# Frank Solutions for Class 10 Maths Chapter 9 Ratio and Proportion

Frank Solutions for Class 10 Maths Chapter 9 Ratio and Proportion are given here. The subject experts at BYJU’S created these solutions to guide you in solving a variety of questions in many methods and managing time in examinations. We, therefore, suggest that students solve Frank Solutions for Class 10 Maths to obtain good marks in the board exam.

Chapter 9, Ratio and Proportion, contains questions related to the ratio in simplest form and the comparison of ratios. The ratio of two quantities of the same kind and in the same unit is the fraction that one quantity is of others. The ratio â€˜aâ€™ to â€˜bâ€™ is the fraction (a/b), written as a:b. Some of the important topics covered in this chapter are: ratio, proportion, compound ratio, duplicate ratio, triplicate ratio, sub-duplicate ratio, sub-triplicate ratio, reciprocal ratio, etc.

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1. Find the ratio of the following in the simplest form:

(i) 5.60 and 2.40

Solution:-

Given numbers can be written as 5.60/2.40

Now, shifting the decimal points, we get,

= 560/240

= 56/24

= 28/12 â€¦ [because diving both by 2]

= 14/6 â€¦ [because diving both by 2]

= 7/3 â€¦ [because diving both by 2]

Therefore, the ratio of the given number is 7: 3.

(ii) 432 and 120

Solution:-

Given numbers can be written as, 432/120

= 432/120

= 216/60 â€¦ [because diving both by 2]

= 108/30 â€¦ [because diving both by 2]

= 54/15 â€¦ [because diving both by 2]

= 18/5 â€¦ [because diving both by 3]

Therefore, the ratio of the given number is 18: 5.

(iii) â‚¹ 5.40 and 180 paise

Solution:-

Given numbers can be written as, 5.40/180

We know that, â‚¹ 1 = 100 paise

So, â‚¹ 5.40 = 540 paise

= 540/180

= 54/18

= 3/1 â€¦ [because diving both by 3]

Therefore, the ratio of the given number is 3: 1.

(iv) a4 + b4 and a3 â€“ b3

Solution:-

The given question can be written as,

= (a4 + b4)/(a3 â€“ b3)

We know that, a4 + b4 = (a – b) (a3 + ab2Â + a2b + b3)

Then, a3 â€“ b3 = (a – b) (a2 + ab + b2)

So, [(a – b) (a3 + ab2 + a2b + b3)]/[(a – b) (a2 + ab + b2)]

= (a3 + ab2 + a2b + b3)/(a2 + ab + b2)

Therefore, the ratio of the given terms is (a3 + ab2 + a2b + b3): (a2 + ab + b2).

v. x2 + 4x + 4 and x2 â€“ x -6

Solution:-

The given question can be written as,

= (x2 + 4x + 4)/(x2 â€“ x â€“ 6)

= (x + 2)2/[(x – 3) (x + 2)]

= (x + 2)/(x – 3)

Therefore, ratio of the given terms is (x + 2): (x – 3)

2. If a: b = 4: 7, find the following

(i) (5a + 2b)/(5a â€“ 2b)

Solution:-

From the question, it is given that,

a: b = 4: 7

a/b = 4/7

(5a + 2b)/(5a â€“ 2b)

Now, divide both numerator and denominator by â€˜bâ€™ we get,

= [(5a/b) + (2b/b)]/[(5a/b) â€“ (2b/b)]

= [(5a/b) + 2]/[(5a/b) – 2]

Now, substitute the value of a and b we get,

= [(5(4/7)) + 2]/[(5(4/7)) – 2]

= ((20/7) + 2)/((20/7) – 2)

= 34/6

= 17/3

(ii) (6a – b)/(a + 3b)

Solution:-

From the question, it is given that,

a: b = 4: 7

a/b = 4/7

(6a – b)/(a + 3b)

Now, divide both numerator and denominator by â€˜bâ€™ we get,

= [(6a/b) – (b/b)]/[(a/b) + (3b/b)]

= [(6a/b) – 1]/[(a/b) + 3]

Now, substitute the value of a and b we get,

= [(6(4/7)) – 1]/[((4/7)) + 3]

= ((24/7) – 1)/((4/7) + 3)

= 17/25

(iii) (5a – 4b)/(2a – 3b)

Solution:-

From the question, it is given that,

a: b = 4: 7

a/b = 4/7

(5a – 4b)/(2a – 3b)

Now, divide both numerator and denominator by â€˜bâ€™ we get,

= [(5a/b) – (4b/b)]/[(2a/b) – (3b/b)]

= [(5a/b) – 4]/[(2a/b) – 3]

Now, substitute the value of a and b we get,

= [(5(4/7)) – 4]/[(2(4/7)) – 3]

= ((20/7) – 4)/((8/7) – 3)

= -8/-13

= 8/13

3. If m: n = 3: 8, find the value of (3m + 2n): (5m + n)

Solution:-

From the question, it is given that,

m: n = 3: 8

m/n = 3/8

(3m + 2n)/(5m + n)

Now, divide both numerator and denominator by â€˜nâ€™ we get,

= [(3m/n) + (2n/n)]/[(5m/n) + (n/n)]

= [(3m/n) + 2]/[(5m/n) + 1]

Now, substitute the value of m and n we get,

= [(3(3/8)) + 2]/[(5(3/8)) + 1]

= ((9/8) + 2)/((15/8) + 1)

= 25/23

Therefore, the value of (3m + 2n): (5m + n) = 25: 23

4. A manâ€™s monthly income is â‚¹ 5,000. He saves every month a minimum of â‚¹ 800. Find the ratio of his:

(i) Annual expenses to annual income.

(ii) Monthly savings to monthly expenses.

Solution:-

From the question, it is given that,

Monthly income of a man = â‚¹ 5,000

Every month man saves â‚¹ 800

(i) Annual expenses to annual income,

Annual income = monthly income Ã— 12

= â‚¹ 5,000 Ã— 12

= â‚¹ 60,000

Then, monthly expenses = â‚¹ 5,000 â€“ 800

= â‚¹ 4,200

Annual expenses = monthly expenses Ã— 12

= â‚¹ 4,200 Ã— 12

= â‚¹ 50,400

Annual expenses/Annual income = 50,400/60,000

= 504/600

= 21/25

Therefore, Annual expenses: annual income = 21: 25

(ii) Monthly savings to monthly expenses,

Monthly expenses = â‚¹ 5,000 – â‚¹ 800 = â‚¹ 4,200

Then, monthly savings/monthly expenses = 800/4,200

= 8/42

= 4/21

Therefore, monthly savings: monthly expenses = 4: 21.

5. If a + b: a â€“ b = 11: 8; find the value of a: b

Solution:-

From the question, it is given that, a + b: a â€“ b = 11: 8.

(a + b)/(a – b) = 11/8

By cross multiplication, we get,

8(a + b) = 11(a – b)

8a + 8b = 11a â€“ 11b

Transposing we get,

11b + 8b = 11a â€“ 8a

19b = 3a

19/3 = a/b

a: b = 19: 3

6. If p: q = 2: 5, q: r = 4: 3, then find p: r

Solution:-

From the question it is given that, p: q = 2: 5, q: r = 4: 3

So, p/q = 2/5

q/r = 4/3

(p/q) Ã— (q/r) = (2/5) Ã— (4/3)

By simplification, we get,

p/r = 8/15

Therefore, the value of p: r = 8: 15

7. If a: e = 5: 12, e: i = 8: 3 and i: u = 9: 16, then find a: u

Solution:-

From the question it is given that, a: e = 5: 12, e: i = 8: 3 and i: u = 9: 16

So, a/e = 5/12

e/i = 8/3

i/u = 9/16

(a/e) Ã— (e/i) Ã— (i/u) = (5/12) Ã— (8/3) Ã— (9/16)

By simplification, we get,

a/u = 10/16

a/u = 5/8

Therefore, the value of a: u = 5: 8

8. Find the compounded ratio of the following:

(i) 15: 16 and 8: 5

Solution:-

The given ratio can be written as,

15/16 and 8/5

= 15/16 Ã— 8/5

= (15 Ã— 8)/(16 Ã— 5)

= (3 Ã— 1)/(2 Ã— 1)

= 3/2

Therefore, the compounded ratio of 15: 16 and 8: 5 is 3: 2.

(ii) (a2 â€“ b2): (a2 + b2) and (a4 â€“ b4): (a + b)4

Solution:-

The given ratio can be written as,

(a2 â€“ b2)/(a2Â + b2) and (a4 â€“ b4)/(a + b)4

= (a2 â€“ b2)/(a2Â + b2) Ã— (a4 â€“ b4)/(a + b)4

We know that, (a2 â€“ b2) = (a + b) (a – b)

= ((a + b) (a – b))/(a2 + b2) Ã— ((a2 + b2) (a2 â€“ b2))

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

= (a – b)2/(a + b)2

Therefore, the compounded ratio is (a – b)2: (a + b)2.

(iii) 3: 5, 7: 9 and 15: 28

Solution:-

The given ratio can be written as,

3/5, 7/9 and 15/28

= 3/5 Ã— 7/9 Ã— 15/28

= (3 Ã— 7 Ã— 15)/(5 Ã— 9 Ã— 28)

= (1 Ã— 1 Ã— 3)/ (1 Ã— 3 Ã— 4)

= (1 Ã— 1 Ã— 1)/(1 Ã— 1 Ã— 4)

= Â¼

Therefore, the compounded ratio is 1: 4.

(iv) âˆš8: 4, 3: âˆš5 and âˆš20: âˆš27

Solution:-

The given ratio can be written as,

âˆš8/4, 3/âˆš5 and âˆš20/âˆš27

= âˆš8/4 Ã— 3/âˆš5 Ã— âˆš20/âˆš27

= 2âˆš2/4 Ã— 3/âˆš5 Ã— 2âˆš5/3âˆš3

= âˆš2/âˆš3

Therefore, the compounded ratio is âˆš2: âˆš3.

(v) (m – n): (m + n), (m + n)2: (m2 + n2) and (m4 â€“ n4): (m2 â€“ n2)2

Solution:-

The given ratio can be written as,

(m – n)/(m + n), (m + n)2/(m2 + n2) and (m4 â€“ n4)/(m2 â€“ n2)2

= (m – n)/(m + n) Ã— (m + n)2/(m2 + n2) Ã— (m4 â€“ n4)/(m2 â€“ n2)2

= (m – n)/(1) Ã— (m + n)/(m2 + n2) Ã— ((m2 + n2) (m2 â€“ n2))/(m2 â€“ n2)2

By simplification, we get,

= 1/1

Therefore, the compounded ratio is 1 : 1.

9. Find the duplicate ratio of the following:

(i) âˆš10 : âˆš14

Solution:-

Given, âˆš10 : âˆš14

= (âˆš10)2: (âˆš14)2

= 10: 14

= 10/14

= 5/7

Therefore, the duplicate ratio is 5: 7.

(ii) 3âˆš2a : 2âˆš3a

Solution:-

Given, 3âˆš2a: 2âˆš3a

= (3âˆš2a)2: (2âˆš3a)2

= 18a: 12a

= 18a/12a

= 3/2

Therefore, the duplicate ratio is 3: 2.

(iii) 2/3: 4/9

Solution:-

Given, 2/3: 4/9

= (2/3)2: (4/9)2

= 4/9: 16/81

= (4/9) Ã— (81/16)

= (4 Ã— 81)/(9 Ã— 16)

= (1 Ã— 9)/(1 Ã— 4)

= 9/4

Therefore, duplicate ratio = 9: 4

(iv) (a + b): (a2 â€“ b2)

Solution:-

Given, (a + b): (a2 â€“ b2)

= (a + b)2 : (a2 â€“ b2)2

= (a + b)2/((a + b)2(a – b)2)

= 1/(a – b)2

Therefore, duplicate ratio = 1: (a – b)2

10. Find the triplicate ratio of the following:

(i) 3: 5

Solution:-

Given, 3: 5

= 33 : 53

= 27: 125

Therefore, the triplicate ratio is 27: 125

(ii) 2âˆš5 : 5âˆš2

Solution:-

Given, 2âˆš5 : 5âˆš2

= (2âˆš5)3: (5âˆš2)3

= (8 Ã— 5âˆš5)/(125 Ã— 2âˆš2)

By simplification,

= 4âˆš5: 25âˆš2

Therefore, the triplicate ratio = 4âˆš5: 25âˆš2

(iii) âˆš15 : âˆš18

Solution:-

Given, âˆš15: âˆš18

= (âˆš15)3: (âˆš18)3

= 15âˆš5: 18 Ã— 3âˆš2

= 5âˆš15: 18âˆš2

Therefore, the triplicate ratio is 5âˆš15: 18âˆš2

(iv) 3âˆš(ab)2: 3âˆš(a2b)

Solution:-

Given, 3âˆš(ab)2: 3âˆš(a2b)

By simplification, we get,

= (3âˆš(ab)2)3: (3âˆš(a2b))3

= ab2 : a2b

= b: a

Therefore, the triplicate ratio is b: a

11. Find the subâ€“duplicate ratio of the following:

(i) x6: y4

Solution:-

Given, x6: y4

= âˆšx6: âˆšy4

= (x6)1/2: (y4)1/2

= x3: y2

Therefore, the sub-duplicate ratio is x3:y2

(ii) 63m2: 28n2

Solution:-

Given, 63m2: 28n2

= âˆš(63m2): âˆš(28n2)

= 3âˆš7m: 2âˆš7n

= 3m: 2n

Therefore, the sub-duplicate ratio is 3m: 2n

(iii) 1/16: 1/36

Solution:-

Given, 1/16: 1/36

= âˆš(1/16): âˆš(1/36)

= Â¼: 1/6

= (Â¼)/(1/6)

= (Â¼) Ã— (6/1)

= 3/2

Therefore, the sub-duplicate ratio is 3: 2

(iv) 9a2/5: 25a2/3

Solution:-

Given, 9a2/5: 25a2/3

= âˆš(9a2/5): âˆš(25a2/3)

= 3a(1/âˆš5): 5a(1/âˆš3)

= 3âˆš3: 5âˆš5

Therefore, the sub-duplicate ratio is 3âˆš3: 5âˆš5.

12. Find the sub â€“ triplicate ratio of the following:

(i) 512: 216

Solution:-

Given, 512: 216

= 3âˆš512 : 3âˆš216

= (83)1/3: (63)1/3

= 8: 6

= 8/6

= 4/3

Therefore, the sub-triplicate ratio is 4: 3.

(ii) m3n6: m6n3

Solution:-

Given, m3n6: m6n3

= 3âˆš(m3n6) : 3âˆš(m6n3)

= (m3n6)1/3 : (m6n3)1/3

= mn2: m2n

= mn2/m2n

= n/m

Therefore, the sub-triplicate ratio is n: m.

(iii) 125a3: 343b6

Solution:-

Given, 125a3: 343b6

= 3âˆš(125a3): 3âˆš(343b6)

= (125a3)1/3: (343b6)1/3

= 5a: 7b2

Therefore, the sub-triplicate ratio is 5a: 7b2.

(iv) 64m3/729n3: 216m3/27n3

Solution:-

Given, 64m3/729n3: 216m3/27n3

= 3âˆš(64m3/729n3): 3âˆš(216m3/27n3)

= (64m3/729n3)1/3: (216m3/27n3)1/3

By simplification, we get,

= 4m/9n: 6m/3n

= (4m/9n) Ã— (3n/6m)

= 2/9

Therefore, the sub-triplicate ratio is 2: 9.

13. Find the reciprocal ratio of the following:

(i) 17/45: 51/27

Solution:-

Given ratio, 17/45: 51/27

The reciprocal of the given ratio is 45/17: 51/27

= (45/17) Ã— (51/27)

= (45/1) Ã— (3/27)

= (45/1) Ã— (1/9)

= 5/1

Therefore, the reciprocal of the ratio is 5: 1

(ii) 1/45: 1/54

Solution:-

Given ratio, 1/45: 1/54

The reciprocal of the given ratio is 45/1: 54/1

= 45: 54

= 45/54

= 5/6

Therefore, the reciprocal of the ratio is 5: 6

(iii) a3b2: a2b3

Solution:-

Given ratio, a3b2: a2b3

The reciprocal of the given ratio is 1/a3b2: 1/a2b3

= (1/a3b2) Ã— (a2b3/1)

= b: a

Therefore, the reciprocal of the ratio is b: a

(iv) 81pq2: 54p2q

Solution:-

Given ratio, 81pq2: 54p2q

The reciprocal of the given ratio 1/81pq2: 1/54p2q

= (1/81pq2) Ã— (54p2q/1)

By simplification, we get,

= 2p/3q

Therefore, the reciprocal of the ratio is 2p: 3q

14. Which of the following ratios is greater?

(i) 3: 5 and 2: 11

Solution:-

The given ratio can be written as 3/5 and 2/11

Then,

3 Ã— 11 > 2 Ã— 5

33 > 10

Therefore, 3: 5 > 2: 11

So, 3: 5 is greater.

(ii) 2: 3 and 13: 19

Solution:-

Given ratio can be written as, 2/3 and 13/19

Then,

2 Ã— 19 < 3 Ã— 13

38 < 39

Therefore, 2: 3 < 13: 19

So, 13: 19 is greater.

(iii) 5: 8 and 7: 10

Solution:-

Given ratio can be written as 5/8 and 7/10

Then,

5 Ã— 10 < 8 Ã— 7

50 < 56

Therefore, 5: 8 < 7: 10

So, 7: 10 is greater.

(iv) (5/2): (15/4) and (5/3): (11/6)

Solution:-

Given ratio, (5/2): (15/4) and (5/3): (11/6).

(5/2): (15/4) = (5/3) Ã— (4/15) = 2/3

(5/3): (11/6) = (5/3) Ã— (6/11) = 10/11

Consider, 2/3 : 10/11

2 Ã— 11 < 3 Ã— 10

22 < 30

2: 3 < 10: 11

5/2 : 15/4 < 5/3 : 11/6

Therefore, 5/3: 11/6 is greater.

15. Two numbers are in the ratio 7: 10. If 8 is added to each number, the ratio

becomes 3: 4. Find the numbers.

Solution:-

From the question, it is given that,

Two numbers are in the ratio 7: 10.

Let us assume the two numbers be 7y and 10y.

Then, (7y + 8)/(10y Ã— 8) = Â¾

28y + 32 = 30x + 24

2y = 8

y = 8/2

y = 4

So, 7y = 7 Ã— 4 = 28

10y = 10 Ã— 4 = 40

Therefore, the two numbers are 28 and 40.