Question 1: Find the fourth proportional to:

i) $1.5, \ 4.5 \ and \ 3.5$

ii) $3a, 6a^2 \ and \ 2ab^2$

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) $a-b \ and \ a^2-b^2$

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) $a-b \ and \ a^3-a^2b$

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 $x$.

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?

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]

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?

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?

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 $x^2+y^2 \ and \ y^2+z^2$ .

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: $pqr(p+q+r)^3=(pq+qr+pr)^3$ .

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.

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: $\frac{(x^2-y^2+z^2)}{(x^{-2}-y^{-2}+z^{-2} )}=y^4$ .

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: $(a-c)b^2:(b-d)cd=(a^2 -b^2-ab) : (c^2-d^2-cd)$

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 $96$ .

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 $\frac{x}{y}+\frac{y}{x} \ and \ \sqrt{x^2+y^2 }$

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:   $mp+nq:q=mr+ns:s$

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: $p:q=r:s$ .

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

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

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: $\frac{ 2x ^3- 3y ^3+ 4z ^3}{ 2a ^3- 3b ^3+ 4c ^3 }=(\frac{2x-3y+4z}{2a-3b+4c})^3$

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.