\displaystyle \textbf{Question 1: } \text{Find the fourth proportional to:}
\displaystyle \text{(i) }1.5,\ 4.5\text{ and }3.5 \qquad \text{(ii) }3a,\ 6a^2\text{ and }2ab^2
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let the fourth proportional be }x.
\displaystyle \therefore \frac{1.5}{4.5}=\frac{3.5}{x}
\displaystyle \Rightarrow x=\frac{3.5\times4.5}{1.5}=10.5
\displaystyle \therefore \text{The fourth proportional is }10.5.
\displaystyle \text{(ii) Let the fourth proportional be }x.
\displaystyle \therefore \frac{3a}{6a^2}=\frac{2ab^2}{x}
\displaystyle \Rightarrow x=\frac{2ab^2\times6a^2}{3a}=4a^2b^2
\displaystyle \therefore \text{The fourth proportional is }4a^2b^2.
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\displaystyle \textbf{Question 2: } \text{Find the third proportional to:}
\displaystyle \text{(i) }2\frac{2}{3}\text{ and }4 \qquad \text{(ii) }a-b\text{ and }a^2-b^2
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let the third proportional be }x.
\displaystyle \therefore 2\frac{2}{3}:4=4:x
\displaystyle \Rightarrow x=\frac{4\times4}{2\frac{2}{3}}=6
\displaystyle \therefore \text{The third proportional is }6.
\displaystyle \text{(ii) Let the third proportional be }x.
\displaystyle \therefore (a-b):(a^2-b^2)=(a^2-b^2):x
\displaystyle \Rightarrow x=\frac{(a^2-b^2)^2}{a-b}
\displaystyle =(a+b)(a^2-b^2)
\displaystyle \therefore \text{The third proportional is }(a+b)(a^2-b^2).
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\displaystyle \textbf{Question 3: } \text{Find the mean proportional between:}
\displaystyle \text{(i) }17.5\text{ and }0.007 \qquad \text{(ii) }6+3\sqrt{3}\text{ and }8-4\sqrt{3} \qquad \text{(iii) }a-b\text{ and }a^3-a^2b
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let the mean proportional be }x.
\displaystyle \therefore 17.5:x=x:0.007
\displaystyle \Rightarrow x^2=17.5\times0.007
\displaystyle \Rightarrow x^2=0.1225
\displaystyle \Rightarrow x=0.35
\displaystyle \text{(ii) Let the mean proportional be }x.
\displaystyle \therefore (6+3\sqrt{3}):x=x:(8-4\sqrt{3})
\displaystyle \Rightarrow x^2=(6+3\sqrt{3})(8-4\sqrt{3})
\displaystyle \Rightarrow x^2=12
\displaystyle \Rightarrow x=2\sqrt{3}
\displaystyle \text{(iii) Let the mean proportional be }x.
\displaystyle \therefore (a-b):x=x:(a^3-a^2b)
\displaystyle \Rightarrow x^2=(a-b)(a^3-a^2b)
\displaystyle =(a-b)\times a^2(a-b)=a^2(a-b)^2
\displaystyle \Rightarrow x=a(a-b)
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\displaystyle \textbf{Question 4: } \text{If }x+5\text{ is the mean proportional between }x+2\text{ and }x+9,\text{ find }x.
\displaystyle \text{Answer:}
\displaystyle \text{Since }x+5\text{ is the mean proportional,}
\displaystyle (x+2):(x+5)=(x+5):(x+9)
\displaystyle \Rightarrow (x+5)^2=(x+2)(x+9)
\displaystyle \Rightarrow x^2+10x+25=x^2+11x+18
\displaystyle \Rightarrow x=7
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\displaystyle \textbf{Question 5: } \text{What least number must be added to each of the numbers }16,\ 7,\ 79\text{ and }43
\displaystyle \text{so that the resulting numbers are in proportion?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required number be }x.
\displaystyle \therefore (16+x):(7+x)=(79+x):(43+x)
\displaystyle \Rightarrow (16+x)(43+x)=(79+x)(7+x)
\displaystyle \Rightarrow x^2+59x+688=x^2+86x+553
\displaystyle \Rightarrow 135=27x
\displaystyle \Rightarrow x=5
\displaystyle \therefore \text{The least number to be added is }5.
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\displaystyle \textbf{Question 6: } \text{What least number must be added to each of the numbers }6,\ 15,\ 20\text{ and }43
\displaystyle \text{to make them proportional?}\hfill\text{[ICSE 2005, 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required number be }x.
\displaystyle \therefore (6+x):(15+x)=(20+x):(43+x)
\displaystyle \Rightarrow (6+x)(43+x)=(20+x)(15+x)
\displaystyle \Rightarrow x^2+49x+258=x^2+35x+300
\displaystyle \Rightarrow 14x=42
\displaystyle \Rightarrow x=3
\displaystyle \therefore \text{The least number to be added is }3.
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\displaystyle \textbf{Question 7: } \text{What number must be added to each of the numbers }16,\ 26\text{ and }40
\displaystyle \text{so that the resulting numbers may be in continued proportion?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required number be }x.
\displaystyle \therefore (16+x):(26+x)=(26+x):(40+x)
\displaystyle \Rightarrow (16+x)(40+x)=(26+x)^2
\displaystyle \Rightarrow x^2+56x+640=x^2+52x+676
\displaystyle \Rightarrow 4x=36
\displaystyle \Rightarrow x=9
\displaystyle \therefore \text{The required number is }9.
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\displaystyle \textbf{Question 8: } \text{What least number must be subtracted from each of the numbers }7,\ 17\text{ and }47
\displaystyle \text{so that the remainders are in continued proportion?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required number be }x.
\displaystyle \therefore (7-x):(17-x)=(17-x):(47-x)
\displaystyle \Rightarrow (7-x)(47-x)=(17-x)^2
\displaystyle \Rightarrow x^2-54x+329=x^2-34x+289
\displaystyle \Rightarrow 20x=40
\displaystyle \Rightarrow x=2
\displaystyle \therefore \text{The least number to be subtracted is }2.
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\displaystyle \textbf{Question 9: } \text{If }y\text{ is the mean proportional between }x\text{ and }z,\text{ show that }
\displaystyle xy+yz \ \text{is the mean proportional between }x^2+y^2\text{ and }y^2+z^2.
\displaystyle \text{Answer:}
\displaystyle \text{Since }y\text{ is the mean proportional between }x\text{ and }z,
\displaystyle x:y=y:z
\displaystyle \Rightarrow y^2=xz
\displaystyle \text{Let the mean proportional between }x^2+y^2\text{ and }y^2+z^2\text{ be }p.
\displaystyle \therefore p^2=(x^2+y^2)(y^2+z^2)
\displaystyle \Rightarrow p^2=(x^2+xz)(xz+z^2)
\displaystyle =x(x+z)\cdot z(x+z)
\displaystyle =xz(x+z)^2
\displaystyle =y^2(x+z)^2
\displaystyle \Rightarrow p=y(x+z)=xy+yz
\displaystyle \therefore xy+yz\text{ is the mean proportional between }x^2+y^2\text{ and }y^2+z^2.
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\displaystyle \textbf{Question 10: } \text{If }q\text{ is the mean proportional between }p\text{ and }r,\text{ show that}
\displaystyle pqr(p+q+r)^3=(pq+qr+pr)^3.
\displaystyle \text{Answer:}
\displaystyle \text{Since }q\text{ is the mean proportional between }p\text{ and }r,
\displaystyle q^2=pr
\displaystyle LHS=pqr(p+q+r)^3
\displaystyle =q^3(p+q+r)^3 \qquad (\because pr=q^2)
\displaystyle =\{q(p+q+r)\}^3
\displaystyle =(pq+q^2+qr)^3
\displaystyle =(pq+pr+qr)^3 \qquad (\because q^2=pr)
\displaystyle =(pq+qr+pr)^3=RHS
\displaystyle \therefore pqr(p+q+r)^3=(pq+qr+pr)^3.
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\displaystyle \textbf{Question 11: } \text{If three quantities are in continued proportion, show that the ratio of }
\displaystyle \text{the first to the third is the duplicate ratio of the first to the second.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three quantities be }x,\ y\text{ and }z.
\displaystyle \text{Since they are in continued proportion,}
\displaystyle x:y=y:z
\displaystyle \Rightarrow y^2=xz
\displaystyle \text{We have to prove that }x:z=x^2:y^2.
\displaystyle x:z=x^2:y^2
\displaystyle \Rightarrow xy^2=x^2z
\displaystyle \text{Using }y^2=xz,\text{ LHS}=xy^2=x(xz)=x^2z=RHS
\displaystyle \therefore x:z=x^2:y^2.
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\displaystyle \textbf{Question 12: } \text{If }y\text{ is the mean proportional between }x\text{ and }z,\text{ prove that}
\displaystyle \frac{x^2-y^2+z^2}{x^{-2}-y^{-2}+z^{-2}}=y^4.
\displaystyle \text{Answer:}
\displaystyle \text{Since }y\text{ is the mean proportional between }x\text{ and }z,
\displaystyle y^2=xz
\displaystyle LHS=\frac{x^2-y^2+z^2}{x^{-2}-y^{-2}+z^{-2}}
\displaystyle =\frac{x^2-xz+z^2}{\frac{1}{x^2}-\frac{1}{xz}+\frac{1}{z^2}}
\displaystyle =\frac{x^2-xz+z^2}{\frac{z^2-xz+x^2}{x^2z^2}}
\displaystyle =x^2z^2
\displaystyle =(xz)^2=(y^2)^2=y^4=RHS
\displaystyle \therefore \frac{x^2-y^2+z^2}{x^{-2}-y^{-2}+z^{-2}}=y^4.
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\displaystyle \textbf{Question 13: } \text{Given four quantities }a,\ b,\ c\text{ and }d\text{ are in proportion, show that}
\displaystyle (a-c)b^2:(b-d)cd=(a^2-b^2-ab):(c^2-d^2-cd).
\displaystyle \text{Answer:}
\displaystyle \text{Given }a,\ b,\ c\text{ and }d\text{ are in proportion.}
\displaystyle \Rightarrow \frac{a}{b}=\frac{c}{d}=k
\displaystyle \Rightarrow a=bk\text{ and }c=dk
\displaystyle \text{Now, }\frac{(a-c)b^2}{(b-d)cd}=\frac{(bk-dk)b^2}{(b-d)(dk)d}
\displaystyle =\frac{k(b-d)b^2}{k(b-d)d^2}=\frac{b^2}{d^2}
\displaystyle \text{Also, }\frac{a^2-b^2-ab}{c^2-d^2-cd}=\frac{b^2k^2-b^2-b^2k}{d^2k^2-d^2-d^2k}
\displaystyle =\frac{b^2(k^2-1-k)}{d^2(k^2-1-k)}=\frac{b^2}{d^2}
\displaystyle \therefore (a-c)b^2:(b-d)cd=(a^2-b^2-ab):(c^2-d^2-cd).
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\displaystyle \textbf{Question 14: } \text{Find two numbers such that the mean proportional between them is }12
\displaystyle \text{and the third proportional to them is }96.
\displaystyle \text{Answer:}
\displaystyle \text{Let the two numbers be }a\text{ and }b.
\displaystyle \text{Since }12\text{ is the mean proportional,}
\displaystyle a:12=12:b
\displaystyle \Rightarrow ab=144
\displaystyle \text{Since }96\text{ is the third proportional,}
\displaystyle a:b=b:96
\displaystyle \Rightarrow b^2=96a
\displaystyle \text{From }ab=144,\ b=\frac{144}{a}
\displaystyle \Rightarrow \left(\frac{144}{a}\right)^2=96a
\displaystyle \Rightarrow \frac{20736}{a^2}=96a
\displaystyle \Rightarrow a^3=216
\displaystyle \Rightarrow a=6
\displaystyle \Rightarrow b=\frac{144}{6}=24
\displaystyle \therefore \text{The two numbers are }6\text{ and }24.
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\displaystyle \textbf{Question 15: } \text{Find the third proportional to }\frac{x}{y}+\frac{y}{x}\text{ and }\sqrt{x^2+y^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the third proportional be }p.
\displaystyle \therefore \left(\frac{x}{y}+\frac{y}{x}\right):\sqrt{x^2+y^2}=\sqrt{x^2+y^2}:p
\displaystyle \Rightarrow p\left(\frac{x}{y}+\frac{y}{x}\right)=\left(\sqrt{x^2+y^2}\right)^2
\displaystyle \Rightarrow p\left(\frac{x^2+y^2}{xy}\right)=x^2+y^2
\displaystyle \Rightarrow p=xy
\displaystyle \therefore \text{The third proportional is }xy.
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\displaystyle \textbf{Question 16: } \text{If }p:q=r:s,\text{ show that }mp+nq:q=mr+ns:s.
\displaystyle \text{Answer:}
\displaystyle \text{Given }p:q=r:s
\displaystyle \Rightarrow \frac{p}{q}=\frac{r}{s}
\displaystyle \text{Multiplying both sides by }m,
\displaystyle \frac{mp}{q}=\frac{mr}{s}
\displaystyle \text{Adding }n\text{ to both sides,}
\displaystyle \frac{mp}{q}+n=\frac{mr}{s}+n
\displaystyle \Rightarrow \frac{mp+nq}{q}=\frac{mr+ns}{s}
\displaystyle \therefore mp+nq:q=mr+ns:s.
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\displaystyle \textbf{Question 17: } \text{If }p+r=mq\text{ and }\frac{1}{q}+\frac{1}{s}=\frac{m}{r},\text{ prove that }p:q=r:s.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{1}{q}+\frac{1}{s}=\frac{m}{r}
\displaystyle \Rightarrow \frac{s+q}{qs}=\frac{m}{r}
\displaystyle \Rightarrow \frac{s+q}{s}=\frac{mq}{r}
\displaystyle \text{Also, }p+r=mq
\displaystyle \Rightarrow \frac{s+q}{s}=\frac{p+r}{r}
\displaystyle \Rightarrow 1+\frac{q}{s}=1+\frac{p}{r}
\displaystyle \Rightarrow \frac{q}{s}=\frac{p}{r}
\displaystyle \Rightarrow p:q=r:s
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\displaystyle \textbf{Question 18: } \text{If }\frac{a}{b}=\frac{c}{d},\text{ prove that each of the following is equal to the given ratio:}
\displaystyle \text{(i) }\frac{5a+4c}{5b+4d}\qquad \text{(ii) }\frac{13a-8c}{13b-8d}\qquad \text{(iii) }\sqrt{\frac{3a^2-10c^2}{3b^2-10d^2}}
\displaystyle \text{(iv) }\left(\frac{8a^3+15c^3}{8b^3+15d^3}\right)^{\frac{1}{3}}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{a}{b}=\frac{c}{d}=k
\displaystyle \Rightarrow a=bk\text{ and }c=dk
\displaystyle \text{(i) }\frac{5a+4c}{5b+4d}=\frac{5bk+4dk}{5b+4d}
\displaystyle =k\left(\frac{5b+4d}{5b+4d}\right)=k
\displaystyle \text{(ii) }\frac{13a-8c}{13b-8d}=\frac{13bk-8dk}{13b-8d}
\displaystyle =k\left(\frac{13b-8d}{13b-8d}\right)=k
\displaystyle \text{(iii) }\sqrt{\frac{3a^2-10c^2}{3b^2-10d^2}}=\sqrt{\frac{3b^2k^2-10d^2k^2}{3b^2-10d^2}}
\displaystyle =\sqrt{k^2\left(\frac{3b^2-10d^2}{3b^2-10d^2}\right)}=k
\displaystyle \text{(iv) }\left(\frac{8a^3+15c^3}{8b^3+15d^3}\right)^{\frac{1}{3}}=\left(\frac{8b^3k^3+15d^3k^3}{8b^3+15d^3}\right)^{\frac{1}{3}}
\displaystyle =\left[k^3\left(\frac{8b^3+15d^3}{8b^3+15d^3}\right)\right]^{\frac{1}{3}}=k
\displaystyle \therefore \text{Each of the given ratios is equal to }\frac{a}{b}\text{ or }\frac{c}{d}.
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\displaystyle \textbf{Question 19: } \text{If }a,\ b,\ c\text{ and }d\text{ are in proportion, prove that:}
\displaystyle \text{(i) }\frac{13a+17b}{13c+17d}=\sqrt{\frac{2ma^2-3nb^2}{2mc^2-3nd^2}}
\displaystyle \text{(ii) }\sqrt{\frac{4a^2+9b^2}{4c^2+9d^2}}=\left(\frac{xa^3-5yb^3}{xc^3-5yd^3}\right)^{\frac{1}{3}}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{a}{b}=\frac{c}{d}=k
\displaystyle \Rightarrow a=bk\text{ and }c=dk
\displaystyle \text{(i) LHS}=\frac{13a+17b}{13c+17d}
\displaystyle =\frac{13bk+17b}{13dk+17d}
\displaystyle =\frac{b(13k+17)}{d(13k+17)}=\frac{b}{d}
\displaystyle \text{RHS}=\sqrt{\frac{2ma^2-3nb^2}{2mc^2-3nd^2}}
\displaystyle =\sqrt{\frac{2m(bk)^2-3nb^2}{2m(dk)^2-3nd^2}}
\displaystyle =\sqrt{\frac{b^2(2mk^2-3n)}{d^2(2mk^2-3n)}}=\frac{b}{d}
\displaystyle \therefore \frac{13a+17b}{13c+17d}=\sqrt{\frac{2ma^2-3nb^2}{2mc^2-3nd^2}}
\displaystyle \text{(ii) LHS}=\sqrt{\frac{4a^2+9b^2}{4c^2+9d^2}}
\displaystyle =\sqrt{\frac{4(bk)^2+9b^2}{4(dk)^2+9d^2}}
\displaystyle =\sqrt{\frac{b^2(4k^2+9)}{d^2(4k^2+9)}}=\frac{b}{d}
\displaystyle \text{RHS}=\left(\frac{xa^3-5yb^3}{xc^3-5yd^3}\right)^{\frac{1}{3}}
\displaystyle =\left(\frac{x(bk)^3-5yb^3}{x(dk)^3-5yd^3}\right)^{\frac{1}{3}}
\displaystyle =\left(\frac{b^3(xk^3-5y)}{d^3(xk^3-5y)}\right)^{\frac{1}{3}}=\frac{b}{d}
\displaystyle \therefore \sqrt{\frac{4a^2+9b^2}{4c^2+9d^2}}=\left(\frac{xa^3-5yb^3}{xc^3-5yd^3}\right)^{\frac{1}{3}}
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\displaystyle \textbf{Question 20: } \text{If }\frac{x}{a}=\frac{y}{b}=\frac{z}{c},\text{ prove that}
\displaystyle \frac{2x^3-3y^3+4z^3}{2a^3-3b^3+4c^3}=\left(\frac{2x-3y+4z}{2a-3b+4c}\right)^3.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k
\displaystyle \Rightarrow x=ak,\ y=bk,\ z=ck
\displaystyle \text{LHS}=\frac{2x^3-3y^3+4z^3}{2a^3-3b^3+4c^3}
\displaystyle =\frac{2(ak)^3-3(bk)^3+4(ck)^3}{2a^3-3b^3+4c^3}
\displaystyle =\frac{k^3(2a^3-3b^3+4c^3)}{2a^3-3b^3+4c^3}=k^3
\displaystyle \text{RHS}=\left(\frac{2x-3y+4z}{2a-3b+4c}\right)^3
\displaystyle =\left(\frac{2ak-3bk+4ck}{2a-3b+4c}\right)^3
\displaystyle =\left(\frac{k(2a-3b+4c)}{2a-3b+4c}\right)^3=k^3
\displaystyle \therefore \frac{2x^3-3y^3+4z^3}{2a^3-3b^3+4c^3}=\left(\frac{2x-3y+4z}{2a-3b+4c}\right)^3.
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