (3 × 14) = 28/42

17/21 = (17 × 2)/ (21 × 2) = 34/ 42

11/14 = (11 × 3)/ (14 × 3) = 33/42

5/7 = (5 × 6)/ (7 × 6) = 30/42

Now writing it in ascending order

28/42, 30/42, 33/42, 34/42

By further simplification

2/3, 5/7, 11/14, 17/21

So we get

2: 3, 5: 7, 11: 14 and 17: 21

9. (i) If A: B = 2: 3, B: C = 4: 5 and C: D = 6: 7, find A: D.

(ii) If x: y = 2: 3 and y: z = 4: 7, find x: y: z.

Solution:

(i) It is given that

A: B = 2: 3, B: C = 4: 5 and C: D = 6: 7

We can write it as

A/ B = 2/3, B/C = 4/5, C/D = 6/7

By multiplication

A/B × B/C × C/D = 2/3 × 4/5 × 6/7

So we get

A/D = 16/35

A: D = 16: 35

(ii) We know that the LCM of y terms 3 and 4 is 12

Now making equals of y as 12

x/y = 2/3 = (2 × 4)/ (3 × 4) = 8/12 = 8: 12

y/z = 4/7 × 3/3 = 12/21 = 12: 21

So x: y: z = 8: 12: 21

10. (i) If A: B = 1/4: 1/5 and B: C = 1/7: 1/6, find A: B: C.

(ii) If 3A = 4B = 6C, find A: B: C

Solution:

(i) We know that

A: B = 1/4 × 5/1 = 5/4

B: C = 1/7 × 6/1 = 6/7

Here the LCM of B terms 4 and 6 is 12

Now making terms of B as 12

A/B = (5 × 3)/ (4 × 3) = 15/12 = 15: 12

B/C = (6 × 2)/ (7 × 2) = 12/14 = 12: 14

So A: B: C = 15: 12: 14

(ii) It is given that

3A = 4B

We can write it as

A/B = 4/3

A: B = 4: 3

Similarly 4B = 6C

We can write it as

B/C = 6/4 = 3/2

B: C = 3: 2

So we get

A: B: C = 4: 3: 2

11. (i) If 3x + 5y/ 3x – 5y = 7/3, find x: y.

(ii) If a: b = 3: 11, find (15a – 3b): (9a + 5b).

Solution:

(i) 3x + 5y/ 3x – 5y = 7/3

By cross multiplication

9x + 15y = 21x – 35y

By further simplification

21x – 9x = 15y + 35y

12x = 50y

So we get

x/y = 50/12 = 25/6

Therefore, x: y = 25: 6

(ii) It is given that

a: b = 3: 11

a/b = 3/11

It is given that

(15a – 3b)/ (9a + 5b)

Now dividing both the numerator and denominator by b

= [15a/b – 3b/b]/ [9a/b + 5b/b]

By further calculation

= [15a/b – 3]/ [9a/b + 5]

Substituting the value of a/ b

= [15 × 3/11 – 3]/ [9 × 3/11 + 5]

So we get

= [45/11 – 3]/ [27/11 + 5]

Taking LCM

= [(45 – 33)/ 11]/ [(27 + 55)/ 11]

= 12/11/ 82/11

We can write it as

= 12/11 × 11/82

= 12/82

= 6/41

Hence, (15a – 3b): (9a + 5b) = 6: 41.

12. (i) If (4x² + xy): (3xy – y²) = 12: 5, find (x + 2y): (2x + y).

(ii) If y (3x – y): x (4x + y) = 5: 12. Find (x² + y²): (x + y)².

Solution:

(i) (4x² + xy): (3xy – y²) = 12: 5

We can write it as

(4x² + xy)/ (3xy – y²) = 12/ 5

By cross multiplication

20x² + 5xy = 36xy – 12y²

20x² + 5xy – 36xy + 12y² = 0

20x² – 31xy + 12y² = 0

Now divide the entire equation by y²

20x²/y² – 31xy/y² + 12y²/y² = 0

So we get

20 (x/y)² – 31 (x/y) + 12 = 0

20 (x/y)² – 15(x/y) – 16 (x/y) + 12 = 0

Taking common terms

5 (x/y) [4 (x/y) – 3] – 4 [4 (x/y) – 3] = 0

Here 4 (x/y) – 3 = 0

4 (x/y) = 3

So we get x/y = ¾

Similarly 5 (x/y) – 4 = 0

5 (x/y) = 4

So we get x/y = 4/5

Now dividing by y

(x + 2y)/ (2x + y) = (x/y + 2)/ (2 x/y + 1)

(a) If x/y = 3/4, then

= (x/y + 2)/ (2 x/y + 1)

Substituting the values

= (3/4 + 2)/ (2 × 3/4 + 1)

By further calculation

= 11/4/ (3/2 + 1)

= 11/4/ 5/2

= 11/4 × 2/5

= 11/10

So we get

(x + 2y): (2x + y) = 11: 10

(b) If x/y = 4/5 then

(x + 2y)/ (2x + y) = [x/y + 2]/ [2 x/y + 1]

Substituting the value of x/y

= [4/5 + 2]/ [2 × 4/5 + 1]

So we get

= 14/5/ [8/5 + 1]

= 14/5/ 13/5

= 14/5 × 5/13

= 14/13

We get

(x + 2y)/ (2x + y) = 11/10 or 14/13

(x + 2y): (2x + y) = 11: 10 or 14: 13

(ii) y (3x – y): x (4x + y) = 5: 12

It can be written as

(3xy – y²)/ (4x² + xy) = 5/12

By cross multiplication

36xy – 12y² = 20x² + 5xy

20x² + 5xy – 36xy + 12y² = 0

20x² – 31xy + 12y² = 0

Divide the entire equation by y²

20x²/y² – 31 xy/y² + 12y²/y² = 0

20(x²/y²) – 31 (xy/y²) + 12 = 0

We can write it as

20(x²/y²) – 15 (x/y) – 16 (x/y) + 12 = 0

Taking common terms

5 (x/y) [4 (x/y) – 3] – 4 [4 (x/y) – 3] = 0

Here

4 (x/y) – 3 = 0

So we get

4 (x/y) = 3

x/y = 3/4

Similarly

5 (x/y) – 4 = 0

So we get

5 (x/y) = 4

x/y = 4/5

(a) x/y = 3/4

We know that

(x² + y²): (x + y)² = (x² + y²)/ (x + y)²

Dividing both numerator and denominator by y²

= (x²/y² + y²/y²)/ [1/y² (x + y)²]

= (x²/ y² + 1) (x/y + 1)²

Substituting the value of x/y

= [(3/4)² + 1]/ [3/4 + 1]²

By further calculation

= (9/16 + 1)/ (7/4)²

So we get

= 25/16/ 49/16

= 25/16 × 16/49

= 25/49

So we get

(x² + y²): (x + y)² = 25: 49

(b) x/y = 4/5

We know that

(x² + y²): (x + y)² = (x² + y²)/ (x + y)²

Dividing both numerator and denominator by y²

= (x²/y² + y²/y²)/ [1/y² (x + y)²]

= (x²/ y² + 1) (x/y + 1)²

Substituting the value of x/y

= [(4/5)² + 1]/ [4/5 + 1]²

By further calculation

= (16/25 + 1)/ (9/5)²

So we get

= 41/25/ 81/25

= 41/25 × 25/81

= 41/81

So we get

(x² + y²): (x + y)² = 41: 81

13. (i) If (x – 9): (3x + 6) is the duplicate ratio of 4: 9, find the value of x.
(ii) If (3x + 1): (5x + 3) is the triplicate ratio of 3: 4, find the value of x.

(iii) If (x + 2y): (2x – y) is equal to the duplicate ratio of 3: 2, find x: y.

Solution:

(i) (x – 9)/ (3x + 6) = (4/9)²

So we get

(x – 9)/ (3x + 6) = 16/81

By cross multiplication

81x – 729 = 48x + 96

81x – 48x = 96 + 729

So we get

33x = 825

x = 825/33 = 25

(ii) (3x + 1)/ (5x + 3) = 3³/ 4³

So we get

(3x + 1)/ (5x + 3) = 27/64

By cross multiplication

64 (3x + 1) = 27 (5x + 3)

192x + 64 = 135x + 81

192x – 135x = 81 – 64

57x = 17

So we get

x = 17/57

(iii) (x + 2y)/ (2x – y) = 3²/ 2²

So we get

(x + 2y)/ (2x – y) = 9/4

By cross multiplication

9 (2x – y) = 4 (x + 2y)

18x – 9y = 4x + 8y

18x = 4x = 8y + 9y

So we get

14x = 17y

x/y = 17/14

x: y = 17: 14

14. (i) Find two numbers in the ratio of 8: 7 such that when each is decreased by 12 ½, they are in the ratio 11: 9.

(ii) The income of a man is increased in the ratio of 10: 11. If the increase in his income is Rs 600 per month, find his new income.

Solution:

(i) Ratio = 8: 7

Consider the numbers as 8x and 7x

Using the condition

Taking LCM

By further calculation

(16x – 25)/ (14x – 25) = 11/9

By cross multiplication

154x – 275 = 144x – 225

154x – 144x = 275 – 225

10x = 50

x = 50/10 = 5

So the numbers are

8x = 8 × 5 = 40

7x = 7 × 5 = 35

(ii) Consider the present income = 10x

Increased income = 11x

So the increase per month = 11x – 10x = x

Here x = Rs 600

New income = 11x = 11 × 600 = Rs 6600

15. (i) A woman reduces her weight in the ratio 7: 5. What does her weight become if originally it was 91 kg.

(ii) A school collected Rs 2100 for charity. It was decided to divide the money between an orphanage and a blind school in the ratio of 3: 4. How much money did each receive?

Solution:

(i) Ratio of original and reduced weight of woman = 7: 5

Consider original weight = 7x

Reduced weight = 5x

Here, the original weight = 91 kg

So the reduced weight = (91 × 5x)/ 7x = 65 kg

(ii) Amount collected for charity = Rs 2100

Here the ratio between an orphanage and a blind school = 3: 4

Sum of ratios = 3 + 4 = 7

We know that

Orphanage schools share = 2100 × 3/7 = Rs 900

Blind schools share = 2100 × 4/7 = Rs 1200

16. (i) The sides of a triangle are in the ratio 7: 5: 3 and its perimeter is 30 cm. Find the lengths of sides.

(ii) If the angles of a triangle are in the ratio 2: 3: 4, find the angles.

Solution:

(i) It is given that

Perimeter of triangle = 30 cm

Ratio among sides = 7: 5: 3

Here the sum of ratios = 7 + 5 + 3 = 15

We know that

Length of first side = 30 × 7/15 = 14 cm

Length of second side = 30 × 5/15 = 10 cm

Length of third side = 30 × 3/15 = 6 cm

Therefore, the sides are 14 cm, 10 cm and 6 cm.

(ii) We know that

Sum of all the angles of a triangle = 180⁰

Here the ratio among angles = 2: 3: 4

Sum of ratios = 2 + 3 + 4 = 9

So we get

First angle = 180 × 2/9 = 40⁰

Second angle = 180 × 3/9 = 60⁰

Third angle = 180 × 4/9 = 80⁰

Hence, the angles are 40⁰, 60⁰ and 80⁰.

17. Three numbers are in the ratio 1/2: 1/3: ¼. If the sum of their squares is 244, find the numbers.

Solution:

It is given that

Ratio of three numbers = 1/2: 1/3: 1/4

= (6: 4: 3)/ 12

= 6: 4: 3

Consider first number = 6x

Second number = 4x

Third number = 3x

So based on the condition

(6x)² + (4x)² + (3x)² = 244

36x² + 16x² + 9x² = 244

So we get

61x² = 244

x² = 244/61 = 4 = 2²

x = 2

Here

First number = 6x = 6 × 2 = 12

Second number = 4x = 4 × 2 = 8

Third number = 3x = 3 × 2 = 6

18. (i) A certain sum was divided among A, B and C in the ratio 7: 5: 4. If B got Rs 500 more than C, find the total sum divided.

(ii) In a business, A invests Rs 50000 for 6 months, B Rs 60000 for 4 months and C Rs 80000 for 5 months. If they together earn Rs 18800 find the share of each.

Solution:

(i) It is given that

Ratio between A, B and C = 7: 5: 4

Consider A share = 7x

B share = 5x

C share = 4x

So the total sum = 7x + 5x + 4x = 16x

Based on the condition

5x – 4x = 500

x = 500

So the total sum = 16x = 16 × 500 = Rs 8000

(ii) 6 months investment of A = Rs 50000

1 month investment of A = 50000 × 6 = Rs 300000

4 months investment of B = Rs 60000

1 month investment of B = 60000 × 4 = Rs 240000

5 months investment of C = Rs 80000

1 month investment of C = 80000 × 5 = Rs 400000

Here the ratio between their investments = 300000: 240000: 400000

= 30: 24: 40

Sum of ratio = 30 = 24 + 40 = 94

Total earnings = Rs 18800

So we get

A share = 30/94 × 18800 = Rs 6000

B share = 24/94 × 18800 = Rs 4800

C share = 40/94 = 18800 = Rs 8000

19. (i) In a mixture of 45 litres, the ratio of milk to water is 13: 2. How much water must be added to this mixture to make the ratio of milk to water as 3: 1?

(ii) The ratio of the number of boys to the numbers of girls in a school of 560 pupils is 5: 3. If 10 new boys are admitted, find how many new girls may be admitted so that the ratio of the number of boys to the number of girls may change to 3: 2.

Solution:

(i) It is given that

Mixture of milk to water = 45 litres

Ratio of milk to water = 13: 2

Sum of ratio = 13 + 2 = 15

Here the quantity of milk = (45 × 13)/ 15 = 39 litres

Quantity of water = 45 × 2/15 = 6 litres

Consider x litre of water to be added, then water = (6 + x) litres

Here the new ratio = 3: 1

39: (6 + x) = 3: 1

We can write it as

39/ (6 + x) = 3/1

By cross multiplication

39 = 18 + 3x

3x = 39 – 18 = 21

x = 21/3 = 7 litres

Hence, 7 litres of water is to be added to the mixture.

(ii) It is given that

Ratio between boys and girls = 5: 3

Number of pupils = 560

So the sum of ratios = 5 + 3 = 8

We know that

Number of boys = 5/8 × 560 = 350

Number of girls = 3/8 × 560 = 210

Number of new boys admitted = 10

So the total number of boys = 350 + 10 = 360

Consider x as the number of girls admitted

Total number of girls = 210 + x

Based on the condition

360: 210 + x = 3: 2

We can write it as

360/ 210 + x = 3/2

By cross multiplication

630 + 3x = 720

3x = 720 – 630 = 90

So we get

x = 90/3 = 30

Hence, 30 new girls are to be admitted.

20. (i) The monthly pocket money of Ravi and Sanjeev are in the ratio 5: 7. Their expenditures are in the ratio 3: 5. If each saves Rs 80 per month, find their monthly pocket money.

(ii) In class X of a school, the ratio of the number of boys to that of the girls is 4: 3. If there were 20 more boys and 12 less girls, then the ratio would have been 2: 1. How many students were there in the class?

Solution:

(i) Consider the monthly pocket money of Ravi and Sanjeev as 5x and 7x

Their expenditure is 3y and 5y respectively.

5x – 3y = 80 …… (1)

7x – 5y = 80 …… (2)

Now multiply equation (1) by 7 and (2) by 5

Subtracting both the equations

35x – 21y = 560

35x – 25y = 400

So we get

4y = 160

y = 40

In equation (1)

5x = 80 + 3 × 40 = 200

x = 40

Here the monthly pocket money of Ravi = 5 × 40 = 200

(ii) Consider x as the number of students in class

Ratio of boys and girls = 4: 3

Number of boys = 4x/7

Number of girls = 3x/7

Based on the problem

(4x/7 + 20): (3x/7 – 12) = 2: 1

We can write it as

(4x + 140)/ 7: (3x – 84)/ 7 = 2: 1

So we get

(4x + 140)/ 7 × 7/ (3x – 84) = 2/1

(4x + 140)/ (3x – 84) = 2/1

6x – 168 = 4x + 140

6x – 4x = 140 + 168

2x = 308

x = 308/2 = 154

Therefore, 154 students were there in the class.

21. In an examination, the ratio of passes to failures was 4: 1. If 30 less had appeared and 20 less passed, the ratio of passes to failures would have been 5: 1. How many students appeared for the examination.

Solution:

Consider the number of passes = 4x

Number of failures = x

Total number of students appeared = 4x + x = 5x

In case 2

Number of students appeared = 5x – 30

Number of passes = 4x – 20

So the number of failures = (5x – 30) – (4x – 20)

By further calculation

= 5x – 30 – 4x + 20

= x – 10

Based on the condition

(4x – 20)/ (x – 10) = 5/1

By cross multiplication

5x – 50 = 4x – 20

5x – 4x = – 20 + 50

x = 30

Number of students appeared = 5x = 5 × 30 = 150

Exercise 7.2

1. Find the value of x in the following proportions:

(i) 10: 35 = x: 42

(ii) 3: x = 24: 2

(iii) 2.5: 1.5 = x: 3

(iv) x: 50 :: 3: 2

Solution:

(i) 10: 35 = x: 42

We can write it as

35 × x = 10 × 42

So we get

x = (10 × 42)/ 35

x = 2 × 6

x = 12

(ii) 3: x = 24: 2

We can write it as

x × 24 = 3 × 2

So we get

x = (3 × 2)/ 24

x = ¼

(iii) 2.5: 1.5 = x: 3

We can write it as

1.5 × x = 2.5 × 3

So we get

x = (2.5 × 3)/ 1.5

x = 5.0

(iv) x: 50 :: 3: 2

We can write it as

x × 2 = 50 × 3

So we get

x = (50 × 3)/ 2

x = 75

2. Find the fourth proportional to

(i) 3, 12, 15

(ii) 1/3, 1/4, 1/5

(iii) 1.5, 2.5, 4.5

(iv) 9.6 kg, 7.2 kg, 28.8 kg

Solution:

(i) 3, 12, 15

Consider x as the fourth proportional to 3, 12 and 15

3: 12 :: 15: x

We can write it as

3 × x = 12 × 15

So we get

x = (12 × 15)/ 3

x = 60

(ii) 1/3, 1/4, 1/5

Consider x as the fourth proportional to 1/3, 1/4 and 1/5

1/3: 1/4:: 1/5: x

We can write it as

1/3 × x = 1/4 × 1/5

So we get

x = 1/4 × 1/5 × 3/1

x = 3/20

(iii) 1.5, 2.5, 4.5

Consider x as the fourth proportional to 1,5, 2.5 and 4.5

1.5: 2.5 :: 4.5: x

We can write it as

1.5 × x = 2.5 × 4.5

So we get

x = (2.5 × 4.5)/ 1.5

x = 7.5

(iv) 9.6 kg, 7.2 kg, 28.8 kg

Consider x as the fourth proportional to 9.6, 7.2 and 28.8

9.6: 7.2 :: 28.8: x

We can write it as

9.6 × x = 7.2 × 28.8

So we get

x = (7.2 × 28.8)/ 9.6

x = 21.6

3. Find the third proportional to

(i) 5, 10

(ii) 0.24, 0.6

(iii) Rs. 3, Rs. 12

(iv) 5 ¼ and 7.

Solution:

(i) Consider x as the third proportional to 5, 10

5: 10 :: 10: x

It can be written as

5 × x = 10 × 10

x = (10 × 10)/ 5 = 20

Hence, the third proportional to 5, 10 is 20.

(ii) Consider x as the third proportional to 0.24, 0.6

0.24: 0.6 :: 0.6: x

It can be written as

0.24 × x = 0.6 × 0.6

x = (0.6 × 0.6)/ 0.24 = 1.5

Hence, the third proportional to 0.24, 0.6 is 1.5.

(iii) Consider x as the third proportional to Rs. 3 and Rs. 12

3: 12 :: 12: x

It can be written as

3 × x = 12 × 12

x = (12 × 12)/ 3 = 48

Hence, the third proportional to Rs. 3 and Rs. 12 is Rs. 48

(iv) Consider x as the third proportional to 5 ¼ and 7

5 ¼: 7 :: 7: x

It can be written as

21/4 × x = 7 × 7

x = (7 × 7 × 4)/ 21 = 28/3 = 9 1/3

Hence, the third proportional to 5 ¼ and 7 is 9 1/3.

4. Find the mean proportion of:

(i) 5 and 80

(ii) 1/12 and 1/75

(iii) 8.1 and 2.5

(iv) (a – b) and (a³ – a²b), a ˃ b

Solution:

(i) Consider x as the mean proportion of 5 and 80

5: x :: x: 80

It can be written as

x² = 5 × 80 = 400

x = √400 = 20

Therefore, mean proportion of 5 and 80 is 20.

(ii) Consider x as the mean proportion of 1/12 and 1/75

1/12: x :: x: 1/75

It can be written as

x² = 1/12 × 1/75 = 1/900

x = √1/900 = 1/30

Therefore, mean proportion of 1/12 and 1/75 is 1/30.

(iii) Consider x as the mean proportion of 8.1 and 2.5

8.1: x :: x: 2.5

It can be written as

x² = 8.1 × 2.5 = 20.25

x = √20.25 = 4.5

Therefore, mean proportion of 8.1 and 2.5 is 4.5.

(iv) Consider x as the mean proportion of (a – b) and (a³ – a²b), a ˃ b

(a – b): x :: (a³ – a²b)

It can be written as

x² = (a – b) (a³ – a²b)

So we get

x² = (a – b) a² (a – b)

x² = a² (a – b)²

Here

x = a (a – b)

Therefore, mean proportion of (a – b) and (a³ – a²b), a ˃ b is a (a – b).

5. If a, 12, 16 and b are in continued proportion find a and b.

Solution:

It is given that

a, 12, 16 and b are in continued proportion

a/12 = 12/16 = 16/b

We know that

a/12 = 12/16

By cross multiplication

16a = 144

a = 144/16 = 9

Similarly

12/16 = 16/b

By cross multiplication

12b = 16 × 16 = 256

b = 256/12 = 64/3 = 21 1/3

Therefore, a = 9 and b = 64/3 or 21 1/3.

6. What number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion?

Solution:

Consider x to be added to 5, 11, 19 and 37 to make them in proportion

5 + x: 11 + x :: 19 + x: 37 + x

It can be written as

(5 + x) (37 + x) = (11 + x) (19 + x)

By further calculation

185 + 5x + 37x + x² = 209 + 11x + 19x + x²

185 + 42x + x² = 209 + 30x + x²

So we get

42x – 30x + x² – x² = 209 – 185

12x = 24

x = 2

Hence, the least number to be added is 2.

7. What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion?

Solution:

Consider x to be subtracted from each term

23 – x, 30 – x, 57 – x and 78 – x are proportional

It can be written as

23 – x: 30 – x :: 57 – x: 78 – x

(23 – x)/ (30 – x) = (57 – x)/ (78 – x)

By cross multiplication

(23 – x) (78 – x) = (30 – x) (57 – x)

By further calculation

1794 – 23x – 78x + x² = 1710 – 30x – 57x + x²

x² – 101x + 1794 – x² + 87x – 1710 = 0

So we get

– 14x + 84 = 0

14x = 84

x = 84/14 = 6

Therefore, 6 is the number to be subtracted from each of the numbers.

8. If k + 3, k + 2, 3k – 7 and 2k – 3 are in proportion, find k.

Solution:

It is given that

k + 3, k + 2, 3k – 7 and 2k – 3 are in proportion

We can write it as

(k + 3) (2k – 3) = (k + 2) (3k – 7)

By further calculation

2k² – 3k + 6k – 9 = 3k² – 7k + 6k – 14

3k² – 7k + 6k – 14 – 2k² + 3k – 6k + 9 = 0

k² – 4k – 5 = 0

k² – 5k + k – 5 = 0

k(k – 5) + 1(k – 5) = 0

(k + 1) (k – 5) = 0

So,

k + 1 = 0 or k – 5 = 0

k = -1 or k = 5

Therefore, the value of k is -1, 5.

9. If x + 5 is the mean proportion between x + 2 and x + 9, find the value of x.

Solution:

It is given that

x + 5 is the mean proportion between x + 2 and x + 9

We can write it as

(x + 5)² = (x + 2) (x + 9)

By further calculation

x² + 10x + 25 = x² + 11x + 18

x² + 10x – x² – 11x = 18 – 25

So we get

– x = – 7

x = 7

Hence, the value of x is 7.

10. What number must be added to each of the numbers 16, 26 and 40 so that the resulting numbers may be in continued proportion?

Solution:

Consider x be added to each number

16 + x , 26 + x and 40 + x are in continued proportion

It can be written as

(16 + x)/ (26 + x) = (26 + x)/ (40 + x)

By cross multiplication

(16 + x) (40 + x) = (26 + x) (26 + x)

On further calculation

640 + 16x + 40x + x² = 676 + 26x + 26x + x²

640 + 56x + x² = 676 + 52x + x²

56x + x² – 52x – x² = 676 – 640

So we get

4x = 36

x = 36/4 = 9

Hence, 9 is the number to be added to each of the numbers.

11. Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.

Solution:

Consider a and b as the two numbers

It is given that 28 is the mean proportional

a: 28 :: 28: b

We get

ab = 28² = 784

Here a = 784/b …… (1)

We know that 224 is the third proportional

a: b :: b: 224

So we get

b² = 224a ….. (2)

Now by substituting the value of a in equation (2)

b² = 224 × 784/b

So we get

b³ = 224 × 784

b³ = 175616 = 56³

b = 56

By substituting the value of b in equation (1)

a = 784/56 = 14

Therefore, 14 and 56 are the two numbers.

12. If b is the mean proportional between a and c, prove that a, c, a² + b² and b² + c² are proportional.

Solution:

It is given that

b is the mean proportional between a and c

We can write it as

b² = a × c

b² = ac ….. (1)

We know that

a, c, a² + b² and b² + c² are in proportion

It can be written as

a/c = (a² + b²)/ (b² + c²)

By cross multiplication

a (b² + c²) = c (a² + b²)

Using equation (1)

a (ac + c²) = c (a² + ac)

So we get

ac (a + c) = a²c + ac²

Here ac (a + c) = ac (a + c) which is true.

Therefore, it is proved.

13. If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between (a² + b²) and (b² + c²).

Solution:

It is given that

b is the mean proportional between a and c

b² = ac …. (1)

Here (ab + bc) is the mean proportional between (a² + b²) and (b² + c²)

(ab + bc)² = (a² + b²) (b² + c²)

Consider LHS = (ab + bc)²

Expanding using formula

= a²b² + b²c² + 2ab²c

Using equation (1)

= a² (ac) + ac (c)² + 2a. ac. c

= a³c + ac³ + 2a²c²

Taking ac as common

= ac (a² + c² + 2ac)

= ac (a + c)²

RHS = (a² + b²) (b² + c²)

Using equation (1)

= (a² + ac) (ac + c²)

Taking common terms out

= a (a + c) c (a + c)

= ac (a + c)²

Hence, LHS = RHS.

14. If y is mean proportional between x and z, prove that xyz (x + y + z)³ = (xy + yz + zx)³.

Solution:

It is given that

y is the mean proportional between x and z

We can write it as

y² = xz …… (1)

Consider

LHS = xyz (x + y + z)³

It can be written as

= xz. y (x + y + z)³

Using equation (1)

= y² y (x + y + z)³

= y³ (x + y + z)³

So we get

= [y (x + y + z)]³

By further calculation

= (xy + y² + yz)³

Using equation (1)

= (xy + yz + zx)³

= RHS

Hence, it is proved.

15. If a + c = mb and 1/b + 1/d = m/c, prove that a, b, c and d are in proportion.

Solution:

It is given that

a + c = mb and 1/b + 1/d = m/c

a + c = mb

Dividing the equation by b

a/b + c/d = m ……. (1)

1/b + 1/d = m/c

Multiplying the equation by c

c/b + c/d = m …… (2)

Using equations (1) and (2)

a/b + c/b = c/b + c/d

So we get

a/b = c/d

Therefore, it is proved that a, b, c and d are in proportion.

16. If x/a = y/b = z/c, prove that

Solution:

It is given that

x/a = y/b = z/c

We can write it as

x = ak, y = bk and z = ck

Therefore, LHS = RHS.

Therefore, LHS = RHS.

Therefore, LHS = RHS.

17. If a/b = c/d = e/f prove that:

(i) (b² + d² + f²) (a² + c² + e²) = (ab + cd + ef)²

Solution:

Consider

a/b = c/d = e/f = k

So we get

a = bk, c = dk, e = fk

(i) LHS = (b² + d² + f²) (a² + c² + e²)

We can write it as

= (b² + d² + f²) (b²k² + d²k² + f²k²)

Taking out the common terms

= (b² + d² + f²) k² (b² + d² + f²)

So we get

= k² (b² + d² + f²)

RHS = (ab + cd + ef)²

We can write it as

= (b. kb + dk. d + fk. f)²

So we get

= (kb² + kd² + kf²)

Taking out common terms

= k² (b² + d² + f²)²

Therefore, LHS = RHS.

Therefore, LHS = RHS.

Therefore, LHS = RHS.

So we get

= bdf (k + 1 + k + 1 + k + 1)³

By further calculation

= bdf (3k + 3)³

= 27 bdf (k + 1)³

RHS = 27 (a + b) (c + d) (e + f)

It can be written as

= 27 (bk + b) (dk + d) (fk + f)

Taking out the common terms

= 27 b (k + 1) d (k + 1) f (k + 1)

So we get

= 27 bdf (k + 1)³

Therefore, LHS = RHS.

18. If ax = by = cz; prove that

Solution:

Consider ax = by = cz = k

It can be written as

x = k/a, y = k/b, z = k/c

19. If a, b, c and d are in proportion, prove that:

(i) (5a + 7v) (2c – 3d) = (5c + 7d) (2a – 3b)

(ii) (ma + nb): b = (mc + nd): d

(iii)(a⁴ + c⁴): (b⁴ + d⁴) = a²c²: b²d²

Solution:

It is given that

a, b, c, d are in proportion

Consider a/b = c/d = k

a = b, c = dk

(i) LHS = (5a + 7b) (2c – 3d)

Substituting the values

= (5bk + 7b) (2dk – 3d)

Taking out the common terms

= k (5b + 7b) k (2d – 3d)

So we get

= k² (12b) (-d)

= – 12 bd k²

RHS = (5c + 7d) (2a – 3b)

Substituting the values

= (5dk + 7d) (2kb – 3b)

Taking out the common terms

= k (5d + 7d) k (2b – 3b)

So we get

= k² (12d) (-b)

= – 12 bd k²

Therefore, LHS = RHS.

(ii) (ma + nb): b = (mc + nd): d

We can write it as

Therefore, LHS = RHS.

(iii)(a⁴ + c⁴): (b⁴ + d⁴) = a²c²: b²d²

We can write it as

Therefore, LHS = RHS.

Therefore, LHS = RHS.

Therefore, LHS = RHS.

Therefore, LHS = RHS.

Therefore, LHS = RHS.

So we get

= d² (1 + k²) + b² (1 + k²)

= (1 + k²) (b² + d²)

RHS = a² + b² + c² + d²

We can write it as

= b²k² + b² + d²k² + d²

Taking out the common terms

= b² (k² + 1) + d² (k² + 1)

= (k² + 1) (b² + d²)

Therefore, LHS = RHS.

20. If x, y, z are in continued proportion, prove that:

(x + y)²/ (y + z)² = x/z.

Solution:

It is given that

x, y, z are in continued proportion

Consider x/y = y/z = k

So we get

y = kz

x = yk = kz × k = k²z

Therefore, LHS = RHS.

21. If a, b, c are in continued proportion, prove that:

Solution:

It is given that

a, b, c are in continued proportion

Consider a/b = b/c = k

So we get

a = bk and b = ck ….. (1)

From equation (1)

a = (ck) k = ck² and b = ck

We know that

Therefore, LHS = RHS.

22. If a, b, c are in continued proportion, prove that:

(iii) a: c = (a² + b²): (b² + c²)

(iv) a²b²c² (a^-4 + b^-4 + c^-4) = b^-2 (a⁴ + b⁴ + c⁴)

(v) abc (a + b + c)³ = (ab + bc + ca)³

(vi) (a + b + c) (a – b + c) = a² + b² + c²

Solution:

It is given that

a, b, c are in continued proportion

So we get

a/b = b/c = k

Therefore, LHS = RHS.

Therefore, LHS = RHS.

(iii) a: c = (a² + b²): (b² + c²)

We can write it as

Therefore, LHS = RHS.

(iv) a²b²c² (a^-4 + b^-4 + c^-4) = b^-2 (a⁴ + b⁴ + c⁴)

Therefore, LHS = RHS.

(v) LHS = abc (a + b + c)³

We can write it as

= ck². ck. c [ck² + ck + c]³

Taking out the common terms

= c³ k³ [c (k² + k + 1)]³

So we get

= c³ k³. c³ (k² + k + 1)³

= c⁶ k³ (k² + k + 1)³

RHS = (ab + bc + ca)³

We can write it as

= (ck². ck + ck. c + c. ck²)³

So we get

= (c²k³ + c²k + c²k²)³

= (c²k³ + c²k² + c²k)³

Taking out the common terms

= [c²k (k² + k + 1)]³

= c⁶k³ (k² + k + 1)³

Therefore, LHS = RHS.

(vi) LHS = (a + b + c) (a – b + c)

We can write it as

= (ck² + ck + c) (ck² – ck + c)

Taking out the common terms

= c (k² + k + 1) c (k² – k + 1)

= c² (k² + k + 1) (k² – k + 1)

So we get

= c² (k⁴ + k² + 1)

RHS = a² + b² + c²

We can write it as

= (ck²)² + (ck)² + (c)²

So we get

= c²k⁴ + c²k² + c²

Taking out the common terms

= c² (k⁴ + k² + 1)

Therefore, LHS = RHS.

23. If a, b, c, d are in continued proportion, prove that:

(ii) (a² – b²) (c² – d²) = (b² – c²)²

(iii) (a + d) (b + c) – (a + c) (b + d) = (b – c)²

(iv) a: d = triplicate ratio of (a – b): (b – c)

Solution:

It is given that

a, b, c, d are in continued proportion

Here we get

a/b = b/c = c/d = k

c = dk, b = ck = dk . k = dk²

a = bk = dk² . k = dk³

Therefore, LHS = RHS.

(ii) LHS = (a² – b²) (c² – d²)

We can write it as

= [(dk³)² – (dk²)²] [(dk)² – d²]

By further calculation

= (d²k⁶ – d²k⁴) (d²k² – d²)

Taking out the common terms

= d²k⁴ (k² – 1) d² (k² – 1)

= d⁴k⁴ (k² – 1)²

RHS = (b² – c²)²

We can write it as

= [(dk²)² – (dk)²]²

By further calculation

= [d²k⁴ – d²k²]²

Taking out the common terms

= [d²k² (k² – 1)]²

= d⁴ k⁴ (k² – 1)²

Therefore, LHS = RHS.

(iii) LHS = (a + d) (b + c) – (a + c) (b + d)

We can write it as

= (dk³ + d) (dk² + dk) – (dk³ + dk) (dk² + d)

Taking out the common terms

= d (k³ + 1) dk (k + 1) – dk (k² + 1) d (k² + 1)

By further simplification

= d²k (k + 1) (k³ + 1) – d²k (k² + 1) (k² + 1)

So we get

= d²k (k⁴ + k³ + k + 1 – k⁴ – 2k² – 1)

= d²k (k³ – 2k² + k)

Taking k as common

= d²k² (k² – 2k + 1)

= d²k² (k – 1)²

RHS = (b – c)²

We can write it as

= (dk² – dk)²

Taking out the common terms

= d²k² (k – 1)²

Therefore, LHS = RHS.

(iv) a: d = triplicate ratio of (a – b): (b – c) = (a – b)³: (b – c)³

We know that

Therefore, LHS = RHS.

(v)

Therefore, LHS = RHS.

Exercise 7.3

1. If a: b :: c: d, prove that

(iii) (2a + 3b) (2c – 3d) = (2a – 3b) (2c + 3d)

(iv) (la + mb): (lc + mb) :: (la – mb): (lc – mb)

Solution:

(i) We know that

If a: b :: c: d we get a/b = c/d

By multiplying 2/5

2a/5b = 2c/5d

By applying componendo and dividendo

(2a + 5b)/ (2a – 5b) = (2c + 5d)/ (2c – 5d)

(ii) We know that

If a: b :: c: d we get a/b = c/d

By multiplying 5/11

5a/11b = 5c/11d

By applying componendo and dividendo

(5a + 11b)/ (5a – 11b) = (5c + 11d)/ (5c – 11d)

Now by applying alternendo

(5a + 11b)/ (5c + 11d) = (5a – 11b)/ (5c – 11d)

(iii) We know that

If a: b :: c: d we get a/b = c/d

By multiplying 2/3

2a/3b = 2c/3d

By applying componendo and dividendo

(2a + 3b)/ (2a – 3b) = (2c + 3d)/ (2c – 3d)

By cross multiplication

(2a + 3b) (2c – 3d) = (2a – 3b) (2c + 3d)

(iv) We know that

If a: b :: c: d we get a/b = c/d

By multiplying l/m

la/mb = lc/md

By applying componendo and dividendo

(la + mb)/ (la – mb) = (lc + md)/ (lc – md)

Now by applying alternendo

(la + mb)/ (lc + md) = (la – mb)/ (lc – md)

So we get

(la + mb): (lc + md) :: (la – mb): (lc – md)

2.

Solution:

Therefore, it is proved.

Therefore, it is proved.

3. If (4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d), prove that a, b, c, d are in proportion.

Solution:

It is given that

(4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d)

We can write it as

Therefore, it is proved that a, b, c, d are in proportion.

4. If (pa + qb): (pc + qd) :: (pa – qb): (pc – qd) prove that a: b :: c: d.

Solution:

It is given that

(pa + qb): (pc + qd) :: (pa – qb): (pc – qd)

We can write it as

Therefore, it is proved that a: b :: c: d.

5. If (ma + nb): b :: (mc + nd): d, prove that a, b, c, d are in proportion.

Solution:

It is given that

(ma + nb): b :: (mc + nd): d

We can write it as

(ma + nb)/ b = (mc + nd)/ d

By cross multiplication

mad + nbd = mbc + nbd

Here mad = mbc

ad = bc

By further calculation

a/b = c/d

Therefore, it is proved that a, b, c, d are in proportion.

6. If (11a² + 13b²) (11c² – 13d²) = (11a² – 13b²) (11c² + 13d²), prove that a: b :: c: d.

Solution:

It is given that

(11a² + 13b²) (11c² – 13d²) = (11a² – 13b²) (11c² + 13d²)

We can write it as

Therefore, it is proved that a: b :: c: d.

7.

Solution:

= 2(a – b)/ (a – b)

= 2

8.

Solution:

= 2(a – b)/ (a – b)

= 2

9.

Solution:

10. Using properties of properties, find x from the following equations:

Solution:

By cross multiplication

8 + 4x = 2 – x

So we get

4x + x = 2 – 8

5x = – 6

x = -6/5

By cross multiplication

49x – 490 = 9x + 36

49x – 9x = 36 + 490

So we get

40x = 526

x = 526/40

x = 263/20

By cross multiplication

50x – 75 = 12x + 1

50x – 12x = 1 + 75

So we get

38x = 76

x = 76/38 = 2

By cross multiplication

81x² – 45 = 36x²

81x² – 36x² = 45

So we get

45x² = 45

x² = 1

x = ± 1

x = 1, – 1

Verification:

(i) If x = 1

Hence, x = 1.

(ii) If x = -1

Here 1/5 ≠ 5/1

x = – 1 is not the solution

Therefore, x = 1.

11. Using properties of proportion solve for x. Given that x is positive.

(i) (ii)

Solution:

By cross multiplication

81x² – 45 = 36x²

81x² – 36x² = 45

So we get

45x² = 45

x² = 1

x = ± 1

x = 1, – 1

Verification:

(i) If x = 1

Hence, x = 1.

(ii)

Solution:

Upon cross-multiplication,

36x² = 25(4x² – 1)

36x² = 100x² – 25

64x² = 25

x² = 25/64

Taking square root on both sides,

x = √25/64

x = ± 5/8

Given that, x is positive

Thus the value of x = 5/8.

12. Solve

Solution:

x = 1/5

13. Solve for x:

Solution:

So we get

3x = a

x = a/3

So we get

x = 3a

Therefore, x = a/3, 3a.

14.

Solution:

It is given that

We get

2ax = x² + 1

x² – 2ax + 1 = 0

Therefore, it is proved.

15.

Solution:

16.

Solution:

It is given that

By cross multiplication

a + b = 5a – 5b

We can write it as

5a – a – 5b – b = 0

4a – 6b = 0

4a = 6b

We get.

a/b = 6/4

a/b = 3/2

∴ a: b = 3: 2

17.

Solution:

It is given that

By further calculation

2x/4 = 2y/3

x/2 = y/3

By cross multiplication

x/y = 2/3

Hence, the required ratio x: y is 2: 3.

18. Using the properties of proportion, solve the following equation for x; given

Solution:

It is given that

By cross multiplication

6x – 6 = 5x + 5

6x – 5x = 5 + 6

x = 11

19.

Solution:

It is given that

If x + y + z ≠ 0

Therefore, it is proved.

Chapter Test

1. Find the compound ratio of

(a + b)²: (a – b)², (a² – b²): (a² + b²) and (a⁴ – b⁴): (a + b)⁴

Solution:

(a + b)²: (a – b)²

(a² – b²): (a² + b²)

(a⁴ – b⁴): (a + b)⁴

We can write it as

2. If (7p + 3q): (3p – 2q) = 43: 2, find p: q.

Solution:

It is given that

(7p + 3q): (3p – 2q) = 43: 2

We can write it as

(7p + 3q)/ (3p – 2q) = 43/2

By cross multiplication

129p – 86q = 14p + 6q

129p – 14p = 6q + 86q

So we get

115p = 92q

By division

p/q = 92/115 = 4/5

Hence, p: q = 4: 5.

3. If a: b = 3: 5, find (3a + 5b): (7a – 2b).

Solution:

It is given that

a: b = 3: 5

We can write it as

a/b = 3/5

Here

(3a + 5b): (7a – 2b)

Now dividing the terms by b

Here

(3a + 5b): (7a – 2b) = 34: 11

4. The ratio of the shorter sides of a right angled triangle is 5: 12. If the perimeter of the triangle is 360 cm, find the length of the longest side.

Solution:

Consider the two shorter sides of a right-angled triangle as 5x and 12x

So the third longest side

= 13x

It is given that

5x + 12x + 13x = 360 cm

By further calculation

30x = 360

We get

x = 360/30 = 12

Here the length of the longest side = 13x

Substituting the value of x

= 13 × 12

= 156 cm

5. The ratio of the pocket money saved by Lokesh and his sister is 5: 6. If the sister saves Rs 30 more, how much more the brother should save in order to keep the ratio of their savings unchanged?

Solution:

Consider 5x and 6x as the savings of Lokesh and his sister.

Lokesh should save Rs y more

Based on the problem

(5x + y)/ (6x + 30) = 5/6

By cross multiplication

30x + 6y = 30x + 150

By further calculation

30x + 6y – 30x = 150

So we get

6y = 150

y = 150/6 = 25

Therefore, Lokesh should save Rs 25 more than his sister.

6. In an examination, the number of those who passed and the number of those who failed were in the ratio of 3: 1. Had 8 more appeared, and 6 less passed, the ratio of passed to failures would have been 2: 1. Find the number of candidates who appeared.

Solution:

Consider the number of passed = 3x

Number of failed = x

So the total candidates appeared = 3x + x = 4x

In the second case

Number of candidates appeared = 4x + 8

Number of passed = 3x – 6

Number of failed = 4x + 8 – 3x + 6 = x + 14

Ratio = 2: 1

Based on the condition

(3x – 6)/ (x + 14) = 2/1

By cross multiplication

3x – 6 = 2x + 28

3x – 2x = 28 + 6

x = 34

Here the number of candidates appeared = 4x = 4 × 34 = 136

7. What number must be added to each of the numbers 15, 17, 34 and 38 to make them proportional?

Solution:

Consider x be added to each number

So the numbers will be

15 + x, 17 + x, 34 + x and 38 + x

Based on the condition

(15 + x)/ (17 + x) = (34 + x)/ (38 + x)

By cross multiplication

(15 + x) (38 + x) = (34 + x) (17 + x)

By further calculation

570 + 53x + x² = 578 + 51x + x²

So we get

x² + 53x – x² – 51x = 578 – 570

2x = 8

x = 4

Hence, 4 must be added to each of the numbers.

8. If (a + 2b + c), (a – c) and (a – 2b + c) are in continued proportion, prove that b is the mean proportional between a and c.

Solution:

It is given that

(a + 2b + c), (a – c) and (a – 2b + c) are in continued proportion

We can write it as

(a + 2b + c)/ (a – c) = (a – c)/ (a – 2b + c)

By cross multiplication

(a + 2b + c) (a – 2b + c) = (a – c)²

On further calculation

a² – 2ab + ac + 2av – 4b² + 2bc + ac – 2bc + c² = a² – 2ac + c²

So we get

a² – 2ab + ac + 2ab – 4b² + 2bc + ac – 2bc + c² – a² + 2ac – c² = 0

4ac – 4b² = 0

Dividing by 4

ac – b² = 0

b² = ac

Therefore, it is proved that b is the mean proportional between a and c.

9. If 2, 6, p, 54 and q are in continued proportion, find the values of p and q.

Solution:

It is given that

2, 6, p, 54 and q are in continued proportion

We can write it as

2/6 = 6/p = p/54 = 54/q

(i) We know that

2/6 = 6/p

By cross multiplication

2p = 36

p = 18

(ii) We know that

p/54 = 54/q

By cross multiplication

pq = 54 × 54

Substituting the value of p

q = (54 × 54)/ 18 = 162

Therefore, the values of p and q are 18 and 162.

10. If a, b, c, d, e are in continued proportion, prove that: a: e = a⁴: b⁴.

Solution:

It is given that

a, b, c, d, e are in continued proportion

We can write it as

a/b = b/c = c/d = d/e = k

d = ek, c = ek², b = ek³ and a = ek⁴

Here

LHS = a/e

Substituting the values

= ek⁴/ e

= k⁴

RHS = a⁴/b⁴

Substituting the values

= (ek⁴)⁴/ (ek³)⁴

So we get

= e⁴k¹⁶/ e⁴k¹²

= k^{16 – 12}

= k⁴

Hence, it is proved that a: e = a⁴: b⁴.

11. Find two numbers whose mean proportional is 16 and the third proportional is 128.

Solution:

Consider x and y as the two numbers

Mean proportion = 16

Third proportion = 128

√xy = 16

xy = 256

Here

x = 256/y ….. (1)

y²/x = 128

Here

x = y²/128 …. (2)

Using both the equations

256/y = y³/ 128

By cross multiplication

y³ = 256 × 128 = 32768

y³ = 32³

y = 32

Substituting the value of y in equation (1)

x = 256/y

So we get

x = 256/32 = 8

Hence, the two numbers are 8 and 32.

12. If q is the mean proportional between p and r, prove that:

Solution:

It is given that

q is the mean proportional between p and r

q² = pr

Here

LHS = p² – 3q² + r²

We can write it as

= p² – 3pr + r²

So we get

= r² – 3pr + p²

Here

LHS = RHS

Therefore, it is proved.

13. If a/b = c/d = e/f, prove that each ratio is

Solution:

It is given that

a/b = c/d = e/f = k

So we get

a = k, c = dk, e = fk

Therefore, it is proved.

= k

Therefore, it is proved.

14. If x/a = y/b = z/c, prove that

Solution:

It is given that

x/a = y/b = z/c = k

So we get

x = ak, y = bk, z = ck

Here

= k³

= k³

Hence, LHS = RHS.

15. If x: a = y: b, prove that

Solution:

We know that

x/a = y/b = k

So we get

x = ak, y = bk

Here

Here LHS = RHS

Therefore, it is proved.

16.

Solution:

Consider

So we get

x = k (b + c – a)

y = k (c + a – b)

z = k (a + b – a)

Here

= k

Therefore, it is proved.

17. If a: b = 9: 10, find the value of

Solution:

It is given that

a: b = 9: 10

So we get

a/b = 9/10

= 5

18. If (3x² + 2y²): (3x² – 2y²) = 11: 9, find the value of

Solution:

It is given that

(3x² + 2y²): (3x² – 2y²) = 11: 9

We can write it as

Here

19.

Solution:

It is given that

20.

Solution:

It is given that

= RHS

21.

Solution:

It is given that

22.

Solution:

It is given that

By cross multiplication

x³ + 3x = 3ax² + a

x³ – 3ax² + 3x – a = 0

Therefore, it is proved.