Class 10: Ratio and Proportion – Miscellaneous Problems – Set 1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1:

(i) lf $2x + 3y : 3x+ 5y = 18 : 29,$ find $x : y$

(ii) If $x:y = 2:3,$ find the value of $3x + 2y : 2x + 5y$

Answer:

(i) $\displaystyle 2x + 3y :3x+ 5y = 18:29$

$\displaystyle \Rightarrow \frac{2x+3y}{3x+ 5y} = \frac{18}{29}$

$\displaystyle \Rightarrow 58x+87y = 54x + 90y$

$\displaystyle \Rightarrow 4x = 3y$

$\displaystyle \Rightarrow \frac{x}{y} = \frac{3}{4}$

$\displaystyle \Rightarrow x:y = 3: 4$

$\displaystyle \text{(ii) } x: y = 2: 3 \hspace{1.0cm} \Rightarrow \frac{x}{y} = \frac{2}{3}$

$\displaystyle 3x+2y: 2x + 5y = \frac{3x+2y}{2x+ 5y}$

$\displaystyle = \frac{3 \frac{x}{y}+2}{2 \frac{x}{y}+5}$

$\displaystyle = \frac{3 \frac{2}{3}+2}{2 \frac{2}{3}+5}$

$\displaystyle = 12 : 19$

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$\displaystyle \text{Question 2: If } a : b= 5 : 3, \text{ find } (5a + 8b) : (6a -7b).$ [ICSE Board 2002]

Answer:

$\displaystyle \text{Let } a : b = 5 : 3 \hspace{1.0cm} \Rightarrow \text{ if } a = 5x , \text{then } b = 3x$

$\displaystyle \text{and } \frac{5a+ 8b}{6a - 7b} = \frac{5 \times 5x+ 8 \times 3x}{6 \times 5x - 7 \times 3x} = \frac{49x}{9x} = 49 : 9$

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Question 3: Two numbers are in the ratio 3 : 5. If 8 is added to each number, the ratio becomes 2 : 3. Find the numbers.

Answer:

Since, the ratio between the numbers is $3 : 5$

Therefore if one number is $3x;$ the other number is $5x$

$\displaystyle \text{Given : } \frac{3x+8}{5x+8} = \frac{2}{3}$

$\Rightarrow 10x + 16 = 9x + 24$

$\Rightarrow x = 8$

$\displaystyle \text{Therefore Nos are } 3x \text{ and } 5x = 3 \times 8 \text{ and } 5 \times 8 = 24 \text{ and } 40$

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Question 4:

(i) What quantity must be added to each term of the ratio 8 : 15 so that it becomes equal to 3:5?

(ii) What quantity must be subtracted from each term of the ratio a : b so that it becomes c : d?

Answer:

(i) Let $x$ be added to each term of the ratio $8 : 15.$

$\displaystyle \text{Given: } \frac{8+x}{15+x} = \frac{3}{5}$

$\displaystyle \Rightarrow 40 + 5x = 45 + 3x$

$\displaystyle \Rightarrow x = 2\frac{1}{2}$

(ii) Let $\displaystyle x$ be subtracted, then :

$\displaystyle \frac{a-x}{b-x} = \frac{c}{d}$

$\displaystyle \Rightarrow ad-dx=bc-cx$

$\displaystyle \Rightarrow cx-dx=bc-ad$

$\displaystyle \Rightarrow x(c-d) = bc - ad$

$\displaystyle \Rightarrow x = \frac{bc - ad}{c-d}$

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Question 5: The work done by $(x - 3)$ men in $(2x + 1)$ days and the work done by $(2x + 1)$ men in $(x + 4)$ days are in the ratio $3 : 10.$ Find the value of $x.$

Answer:

Assuming that all the men do the same amount of work in one day and one day work of each man = 1 unit; we get :

Amount of work done by $(x - 3)$ men in $(2x + 1)$ days

= amount of work done by $(x - 3) (2x + 1)$ men in one day

$= (x -3) (2x + 1)$ units of work.

Similarly, amount of work done by $(2x + 1)$ men in $(x + 4)$ days.

= amount of work done by $(2x + 1) (x + 4)$ men in one day.

$= (2x + 1) (x + 4)$ units of work.

According to the given statement:

$\displaystyle \frac{(x - 3) (2x + 1)}{(2x + 1) (x + 4)} = \frac{3}{10}$

$\displaystyle \Rightarrow \frac{2x^2+x - 6x - 3}{2x^2+8x+x+4} = \frac{3}{10}$

$\displaystyle \Rightarrow \frac{2x^2 - 5x - 3}{2x^2+9x+4} = \frac{3}{10}$

$\displaystyle \Rightarrow 20x^2 - 50 x - 30 = 6x^2 + 27x + 12$

$\displaystyle \Rightarrow 14x^2 - 77x - 42 = 0$

$\displaystyle \Rightarrow 2x^2 - 11x - 6=0$

$\displaystyle \Rightarrow ( x - 6) ( 2x+ 1) = 0$

$\displaystyle \Rightarrow x = 6 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = \frac{-1}{2}$

$\displaystyle x = \frac{-1}{2} \text{ is not possible as it will make no. of men } (x - 3) \text{ negative. }$

Therefore $x = 6$

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Question 6: When the fare of a certain journey by an airliner was increased in the ratio 5:7 the cost of the ticket for the journey became Rs. 1421. Find the increase in the fare.

Answer:

According to the given statement :

$\displaystyle \text{The original fair : Increased fair } = 5 :7$

$\displaystyle \Rightarrow 7 \times \text{The original fare} = 5 \times \text{Increased fair}$

$\displaystyle \Rightarrow 7 \times \text{The original fare }= 5 \times 1421$

$\displaystyle \Rightarrow \text{The original fare} = \frac{5 \times 1421}{7} = \text{ Rs. } 1015$

$\displaystyle \text{Therefore increase in the fare }= ( 1421 - 1015 ) = \text{ Rs. } 406$

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Question 7: In a regiment, the ratio of number of officers to the number of soldiers was 3 : 31 before a battle. In the battle 6 officers and 22 soldiers were killed. The ratio between the number of officers and the number of soldiers now is 1 : 13. Find the number of officers and soldiers in the regiment before the battle. [ICSE Board 1992]

Answer:

Before the battle:

Let the number of officers be $3x \Rightarrow$ the number of soldiers $= 31x$

After the battle:

The number of officers $= 3x - 6$ and the number of soldiers $= 31x - 22$

Given

$\displaystyle \frac{3x-6}{31x-22} = \frac{1}{13}$

$\Rightarrow x = 7$

Therefore the number of officers before the battle $= 3x = 3 \times 7 = 21$

And the number of soldiers before the battle $= 31 x = 31 \times 7 = 217$

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$\displaystyle \text{Question 8: If } \frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} \text{ and } a + b + c = 0; \\ \\ \text{ show that each given ratio is equal to } -1.$

Answer:

$\displaystyle \text{Since } a + b + c = 0 \Rightarrow a+b = - c, b+c = - a, \text{ and } c+a = - b$

$\displaystyle \therefore \frac{a}{b+c} = \frac{a}{-a} = -1 ;$

$\displaystyle \frac{b}{c+a} = \frac{b}{-b} = - 1$

$\displaystyle \text{and } \frac{c}{a+b} = \frac{c}{-c} = -1$

$\\$

$\displaystyle \text{Question 9: If } \frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} \text{ and } a + b + c \neq 0; \\ \\ \text{ show that each given ratio is equal to } \frac{1}{2}.$

Answer:

$\displaystyle \frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} = \frac{a+b+c}{(b+c) + ( c+a) + ( a+b)} = \frac{a+b+c}{2(a+b+c)} =\frac{1}{2}$

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Question 10: Find the compound ratio of :

$\displaystyle \text{(i) } 3a : 2b , 2m : n \text{ and } 4x: 3y$

$\displaystyle \text{(ii) } a-b : a+b, (a+b)^2 \text{ and } a^4 - b^4 : ( a^2 - b^2)^2$

Answer:

$\displaystyle \text{(i) Required compound ratio } = (3a \times 2m \times 4x) : (2b \times n \times 3y) \\ \\ { \hspace{5.5cm} = \frac{24amx}{6bny} = 4amx: bny }$

$\displaystyle \text{(ii) Required compound ratio } \\ \\ = [ (a-b) \cdot (a+b)^2 \cdot (a^4 - b^4)] : [(a+b) \cdot (a^2+b^2) \cdot (a^2-b^2)^2]$

$\displaystyle = \frac{(a-b)(a+b)^2(a^2+b^2)(a^2-b^2)}{(a+b)(a^2+b^2)(a^2-b^2) ( a+b)(a-b)}$

$\displaystyle = 1: 1$

$\\$

Question 11: Find the ratio compounded of the duplicate ratio of 5 : 6, the reciprocal ratio of 25 : 42 and the sub-triplicate ratio of 216 : 343.

Answer:

$\displaystyle \text{Since, duplicate ratio of } 5 : 6 = 5^2 : 6^2 = 25 : 36$

$\displaystyle \text{Reciprocal ratio of } 25 :42 = \frac{1}{25}: \frac{1}{42} =42: 25$

$\displaystyle \text{And sub-triplicate ratio of } 216 : 343 = \sqrt[3]{216} : \sqrt[3]{343} = 6: 7$

$\displaystyle \text{Therefore, the required compounded ratio } \\ \\ = ( 25 \times 42 \times 6) : (36 \times 25 \times 7) = \frac{ 25 \times 42 \times 6}{36 \times 25 \times 7} = 1: 1$

$\\$

Question 12: Find:

(i) the fourth proportional to 3, 6 and 4.5

(ii) the mean proportional between 6.25 and 0.16

(iii) the third proportional to 1.2 and 1.8

Answer:

(i) Let the fourth proportional to $3, 6$ and $4.5$ be $x.$

$3 : 6 = 4.5 : x \Rightarrow 3 \times x = 6 \times 4.5 \Rightarrow x = 9$

(ii) Let the mean proportional between $6.25$ and $0.16$ be be $x.$

$\Rightarrow 6.25, x \text{ and } 0.16 \text{ are in continued proportion. }$

$\Rightarrow 6.25: x = x : 0.16$

$\Rightarrow x \times x =6.25 \times 0.16 \Rightarrow x^2=1$

$\\$

Question 13: Quantities $a, 2, 10,$ and $b$ are in continues proportion; find the values of a and b.

Answer:

$\displaystyle a , 2, 10 \text{ and } b \text{ are in continued proportion.}$

$\displaystyle \Rightarrow \frac{a}{2} = \frac{2}{10} = \frac{10}{b}$

$\displaystyle \Rightarrow \frac{a}{2} = \frac{2}{10} \hspace{0.5cm} \text { and } \hspace{0.5cm} \frac{2}{10} = \frac{10}{b}$

$\displaystyle \Rightarrow a = \frac{4}{10} =0.4 \hspace{0.5cm} \text { and } \hspace{0.5cm} b = \frac{100}{2} = 50$

$\\$

Question 14: What number should be subtracted from each of the numbers 23, 30, 57 and 78; so that the remainders are in proportion. [ICSE Board 2004 ]

Answer:

Let the number subtracted be $x .$

$\displaystyle \therefore ( 23-x) : ( 30-x) :: ( 57-x): (78-x)$

$\displaystyle \Rightarrow \frac{23-x}{30-x} = \frac{57-x}{78-x}$

$\displaystyle \Rightarrow 1794 - 101 x + x^2 = 1710 - 87 x + x^2$

$\displaystyle \Rightarrow 14 x = 84$

$\displaystyle \Rightarrow x = 6$

$\\$

Question 15: What should be added to each of the numbers 13, 17 and 22 so that the resulting numbers are in continued proportion.

Answer:

Let the number added be $x .$

$\displaystyle \therefore 13+x, 17+x \text{ and } 22+x \text{ are in continued proportion. }$

$\displaystyle \Rightarrow \frac{13+x}{17+x} = \frac{17+x}{22+x}$

$\displaystyle \Rightarrow (13+x)(22+x) = ( 17+x)^2$

$\displaystyle \Rightarrow 286 + 35 x + x^2 = 289 + 34 x + x^2$

$\displaystyle \Rightarrow x = 3$

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Question 16: If $(a^2 + c^2) , ( ab + cd)$ and $( b^2 +d^2)$ are in continued proportion: prove that $a, b, c$ and $d$ are in proportion.

Answer:

Given $(a^2 + c^2) , ( ab + cd)$ and $( b^2 +d^2)$ are in continued proportion.

$\displaystyle \therefore \frac{a^2 + c^2 }{ab + cd} = \frac{ab+cd}{b^2 + d^2}$

$\displaystyle \Rightarrow ( a^2 + c^2)(b^2 + d^2) = ( ab + cd)^2$

$\displaystyle \Rightarrow a^2 b^2 + a^2 d^2 + b^2 c^2 + c^2 d^2 = a^2 b^2 + 2 abcd + c^2 d^2$

$\displaystyle \Rightarrow a^2 d^2 + b^2 c^2 - 2 abcd = 0$

$\displaystyle \Rightarrow ( ad - cd)^2 = 0$

$\displaystyle \Rightarrow ad - bc = 0$

$\displaystyle \Rightarrow ad = bc$

$\displaystyle \Rightarrow \frac{a}{b} = \frac{c}{d}$

Therefore $a, b, c,$ and $d$ are in continued proportion.

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$\displaystyle \text{Question 17: If }p:q :: q:r, \text{ prove that } p:r = p^2:q^2.$

Answer:

$\displaystyle p:q :: q:r \Rightarrow q^2 = pr$

$\displaystyle \therefore p^2 : q^2 = \frac{p^2}{q^2} = \frac{p^2}{pr} = \frac{p}{r} = p:r$

$\\$

Question 18: If $a \neq b$ and $a : b$ is the duplicate ratio of $a+c$ and $b+ c,$ prove that $c$ is the mean proportion between $a$ and $b.$

Answer:

$\displaystyle \text{Given: } \frac{a}{b} = \frac{(a+c)^2}{(b+c)^2}$

$\displaystyle \Rightarrow a(b^2 + c^2 + 2bc) = b ( a^2 + c^2 + 2ac)$

$\displaystyle \Rightarrow ab^2 + ac^2 + 2abc = a^2b + bc^2+ 2abc$

$\displaystyle \Rightarrow ac^2 - bc^2 = a^2 b - ab^2$

$\displaystyle \Rightarrow c^2 ( a - b) = ab ( a-b)$

$\displaystyle \Rightarrow c^2 = ab$

$\displaystyle \Rightarrow c$ is mean proportion between $a \text{ and } b$

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$\displaystyle \text{Question 19: If } a+c = mb \text{ and } \frac{1}{b}+\frac{1}{d}= \frac{m}{c}, \text{ prove that } a, b, c \text{ and } d \\ \\ \text{ are in proportion. }$

Answer:

$\displaystyle \frac{1}{b}+\frac{1}{d}= \frac{m}{c}$

$\displaystyle \Rightarrow \frac{d+b}{bd} = \frac{m}{c}$

$\displaystyle \Rightarrow cd + cb = mbd$

$\displaystyle \Rightarrow cd + bc = (a+c) d$

$\displaystyle \Rightarrow cd + bc = ad + cd$

$\displaystyle \Rightarrow bc = ad$

$\displaystyle \Rightarrow \frac{a}{b} = \frac{c}{d}$

$\displaystyle \Rightarrow a, b, c$ and $d$ are in proportion

$\\$

Question 20: If q is the mean proportional between p and r, prove that:

$\displaystyle p^2 - q^2 + r^2 = q^4 \Big( \frac{1}{p^2} - \frac{1}{q^2} + \frac{1}{r^2} \Big)$

Answer:

Since $q$ is the mean proportional between $p$ and $r \Rightarrow q^2 = pr$

$\displaystyle RHS = q^4 \Big( \frac{1}{p^2} - \frac{1}{q^2} + \frac{1}{r^2} \Big)$

$\displaystyle = \frac{q^4}{p^2} - \frac{q^4}{q^2} + \frac{q^4}{r^2}$

$\displaystyle = \frac{q^4}{p^2} - q^2 + \frac{q^4}{r^2}$

$\displaystyle = \frac{p^2r^2}{p^2} - q^2 + \frac{p^2r^2}{r^2}$

$\displaystyle = r^2 - q^2 + p^2 = LHS$

$\\$

$\displaystyle \text{Question 21: If } \frac{a}{b} = \frac{c}{d}, \text{ prove that each given ratio } \Big( \frac{a}{b} \text{ and } \frac{c}{d} \Big) \text{ is equal to: }$

$\displaystyle \text{(i) } \frac{3a-5c}{3b-5d} \hspace{1.0cm} \text{(ii) } \sqrt{ \frac{2a^2+9c^2}{2b^2+9d^2} } \hspace{1.0cm} \text{(iii) } \Big( \frac{5a^3 - 13c^3}{5b^3 - 13d^3} \Big)^{\frac{1}{3}}$

Answer:

$\displaystyle \text{Let } \frac{a}{b} = \frac{c}{d} = k \Rightarrow a = bk \text{ and } c = dk$

$\displaystyle \text{(i) } \frac{3a-5c}{3b-5d} = \frac{3(bk) - 5(dk)}{3b-5d} = \frac{k(3b-5d)}{3b-5d} = k = \text{each given ratio }$

$\\$

$\displaystyle \text{(ii) } \sqrt{ \frac{2a^2+9c^2}{2b^2+9d^2} } = \sqrt{ \frac{2(bk)^2+9(dk)^2}{2b^2+9d^2} } = \sqrt{ \frac{k^2 ( 2b^2+9d^2)}{2b^2+9d^2} } = k = \text{each given ratio }$

$\\$

$\displaystyle \text{(iii) } \Bigg( \frac{5a^3 - 13c^3}{5b^3 - 13d^3} \Bigg)^{\frac{1}{3}} = \Bigg( \frac{5(bk)^3 - 13(dk)^3}{5b^3 - 13d^3} \Bigg)^{\frac{1}{3}} = \Bigg( \frac{k^3 ( 5b^3 - 13d^3)}{5b^3 - 13d^3} \Bigg)^{\frac{1}{3}} \\ \\ \\ { \hspace{3.0cm} = (k^3)^{\frac{1}{3}} = k = \text{each given ratio } }$

$\\$

Question 22: If a, b, c and d are in proportion, prove that :

$\displaystyle \text{(i) } \frac{a-b}{c-d} = \sqrt{\frac{3a^2+ 8b^2}{3c^2+8d^2} } \hspace{1.0cm} \text{(ii) } \frac{5a^2+12c^2}{5b^2+12d^2} = \sqrt{ \frac{3a^4 - 7c^4}{3b^4 - 7d^4} }$

Answer:

$\displaystyle \text{Let } \frac{a}{b} = \frac{c}{d} = k \Rightarrow a = bk \text{ and } c = dk$

$\displaystyle \text{(i) LHS } = \frac{a-b}{c-d} = \frac{bk-b}{dk-d} = \frac{b(k-1)}{d(k-1)} = \frac{b}{d}$

$\displaystyle \text{RHS } = \sqrt{\frac{3a^2+ 8b^2}{3c^2+8d^2} } = \sqrt{\frac{3(bk)^2+ 8b^2}{3(dk)^2+8d^2} } = \sqrt{\frac{b^2(3k^2+8)}{d^2(3k^2+8)} } = \sqrt{ \frac{b^2}{d^2} } = \frac{b}{d}$

Hence LHS = RHS. Hence proved.

$\\$

$\displaystyle \text{(ii) LHS } = \frac{5a^2+12c^2}{5b^2+12d^2} = \frac{5(bk)^2+12(dk)^2}{5b^2+12d^2} = \frac{k^4(5b^2+12d^2)}{5b^2+12d^2} = k^2$

$\displaystyle \text{RHS } = \sqrt{ \frac{3a^4 - 7c^4}{3b^4 - 7d^4} } = \sqrt{ \frac{3(bk)^4 - 7(dk)^4}{3b^4 - 7d^4} } = \sqrt{ \frac{k^4(3b^4 - 7d^4)}{3b^4 - 7d^4} } = k^2$

Hence LHS = RHS. Hence proved.

$\\$

Question 23: 6 is the mean proportion between two numbers $x$ and $y$ and 48 is third proportion to $x$ and $y.$ Find the number. [ICSE Board 2011]

Answer:

Since, $6$ is mean proportional between $x$ and $y.$

$\Rightarrow x:6=6:y \hspace{1.0cm} \Rightarrow xy=36 \text{ ... ... ... ... ... i) }$

and, $48$ is third proportional to $x$ and $y$

$\Rightarrow x:y= y:48 \hspace{1.0cm} \Rightarrow y^2 = 48 x \text{ ... ... ... ... ... ii) }$

$\displaystyle \text{From eq. i); } xy = 3 \hspace{1.0cm} \Rightarrow x = \frac{36}{y}$

$\displaystyle \text{Substituting } x = \frac{36}{y} \text{ in eq. ii), we get :}$

$\displaystyle y^2 = 48 \times \frac{36}{y} \hspace{1.0cm} \Rightarrow y^3 = 36 \times 48 \text{ and } y = 12$

$\displaystyle \text{Therefore } x = \frac{36}{y} = \frac{36}{12} = 3$

The required nos. are $3$ and $12.$

$\\$

$\displaystyle \text{Question 24: If } \frac{8x+13y}{8x-13y} = \frac{9}{7}, \text{ find x: y }$

Answer:

Applying componendo and dividendo:

$\displaystyle \frac{8x+13y}{8x-13y} = \frac{9}{7}$

$\displaystyle \Rightarrow \frac{8x+13y + 8x-13y}{8x+13y- 8x+13y} = \frac{9+7}{9-7}$

$\displaystyle \Rightarrow \frac{16x}{26y} = \frac{16}{2}$

$\displaystyle \Rightarrow \frac{x}{y} = 8 \times \frac{26}{16} = \frac{13}{1}$

$\displaystyle \Rightarrow x:y = 13: 1$

$\\$

$\displaystyle \text{Question 25: If } a: b = c: d, \text{ show that: } 3a +2b:3a-2b =3c +2d:3c -2d.$

Answer:

$a: b = c: d$

$\displaystyle \Rightarrow \frac{a}{b} = \frac{c}{d}$

$\displaystyle \Rightarrow \frac{3a}{2b} = \frac{3c}{2d}$

Applying componendo and dividendo:

$\displaystyle \Rightarrow \frac{3a+2b}{3a-2b} = \frac{3c+2d}{3c-2d}$

$\displaystyle \Rightarrow 3a +2b:3a-2b =3c +2d:3c -2d$

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$\displaystyle \text{Question 26: If } \frac{8a-5b}{8c-5d} = \frac{8a+5b}{8c+5d}, \text{ prove that } \frac{a}{b} = \frac{c}{d}$

Answer:

$\displaystyle \frac{8a-5b}{8c-5d} = \frac{8a+5b}{8c+5d}$

$\displaystyle \Rightarrow \frac{8a-5b}{8a+5b} = \frac{8c-5d}{8c+5d}$

Applying componendo and dividendo:

$\displaystyle \Rightarrow \frac{8a-5b + 8a+5b}{8a-5b-8a-5b} = \frac{8c-5d+ 8c+5d}{8c-5d-8c-5d}$

$\displaystyle \Rightarrow \frac{16a}{-10b} = \frac{16c}{-10d}$

$\displaystyle \Rightarrow \frac{a}{b} = \frac{c}{d}$

$\\$

$\displaystyle \text{Question 27: If } p = \frac{4xy}{x+y}, \text{ find the value of } \frac{p+2x}{p-2x} + \frac{p+2y}{p-2y}$

Answer:

$\displaystyle p = \frac{4xy}{x+y}$ $\displaystyle \Rightarrow \frac{p}{2x} = \frac{2y}{x+y}$

Now apply componendo and dividendo

$\displaystyle \Rightarrow \frac{p+2x}{p-2x} = \frac{2y+ x + y}{2y - x - y}$

$\displaystyle \Rightarrow \frac{p+2x}{p-2x} = \frac{x+3y}{y-x}$

Again

$\displaystyle p = \frac{4xy}{x+y}$ $\displaystyle \Rightarrow \frac{p}{2y} = \frac{2x}{x+y}$

Now apply componendo and dividendo

$\displaystyle \Rightarrow \frac{p+2y}{p-2y} = \frac{2x+ x + y}{2x - x - y}$

$\displaystyle \Rightarrow \frac{p+2y}{p-2y} = \frac{3x+y}{x-y}$

$\displaystyle \therefore \frac{p+2x}{p-2x} + \frac{p+2y}{p-2y} = \frac{x+3y}{y-x} + \frac{3x+y}{x-y} \\ \\ \\ { \hspace{3.5cm} = \frac{x+3y}{y-x} - \frac{3x+y}{y-x} = \frac{x+3y-3x-y}{y-x} = 2 }$

$\\$

Question 28: If $a: b = c : d;$ prove that:

$(a^2 + ac + c^2) : (a^2 - ac + c^2) = (b^2 + bd + d^2) : (b^2 - bd + d^2)$

Answer:

$\displaystyle \text{Let } \frac{a}{b} = \frac{c}{d} = k \Rightarrow a = bk \text{ and } c = dk$

$\displaystyle \therefore (a^2 + ac + c^2) : (a^2 - ac + c^2) = \frac{a^2 + ac + c^2}{a^2 - ac + c^2}$

$\displaystyle = \frac{(bk)^2 + (bk)(dk) + (dk)^2}{(bk)^2 - (bk)(dk) + (dk)^2}$

$\displaystyle = \frac{k^2( b^2 + bd + d^2)}{k^2 ( b^2 - bd + d^2)}$

$\displaystyle = (b^2 + bd + d^2) : (b^2 - bd + d^2)$

Hence proved.

$\\$

Question 29: If $x, y$ and $z$ are in continued proportion, prove that :

Answer:

$x, y$ and $z$ are in continued proportion

$\displaystyle \Rightarrow \frac{x}{y} = \frac{y}{z} = k \hspace{1.0cm} \Rightarrow x = yk , y = zk \text{ and } x = yk = ( xk)k = zk^2$

$\displaystyle \therefore x^2 - y^2 : x^2 + y^2 = \frac{x^2 - y^2 }{x^2 + y^2} = \frac{y^2k^2- y^2}{y^2k^2 + y^2} = \frac{y^2(k^2-1)}{y^2(k^2+1)} = \frac{k^2 - 1}{k^2+1}$

Also,

$\displaystyle x-z : x+ z = \frac{x-z}{x+z} = \frac{zk^2-z}{zk^2+z} = \frac{z(k^2-1)}{z(k^2+1)} = \frac{k^2 - 1}{k^2 +1}$

Therefore $x^2 - y^2 : x^2 + y^2 = x-z : x+ z$

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Question 30: Using the properties of proportion, solve the following equation for $x :$

$\displaystyle \frac{x^3+3x}{3x^2+1}= \frac{341}{91}$

Answer:

Applying componendo and dividendo, we get :

$\displaystyle \frac{x^3+3x+3x^2+1}{x^3+3x-3x^2-1} = \frac{341+91}{341-91}$

$\displaystyle \Rightarrow \frac{(x+1)^3}{(x-1)^3} = \frac{432}{250} = \frac{216}{125} = \Big(\frac{6}{5} \Big)^3$

$\displaystyle \Rightarrow \frac{x+1}{x-1} = \frac{6}{5}$

Again, applying componendo and dividendo, we get :

$\displaystyle \frac{x+1+x-1}{x+1-x+1} = \frac{6+5}{6-5}$

$\displaystyle \Rightarrow \frac{2x}{2}= \frac{11}{1}$

$\displaystyle \Rightarrow x = 11$

$\\$

$\displaystyle \text{Question 31: If } x = \frac{\sqrt{3a+2b}+ \sqrt{3a-2b}}{\sqrt{3a+2b}-\sqrt{3a-2b}} , \text{ prove that: } bx^2 - 3ax + b = 0.$

Answer:

Given:

$\displaystyle \frac{x}{1} = \frac{\sqrt{3a+2b}+ \sqrt{3a-2b}}{\sqrt{3a+2b}-\sqrt{3a-2b}}$

Applying componendo and dividendo, we get :

$\displaystyle \Rightarrow \frac{x+1}{x-1} = \frac{\sqrt{3a+2b}+ \sqrt{3a-2b}+ \sqrt{3a+2b}-\sqrt{3a-2b}}{\sqrt{3a+2b}+ \sqrt{3a-2b} - \sqrt{3a+2b}+\sqrt{3a-2b}}$