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Question 1: Find the fourth proportional to:

i) 1.5, \ 4.5 \ and \ 3.5

ii) 3a, 6a^2 \ and \ 2ab^2  

Answer:

i) Let the 4^{th}   proportion be x

Therefore

\frac{1.5}{4.5} = \frac{3.5}{x} \Rightarrow x = \frac{3.5 \times 4.5}{1.5} = 10.5

Hence x = 10.5

ii) Let the 4^{th}  proportion be x

Therefore

\frac{3a}{6a^2} = \frac{2ab^2}{x} \Rightarrow x = \frac{2ab^2 \times 6a^2}{3a} = 4a^2b^2

Hence x = 4a^2b^2

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Question 2: Find the third proportional to:

i) 2 \frac{2}{3}  \ and \ 4     

ii) a-b \ and \ a^2-b^2    

Answer:

i) Let the 3^{rd}  proportion be x

Therefore

 2 \frac{2}{3} : 4 = 4 : x \Rightarrow x = \frac{4 \times 4}{2 \frac{2}{3}} = 6

Hence x = 6

ii) Let the 3^{rd}  proportion be x

Therefore

a-b : a^2-b^2 = a^2-b^2: x \Rightarrow x = \frac{(a^2-b^2) \times (a^2-b^2)}{(a-b)} = (a+b)(a^2-b^2)

Hence x = (a+b)(a^2-b^2)

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Question 3: Find the mean proportional between:

i) 17.5 \ and \ 0.007      

ii) 6+ 3\sqrt{3} \ and \ 8-4\sqrt{3}      

iii) a-b \ and \ a^3-a^2b      

Answer:

i) Let the mean proportional be x

Therefore 17.5 : x = x:0.007 \Rightarrow x^2= 17.5 \times 0.007 

 \Rightarrow x = 0.35

ii) Let the mean proportional be x

Therefore  6+ 3\sqrt{3} : x = x : 8-4\sqrt{3} \Rightarrow x^2 = (6+ 3\sqrt{3}) \times (8-4\sqrt{3})

 \Rightarrow x = 2\sqrt{3}

iii) Let the mean proportional be x

Therefore a-b : x= x: a^3-a^2b   \Rightarrow x^2 = (a-b) \times (a^3-a^2b )

 \Rightarrow x = a(a-b)

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Question 4: If x+5  is the means proportion between  x+2 \ and \ x+9   ; find the value of x .

Answer:

Given x+5  is the means proportion between  x+2 \ and \ x+9  

Therefore

(x+2) :(x+5)= (x+5):(x+9)

(x+5)^2 = (x+2)(x+9)

\Rightarrow x^2+25+10x = x^2+11x+18

\Rightarrow x = 7

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Question 5: what least number must be added to each of the numbers 16, 7, 79 \ and \ 43       so that the resulting numbers are in proportion?

Answer:

Let the number added be x

Therefore (16+x): (7+x) = (79+x): (43+x)

\Rightarrow (16+x) \times (43+x) =  (79+x) \times (7+x)

\Rightarrow x^2+59x+688 = x^2+ 86x +553

\Rightarrow x = 5

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Question 6: What least number must be added to each of the numbers 6, 15, 20, \ and \ 43      to make them proportional. [2005, 2013]

Answer:

Let the number added be x

Therefore (6+x): (15+x) = (20+x): (43+x)

\Rightarrow (6+x) \times (43+x) =  (20+x) \times (15+x)

\Rightarrow x^2+49x+258 = x^2+ 35x +300

\Rightarrow x = 3

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Question 7: What number must be added to each of the numbers 16, 26 \ and \ 40      so that the resulting numbers may be in continued proportion?

Answer:

Let the number added be x

Therefore (16+x): (26+x) = (26+x): (40+x)

\Rightarrow (16+x) \times (40+x) =  (26+x) \times (26+x)

\Rightarrow x^2+56x+640 = x^2+ 52x +676

\Rightarrow x = 9

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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?

Answer:

Let the number subtracted be x

Therefore (7-x): (17-x) = (17-x): (47-x)

\Rightarrow (7-x) \times (47-x) =  (17-x) \times (17-x)

\Rightarrow x^2-54x+329 = x^2-34x +289

\Rightarrow x = 2

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Question 9: If y  is the mean proportional between x \ and \ z ; show that xy+yz is the mean proportional between x^2+y^2 \ and \ y^2+z^2 .

Answer:

Since y  is the mean proportional between x \ and \ z

 x:y = y : z \Rightarrow y^2 = xz

Let the mean proportional between x^2+y^2 \ and \ y^2+z^2 by p

Therefore  (x^2+y^2 ):  p = p : (x^2+z^2)

 \Rightarrow p^2 = (x^2+y^2 ) \times (y^2+z^2)

 \Rightarrow p^2= (x^2+xz ) \times (xz+z^2)

  = xz(x+z)(x+z) = y^2(x+z)^2

 \Rightarrow p = y(x+z) = yx+yz . Hence proved.

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Question 10: If q is the mean proportional between p \ and \ r . show that: pqr(p+q+r)^3=(pq+qr+pr)^3 .

Answer:

Given q is the mean proportional between p \ and \ r

Therefore q^2 = pr

LHS = pqr(p+q+r)^3 = q^3(p+q+r)^3 

= [q(p+q+r)]^3 = (pq+qr+pr)^3 = RHS . Hence proved.

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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.

Answer:

Let the three quantities by x, \ y \ and \ z

If they are in proportion, then we have x:y =y: z \Rightarrow y^2 = xz

Now we have to prove that x: z = x^2 : y^2

\Rightarrow x \times y^2 = x^2 \times z 

Substituting y^2 = xz \Rightarrow LHS = x^2z = RHS . Hence proved.

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Question 12: If y is the mean proportional between x \ and \ z ,  prove that: \frac{(x^2-y^2+z^2)}{(x^{-2}-y^{-2}+z^{-2} )}=y^4 .

Answer:

Given y is the mean proportional between x \ and \ z

Therefore y^2 = xz

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

 = \frac{(x^2-y^2+z^2)}{\frac{1}{x^2}-\frac{1}{y^2} + \frac{1}{z^2}}

 = \frac{(x^2-xz+z^2)}{\frac{1}{x^2}-\frac{1}{xz} + \frac{1}{z^2}}

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

 =x^2z^2 = (xy)^2 = (y^2)^2 = y^4  = RHS . Hence proved.

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Question 13: Give four quantities a, b, c \ and \ d are in proportion. Show that: (a-c)b^2:(b-d)cd=(a^2 -b^2-ab) : (c^2-d^2-cd)

Answer:

Given a, b, c \ and \ d are in proportion

 \Rightarrow \frac{a}{b}=\frac{c}{d} = k

 \Rightarrow a = bk \ and \ a = bk \ and \ c = dk

To prove (a-c)b^2:(b-d)cd=(c^2-d^2-cd)

 LHS = \frac{(a-c)b^2}{(b-d)cd} = \frac{(bk-dk)b^2}{(b-d)d^2k} = \frac{b^2}{d^2}

RHS   = \frac{a^2 -b^2-ab}{c^2-d^2-cd} =  \frac{b^2k^2 -b^2-bkd}{d^2k^2-d^2-kd^2}

 =\frac{b^2(k^2-1-k)}{d^2(k^2-1-k)} = \frac{b^2}{d^2}

Hence LHS =  RHS . Hence proved.

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

Answer:

Let the two numbers be a \ and \ b

Therefore a: 12 = 12: b \Rightarrow ab = 144

If 96 is the third proportion

 \Rightarrow a : b = b : 96 

 \Rightarrow b^2 = 96a

 \Rightarrow (\frac{144}{a})^2 = 96a

 \Rightarrow a ^3 = 216 \ or \ a = 6 \ and \  b = \frac{144}{6} = 24.

Hence a = 6 \ and \ b = 24.

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Question 15: Find the third proportional to \frac{x}{y}+\frac{y}{x} \ and \  \sqrt{x^2+y^2 } 

Answer:

Let the third proportion by p

Therefore \frac{x}{y}+\frac{y}{x} :  \sqrt{x^2+y^2 } =  \sqrt{x^2+y^2 } : p

 \Rightarrow p(\frac{x}{y}+\frac{y}{x}) = (\sqrt{x^2+y^2 })^2

 \Rightarrow p(\frac{x^2+y^2}{xy} )= x^2+y^2

 \Rightarrow p = xy

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Question 16: If p:q=r:s ; then show that:   mp+nq:q=mr+ns:s

Answer:

Given p:q=r:s

Therefore \frac{p}{q} = \frac{r}{s}

Multiplying both sides by m

\frac{mp}{q} = \frac{mr}{s}

Adding n to both sides

\frac{mp}{q}+n = \frac{mr}{s}+n 

\Rightarrow \frac{mp+nq}{q} = \frac{mr+sn}{s} 

or mp+nq:q=mr+ns:s

Hence proved.

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Question 17: If p+r=mq \ and \  \frac{1}{q}+\frac{1}{s}=\frac{m}{r} ; then prove that: p:q=r:s .

Answer:

Given  \frac{1}{q}+\frac{1}{s}=\frac{m}{r}

  \frac{1}{q}+\frac{1}{s}=\frac{m}{r}

  \frac{s+q}{qs} = \frac{m}{r}

  \Rightarrow \frac{s+q}{s} = \frac{mq}{r}

Given   p+r=mq

  \Rightarrow \frac{s+q}{s} = \frac{p+r}{r}

  \Rightarrow  1+ \frac{q}{s} = 1+ \frac{p}{r}

or p:q=r:s

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Question 18: If \frac{a}{b}=\frac{c}{d} , prove that each of the given ratio is equal to:

i) \frac{5a+4c}{5b+4d}

ii) \frac{13a-8c}{13b-8d}

iii) \sqrt{\frac{3a^2-10c^2}{3b^2-10d^2}}

iv) (\frac{8a ^3+ 15c ^3}{ 8b ^3+ 15d ^3 })^\frac{1}{3} 

Answer:

Given \frac{a}{b}=\frac{c}{d} = k

 a = bk \ and \ c = dk 

i) \frac{5a+4c}{5b+4d} = \frac{5(bk) +4(dk)}{5b+4d} = k\frac{5b +4d}{5b+4d}= k

ii) \frac{13a-8c}{13b-8d}  = \frac{13(bk)-8(dk)}{13b-8d}=k\frac{13b-8d}{13b-8d}  = k

iii) \sqrt{\frac{3a^2-10c^2}{3b^2-10d^2}} = \sqrt{\frac{3(bk)^2-10(dk)^2}{3b^2-10d^2}} =  k\sqrt{\frac{3b^2-10d^2}{3b^2-10d^2}} = k  

iv) (\frac{8a ^3+ 15c ^3}{ 8b ^3+ 15d ^3 })^\frac{1}{3} = (\frac{8(bk) ^3+ 15(dk) ^3}{ 8b ^3+ 15d ^3 })^\frac{1}{3} = (k^3(\frac{8b^3+ 15d^3}{ 8b ^3+ 15d ^3 }))^\frac{1}{3} = k

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

i) \frac{13a+17b}{13c+17d}=\sqrt{\frac{2ma ^2- 3nb ^2}{ 2mc ^2- 3nd ^2 }}

ii)  \sqrt {\frac{ 4a ^2+ 9b ^2}{ 4c ^2+ 9d ^2 }}=(\frac{ xa ^3-5yb^3}{ xc ^3-5yd^3 })^\frac{1}{3} 

Answer:

Given \frac{a}{b}=\frac{c}{d} = k

 a = bk \ and \ c = dk 

i) \frac{13a+17b}{13c+17d}=\sqrt{\frac{2ma ^2- 3nb ^2}{ 2mc ^2- 3nd ^3 }}

LHS = \frac{13a+17b}{13c+17d} = \frac{13bk+17b}{13dk+17d} = \frac{b(13k+17)}{d(13dk+17)} = \frac{b}{d} 

RHS = \sqrt{\frac{2ma ^2- 3nb ^2}{ 2mc ^2- 3nd ^3 }} = \sqrt{\frac{2m(bk)^2- 3nb ^2}{ 2m(dk)^2- 3nd ^2 }} = \sqrt{\frac{b^2(2mk^2- 3n)}{d^2( 2mk^2- 3n)}} = \frac{b}{d} 

Hence LHS = RHS.

ii)  \sqrt {\frac{ 4a ^2+ 9b ^2}{ 4c ^2+ 9d ^2 }}=(\frac{ xa ^3-5yb^3}{ xc ^3-5yd^3 })^\frac{1}{3} 

LHS = \sqrt {\frac{4a^2+ 9b^2}{ 4c^2+ 9d^2}} = \sqrt {\frac{ 4(bk)^2+ 9b ^2}{4(dk)^2+ 9d ^2}}=  \sqrt {\frac{ b^2(4k^2+ 9)}{d^2(4k^2+ 9)}} = \frac{b}{d} 

RHS = (\frac{ xa ^3-5yb^3}{ xc ^3-5yd^3 })^\frac{1}{3}= (\frac{ x(bk)^3-5yb^3}{ x(dk)^3-5yd^3 })^\frac{1}{3} =  (\frac{ b^3(xk^3-5y)}{ d^3(xk^3-5y)})^\frac{1}{3} =  \frac{b}{d} 

Hence LHS = RHS.

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Question 20: If   \frac{x}{a}=\frac{y}{b}=\frac{z}{c} , prove that: \frac{ 2x ^3- 3y ^3+ 4z ^3}{ 2a ^3- 3b ^3+ 4c ^3 }=(\frac{2x-3y+4z}{2a-3b+4c})^3

Answer:

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

Therefore x = ak, \ y = bk, \ and \ z = ck

LHS = \frac{ 2x ^3- 3y ^3+ 4z ^3}{ 2a ^3- 3b ^3+ 4c ^3 }

= \frac{ 2(ak)^3- 3(bk)^3+ 4(ck)^3}{ 2a ^3- 3b ^3+ 4c ^3 }

= \frac{ k^3(2a^3- 3b^3+ 4c)^3}{ 2a^3- 3b^3+ 4c^3 }

=k^3

RHS = (\frac{2x-3y+4z}{2a-3b+4c})^3

= (\frac{2(ak)-3(bk)+4(ck)}{2a-3b+4c})^3

=k^3

Hence LHS = RHS.

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