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 course in 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 of the prescribed board.

PDF Solution for ML Aggarwal Class 10 Maths Chapter 7 :-Click Here to Download PDF

 

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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 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

[4 (x/y) – 3] [5 (x/y) – 4] = 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

[4 (x/y) – 3] [5 (x/y) – 4] = 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

[8x – 25/2]/ [7x – 25/2] = 11/9

Taking LCM

[(16x – 25)/ 2]/ [(14x – 25)/ 2] = 11/9

By further calculation

[(16x – 25) × 2]/ [2 (14x – 25)] = 11/9

(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 original weight = 91 kg

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

(ii) Amount collected for charity = Rs 2100

Here the ratio between 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 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

No. 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 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 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion, find the value of x.

Solution:

It is given that

2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion

We can write it as

(2x – 1) (15x – 9) = (5x – 6) (6x + 2)

By further calculation

30x2 – 18x – 15x + 9 = 30x2 + 10x – 36x – 12

30x2 – 33x + 9 = 30x2 – 26x – 12

30x2 – 33x – 30x2 + 26x = – 12 – 9

So we get

-7x = – 21

x = -21/-7 = 3

Therefore, the value of x is 3.

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 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 equation (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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 1

Solution:

It is given that

x/a = y/b = z/c

We can write it as

x = ak, y = bk and z = ck

ML Aggarwal Solutions for Class 10 Chapter 7 Image 2

Therefore, LHS = RHS.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 3

Therefore, LHS = RHS.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 4

ML Aggarwal Solutions for Class 10 Chapter 7 Image 5

Therefore, LHS = RHS.

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

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 6

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.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 7

Therefore, LHS = RHS.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 8

ML Aggarwal Solutions for Class 10 Chapter 7 Image 9

Therefore, LHS = RHS.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 10

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 11

Solution:

Consider ax = by = cz = k

It can be written as

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 12

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 13

ML Aggarwal Solutions for Class 10 Chapter 7 Image 14

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 15

ML Aggarwal Solutions for Class 10 Chapter 7 Image 16

Therefore, LHS = RHS.

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

We can write it as

ML Aggarwal Solutions for Class 10 Chapter 7 Image 17

Therefore, LHS = RHS.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 18

Therefore, LHS = RHS.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 19

ML Aggarwal Solutions for Class 10 Chapter 7 Image 20

Therefore, LHS = RHS.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 21

ML Aggarwal Solutions for Class 10 Chapter 7 Image 22

Therefore, LHS = RHS.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 23

Therefore, LHS = RHS.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 24

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 25

ML Aggarwal Solutions for Class 10 Chapter 7 Image 26

Therefore, LHS = RHS.

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 27

Solution:

It is given that

a, b, c are in continued proportion

ML Aggarwal Solutions for Class 10 Chapter 7 Image 28

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 29

ML Aggarwal Solutions for Class 10 Chapter 7 Image 30

Therefore, LHS = RHS.

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 31

(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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 32

ML Aggarwal Solutions for Class 10 Chapter 7 Image 33

Therefore, LHS = RHS.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 34

ML Aggarwal Solutions for Class 10 Chapter 7 Image 35

Therefore, LHS = RHS.

(iii) a: c = (a2 + b2): (b2 + c2)

We can write it as

ML Aggarwal Solutions for Class 10 Chapter 7 Image 36

ML Aggarwal Solutions for Class 10 Chapter 7 Image 37

Therefore, LHS = RHS.

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 38

ML Aggarwal Solutions for Class 10 Chapter 7 Image 39

ML Aggarwal Solutions for Class 10 Chapter 7 Image 40

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:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 41

(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)

ML Aggarwal Solutions for Class 10 Chapter 7 Image 42

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 43

ML Aggarwal Solutions for Class 10 Chapter 7 Image 44

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 45

Therefore, LHS = RHS.

(v)

ML Aggarwal Solutions for Class 10 Chapter 7 Image 46

ML Aggarwal Solutions for Class 10 Chapter 7 Image 47

ML Aggarwal Solutions for Class 10 Chapter 7 Image 48

Therefore, LHS = RHS.

Exercise 7.3

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 49

(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.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 50

Solution:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 51

Therefore, it is proved.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 52

ML Aggarwal Solutions for Class 10 Chapter 7 Image 53

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 54

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 55

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 56

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

7. If (a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d), prove that a: b:: c: d.

Solution:

It is given that

(a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d)

We can write it as

ML Aggarwal Solutions for Class 10 Chapter 7 Image 57

ML Aggarwal Solutions for Class 10 Chapter 7 Image 58

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

8.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 59

Solution:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 60

ML Aggarwal Solutions for Class 10 Chapter 7 Image 61

= 2(a – b)/ (a – b)

= 2

9.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 62

Solution:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 63

ML Aggarwal Solutions for Class 10 Chapter 7 Image 64

= 2(a – b)/ (a – b)

= 2

10.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 65

Solution:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 66

ML Aggarwal Solutions for Class 10 Chapter 7 Image 67

ML Aggarwal Solutions for Class 10 Chapter 7 Image 68

11.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 69

Solution:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 70

ML Aggarwal Solutions for Class 10 Chapter 7 Image 71

By cross multiplication

36x × 25 = 16 (36x + 1)

900x = 576x + 16

900x – 576x = 16

So we get

324x = 16

x = 16/324

x = 4/81

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 72

Solution:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 73

By cross multiplication

8 + 4x = 2 – x

So we get

4x + x = 2 – 8

5x = – 6

x = -6/5

ML Aggarwal Solutions for Class 10 Chapter 7 Image 74

ML Aggarwal Solutions for Class 10 Chapter 7 Image 75

By cross multiplication

49x – 490 = 9x + 36

49x – 9x = 36 + 490

So we get

40x = 526

x = 526/40

x = 263/20

ML Aggarwal Solutions for Class 10 Chapter 7 Image 76

ML Aggarwal Solutions for Class 10 Chapter 7 Image 77

ML Aggarwal Solutions for Class 10 Chapter 7 Image 78

ML Aggarwal Solutions for Class 10 Chapter 7 Image 79

By cross multiplication

50x – 75 = 12x + 1

50x – 12x = 1 + 75

So we get

38x = 76

x = 76/38 = 2

ML Aggarwal Solutions for Class 10 Chapter 7 Image 80

ML Aggarwal Solutions for Class 10 Chapter 7 Image 81

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 82

Hence, x = 1.

(ii) If x = -1

ML Aggarwal Solutions for Class 10 Chapter 7 Image 83

Here 1/5 ≠ 5/1

x = – 1 is not the solution

Therefore, x = 1.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 84

13. Using properties of proportion solve for x. Give that x is positive.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 85

Solution:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 86

ML Aggarwal Solutions for Class 10 Chapter 7 Image 87

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 88

Hence, x = 1.

14. Solve

ML Aggarwal Solutions for Class 10 Chapter 7 Image 89

Solution:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 90

x = 1/5

15. Solve for x:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 91

Solution:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 92

ML Aggarwal Solutions for Class 10 Chapter 7 Image 93

So we get

3x = a

x = a/3

ML Aggarwal Solutions for Class 10 Chapter 7 Image 94

So we get

x = 3a

Therefore, x = a/3, 3a.

16.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 95

Solution:

It is given that

ML Aggarwal Solutions for Class 10 Chapter 7 Image 96

We get

2ax = x2 + 1

x2 – 2ax + 1 = 0

Therefore, it is proved.

17. ML Aggarwal Solutions for Class 10 Chapter 7 Image 97

Solution:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 98

ML Aggarwal Solutions for Class 10 Chapter 7 Image 99

18.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 100

Solution:

It is given that

ML Aggarwal Solutions for Class 10 Chapter 7 Image 101

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

19.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 102

Solution:

It is given that

ML Aggarwal Solutions for Class 10 Chapter 7 Image 103

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.

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 104

Solution:

It is given that

ML Aggarwal Solutions for Class 10 Chapter 7 Image 105

By cross multiplication

6x – 6 = 5x + 5

6x – 5x = 5 + 6

x = 11

21.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 106

Solution:

It is given that

ML Aggarwal Solutions for Class 10 Chapter 7 Image 107

ML Aggarwal Solutions for Class 10 Chapter 7 Image 108

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)

(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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 109

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 110

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 111

= 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

= k16 – 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:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 112

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 113

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 114

Solution:

It is given that

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

So we get

a = k, c = dk, e = fk

ML Aggarwal Solutions for Class 10 Chapter 7 Image 115

Therefore, it is proved.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 116

ML Aggarwal Solutions for Class 10 Chapter 7 Image 117

= k

Therefore, it is proved.

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 118

Solution:

It is given that

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

So we get

x = ak, y = bk, z = ck

Here

ML Aggarwal Solutions for Class 10 Chapter 7 Image 119

= k3

ML Aggarwal Solutions for Class 10 Chapter 7 Image 120

= k3

Hence, LHS = RHS.

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 121

Solution:

We know that

x/a = y/b = k

So we get

x = ak, y = bk

Here

ML Aggarwal Solutions for Class 10 Chapter 7 Image 122

ML Aggarwal Solutions for Class 10 Chapter 7 Image 123

ML Aggarwal Solutions for Class 10 Chapter 7 Image 124

Here LHS = RHS

Therefore, it is proved.

16.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 125

Solution:

Consider

ML Aggarwal Solutions for Class 10 Chapter 7 Image 126

So we get

x = k (b + c – a)

y = k (c + a – b)

z = k (a + b – a)

Here

ML Aggarwal Solutions for Class 10 Chapter 7 Image 127

ML Aggarwal Solutions for Class 10 Chapter 7 Image 128

= k

Therefore, it is proved.

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 129

Solution:

It is given that

a: b = 9: 10

So we get

a/b = 9/10

ML Aggarwal Solutions for Class 10 Chapter 7 Image 130

= 5

ML Aggarwal Solutions for Class 10 Chapter 7 Image 131

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

ML Aggarwal Solutions for Class 10 Chapter 7 Image 132

Solution:

It is given that

(3x2 + 2y2): (3x2 – 2y2) = 11: 9

We can write it as

ML Aggarwal Solutions for Class 10 Chapter 7 Image 133

Here

ML Aggarwal Solutions for Class 10 Chapter 7 Image 134

ML Aggarwal Solutions for Class 10 Chapter 7 Image 135

19.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 136

Solution:

It is given that

ML Aggarwal Solutions for Class 10 Chapter 7 Image 137

ML Aggarwal Solutions for Class 10 Chapter 7 Image 138

20.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 139

Solution:

It is given that

ML Aggarwal Solutions for Class 10 Chapter 7 Image 140

ML Aggarwal Solutions for Class 10 Chapter 7 Image 141

= RHS

21.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 142

Solution:

It is given that

ML Aggarwal Solutions for Class 10 Chapter 7 Image 143

ML Aggarwal Solutions for Class 10 Chapter 7 Image 144

ML Aggarwal Solutions for Class 10 Chapter 7 Image 145

22.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 146

Solution:

It is given that

ML Aggarwal Solutions for Class 10 Chapter 7 Image 147

By cross multiplication

x3 + 3x = 3ax2 + a

x3 – 3ax2 + 3x – a = 0

Therefore, it is proved.

23.

ML Aggarwal Solutions for Class 10 Chapter 7 Image 148

Solution:

ML Aggarwal Solutions for Class 10 Chapter 7 Image 149

ML Aggarwal Solutions for Class 10 Chapter 7 Image 150

Therefore, it is proved.

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