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 ‘a’ to ‘b’ 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 … [because diving both by 2]
= 14/6 … [because diving both by 2]
= 7/3 … [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 … [because diving both by 2]
= 108/30 … [because diving both by 2]
= 54/15 … [because diving both by 2]
= 18/5 … [because diving both by 3]
Therefore, the ratio of the given number is 18: 5.
(iii) ₹ 5.40 and 180 paise
Solution:-
Given numbers can be written as, 5.40/180
We know that, ₹ 1 = 100 paise
So, ₹ 5.40 = 540 paise
= 540/180
= 54/18
= 3/1 … [because diving both by 3]
Therefore, the ratio of the given number is 3: 1.
(iv) a4 + b4 and a3 – b3
Solution:-
The given question can be written as,
= (a4 + b4)/(a3 – b3)
We know that, a4 + b4 = (a – b) (a3 + ab2Â + a2b + b3)
Then, a3 – 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 – x -6
Solution:-
The given question can be written as,
= (x2 + 4x + 4)/(x2 – x – 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 – 2b)
Solution:-
From the question, it is given that,
a: b = 4: 7
a/b = 4/7
(5a + 2b)/(5a – 2b)
Now, divide both numerator and denominator by ‘b’ we get,
= [(5a/b) + (2b/b)]/[(5a/b) – (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 ‘b’ 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 ‘b’ 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 ‘n’ 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 man’s monthly income is ₹ 5,000. He saves every month a minimum of ₹ 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 = ₹ 5,000
Every month man saves ₹ 800
(i) Annual expenses to annual income,
Annual income = monthly income × 12
= ₹ 5,000 × 12
= ₹ 60,000
Then, monthly expenses = ₹ 5,000 – 800
= ₹ 4,200
Annual expenses = monthly expenses × 12
= ₹ 4,200 × 12
= ₹ 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 = ₹ 5,000 – ₹ 800 = ₹ 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 – b = 11: 8; find the value of a: b
Solution:-
From the question, it is given that, a + b: a – b = 11: 8.
(a + b)/(a – b) = 11/8
By cross multiplication, we get,
8(a + b) = 11(a – b)
8a + 8b = 11a – 11b
Transposing we get,
11b + 8b = 11a – 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) × (q/r) = (2/5) × (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) × (e/i) × (i/u) = (5/12) × (8/3) × (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 × 8/5
= (15 × 8)/(16 × 5)
= (3 × 1)/(2 × 1)
= 3/2
Therefore, the compounded ratio of 15: 16 and 8: 5 is 3: 2.
(ii) (a2 – b2): (a2 + b2) and (a4 – b4): (a + b)4
Solution:-
The given ratio can be written as,
(a2 – b2)/(a2 + b2) and (a4 – b4)/(a + b)4
= (a2 – b2)/(a2 + b2) × (a4 – b4)/(a + b)4
We know that, (a2 – b2) = (a + b) (a – b)
= ((a + b) (a – b))/(a2 + b2) × ((a2 + b2) (a2 – 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 × 7/9 × 15/28
= (3 × 7 × 15)/(5 × 9 × 28)
= (1 × 1 × 3)/ (1 × 3 × 4)
= (1 × 1 × 1)/(1 × 1 × 4)
= ¼
Therefore, the compounded ratio is 1: 4.
(iv) √8: 4, 3: √5 and √20: √27
Solution:-
The given ratio can be written as,
√8/4, 3/√5 and √20/√27
= √8/4 × 3/√5 × √20/√27
= 2√2/4 × 3/√5 × 2√5/3√3
= √2/√3
Therefore, the compounded ratio is √2: √3.
(v) (m – n): (m + n), (m + n)2: (m2 + n2) and (m4 – n4): (m2 – n2)2
Solution:-
The given ratio can be written as,
(m – n)/(m + n), (m + n)2/(m2 + n2) and (m4 – n4)/(m2 – n2)2
= (m – n)/(m + n) × (m + n)2/(m2 + n2) × (m4 – n4)/(m2 – n2)2
= (m – n)/(1) × (m + n)/(m2 + n2) × ((m2 + n2) (m2 – n2))/(m2 – n2)2
By simplification, we get,
= 1/1
Therefore, the compounded ratio is 1 : 1.
9. Find the duplicate ratio of the following:
(i) √10 : √14
Solution:-
Given, √10 : √14
= (√10)2: (√14)2
= 10: 14
= 10/14
= 5/7
Therefore, the duplicate ratio is 5: 7.
(ii) 3√2a : 2√3a
Solution:-
Given, 3√2a: 2√3a
= (3√2a)2: (2√3a)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) × (81/16)
= (4 × 81)/(9 × 16)
= (1 × 9)/(1 × 4)
= 9/4
Therefore, duplicate ratio = 9: 4
(iv) (a + b): (a2 – b2)
Solution:-
Given, (a + b): (a2 – b2)
= (a + b)2 : (a2 – 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) 2√5 : 5√2
Solution:-
Given, 2√5 : 5√2
= (2√5)3: (5√2)3
= (8 × 5√5)/(125 × 2√2)
By simplification,
= 4√5: 25√2
Therefore, the triplicate ratio = 4√5: 25√2
(iii) √15 : √18
Solution:-
Given, √15: √18
= (√15)3: (√18)3
= 15√5: 18 × 3√2
= 5√15: 18√2
Therefore, the triplicate ratio is 5√15: 18√2
(iv) 3√(ab)2: 3√(a2b)
Solution:-
Given, 3√(ab)2: 3√(a2b)
By simplification, we get,
= (3√(ab)2)3: (3√(a2b))3
= ab2 : a2b
= b: a
Therefore, the triplicate ratio is b: a
11. Find the sub–duplicate ratio of the following:
(i) x6: y4
Solution:-
Given, x6: y4
= √x6: √y4
= (x6)1/2: (y4)1/2
= x3: y2
Therefore, the sub-duplicate ratio is x3:y2
(ii) 63m2: 28n2
Solution:-
Given, 63m2: 28n2
= √(63m2): √(28n2)
= 3√7m: 2√7n
= 3m: 2n
Therefore, the sub-duplicate ratio is 3m: 2n
(iii) 1/16: 1/36
Solution:-
Given, 1/16: 1/36
= √(1/16): √(1/36)
= ¼: 1/6
= (¼)/(1/6)
= (¼) × (6/1)
= 3/2
Therefore, the sub-duplicate ratio is 3: 2
(iv) 9a2/5: 25a2/3
Solution:-
Given, 9a2/5: 25a2/3
= √(9a2/5): √(25a2/3)
= 3a(1/√5): 5a(1/√3)
= 3√3: 5√5
Therefore, the sub-duplicate ratio is 3√3: 5√5.
12. Find the sub – triplicate ratio of the following:
(i) 512: 216
Solution:-
Given, 512: 216
= 3√512 : 3√216
= (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
= 3√(m3n6) : 3√(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
= 3√(125a3): 3√(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
= 3√(64m3/729n3): 3√(216m3/27n3)
= (64m3/729n3)1/3: (216m3/27n3)1/3
By simplification, we get,
= 4m/9n: 6m/3n
= (4m/9n) × (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) × (51/27)
= (45/1) × (3/27)
= (45/1) × (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) × (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) × (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 × 11 > 2 × 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 × 19 < 3 × 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 × 10 < 8 × 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) × (4/15) = 2/3
(5/3): (11/6) = (5/3) × (6/11) = 10/11
Consider, 2/3 : 10/11
2 × 11 < 3 × 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 × 8) = ¾
28y + 32 = 30x + 24
2y = 8
y = 8/2
y = 4
So, 7y = 7 × 4 = 28
10y = 10 × 4 = 40
Therefore, the two numbers are 28 and 40.
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