# Selina ICSE Class 10 Maths Solutions Chapter 7 Ratio And Proportion Including Properties Uses

Question 1. If a:b = 5:3, find:

Question 2. If x: y = 4: 7, find the value of (3π₯ + 2π¦): (5π₯ + π¦).
Solution:

It is given that,
π₯: π¦ = 4: 7
π₯/π¦ = 4/7
Divide each term by π¦.

Question 3. If a:b = 3:8,

Question 4. If (a β b): (a + b) = 1: 11, find the ratio (5a + 4b + 15): (5a β 4b + 3).
Solution:

It is given that,

Question 5. Find the number which bears the same ratio to 7/33 that 8/21 does to 4/9.
Solution:

Let us assumed that,
The required fraction be π₯/π¦

Question 7. Find π₯/π¦, when π₯2 + 6π¦2 = 5π₯π¦
Solution:

It is given that,
π₯2 + 6π¦2 = 5π₯π¦
By dividing both the sides by π¦2

Let us assumed that, π₯/π¦ = π
π2 + 6 = 5π
π2 β 5π + 6 = 0
π2 β 2π β 3π + 6 = 0
π(π β 2) β 3(π β 2) = 0
(π β 3)(π β 2) = 0
π β 3 = 0          π β 2 = 0
π = 3                π = 2
Hence, the value of π₯/π¦ is 2 and 3.

Question 8. If the ratio between 8 and 11 is the same as the ratio of 2π₯ β π¦ to π₯ + 2π¦, find the value of 7π₯/9π¦.
Solution:

It is given that,

Question 9. Divide Rs. 1290 into A, B and C such that A is 2/5 of B and B:C = 4:3.
Solution:

It is given that,

Question 10. A school has 630 students. The ratio of the number of boys to the number of girls is 3 : 2. This ratio changes to 7 : 5 after the admission of 90 new students. Find the number of newly admitted boys.
Solution:

Let us assumed that,
Number of boys are 3π₯
and number of girls are 2π₯
3π₯ + 2π₯ = 630
5π₯ = 630
π₯ = 630/5
π₯ = 126
Number of boys = 3 Γ 126 = 378
Number of girls = 2 Γ 126 = 252
According to question,
Admission of 90 new students,
Total number of students = 630 + 90
Total number of students = 720
Let us assumed that,
Number of boys are 7π₯
and number of girls are 5π₯
7π₯ + 5π₯ = 720
12π₯ = 720
π₯ = 720/12
π₯ = 60
Number of boys = 7 Γ 60 = 420
Number of girls = 5 Γ 60 = 300
Hence, the number of new students boys are 420β 378 = 42.

Question 11. What quantity must be subtracted from each term of the ratio 9: 17 to make it equal to 1: 3
Solution:
Let us assumed that,
π₯ be deducted from each term of the 9:17 ratio.
9βπ₯/17β π₯ = 1/3
By cross-multiplication,
3(9 β π₯) = 17 β π₯
27 β 3π₯ = 17 β π₯
β3π₯ + π₯ = 17 β 27
β2π₯ = β10
π₯ = 10/2
π₯ = 5
Hence, subtract the required number from the total 5.

Question 12. 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 every month, find their monthly pocket money.
Solution:

It is given that,
The pocket money of Ravi and Sanjeev are in the Ratio 5:7
The pocket expenditures of Ravi and Sanjeev are in the Ratio 3:5
And each of them saves Rs. 80.
Let us assumed that,
Pocket money of Ravi and Sanjeev are in the Ratio 5k and 7k
Pocket expenditures of Ravi and Sanjeev are in the Ratio 3m and 5m
5π β 3π = 80___________(i)
7π β 5π = 80___________(ii)
From equation (i) we get the value of k
5π β 3π = 80

Question 13. The work done by (π₯β 2) men in (4π₯ + 1) days and the work done by (4π₯ + 1) men in (2π₯β 3) days are in the ratio 3: 8. Find the value of π₯.
Solution:

It is given that,
Number of men = (π₯β 2)
Number of days = (4π₯ + 1)
Amount of work done by = (π₯β 2)(4π₯ + 1)units of work
Similarly,
Number of men = (4π₯ + 1)
Number of days = (2π₯ β 3)
Amount of work done by = (4π₯ + 1)(2π₯ β 3)units of work
According to question,

By Cross-multiply,
8(π₯ β 2) = 3(2π₯ β 3)
8π₯ β 16 = 6π₯ β 9
8π₯ β 6π₯ = β9 + 16
2π₯ = 7
π₯ = 7/2
π₯ = 3.5
Hence, the value of π₯ is 3.5.

Question 14. The bus fare between two cities is increased in the ratio 7: 9. Find the increase in the fare, if:
(i) the original fare is Rs 245;
(ii) the increased fare is Rs 207.
Solution:

It is given that,
Bus fare between two cities is increased in the ratio 7:9.
New bus fare = 9/7 Γ Original bus fare
(i) New bus fare = 9/7 Γ 245
New bus fare = Rs. 315
Increase in fare = Rs. 315 β Rs. 245
Increase in fare = Rs. 70
(ii) Rs. 207 = 9/7 Γ Original bus fare
Rs. 207 Γ 7/9 = Original bus fare
Original bus fare = Rs. 161
Increase in fare = Rs. 207 β Rs. 161
Increase in fare = Rs. 46

Question 15. By increasing the cost of entry ticket to a fair in the ratio 10: 13, the number of visitors to the fair has decreased in the ratio 6: 5. In what ratio has the total collection increased or decreased?
Solution:

Let us assumed that,
The initial cost of the entry ticket is 10π₯ and present cost be 13π₯.
The initial number of the visitors are 6π¦ and at present number of visitor be 5π¦.
Total collection = 10π₯ Γ 6π¦ = 60π₯π¦
Total collection = 13π₯ Γ 5π¦ = 65π₯π¦
Ratio of total collection = 60π₯π¦ βΆ 65π₯π¦ = 12 βΆ 13
Hence, the total collection has increased in the ratio 12: 13.

Question 16. In a basket, the ratio between the number of oranges and the number of apples is 7: 13. If 8 oranges and 11 apples are eaten, the ratio between the number of oranges and the number of apples
becomes 1: 2. Find the original number of oranges and the original number of apples in the basket.
Solution:

Let us assumed that,
The original number of oranges and apples be 7π₯ and 13π₯.
According to the question,
7π₯β 8/13π₯ β11 = 1/2
By cross-multiplication,
2(7π₯ β 8) = 13π₯ β 11
14π₯ β 16 = 13π₯ β 11
14π₯ β 13π₯ = β11 + 16
π₯ = 5
Hence, the original number of oranges is 35 and apples are 65 respectively.

Question 17. In a mixture of 126 kg of milk and water, milk and water are in ratio 5 : 2. How much water must be added to the mixture to make this ratio 3 : 2?
Solution:

It is given that,
The ratio of quantity of milk and water is 5:2
Total mixture is 126 kg
Quantity of milk = 126 Γ 5/7
Quantity of milk = 90kg.
Quantity of milk = 126 Γ 2/7
Quantity of milk = 36kg.
Change Ratio of mixture is 3:2
Let the quantity of water be added be π₯kg

Question 18. (A) If A: B = 3: 4 and B: C = 6: 7, find:
(i) A: B: C
(ii) A: C
(B) If A : B = 2 : 5 and A : C = 3 : 4, find:
(i) A : B : C
Solution:

(A) A: B = 3: 4 and B: C = 6: 7
(i) A: B: C
It is given that,

Question 19 (i). If 3A = 4B = 6C; find A: B: C.
Solution:

It is given that,
3π΄ = 4π΅ = 6πΆ
From 3π΄ = 4π΅ we get,

Question 19 (ii). If 2a = 3b and 4b = 5c, find: a : c.
Solution:

It is given that,
2a = 3b and 4b = 5c
2a = 3b

Question 20. Find the compound ratio of:
(π)3: 5 πππ 8: 15
(ππ) 2: 3, 9: 14 πππ 14: 27
(πππ) 2π: 3π, ππ: π₯2 πππ π₯: π
(iv) β2: 1, 3: β5 πππ β20: 9
Solution:

(π) 3: 5 πππ 8: 15
It is given that,
3: 5 πππ 8: 15
Compound ratio = 3 Γ 8: 5 Γ 15

Question 21. Find duplicate ratio of:
(π) 3: 4 (ππ) 3β3 βΆ 2β5
Solution:

(i) Duplicate ratio of 3: 4
= 3 Γ 2: 4 Γ 2
= 9: 16
Hence, the duplicate ratio is 9: 16
(ii) Duplicate ratio of 3β3: 2β5
= (3β3)2: (2β5)Β²
= 3β3 Γ 3β3: 2β5 Γ 2β5
= 9 Γ 3: 4 Γ 5
= 27: 20
Hence, the duplicate ratio is 27: 20

Question 22. Find the triplicated of:
(i) 1: 3 (ii) π/2 : π/3
Solution:

(i) Triplicated ratio of 1: 3
= (1)3: (3)3
= 1: 27
Hence, the triplicated ratio is 1: 27.

Question 23. Find sub-duplicate ratio of:
(i) 9: 16
(ii) (π₯β π¦)4: (π₯ + π¦)6
Solution:

(i) 9: 16
Sub-duplicate ratio of 9: 16
= β9: β16
= 3: 4
Hence, the sub-duplicate ratio is 3:4.
(ii) (π₯β π¦)4: (π₯ + π¦)6
Sub-duplicate ratio of 9: 16

Question 24. Find the sub-triplicate ratio of:
(i) 64: 27
(ii) π₯3: 125π¦3
Solution:

(i) 64: 27
Sub-triplicate ratio of 64 : 27
= β64 3
: β27 3
= 4: 3
Hence, the sub-triplicate ratio is 4: 3.
(ii) π₯3: 125π¦3
Sub-triplicate ratio of π₯Β³: 125π¦Β³
= 3βπ₯3: 3β(5π¦)3
= π₯: 5π¦
Hence, the sub-triplicate ratio is π₯: 5π¦.

Question 25. Find the reciprocal ratio of:
(i) 5:8
(ii) π₯/3 : π¦/7
Solution:

Question 26. If (π₯ + 3): (4π₯ + 1) is the duplicate ratio of 3: 5, find the value of π₯.
Solution:

It is given that,
(π₯ + 3): (4π₯ + 1)
The duplicate Ratio = 3: 5

Question 27. If π: π is the duplicate ratio of π + π₯: π + π₯; show that π₯2 = mn.
Solution:

It is given that,

By cross-multiplication
π(π2 + 2ππ₯ + π₯2) = π(π2 + 2ππ₯ + π₯2)
ππ2 + 2πππ₯ + ππ₯2 = ππ2 + 2πππ₯ + ππ₯2
ππ2 + ππ₯2 = ππ2 + ππ₯2
ππ2 β ππ2 = ππ₯2 β ππ₯2
ππ(π β π) = π₯2(π β π)
ππ = π₯2
π₯2 = ππ
Hence proved

Question 28. If (3π₯β 9): (5π₯ + 4) is the triplicate ratio of 3: 4, find the value of π₯.
Solution:

It is given that,
(3π₯β 9): (5π₯ + 4)
Triplicate Ratio = 3: 4

By cross-multiplication,
64(π₯ β 3) = 9(5π₯ + 4)
64π₯ β 192 = 45π₯ + 36
64π₯ β 45π₯ = 192 + 36
19π₯ = 228
π₯ = 228/19
π₯ = 12
Hence, the value of π₯ is 12.

Question 29. Find the ratio compounded of the reciprocal ratio of 15: 28, the sub-duplicate ratio of 36: 49 and the triplicate ratio of 5: 4.
Solution:
Reciprocal ratio of 15: 28
= 28: 15
Sub-duplicate ratio of 36: 49
= β36: β49
= 6: 7
Triplicate ratio of 5: 4
= 53: 43
= 125: 64

Question 30 (a). If π2 = ππ, show that π: π is the duplicate ratio of (π + π): (π + π).
Solution:

It is given that,
π2 = ππ
Duplicate ratio of (π + π): (π + π) = (π + π)2: (π + π)2
We know that,
(π + π)2 = π2 + 2ππ + π2
= (π2 + 2ππ + π2): (π2 + 2ππ + π2)
Put the value of π2
= (π2 + 2ππ + ππ): (π2 + 2ππ + ππ)
= π(π + 2π + π): π(π + 2π + π)

Question 30 (b). If (π β π₯): (π β π₯) be the duplicate ratio of π: π then show that: 1/π + 1/π = 1/π₯
Solution:

It is given that,

### Exercise 7 B

Question 1. Find the fourth proportional to:
(i) 1.5, 4.5 πππ 3.5
(ii) 3π, 6π2 πππ 2ππ2
Solution:

(i) Let us assumed that,
The fourth proportional to 1.5, 4.5 and 3.5 be π₯.
1.5: 4.5 = 3.5: π₯
1.5 Γ π₯ = 3.5 Γ 4.5
1.5 Γ π₯ = 15.75
π₯ = 10.5
Hence, the fourth proportional to 1.5, 4.5 and 3.5 is 10.5.
(ii) Let us assumed that,
The fourth proportional to 3π, 6π2 and 2ππ2 be π₯.
3π: 6π2 = 2ππ2: π₯
3π Γ π₯ = 2ππ2 Γ 6π2
3π Γ π₯ = 12π3π2
π₯ = 12π3π2/3π
π₯ = 4π2π2
Hence, the fourth proportional to 3π, 6π2 and 2ππ2 is 4π2π2.

Question 2. Find the third proportional to:
(π) 2 (2/3) πππ 4
(ππ) π β π πππ π2 β π2
Solution:

(i) Let us assumed that,
The third proportional to 2 (2/3) πππ 4 be π₯
We know that,
Given ratios are in continue proportion

(ii) Let us assumed that,
The third proportional to π β π πππ π2 β π2 be π₯
We know that,
Given ratios are in continue proportion
π β π: π2 β π2 = π2 β π2: π₯

Question 3. Find the mean proportional between:
(i) 6 + 3β3 πππ 8 β 4β3
(ii) π β π πππ π3 β π2π
Solution:

(i) Let us assumed that,
The mean proportional between 6 + 3β3 πππ 8 β 4β3 be π₯.
6 + 3β3, π₯ πππ 8β 4β3
6 + 3β3 βΆ π₯ = π₯ βΆ 8 β 4β3
π₯ Γ π₯ = (6 + 3β3)(8β 4β3)
π₯2 = 48 + 24β3 β 24β3β 36
π₯2 = 12
π₯ = 2β3
Hence, the mean proportional between 6 + 3β3 πππ 8 β 4β3 is 2β3.
(ii) Let us assumed that,
The mean proportional between πβ π πππ π3β π2π be π₯.
πβ π, π₯ πππ π3β π2π
π β π: π₯ = π₯: π3β π2π
π₯ Γ π₯ = (π β π) (π3β π2π)
π₯2 = (π β π)π2(π β π)
π₯ = [π(πβ π)]
π₯ = π(π β π)
Hence, the mean proportional between π β π πππ π3 β π2π is π(π β π)

Question 4. If π₯ + 5 is the mean proportional between π₯ + 2 and π₯ + 9; find the value of π₯.
Solution:

It is given that,
π₯ + 5 is the mean proportional between π₯ + 2 and π₯ + 9.
(π₯ + 2), (π₯ + 5) and (π₯ + 9)
(π₯ + 2): (π₯ + 5) = (π₯ + 5): (π₯ + 9)

By cross multiply,
(π₯ + 2)(π₯ + 9) = (π₯ + 5)2
π₯2 + 2π₯ + 9π₯ + 18 = π₯2 + 25 + 10π₯
11π₯ β 10π₯ = 25 β 18
π₯ = 7
Hence, the value of π₯ is 7.

Question 5. If π₯2, 4 and 9 are in continued proportion, find x.
Solution:

It is given that,
π₯2, 4 and 9 are in continued proportion
π₯2: 4 = 4: 9

Question 6. What least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional?
Solution:

Let us assumed that,
The number added be π₯.
(6 + π₯): (15: π₯): : (20 + π₯): (43 + π₯)

By cross multiply
(6 + π₯)(43 + π₯) = (20 + π₯)(15 + π₯)
258 + 6π₯ + 43π₯ + π₯2 = 300 + 20π₯ + 15π₯ + π₯2
258 + 49π₯ = 300 + 35π₯
49π₯ β 35π₯ = 300 β 258
14π₯ = 42
π₯ = 42/14
π₯ = 3
Hence, the value of π₯ is 3.

Question 7 (i). If a, b, c are in continued proportion.

Question 7 (ii). If π, π, π are in continued proportion and π(π β π) = 2π. Prove that: π β π = 2(π+π)/π
Solution:

It is given that,
a, b and c are in continued proportion,
π/π = π/π
By cross multiplication,
ππ = π2
Put the value of ac in given equation,
π(π β π) = 2π
ππ β ππ = 2π
ππ β π2 = 2π
π(π β π) = 2π
π β π = 2
Now,
In the LHS side it is given that,

Question 8. What least number must be subtracted from each of the numbers 7, 17 and 47 so that the remainders are in continued proportion?
Solution:

It is given that,
7, 17 and 47 so that the remainders are in continued proportion
Let us assumed that,
The subtracted number be π₯
(7 β π₯): (17 β π₯): : (17 β π₯): (47 β π₯)

By cross multiplication,
(7 β π₯)(47 β π₯) = (17 β π₯)(17 β π₯)
329 β 47π₯ β 7π₯ + π₯2 = 289 β 34π₯ + π₯2
329 β 54π₯ = 289 β 34π₯
329 β 54π₯ + 34π₯ = 289 β 329
β20π₯ = β40
π₯ = 40/20
π₯ = 2
Hence, the number which should be subtracted is 2.

Question 9. If y is the mean proportional between x and z; show that π₯π¦ + π¦π§ is the mean proportional between π₯2 + π¦2 and π¦2 + π§2.
Solution:

It is given that,
The mean proportion between π₯ and π§ is π¦
π¦2 = π₯π§
Need to prove,
π₯π¦ + π¦π§ is the mean proportional between π₯2 + π¦2 and π¦2 + π§2
(π₯π¦ + π¦π§)2 = (π₯2 + π¦2)(π¦2 + π§2)
Now,
In the LHS side it is given that,
(π₯π¦ + π¦π§)2 = [π¦(π₯ + π§)2]
(π₯π¦ + π¦π§)2 = π¦2(π₯ + π§)2
(π₯π¦ + π¦π§)2 = π₯π§(π₯ + π§)2
In the RHS side it is given that,
(π₯2 + π¦2)(π¦2 + π§2) = (π₯2 + π₯π¦)(π₯π¦ + π§2)
(π₯2 + π¦2)(π¦2 + π§2) = π₯(π₯ + π§)π§(π₯ + π§)
(π₯2 + π¦2)(π¦2 + π§2) = π₯π§(π₯ + π§)2
LHS = RHS
Hence proved

Question 10. If q is the mean proportional between p and r, show that: πππ(π + π + π)3 = (ππ + ππ + ππ)3.
Solution:

It is given that,
The mean proportional between π and π is π
π2 = ππ
Now,
In the LHS side it is given that,
πππ(π + π + π)3 = ππ2(π + π + π)3
πππ(π + π + π)3 = π3(π + π + π)3
πππ(π + π + π)3 = [π(π + π + π)3
πππ(π + π + π)3 = (ππ + π2 + ππ)3
(ππ + π2 + ππ)3 = (ππ + π2 + ππ)3
LHS = RHS
Hence proved

Question 11. If three quantities are in continued proportion; show that the ratio of the first to the third is the duplicate ratio of the first to the second.
Solution:

Let us assumed that,
π₯, π¦ and π§ be the three quantities which are in continued proportion
π¦2 = π₯π§_(1)
Now,
We need to prove that,
π₯: π§ = π₯2: π¦2
π₯π¦2 = π₯2π§
In the LHS side it is given that,
π₯π¦2 = π₯(π₯π§) = π₯2π§
LHS = RHS
Hence proved.

Question 12. If y is the mean proportional between π₯ and π§,

Question 13. Given four quantities a, b, c and d are in proportion. Show that:
(π β π)π2: (π β π)ππ = (π2 β π2 β ππ): (π2 β π2 β ππ)
Solution:

It is given that,
a, b, c and d are proportion
Let us assumed that,
π/π = π/π = π
By cross-multiplication,
π = ππ πππ π = ππ
Now
In the LHS side it is given that,

Question 14. Find two numbers such that the mean proportional between them is 12 and the third proportional to them is 96.
Solution:

Let us assumed that,
a and b are the two numbers whose mean proportional is 12.
ππ = (12)2
ππ = 144
π = 144/π __(1)
It is given that,
The third proportional is 96
π: π: : π: 96
π2 = 96π
Put the value of b

Question 15. Find the third proportional of

Question 16. If π: π = π: π ; then show that: ππ + ππ: π = ππ + ππ : π .
Solution:

It is given that,

Question 17. If  and

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

### Exercise 7C

Question 1. If π: π = π: π, prove that:
(π) 5π + 7π: 5πβ 7π = 5π + 7π: 5πβ 7π.
(ππ) (9π + 13π) (9πβ 13π) = (9π + 13π)(9πβ 13π).
(πππ) π₯π + π¦π: π₯π + π¦π = π βΆ π.
Solution:

Question 2. If π: π = π: π, prove that: (6π + 7π)(3πβ 4π) = (6π + 7π)(3πβ 4π).
Solution:

It is given that,

Question 5. If (7π + 8π)(7πβ 8π) = (7πβ 8π)(7π + 8π), prove that π: π = π: π.
Solution:

It is given that,

Question 7. If (π + π + π + π) (πβ πβ π + π) = (π + πβ πβ π)(πβ π + πβ π), prove that π: π = π: π.
Solution:

It is given that,
(π + π + π + π)(πβ πβ π + π) = (π + πβ πβ π)(πβ π + πβ π)

Question 9. If (π2 + π2) (π₯2 + π¦2) = (ππ₯ + ππ¦)2; prove that: π/π₯ = π/π¦.
Solution:

It is given that,
(π2 + π2) (π₯2 + π¦2) = (ππ₯ + ππ¦)2
π2π₯2 + π2π¦2 + π2π₯2 + π2π¦2 = π2π₯2 + 2ππ₯ππ¦+π2π¦2
π2π¦2 + π2π₯2 = 2ππ₯ππ¦
π2π¦2 + π2π₯2 β 2ππ₯ππ¦ = 0
(ππ¦ β ππ₯)2 = 0
ππ¦ β ππ₯ = 0
ππ¦ = ππ₯
π/π₯ = π/π¦
Hence proved

Question 10. If a, b and c are in continued proportion, prove that:

Question 11. Using properties of proportion, solve for :

### Exercise 7D

Question 1. If π: π = 3: 5, find: (10π + 3π): (5π + 2π)
Solution:

It is given that,
π: π = 3: 5

Question 2. If 5π₯ + 6π¦: 8π₯ + 5π¦ = 8: 9, find π₯: π¦.
Solution:

It is given that

Hence, the value of π₯: π¦ is 14: 19.

Question 3. If (3π₯β 4π¦): (2π₯β 3π¦) = (5π₯β 6π¦): (4π₯β 5π¦), find π₯: π¦.
Solution:

It is given that,
(3π₯β 4π¦): (2π₯β 3π¦) = (5π₯β 6π¦): (4π₯β 5π¦)

Question 4. Find the:
(i) duplicate ratio of 2β2: 3β5
(ii) triplicate ratio of 2π: 3π
(iii) sub-duplicate ratio of 9π₯2π4: 25π¦6π2
(iv) sub-triplicate ratio of 216: 343
(v) reciprocal ratio of 3: 5
(vi) ratio compounded of the duplicate ratio of 5: 6, the reciprocal ratio of 25: 42 and the sub duplicate ratio of 36: 49.
Solution:

(i) Duplicate ratio of 2β2: 3β5
(2β2)2: (3β5)2
(2 Γ 2 Γ β2 Γ β2): (3 Γ 3 Γ β5 Γ β5)
8: 45
Hence, the duplicate ratio of 2β2: 3β5 is 8: 45.
(ii) Triplicate ratio of 2π: 3π
(2π)3: (3π)3
(2 Γ 2 Γ 2 Γ π Γ π Γ π): (3 Γ 3 Γ 3 Γ π Γ π Γ π )
8π3: 27π3
Hence, the triplicate ratio of 2π: 3π is 8π3: 27π3.
(iii) Sub-duplicate ratio of 9π₯2π4: 25π¦6π2

Hence, the Sub-triplicate ratio of 216: 343 is 6: 7.
(v) Reciprocal ratio of 3: 5
5: 3
Hence, the reciprocal ratio of 3: 5 is 5: 3.
(vi) ratio compounded of the duplicate ratio of 5: 6, the reciprocal ratio of 25: 42 and the sub-duplicate
ratio of 36: 49.
Duplicate ratio of 5: 6
(5)2: (6)2
25: 36
Reciprocal ratio of 25: 42
42: 25
Sub-duplicate ratio of 36: 49
β36: β49
6: 7

Question 5. Find the value of x, if:
(i) (2π₯ + 3): (5π₯β 38) is the duplicate ratio of β5: β6
(ii) (2π₯ + 1): (3π₯ + 13) is the sub-duplicate ratio of 9: 25.
(iii) (3π₯β 7): (4π₯ + 3) is the sub-triplicate ratio of 8: 27.
Solution:

(i) (2π₯ + 3): (5π₯β 38) is the duplicate ratio of β5: β6 It is given that,
Duplicate ratio of β5: β6
Duplicate ratio of (β5)2: (β6)2
The duplicate ratio is 5: 6

By cross-multiplication,
6(2π₯ + 3) = 5(5π₯ β 38)
12π₯ + 18 = 25π₯ β 190
12π₯ β 25π₯ = β190 β 18
β13π₯ = β208
13π₯ = 208
π₯ = 208/13
π₯ = 16
Hence, the value of π₯ is 16
(ii) (2π₯ + 1): (3π₯ + 13) is the sub-duplicate ratio of 9: 25.
It is given that,
Sub-duplicate ratio of 9: 25
Sub-duplicate ratio of β9: β25
The sub-duplicate ratio is 3: 5

By cross-multiplication,
5(2π₯ + 1) = 3(3π₯ + 13)
10π₯ + 5 = 9π₯ + 39
10π₯ β 9π₯ = +39 β 5
π₯ = 39 β 5
π₯ = 34
Hence, the value of π₯ is 34.
(iii) (3π₯β 7): (4π₯ + 3) is the sub-triplicate ratio of 8: 27.
It is given that,
Sub-duplicate ratio of 8: 27
Sub-duplicate ratio of 3β8 : 3β27
The sub-duplicate ratio is 2: 3

By cross-multiplication,
3(3π₯ β 7) = 2(4π₯ + 3)
9π₯ β 21 = 8π₯ + 6
9π₯ β 8π₯ = 6 + 21
π₯ = 27
Hence, the value of π₯ is 27.

Question 6. What quantity must be added to each term of the ratio π₯: π¦ so that it may become equal to π: π?
Solution:

Let us assumed that,
The quantity which is to be added be a.
π₯+π/π¦+π = π/π
By cross multiplication,
π(π₯ + π) = π(π¦ + π)
ππ₯ + ππ = ππ¦ + ππ
ππ β ππ = ππ¦ β ππ₯
π(π β π) = ππ¦ β ππ₯

Question 7. A woman reduces her weight in the ratio 7: 5. What does her weight become if originally it was 84 kg?
Solution:

It is given that,
Original weight is 84kg
Let us assumed that,
The reduced weight is π₯
84: π₯ = 7: 5
84/π₯ = 7/5
By cross multiplication,
84 Γ 5 = 7 Γ π₯
84Γ5/7 = π₯
12 Γ 5 = π₯
60 = π₯
π₯ = 60
Hence, reduced weight us 60 kg.

Question 8. If 15(2π₯2β π¦2) = 7π₯π¦, find π₯: π¦; if x and y both are positive.
Solution:

It is given that,
15(2π₯2β π¦2) = 7π₯π¦
By substitution,

Question 9. Find the:
(i) fourth proportional to 2π₯π¦, π₯2 and π¦2.
(ii) third proportional to π2β π2 and π + π.
(iii) mean proportional to (π₯β π¦) and (π₯3β π₯2π¦).
Solution:

(i) Let us assumed that,
Fourth proportional to 2π₯π¦, π₯2 and π¦2 be π.
2π₯π¦: π₯2 = π¦2: π

Question 10. Find two numbers such that the mean proportional between them is 14 and third proportional to them is 112.
Solution:
It is given that,
Mean proportional between numbers is 14
Let us assumed that,
The two number are π and π
Now,
π: 14 = 14: π
π/14 = 14/π
By cross multiplication,
π Γ π = 14 Γ 14
ππ = 196
π = 196/π__________________________(1)
The third proportional to π and π is 112.
π: π = π: 112
π/π = π/112
π Γ 112 = π Γ π
π2 = π112 __________________________(2)

Put the value of a in above equation,
π2 = 196/π Γ 112
π2 Γ π = 196 Γ 112
π3 = 196 Γ 112
π3 = 21952
π = 28
Put the value of π in equation 2
π = 196/28
π = 7
Hence, the required two numbers are 7 and 28.

Question 11. If x and y be unequal and π₯: π¦ is the duplicate ratio of π₯ + π§ and π¦ + π§, prove that z is mean proportional between x and y.
Solution:

It is given that,

By cross multiplication,
π₯(π¦2 + 2π¦π§ + π§2) = π¦(π₯2 + 2π₯π§ + π§2)
π₯π¦2 + 2π₯π¦π§ + π₯π§2 = π¦π₯2 + 2π₯π¦π§ + π¦π§2
π₯π¦2 + π₯π§2 = π¦π₯2 + π¦π§2
π₯π¦2 β π₯2π¦ = π¦π§2 β π₯π§2
π₯π¦(π¦ β π₯) = π§2(π¦ β π₯)
π₯π¦ = π§2
Hence, the mean proportional between π₯ and π¦ is π§

Question 13. If (4π + 9π)(4πβ 9π) = (4πβ 9π)(4π + 9π), prove that: π: π = π: π.
Solution:

It is given that,
(4π + 9π)(4πβ 9π) = (4πβ 9π)(4π + 9π)

Question 15. There are 36 members in a student council in a school and the ratio of the number of boys to the number of girls is 3: 1. How any more girls should be added to the council so that the ratio of the number of boys to the number of girls may be 9: 5?
Solution:

It is given that,
Ratio of number of boys to the number of girls = 3: 1
Let us assumed that,
The number of boys be 3π₯
Number of girls be π₯
3π₯ + π₯ = 36
4π₯ = 36
π₯ = 9
Number of boys = 3π₯ = 27
Number of girls = π₯ = 9
Let us assumed that,
Number of girls be added to the council = π
We have:
27/9+π = 9/π
By Cross multiplication,
27 Γ 5 = 9(9 + π)
135 = 81 + 9π
β9π = 81 β 135
β9π = β54
π = 6
Hence, the council has added 6 girls.

Question 16. If 7π₯β 15π¦ = 4π₯ + π¦, find the value of π₯: π¦. Hence, use componendo and dividend to find the values of:

Question 21. Using componendo and dividend find the value of

By cross multiplication,
(π₯ + 1)2(π β 1) = (π₯ β 1)2(π + 1)
(π₯2 + 2π₯ + 1)(π β 1) = (π₯2 β 2π₯ + 1)(π + 1)
π(π₯2 + 2π₯ + 1) β (π₯2 + 2π₯ + 1) = π(π₯2 β 2π₯ + 1) + (π₯2 β 2π₯ + 1)
ππ₯2 + 2ππ₯ + π β π₯2 β 2π₯ β 1 = ππ₯2 β 2ππ₯ + π + π₯2 β 2π₯ + 1
2ππ₯ β π₯2 β 1 = β2ππ₯ + π₯2 + 1
2ππ₯ β π₯2 β 1 + 2ππ₯ β π₯2 β 1 = 0
4ππ₯ β 2π₯2 β 2 = 0
2(2ππ₯ β π₯2 β 1) = 0
2ππ₯ β π₯2 β 1 = 0
Hence proved.

Question 27. Simplification:- (2π₯2 β 5π¦2): π₯π¦ = 1: 3
Solution:

It is given that,
(2π₯2 β 5π¦2): π₯π¦ = 1: 3

By cross multiplication,
3(2π3 β 5) = π
6π3 β 15 = π
6π3 β π β 15 = 0
By splitting the middle term,
6π3 + 9π β 10π β 15 = 0
3π(2π3 + 3) β 5(2π + 3) = 0
(3π β 5)(2π + 3) = 0
(3π β 5) = 0 (2π + 3) = 0
3π β 5 = 0            2π + 3 = 0
3π = 5                   2π = β3
π = 5/3                  π = β 3/2
a canβt be negative so, π = 5/3
π₯/π¦ = π
π₯/π¦ = 5/3
π₯: π¦ = 5: 3
Hence, the ratio of π₯: π¦ is 5: 3

Question 30. If a, b and c are in continued proportion, prove that: π: π = (π2 + π2): (π2 + π2)
Solution:

It is given that,
a, b and c are in continued proportion
π: π = π: π
Let us assumed that,
π/π = π/π = π
By cross multiplication,
π = ππ, π = ππ
π = ππ Γ π
π = ππ2
Now,
In the LHS side it is given that,