Question 1: If a:b=3:5 , find: (10a+3b):(5a+2b)

Answer:

Given a:b=3:5

\Rightarrow a = \frac{3}{5} b 

\frac{(10a+3b)}{(5a+2b)} = \frac{(10 .\frac{3}{5} b+3b)}{(5 .\frac{3}{5} b+2b)}  = \frac{9}{5}

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Question 2: If 5x+6y:8x+5y=8:9  , find: x:y 

Answer:

Given 5x+6y:8x+5y=8:9

9(5x+6y)=8(8x+5y)

45x +54y=64x+40y

14y = 19x

\Rightarrow x:y = 14:19

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Question 3: If (3x-4y):(2x-3y)=(5x-6y):(4x-5y) , find: x:y 

Answer:

(3x-4y):(2x-3y)=(5x-6y):(4x-5y)

(3x-4y)(4x-5y) = (5x-6y)(2x-3y)

12x^2-16xy-15xy+20y^2 = 10x^2-12xy-15xy+18y^2

12x^2-31xy+20y^2 = 10x^2-27xy+18y^2

2x^2-4xy+2y^2 = 0

x^2-2xy+y^2 = 0

(x-y)^2 = 0 

or x-y = 0 

or x:y = 1:1

Note: You could have also done this by applying componendo and dividendo.

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Question 4: Find the:

i) Duplicate ratio of 2\sqrt{2}:3\sqrt{5} 

ii) Triplicate ratio of 2a:3b 

iii) Sub-duplicate ratio of 9x^2 a^4:25y^6 b^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 .

Answer:

i) Duplicate ratio of 2\sqrt{2}:3\sqrt{5}  = (2\sqrt{2})^2:(3\sqrt{5})^2 = 8:45  

ii) Triplicate ratio of 2a:3b  = (2a)^3:(3b)^3 = 8a^3:81b^3

iii) Sub-duplicate ratio of 9x^2 a^4:25y^6 b^2 =  \sqrt{9x^2 a^4}:\sqrt{25y^6 b^2} = 3xa^2:5y^3b

iv) Sub-triplicate ratio of 216:343 = \sqrt[3]{216}:\sqrt[3]{343}  

v) Reciprocal ratio of 3:5 = 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  = 25:36

Reciprocal ratio of 25:42 = 42:25

Sub-duplicate ratio of 36:49 = 6:7

Compound ratio \frac{25}{36} \times \frac{42}{25} \times \frac{6}{7} = \frac{1}{1} \ or \  1:1

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Question 5: Find the value of x , if:

i) (2x+3):(5x-38) is the duplicate ratio of \sqrt{5}:\sqrt{6}  .

ii) (2x+1):(3x+13) is the sub-duplicate ratio of 9:25 .

iii) (3x-7):(4x+3) is the sub-triplicate ratio of 8:27 .

Answer:

i) (2x+3):(5x-38) is the duplicate ratio of \sqrt{5}:\sqrt{6} 

\frac{2x+3}{5x-38} = \frac{5}{6} 

12x+18 = 25x-190 

13x = 208 \Rightarrow x = 16  

ii) (2x+1):(3x+13) is the sub-duplicate ratio of 9:25 .

\frac{2x+1}{3x+13} = \frac{\sqrt{9}}{\sqrt{25}}

\frac{2x+1}{3x+13} = \frac{3}{5} 

10x+5 = 9x+39

x = 34

iii) (3x-7):(4x+3) is the sub-triplicate ratio of 8:27

(3x-7):(4x+3) = \frac{\sqrt[3]{8}}{\sqrt[3]{27}} = \frac{2}{3}

9x-21 = 8x+6

or x = 27

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Question 6: What quantity must be added to each term of the ratio x:y so that it may become equal to c:d ?

Answer:

Let us add quantity a

Therefore

\frac{x+a}{y+a} = \frac{c}{d}

dx+da = cy+ca

(d-c)a = cy-dx

or a = \frac{cy-dx}{d-c}

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Question 7: Two numbers are in the ratio 5:7 . If 3 is subtracted from each of them, the ratio between them becomes 2:3 find the numbers.

Answer:

Let the two numbers be x \ and \ y 

Given

\frac{x}{y} = \frac{5}{7}

\Rightarrow x = \frac{5}{7} y

Therefore if  3 is subtracted from each of them, then

\frac{x-3}{y-3} = \frac{2}{3}

3x-9 = 2y-6

Substituting

3 \frac{5}{7} y - 9 = 2y-6

15y-63 = 14y -42

y = 21

and x = \frac{5}{7} y = 15

Hence x = 15 \ and \ y = 21

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Question 8: If 15(2x^2-y^2 )=7xy , find x:y ; if x \ and \  y both are positive.

Answer:

Given 15(2x^2-y^2 )=7xy 

30x^2-15y^2=7xy

Dividing both sides by  xy , we get

30 \frac{x}{y} - 15 \frac{y}{x} = 7

Let  \frac{x}{y} = a

30a - \frac{15}{a} =7

30a^2-7a-15=0

(6a-5)(5a+3)=0

\Rightarrow a= \frac{5}{6} \ or \ \frac{-3}{5} ( not possible as both x \ and \ y are positive)

Hence \frac{x}{y} = \frac{5}{6} \ or\  x:y=5:6

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Question 9: Find the:

i) Fourth Proportional to 2xy, x^2 \ and \ y^2 .

ii) Third proportional to a^2-b^2 \ and \ a+b

iii) Mean proportional to (x-y) \ and \ (x^3-x^2 y)

Answer:

i) Fourth Proportional to 2xy, x^2 \ and \ y^2 .

Let the fourth proportional be a

Therefore 2xy: x^2 = y^2 : a

 \Rightarrow a = \frac{y^2. x^2}{2xy} = \frac{xy}{2}

ii) Third proportional to a^2-b^2 \ and \ a+b

Let the third proportional be x

Therefore (a^2-b^2): (a+b) = (a+b) : x

 \Rightarrow \frac{a^2-b^2}{a+b} = \frac{a+b}{x}

Simplifying x = \frac{a+b}{a-b}

iii) Mean proportional to (x-y) \ and \ (x^3-x^2 y)

Let a  be the mean proportion.

Therefore (x-y) : a = a:  (x^3-x^2 y)

\Rightarrow \frac{x-y}{a} = \frac{a}{x^3-x^2 y}

Simplifying

a = x(x-y)

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Question 10: Find two numbers such that the mean proportional between them is 14 and third proportional to them is 112 .

Answer:

Let the two numbers be a \ and \ b .

Given, mean proportional between them is 14

Therefore a:14=14:b

or ab = 196 … … … … … … … … i)

Also given that third proportional to them is 112

Therefore a:b = b: 112

or b^2 = 112a   … … … … … … … … ii)

Solving i) and ii)

b^2 = 112 \Big( \frac{196}{b} \Big)

b^3 = 112. 196  \ or \  b = 28

Substituting back in i),

a = \frac{196}{28} = 7

Hence a = 7 \ and \  b = 28

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Question 11: If x \ and \ y be unequal and x:y  is the duplicate ratio of x+z \ and \ y+z , prove that z is mean proportional between x \ and \ y .

Answer:

Given

\frac{x}{y} = \frac{(x+z)^2}{(y+z)^2}

x(y^2+2yz+z^2) = y(x^2+2xz+z^2)

xy^2+2xyz+xz^2 = yx^2+2xyz+yz^2

xy(y-x)=z^2(y-x)

xy=z^2

or x:z=z:y

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Question 12: If q is the proportional between p \ and \ r   prove that:

\frac{p^3+q^3+r^3}{p^2 q^2 r^2 } = \frac{1}{p^3} + \frac{1}{q^3} + \frac{1}{r^3} 

Answer:

Given q is the proportional between p \ and \ r  

or q^2 = pr

LHS   = \frac{p^3+q^3+r^3}{p^2 q^2 r^2 }  

Substituting   q^2 = pr  

 = \frac{p^3+pqr+r^3}{p^2 q^2 r^2 }  

 = \frac{1}{r^3}+ \frac{q}{p^2r^2} + \frac{1}{p^3}

 = \frac{1}{p^3} + \frac{1}{q^3} + \frac{1}{r^3} = RHS

Hence proved.

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Question 13: If a, b \ and \ c are in continued proportion. Prove that: a:c=(a^2+b^2 ):(b^2+c^2)

Answer:

Given a:b=b:c

or \frac{a}{b} = \frac{b}{c} = k

Therefore a = bk \ and \ b= ck

Now, LHS = \frac{a}{c}= \frac{ck^2}{c} =k^2

RHS = \frac{a^2+b^2}{b^2+c^2}

 =  \frac{(bk)^2+b^2}{(ck)^2+c^2}

 = \frac{(ckk)^2+(ck)^2}{(ck)^2+c^2}

 = \frac{c^2k^2(k^2+1)}{c^2(k^2+1)}

 =k^2

LHS = RHS

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Question 14: If x= \frac{2ab}{a+b} , find the value of: \frac{x+a}{x-a} + \frac{x+b}{x-b} .

Answer:

Given x= \frac{2ab}{a+b} 

Therefore

\frac{x}{a} = \frac{2b}{a+b} \ and \ \frac{x}{b} = \frac{2a}{a+b} 

Applying componendo and dividendo

\frac{x+a}{x-a} = \frac{2b+a+b}{2b-a-b} 

\Rightarrow \frac{x+a}{x-a} = \frac{3b+a}{b-a} 

Similarly Applying componendo and dividendo

\frac{x+b}{x-b} = \frac{2a+a+b}{2a-a-b} 

\Rightarrow \frac{x+b}{x-b} = \frac{3a+b}{a-b} 

Adding   \frac{x+a}{x-a} + \frac{x+b}{x-b} = \frac{3b+a}{b-a} + \frac{3a+b}{a-b} 

or  \frac{x+a}{x-a} + \frac{x+b}{x-b} = \frac{2a-2b}{a-b} =2  

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Question 15: If (4a+9b)(4c-9d)=(4a-9b)(4c+9d) , prove that: a:b=c:d .

Answer:

Given (4a+9b)(4c-9d)=(4a-9b)(4c+9d)

or \frac{(4a+9b)}{(4a-9b)} = \frac{(4c+9d)}{(4c-9d)}

Applying componendo and dividendo

\frac{(4a+9b)+(4a-9b)}{(4a+9b)-(4a-9b)} = \frac{(4c+9d)+(4c-9d)}{(4c+9d)-(4c-9d)}

\frac{8a)}{18b} = \frac{8c}{18d}

or a:b=c:d

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Question 16: If \frac{a}{b} = \frac{c}{d} , show that (a+b):(c+d)=\sqrt{a^2+b^2 }:\sqrt{c^2+d^2} 

Answer:

Given \frac{a}{b} = \frac{c}{d} = k

therefore a = bk \ and \ c = dk 

LHS = \frac{a+b}{c+d} 

= \frac{bk+b}{dk+d} = \frac{b}{d}

RHS = \frac{\sqrt{a^2+b^2 }}{\sqrt{c^2+d^2}}

= \frac{\sqrt{(bk)^2+b^2 }}{\sqrt{(dk)^2+d^2}}

= \frac{\sqrt{b^2 }}{\sqrt{d^2}} = \frac{b}{d}

Hence, LHS = RHS

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Question 17: If \frac{x}{a} = \frac{y}{b} = \frac{z}{c} , prove that: \frac{ax-by}{(a+b)(x-y)} + \frac{by-cz}{(b+c)(y-z)} + \frac{cz-ax}{(c+a)(z-x)}=3

Answer:

Given \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k

Therefore

x = ak \ and \ y=bk \ and \ z=ck 

LHS  = \frac{ax-by}{(a+b)(x-y)} + \frac{by-cz}{(b+c)(y-z)} + \frac{cz-ax}{(c+a)(z-x)}

= \frac{a(ak)-b(bk)}{(a+b)(ak-bk)} + \frac{b(bk)-c(ck)}{(b+c)(bk-ck)} + \frac{c(ck)-a(ak)}{(c+a)(ck-ak)}

= \frac{k(a^2-b^2)}{k(a^2-b^2)} + \frac{k(b^2-c^2)}{k(b^2-c^2)} + \frac{k(c^2-a^2)}{k(c^2-a^2)} = 3  

Hence LHS = RHS

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Question 18: There are 36  members in a student’s council in a school and the ratio of the number of boys to the numbers of girls is 3:1 . How many more girls should be added to the council so that the ratio of number of boys to the number of girls may be 9:5  .

Answer:

Let the number of girls = x

The the number of boys = 3x

Therefore 3x+x = 36 \Rightarrow x = 9

Therefore Boys = 27  and Girls = 9

Let n girls be added

Therefore \frac{27}{9+n} = \frac{9}{5}

or 135 = 81 +9n

or n = 6

Hence 6 more girls should be added.

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Question 19: If \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b}  , prove that: ax+by+cz=0

Answer:

Given \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b} = k

x = k(b-c) \ and \ y = k(c-a) \ and \ z = k(a-b) 

LHS = ax+by+cz

= ak(b-c)+bk(c-a) + ck(a-b)

=k(ab-ac+bc-ab+ca-bc) = 0  

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Question 20: If 7x-15y=4x+y , find the value of x:y . hence, use componendo and dividendo to find the value of:

i) \frac{9x+5y}{9x-5y}     ii) \frac{3x^2+2y^2}{3x^2-2y^2} 

Answer:

Given 7x-15y=4x+y

3x = 16y  

or \frac{x}{y} = \frac{16}{3}  

i) \frac{9x+5y}{9x-5y} 

Divide Numerator and Denominator by y , we get

= \frac{9 \times \frac{x}{y}+5}{9 \times \frac{x}{y}-5} 

= \frac{9 \times \frac{16}{3}+5}{9 \times \frac{16}{3}-5} 

= \frac{53}{43}  

ii) \frac{3x^2+2y^2}{3x^2-2y^2} 

Divide Numerator and Denominator by y^2 , we get

= \frac{3 \times (\frac{x}{y})^2+2}{3 \times (\frac{x}{y})^2-2} 

= \frac{262}{250}=\frac{131}{125}  

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Question 21: If \frac{4m+3n}{4m-3n} = \frac{7}{4}   , use properties of proportion to find:

i) m:n     ii) \frac{2m^2-11n^2}{2m^2+11n^2 } .

Answer:

i) Given \frac{4m+3n}{4m-3n} = \frac{7}{4}

Applying componendo and dividendo

= \frac{4m+3n+4m-3n}{4m+3n-4m+3n} = \frac{7+4}{7-4}

= \frac{8m}{6n} = \frac{11}{3}

= \frac{m}{n} = \frac{11 \times 6}{3 \times 8} = \frac{11}{4}

ii) \frac{2m^2-11n^2}{2m^2+11n^2 } 

Applying componendo and dividendo

= \frac{2m^2-11n^2+2m^2+11n^2}{2m^2-11n^2-2m^2-11n^2 } 

= \frac{4m^2}{-22n^2 } 

= \frac{4}{-22} \Big( \frac{m}{n} \Big)^2 = \frac{4}{-22} \Big( \frac{11}{4} \Big)^2 = - \frac{11}{8}

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Question 22: If x, y \ and \ z are in continued proportion, prove that: \frac{(x+y)^2}{(y+z)^2} = \frac{x}{y} .    [2010]

Answer:

If x, y \ and \ z are in continued proportion, then

\frac{x}{y} = \frac{y}{z} \Rightarrow x = \frac{y^2}{z}

Applying componendo and dividendo

\frac{x+y}{x-y} = \frac{y+z}{y-z}

\Rightarrow \frac{x+y}{y+z} = \frac{x-y}{y-z}

Squaring both sides

\Rightarrow \frac{(x+y)^2}{(y+z)^2} =\Big( \frac{x-y}{y-z} \Big)^2

\Rightarrow \frac{(x+y)^2}{(y+z)^2} = \Big( \frac{x-y}{y-z} \Big)^2

Substituting

\Rightarrow \frac{(x+y)^2}{(y+z)^2} = \Big( \frac{\frac{y^2}{z}-y}{y-z} \Big)^2

\Rightarrow \frac{(x+y)^2}{(y+z)^2} = \Big( \frac{y^2-yz}{z(y-z)} \Big)^2= \frac{y^2}{z^2} = \frac{zx}{z^2} = \frac{x}{z}

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Question 23: Given x= \frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 }}  . Use componendo and dividendo to prove that: x^2= \frac{2a^2 x}{x^2+1} .   [2010]

Answer:

Given x= \frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 }} 

Applying componendo and dividendo

\frac{x+1}{x-1} = \frac{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})+(\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 })}{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})-(\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 })} 

Simplifying

\frac{x+1}{x-1} = \frac{\sqrt{a^2+b^2}}{\sqrt{a^2-b^2 }}

Square both sides

\frac{x^2+1+2x}{x^2-2x+1} = \frac{a^2+b^2}{a^2-b^2}

Applying componendo and dividendo

\frac{x^2+1+2x+x^2-2x+1}{x^2+1+2x-x^2+2x-1} = \frac{a^2+b^2+a^2-b^2}{a^2+b^2-a^2+b^2}

\frac{2(x^2+1)}{4x} = \frac{2a^2}{2b^2}

\frac{x^2+1}{2x} = \frac{a^2}{b^2}

Simplifying

b^2 = \frac{2a^2x}{x^2+1}

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Question 24: If \frac{x^2+y^2}{x^2-y^2} =2 \frac{1}{8}  , find:

i) \frac{x}{y}      ii) \frac{x^3+y^3}{x^3-y^3 }     [2014]

Answer:

i) Given \frac{x^2+y^2}{x^2-y^2} =2 \frac{1}{8} = \frac{17}{8} 

Applying componendo and dividendo

\frac{x^2+y^2+x^2-y^2}{x^2+y^2-x^2+y^2} = \frac{17+8}{17-8} 

\frac{2x^2}{2y^2} = \frac{25}{9} 

Simplifying, we get

\frac{x}{y} = \frac{5}{3} 

ii) \frac{x^3+y^3}{x^3-y^3 } 

Applying componendo and dividendo

\frac{x^3+y^3+x^3-y^3 }{x^3+y^3-x^3+y^3 } 

\frac{2x^3}{2y^3} = \Big( \frac{x}{y} \Big)^3 = \Big( \frac{5}{3} \Big)^3 = \frac{125}{9}  

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Question 25: Using componendo and dividendo, find the value of x : \frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}} = 9     [2011]

Answer:

Given \frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}} =9 

Applying componendo and dividendo

\frac{(\sqrt{3x+4}+\sqrt{3x-5})+(\sqrt{3x+4}-\sqrt{3x-5})}{(\sqrt{3x+4}+\sqrt{3x-5})-(\sqrt{3x+4}-\sqrt{3x-5})} = \frac{9+1}{9-1}  

\frac{2\sqrt{3x+4}}{2\sqrt{3x-5}} = \frac{10}{8}  

Simplifying

\frac{\sqrt{3x+4}}{\sqrt{3x-5}} = \frac{5}{4}  

Square both sides

\frac{3x+4}{3x-5} = \frac{25}{14}  

42x+56 = 75x-125   

Simplifying we get x = 7

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Question 26: If x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}  , using properties of proportion show that:  x^2-2ax+1   [2012]

Answer:

Given x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}

Applying componendo and dividendo

\frac{x+1}{x-1} = \frac{(\sqrt{a+1}+\sqrt{a-1})+(\sqrt{a+1}-\sqrt{a-1})}{(\sqrt{a+1}+\sqrt{a-1})-(\sqrt{a+1}-\sqrt{a-1})}

Simplify

\frac{x+1}{x-1} = \frac{\sqrt{a+1}}{\sqrt{a-1}}

Now square both sides

\frac{x^2+1+2x}{x^2-2x+1} = \frac{a+1}{a-1}

Simplifying

x^2+1 = 2ax

or x^2-2ax+1 = 0