# Frank Solutions for Chapter 9 Ratio and Proportion Class 10 ICSE Mathematics

**Exercise 9**

** Find the ratio of the following in the simplest form:**

**(i) 5.60 and 2.40**

**(ii) 432 and 120**

**(iii) ₹** **5.40 and 180 paise**

**(iv) a ^{4} + b^{4} and a^{3} – b^{3}**

**(v) x ^{2} + 4x + 4 and x^{2} – x -6**

**Answer**

**(i)** 5.60 and 2.40

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, ratio of the given number is 7: 3.

**(ii)** 432 and 120

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, ratio of the given number is 18: 5.

**(iii)** ₹ 5.40 and 180 paise

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, ratio of the given number is 3: 1.

**(iv)** a^{4} + b^{4} and a^{3} – b^{3}

Given question can be written as,

= (a^{4} + b^{4})/(a^{3} – b^{3})

We know that, a^{4} + b^{4} = (a – b) (a^{3} + ab^{2} + a^{2}b + b^{3})

Then, a^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2})

So, [(a – b) (a^{3} + ab^{2} + a^{2}b + b^{3})]/[(a – b) (a^{2} + ab + b^{2})]

= (a^{3} + ab^{2} + a^{2}b + b^{3})/(a^{2} + ab + b^{2})

Therefore, ratio of the given terms is (a^{3} + ab^{2} + a^{2}b + b^{3}): (a^{2} + ab + b^{2}).

**(v)** x^{2} + 4x + 4 and x^{2} – x - 6

Given question can be written as,

= (x^{2} + 4x + 4)/(x^{2} – 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)**

**(ii) (6a – b)/(a + 3b)**

**(iii) (5a – 4b)/(2a – 3b)**

**Answer**

From the question it is given that,

a: b = 4: 7

⇒ a/b = 4/7

**(i)** (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)

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)

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)**

**Answer**

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.**

**Answer**

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**

**Answer**

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**

**Answer**

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**

**Answer**

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**

**(ii) (a ^{2} – b^{2}): (a^{2} + b^{2}) and (a^{4} – b^{4}): (a + b)^{4}**

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

**(iv) √8: 4, 3: √5 and √20: √27**

**(v) (m – n): (m + n), (m + n) ^{2}: (m^{2} + n^{2}) and (m^{4} – n^{4}): (m^{2} – n^{2})^{2}**

**Answer**

**(i)** 15: 16 and 8: 5

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)** (a^{2} – b^{2}): (a^{2} + b^{2}) and (a^{4} – b^{4}): (a + b)^{4}

Given ratio can be written as,

(a^{2} – b^{2})/(a^{2} + b^{2}) and (a^{4} – b^{4})/(a + b)^{4}

= (a^{2} – b^{2})/(a^{2} + b^{2}) × (a^{4} – b^{4})/(a + b)^{4}

We know that, (a^{2} – b^{2}) = (a + b) (a – b)

= ((a + b) (a – b))/(a^{2} + b^{2}) × ((a^{2} + b^{2}) (a^{2} – b^{2}))

= ((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

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

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}: (m^{2} + n^{2}) and (m^{4} – n^{4}): (m^{2} – n^{2})^{2}

Given ratio can be written as,

(m – n)/(m + n), (m + n)^{2}/(m^{2} + n^{2}) and (m^{4} – n^{4})/(m^{2} – n^{2})^{2}

= (m – n)/(m + n) × (m + n)^{2}/(m^{2} + n^{2}) × (m^{4} – n^{4})/(m^{2} – n^{2})^{2}

= (m – n)/(1) × (m + n)/(m^{2} + n^{2}) × ((m^{2} + n^{2}) (m^{2} – n^{2}))/(m^{2} – n^{2})^{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**

**(ii) 3√2a : 2√3a**

**(iii) 2/3: 4/9**

**(iv) (a + b): (a ^{2} – b^{2})**

**Answer**

**(i) **√10 : √14

Given, √10 : √14

= (√10)^{2}: (√14)^{2}

= 10: 14

= 10/14

= 5/7

Therefore, duplicate ratio is 5: 7.

**(ii) **3√2a : 2√3a

Given, 3√2a: 2√3a

= (3√2a)^{2}: (2√3a)^{2}

= 18a: 12a

= 18a/12a

= 3/2

Therefore, duplicate ratio is 3: 2.

**(iii)** 2/3: 4/9

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): (a^{2} – b^{2})

Given, (a + b): (a^{2} – b^{2})

= (a + b)^{2} : (a^{2} – b^{2})^{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

(i) 3: 5

**(ii) 2√5 : 5√2**

**(iii) √15 : √18**

**(iv) ∛(ab) ^{2}: ∛(a^{2}b)**

**Answer**

**(i)** 3: 5

Given, 3: 5

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

= 27: 125

Therefore, triplicate ratio is 27: 125

**(ii)** 2√5 : 5√2

Given, 2√5 : 5√2(ii) 2√5 : 5√2

= (2√5)^{3}: (5√2)^{3}

= (8 × 5√5)/(125 × 2√2)

By simplification,

= 4√5: 25√2

Therefore, triplicate ratio = 4√5: 25√2

**(iii)** √15 : √18

Given, √15: √18

= (√15)^{3}: (√18)^{3}

= 15√5: 18 × 3√2

= 5√15: 18√2

Therefore, triplicate ratio is 5√15: 18√2

**(iv)** ∛(ab)^{2}: ∛(a^{2}b)

Given, ∛(ab)^{2}: ∛(a^{2}b)

By simplification we get,

= (∛(ab)^{2})^{3}: (∛(a^{2}b))^{3}

= ab^{2} : a^{2}b

= b: a

Therefore, triplicate ratio is b : a

**11.** **Find the sub – duplicate ratio of the following:**

**(i) x ^{6}: y^{4}**

**(ii) 63m ^{2}: 28n^{2}**

**(iii) 1/16: 1/36**

**(iv) 9a ^{2}/5: 25a^{2}/3**

**Answer**

**(i)** x^{6}: y^{4}

Given, x^{6}: y^{4}

= √x^{6}: √y^{4}

= (x^{6})^{1/2}: (y^{4})^{1/2}

= x^{3}: y^{2}

Therefore, sub – duplicate ratio is x^{3}:y^{2}

**(ii)** 63m^{2}: 28n^{2}

Given, 63m^{2}: 28n^{2}

= √(63m^{2}): √(28n^{2})

= 3√7m: 2√7n

= 3m: 2n

Therefore, sub – duplicate ratio is 3m: 2n

1/16: 1/36

**(iii)** Given, 1/16: 1/36

= √(1/16): √(1/36)

= ¼: 1/6

= (¼)/(1/6)

= (¼) × (6/1)

= 3/2

Therefore, sub – duplicate ratio is 3: 2

**(iv)** 9a^{2}/5: 25a^{2}/3

Given, 9a^{2}/5: 25a^{2}/3

= √(9a^{2}/5): √(25a^{2}/3)

= 3a(1/√5): 5a(1/√3)

= 3√3: 5√5

Therefore, sub–duplicate ratio is 3√3: 5√5.

**12. Find the sub – triplicate ratio of the following:**

**(i) 512: 216**

**(ii) m ^{3}n^{6}: m^{6}n^{3}**

**(iii) 125a ^{3}: 343b^{6}**

**(iv) 64m ^{3}/729n^{3}: 216m^{3}/27n^{3}**

**Answer**

**(i)** 512: 216

Given, 512: 216

= ∛512 : ∛216

= (8^{3})^{1/3}: (6^{3})^{1/3}

= 8: 6

= 8/6

= 4/3

Therefore, sub – triplicate ratio is 4: 3.

**(ii)** m^{3}n^{6}: m^{6}n^{3}

Given, m^{3}n^{6}: m^{6}n^{3}

= ∛(m^{3}n^{6}) : ∛(m^{6}n^{3})

= (m^{3}n^{6})^{1/3} : (m^{6}n^{3})^{1/3}

= mn^{2}: m^{2}n

= mn^{2}/m^{2}n

= n/m

Therefore, sub – triplicate ratio is n: m.

**(iii)** 125a^{3}: 343b^{6}

Given, 125a^{3}: 343b^{6}

= ∛(125a^{3}): ∛(343b^{6})

= (125a^{3})^{1/3}: (343b^{6})^{1/3}

= 5a: 7b^{2}

Therefore, sub – triplicate ratio is 5a: 7b^{2}.

**(iv)** 64m^{3}/729n^{3}: 216m^{3}/27n^{3}

Given, 64m^{3}/729n^{3}: 216m^{3}/27n^{3}

= ∛(64m^{3}/729n^{3}): ∛(216m^{3}/27n^{3})

= (64m^{3}/729n^{3})^{1/3}: (216m^{3}/27n^{3})^{1/3}

By simplification we get,

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

= (4m/9n) × (3n/6m)

= 2/9

Therefore, sub – triplicate ratio is 2: 9.

**13.** **Find the reciprocal ratio of the following:**

**(i) 17/45: 51/27**

**(ii) 1/45: 1/54**

**(iii) a ^{3}b^{2}: a^{2}b^{3}**

**(iv) 81pq ^{2}: 54p^{2}q**

**Answer**

**(i)** 17/45: 51/27

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, reciprocal of the ratio is 5: 1

**(ii)** 1/45: 1/54

Given ratio, 1/45: 1/54

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

= 45: 54

= 45/54

= 5/6

Therefore, reciprocal of the ratio is 5: 6

**(iii)** a^{3}b^{2}: a^{2}b^{3}

Given ratio, a^{3}b^{2}: a^{2}b^{3}

The reciprocal of the given ratio is 1/a^{3}b^{2}: 1/a^{2}b^{3}

= (1/a^{3}b^{2}) × (a^{2}b^{3}/1)

= b: a

Therefore, reciprocal of the ratio is b: a

**(iv)** 81pq^{2}: 54p^{2}q

Given ratio, 81pq^{2}: 54p^{2}q

The reciprocal of the given ratio 1/81pq^{2}: 1/54p^{2}q

= (1/81pq^{2}) × (54p^{2}q/1)

By simplification we get,

= 2p/3q

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

**14. Which of the following ratios is greater?**

**(i) 3: 5 and 2: 11**

**(ii) 2: 3 and 13: 19**

**(iii) 5: 8 and 7: 10**

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

**Answer**

**(i)** 3: 5 and 2: 11

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

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

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)

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.**

**Answer**

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.

**16. Two positive numbers are in the ratio 3 : 4 and sum of their squares is 1225. Find the numbers. **

**Answer**

Let the two numbers be 3x and 4x, since the ratio between them is 3 : 4.

**17. Two numbers are in the ratio 5 : 7 and the difference of their squares is 600. Find the numbers. **

**Answer**

Let the two numbers be 5x and 7x, since the ration between then is 5 : 7.

**18. What quantity must be subtracted from each term of the ratio 39: 89 to make it equal to 2 : 5 ?**

**Answer**

Let x be subtracted from each term such that:

**19. What quantity must be added to each term of the ratio 19: 51 to make it equal to 3 : 7 ?**

**Answer**

Let x be added from each term such that

**20. What quantity must be added to each term of the ratio (p + q) : (p – q) to make it equal to (p + q) ^{2} : (p – q)^{2}? **

**Answer**

Let x be added from each term such that

**21. If (3x – 4) : (2x + 5) is the duplicate ratio of 3 : 4, find x.**

**Answer**

(3x – 4)/(2x + 5) = (3)^{2}/(4)^{2}

**22. If (5x + 3) : (3x + 1) is the triplicate ratio of 4 : 3, find x. **

**Answer**

(5x + 3)/(3x + 1) = (4)^{3}/(3)^{3 }

**23. If r ^{2} = pq, Show that p : q is the duplicate ration of (p + r) : (q + r) **

**Answer**

p/q = (p + r)^{2}/(q + r)^{2}

**24. A sum of money is divided in the ratio 2 : 3. If the larger portion is Rs 7, 47,300. Find the sum distributed. **

**Answer**

2/3 = x/(7,47,300)

**25. Divide a number into two parts in the ratio 5 : 7 so that smaller part is 60. Find the number.**

**Answer**

5/7 = 60/x

⇒ 5x = 420

⇒ x = 84

Number = 84 + 60 = 144

**26. A bag contains Rs 1800 in the form of Re 1, Rs 2 and Rs 5 coins. The ratio of the number of the respective coins is 3 : 7 : 11. Find the total number of coins in the bag. **

**Answer**

Let the number of Re 1, Rs 2 and Rs 5 coins be 3x, 7x and 11x

**27. The present age of two persons are in the ratio 4 : 3. /nine years hence their ages will be in the ratio 23 : 18. Find their present ages. **

**Answer**

Let present ages be 4x and 3x.

**28. The ratio of the pocket money saved by a boy and his sister is 5 : 4. If the brother saves Rs 100 more, how much more should the sister save in order to keep the ratio of their savings unchanged ?**

**Answer**

Let money saved by the boy and her sister be 5x and 4x and let the sister need to save Rs y more

**29. A cistern of milk contains a mixture of milk and water in the ration 11 : 4. If the cistern contains 60 litres of milk, how much more water must be added to make he ratio of milk to water as 11: 6 ?**

**Answer**

Quantity of milk = 60 × 11/15 = 44 litres

**30. In an examination the ratio of the number of successful candidates to unsuccessful candidates is 7 : 5. Had 30 more appear and 10 more passed, the ratio of successful candidates to unsuccessful candidates would have been 4 : 3. Find the number of candidates who appeared in the examination originally. **

**Answer**

Let successful candidates be 7x and unsuccessful candidates be 5x.

**Exercise 9.2 **

**1. Find the value of the unknown in the following proportions: **

**(i) 5 : 12 : : 15 : x **

**(ii) 3 : 4 : : p : 12 **

**(iii) 1/2 : m : : 14/9 : 4/3 **

**(iv) a : 9/2 : : 7/2 : 11/2 **

**Answer**

**(i)** 5 : 12 : : 15 : x

**2. Find the fourth proportion to the following:**

**(i) 3, 5 and 15 **

**(ii) 0.7, 4.9 and 1.6 **

**(iii) (p ^{2}q – qr^{2}).(pqr – pr^{2}) and (pq^{2 }– pr^{2}) **

**(iv) (x ^{2} – y^{2}), (x^{3} + y^{3}) and (x^{3} – xy^{2} + x^{2}y – y^{3}) **

**Answer**

**(i)** 3, 5 and 15

**3. Find the third proportion to the following:**

**(i) 3 and 15 **

**(ii) 16x ^{2} and 24x **

**(iii) (x – y) and m(x – y) **

**(iv) 9/25 and 18/25 **

**Answer**

**(i)** 3 and 15

**4. Find the mean proportion of the following: **

**(i) 24 and 6 **

**(ii) 0.09 and 0.25 **

**(iii) ab ^{3} and a^{3}b**

**(iv) 28/3 and 175/27 **

**Answer**

**(i) **24 and 6

**5. If x, 12 and 16 are in continued proportion, find x.**

**Answer**

Since x, 12 and 16 are in continued proportion

**6. If 1/12, x and 1/75 are in continued proportion, find x.**

**Answer**

Since 1/12, x and 1/75 are in continued proportion

**7. If y is the mean proportional between x and z, show that :**

**xyz(x + y + z) ^{3} = (xy + yz + xz)^{3}**

**Answer**

Since y is the mean proportion between x and z

**8. If y is the mean proportional between x and y; show that y(x + z) is the mean proportional between x ^{2} + y^{2} and y^{2} + z^{2}.**

**Answer**

Since y is the mean proportion between x and z.

**9. 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.**

**Answer**

Let x, y and z are the three quantities which are in continued proportion.

**10. Given four quantities p, q, r and s are in proportion, show that q ^{2}(p – r): rs(q – s) = (p^{2} – r^{2} – pq) : (r^{2} – s^{2} – rs).**

**Answer**

p, q, r and s are in proportion then, p : q : : r : s

**11. If a : b = c : d; then show that (ax + by) : b = (cx + dy) : d**

**Answer**

a/b = c/d ⇒ ax/b = cx/d

**12. If (a + c) : b = 5 : 1 and (bc + cd) = 5 : 1, then prove that a: b = c : d**

**Answer**

(a + c)/b = 5/1

**13. Find the smallest number that must be subtracted from each of the number 20, 29, 84 and 129 so that they are in proportion.**

**Answer**

Let x be subtracted from each number so that 20 – x, 29 – x, 84 – x and 129 – x are in proportion.

**14. Find two numbers mean proportional is 12 and the third proportional is 324.**

**Answer**

Let a and b be the two numbers, whose mean proportional is 12.

**15. Find two numbers whose mean proportional is 18 and the third proportional is 486.**

**Answer**

Let a and b be the two numbers, whose mean proportional is 18.

**Exercise 9.3 **

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

**(i) 2a + 7b : 2a – 7b = 2c + 7d : 2c - 7d **

**(ii) 7a + 11b : 7a – 11b = 7c + 11d : 7c – 11d**

**(iii) (4 + 9b)/(4c + 9d) = (4a – 9b)/(4c – 9d)**

**(iv) (ax + by) : (cx + dy) = (ax – by) : (cx – dy) **

**Answer**

**(i) **2a + 7b : 2a – 7b = 2c + 7d : 2c – 7d

**2. If (7a + 12b)/(7c + 12d) = (7a – 12b)/(7c – 12d), then prove that a/b = c/d.**

**Answer**

(7a + 12b)/(7c + 12d) = (7a – 12b)/(7c – 12d)

**3. If (7m + 8n)(7p – 8q) = (7m – 8n)(7p + 8q), then prove that m : n = p : q**

**Answer**

(7m + 8n)(7p – 8q) = (7m – 8n)(7p + 8q)

^{2 }+ c^{2})/(b^{2}+ d^{2}) = ac/bd**Answer**

a/b = c/d

⇒ a = bc/d

**5. If a : b = c : d, then prove that (a ^{2} + ab + b^{2})/(a^{2} – ab + b^{2}) = (c^{2} + cd + d^{2})/(c^{2} – cd + d^{2})**

**Answer**

a/b = c/d

⇒ a = bc/d

**6. If a : b : : c : d : : e : f, then prove that (ae + bf)/(ae – bf) = (ce + df)/(ce – df)**

**Answer**

a/b = c/d = e/f

**7. If p, q, r and s are in continued proportion, then prove that (p ^{3} + q^{3} + r^{3}) : (q^{3} + r^{3} + s^{3}) : : p : s**

**Answer**

p/q = q/r = r/s = k

**8. If u, v, w and x are in continued proportion, then prove that (2u + 3x) : (3u + 4x) : : (2u ^{3} + 3v^{3}) : (3u^{3 }+ 4v^{3})**

**Answer**

u/v = v/w = w/x = a

**9. If p, q and r in continued proportion, then prove the following:**

**(i) (p ^{2} – q^{2})(q^{2} + r^{2}) = (q^{2} – r^{2})(p^{2} + q^{2}) **

**(ii) (p + q + r)(p – q + r) = p ^{2} + q^{2 }+ r^{2} **

**(iii) (pqr) ^{2 }(1/p^{4} + 1/q^{4} + 1/r^{4}) = (p^{4} + q^{4} + r^{4})/q^{2}**

**(iv) p ^{2 }- q^{2} + r^{2} = q^{4} (1/p^{2} – 1/q^{2} + 1/r^{2}) **

**Answer**

**(i)** (p^{2} – q^{2})(q^{2} + r^{2}) = (q^{2} – r^{2})(p^{2} + q^{2})

**10. If a, b, c and d are in continued proportion, then prove that**

**(i) ad(c ^{2} + d^{2}) = c^{3}(b + d)**

**(iii) (a + d)(b + c) – (a + c)(b + d) – (b – c) ^{2} **

**Answer**

**(i)** ad(c^{2} + d^{2}) = c^{3}(b + d)

**11. If x/(b + c – a) = y/(c + a – b) = z/(a + b – c), then prove that each ratio is equal to the ratio of (x + y + z)/(a + b + c).**

**Answer**

x/(b + c – a) = y/(c + a – b) = z/(a + b – c) = k

**12. If a/(b + c) = b/(c + a) = c/(a + b), then prove that a(b – c) + b(c – a) + c(a – b) = 0**

**Answer**

a/(b + c) = b(c + a) = c/(a + b) = k

**13. If x =****then prove that x ^{3} – 3mx^{2} + 3x = m**

**Answer**

^{2}– b^{2})/ab**Answer**

x = pab/(a + b) ⇒ x/pa = b/(a + b)

**15. Show that the value of x is 11, when (x ^{3 }+ 3x)/(3x^{2} + 1) = 341/91**

**Answer**

(x^{3 }+ 3x)/(3x^{2} + 1) = 341/91