**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.

9βπ₯/17β π₯ = 1/3

By cross-multiplication,

3(9 β π₯) = 17 β π₯

27 β 3π₯ = 17 β π₯

β3π₯ + π₯ = 17 β 27

β2π₯ = β10

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₯ = β9 + 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¦ βΆ 65π₯π¦ = 12 βΆ 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₯ β11 = 1/2

By cross-multiplication,

2(7π₯ β 8) = 13π₯ β 11

14π₯ β 16 = 13π₯ β 11

14π₯ β 13π₯ = β11 + 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΅ = 6πΆ

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 πππ 8: 15

(ππ) 2: 3, 9: 14 πππ 14: 27

(πππ) 2π: 3π, ππ: π₯2 πππ π₯: π

(iv) β2: 1, 3: β5 πππ β20: 9

Solution:

(π) 3: 5 πππ 8: 15

It is given that,

3: 5 πππ 8: 15

Compound ratio = 3 Γ 8: 5 Γ 15

**Question **21. Find duplicate ratio of:

(π) 3: 4 (ππ) 3β3 βΆ 2β5

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 3β3: 2β5

= (3β3)2: (2β5)Β²

= 3β3 Γ 3β3: 2β5 Γ 2β5

= 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

= β9: β16

= 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

= β64 3

: β27 3

= 4: 3

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

(ii) π₯^{3}: 125π¦^{3}

Sub-triplicate ratio of π₯Β³: 125π¦Β³

= 3βπ₯^{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} + 2ππ₯ + π₯^{2}) = π(π^{2} + 2ππ₯ + π₯^{2})

ππ^{2} + 2πππ₯ + ππ₯^{2} = ππ^{2} + 2πππ₯ + ππ₯^{2}

ππ^{2} + ππ₯^{2} = ππ^{2} + ππ₯^{2}

ππ^{2} β ππ^{2} = ππ₯^{2} β ππ₯^{2}

ππ(π β π) = π₯^{2}(π β π)

ππ = π₯^{2}

π₯^{2} = ππ

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

= β36: β49

= 6: 7

Triplicate ratio of 5: 4

= 5^{3}: 4^{3}

= 125: 64

**Question **30 (a). If π^{2} = ππ, show that π: π is the duplicate ratio of (π + π): (π + π).

Solution:

It is given that,

π^{2} = ππ

Duplicate ratio of (π + π): (π + π) = (π + π)^{2}: (π + π)^{2}

We know that,

(π + π)^{2} = π^{2} + 2ππ + π^{2}

= (π^{2} + 2ππ + π^{2}): (π^{2} + 2ππ + π^{2})

Put the value of π^{2}

= (π^{2} + 2ππ + ππ): (π^{2} + 2ππ + ππ)

= 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 πππ 3.5

(ii) 3π, 6π^{2} πππ 2ππ^{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 2ππ^{2} be π₯.

3π: 6π^{2} = 2ππ^{2}: π₯

3π Γ π₯ = 2ππ^{2} Γ 6π^{2}

3π Γ π₯ = 12π3π^{2}

π₯ = 12π^{3}π^{2}/3π

π₯ = 4π^{2}π^{2}

Hence, the fourth proportional to 3π, 6π^{2} and 2ππ^{2} is 4π2π^{2}.

**Question **2. Find the third proportional to:

(π) 2 (2/3) πππ 4

(ππ) π β π πππ π^{2} β π^{2}

Solution:

(i) Let us assumed that,

The third proportional to 2 (2/3) πππ 4 be π₯

We know that,

Given ratios are in continue proportion

(ii) Let us assumed that,

The third proportional to π β π πππ 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 + 3β3 πππ 8 β 4β3

(ii) π β π πππ π^{3} β π^{2}π

Solution:

(i) Let us assumed that,

The mean proportional between 6 + 3β3 πππ 8 β 4β3 be π₯.

6 + 3β3, π₯ πππ 8β 4β3

6 + 3β3 βΆ π₯ = π₯ βΆ 8 β 4β3

π₯ Γ π₯ = (6 + 3β3)(8β 4β3)

π₯^{2} = 48 + 24β3 β 24β3β 36

π₯^{2} = 12

π₯ = 2β3

Hence, the mean proportional between 6 + 3β3 πππ 8 β 4β3 is 2β3.

(ii) Let us assumed that,

The mean proportional between πβ π πππ π^{3}β π^{2}π be π₯.

πβ π, π₯ πππ π^{3}β π^{2}π

π β π: π₯ = π₯: π^{3}β π^{2}π

π₯ Γ π₯ = (π β π) (π^{3}β π^{2}π)

π₯^{2} = (π β π)π^{2}(π β π)

π₯ = [π(πβ π)]

π₯ = π(π β π)

Hence, the mean proportional between π β π πππ π^{3} β π^{2}π 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,

ππ = π^{2}

Put the value of ac in given equation,

π(π β π) = 2π

ππ β ππ = 2π

ππ 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

β20π₯ = β40

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: πππ(π + π + π)^{3} = (ππ + ππ + ππ)^{3}.

Solution:

It is given that,

The mean proportional between π and π is π

π^{2 }= ππ

Now,

In the LHS side it is given that,

πππ(π + π + π)^{3} = ππ^{2}(π + π + π)^{3}

πππ(π + π + π)^{3} = π^{3}(π + π + π)^{3}

πππ(π + π + π)^{3} = [π(π + π + π)^{3}

πππ(π + π + π)^{3} = (ππ + π^{2} + ππ)^{3}

(ππ + π^{2} + ππ)^{3 }= (ππ + π^{2} + ππ)^{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₯

^{2}π§

In the LHS side it is given that,

π₯π¦

^{2}= π₯(π₯π§) = π₯

^{2}π§

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)ππ = (π^{2} β **π ^{2}** β ππ): (π

^{2}β π

^{2}β ππ)

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 = ππ πππ π = ππ

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.

ππ = (12)^{2}

ππ = 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: ππ + ππ: π = ππ + ππ : 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: 5πβ 7π = 5π + 7π: 5πβ 7π.

(ππ) (9π + 13π) (9πβ 13π) = (9π + 13π)(9πβ 13π).

(πππ) π₯π + π¦π: π₯π + π¦π = π βΆ π.

Solution:

**Question **2. If π: π = π: π, prove that: (6π + 7π)(3πβ 4π) = (6π + 7π)(3πβ 4π).

Solution:

It is given that,

**Question **5. If (7π + 8π)(7πβ 8π) = (7πβ 8π)(7π + 8π), prove that π: π = π: π.

Solution:

It is given that,

**Question **7. If (π + π + π + π) (πβ πβ π + π) = (π + πβ πβ π)(πβ π + πβ π), prove that π: π = π: π.

Solution:

It is given that,

(π + π + π + π)(πβ πβ π + π) = (π + πβ πβ π)(πβ π + πβ π)

**Question **9. If (π^{2} + π^{2}) (π₯^{2} + π¦^{2}) = (ππ₯ + ππ¦)^{2}; prove that: π/π₯ = π/π¦.

Solution:

It is given that,

(π^{2} + π^{2}) (π₯^{2} + π¦^{2}) = (ππ₯ + ππ¦)^{2}

π^{2}π₯^{2} + π^{2}π¦^{2} + π^{2}π₯^{2} + π^{2}π¦^{2} = π2π₯^{2} + 2ππ₯ππ¦+π^{2}π¦^{2}

π^{2}π¦^{2} + π^{2}π₯^{2} = 2ππ₯ππ¦

π^{2}π¦^{2} + π^{2}π₯^{2} β 2ππ₯ππ¦ = 0

(ππ¦ β ππ₯)^{2} = 0

ππ¦ β ππ₯ = 0

ππ¦ = ππ₯

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 2β2: 3β5

(ii) triplicate ratio of 2π: 3π

(iii) sub-duplicate ratio of 9π₯^{2}π^{4}: 25π¦^{6}π^{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 2β2: 3β5

(2β2)^{2}: (3β5)^{2}

(2 Γ 2 Γ β2 Γ β2): (3 Γ 3 Γ β5 Γ β5)

8: 45

Hence, the duplicate ratio of 2β2: 3β5 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₯^{2}π^{4}: 25π¦^{6}π^{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

β36: β49

6: 7

**Question **5. Find the value of x, if:

(i) (2π₯ + 3): (5π₯β 38) is the duplicate ratio of β5: β6

(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 β5: β6 It is given that,

Duplicate ratio of β5: β6

Duplicate ratio of (β5)^{2}: (β6)^{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₯ = β190 β 18

β13π₯ = β208

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 β9: β25

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 3β8 : 3β27

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)

ππ₯ + ππ = ππ¦ + ππ

ππ β ππ = ππ¦ β ππ₯

π(π β π) = ππ¦ β ππ₯

**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₯

84Γ5/7 = π₯

12 Γ 5 = π₯

60 = π₯

π₯ = 60

Hence, reduced weight us 60 kg.

**Question **8. If 15(2π₯^{2}β π¦^{2}) = 7π₯π¦, find π₯: π¦; if x and y both are positive.

Solution:

It is given that,

15(2π₯^{2}β 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^{2}β π^{2} and π + π.

(iii) mean proportional to (π₯β π¦) and (π₯^{3}β π₯^{2}π¦).

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

ππ = 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

π2 = π112 __________________________(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₯^{2}π¦ = 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)(4πβ 9π) = (4πβ 9π)(4π + 9π), prove that: π: π = π: π.

Solution:

It is given that,

(4π + 9π)(4πβ 9π) = (4πβ 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

β9π = 81 β 135

β9π = β54

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)

ππ₯^{2} + 2ππ₯ + π β π₯^{2} β 2π₯ β 1 = ππ₯^{2} β 2ππ₯ + π + π₯^{2} β 2π₯ + 1

2ππ₯ β π₯^{2} β 1 = β2ππ₯ + π₯^{2} + 1

2ππ₯ β π₯^{2} β 1 + 2ππ₯ β π₯^{2} β 1 = 0

4ππ₯ β 2π₯^{2} β 2 = 0

2(2ππ₯ β π₯^{2} β 1) = 0

2ππ₯ 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 = β3

π = 5/3 π = β 3/2

a canβt 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 = ππ, π = ππ

π = ππ Γ π

π = ππ

^{2}

Now,

In the LHS side it is given that,