\displaystyle \textbf{Question 1: } \text{If }a:b=5:3,\text{ find }\frac{5a-3b}{5a+3b}.
\displaystyle \text{Answer:}
\displaystyle \frac{5a-3b}{5a+3b}=\frac{5\left(\frac{a}{b}\right)-3}{5\left(\frac{a}{b}\right)+3}
\displaystyle =\frac{5\left(\frac{5}{3}\right)-3}{5\left(\frac{5}{3}\right)+3}=\frac{25-9}{25+9}=\frac{8}{17}
\\

\displaystyle \textbf{Question 2: } \text{If }x:y=4:7,\text{ find }(3x+2y):(5x+y).
\displaystyle \text{Answer:}
\displaystyle \frac{3x+2y}{5x+y}=\frac{3\left(\frac{x}{y}\right)+2}{5\left(\frac{x}{y}\right)+1}
\displaystyle =\frac{3\left(\frac{4}{7}\right)+2}{5\left(\frac{4}{7}\right)+1}=\frac{12+14}{20+7}=\frac{26}{27}
\\

\displaystyle \textbf{Question 3: } \text{If }a:b=3:8,\text{ find }\frac{4a+3b}{6a-b}.
\displaystyle \text{Answer:}
\displaystyle \frac{4a+3b}{6a-b}=\frac{4\left(\frac{a}{b}\right)+3}{6\left(\frac{a}{b}\right)-1}
\displaystyle =\frac{4\left(\frac{3}{8}\right)+3}{6\left(\frac{3}{8}\right)-1}=\frac{12+24}{18-8}=\frac{18}{5}
\\

\displaystyle \textbf{Question 4: } \text{If }(a-b):(a+b)=1:11,\text{ find }(5a+4b+15):(5a-4b+3).
\displaystyle \text{Answer:}
\displaystyle \frac{a-b}{a+b}=\frac{1}{11}
\displaystyle \Rightarrow 11a-11b=a+b
\displaystyle \Rightarrow 10a=12b
\displaystyle \Rightarrow \frac{a}{b}=\frac{6}{5}
\displaystyle \frac{5a+4b+15}{5a-4b+3}=\frac{5\left(\frac{a}{b}\right)+4+\frac{15}{b}}{5\left(\frac{a}{b}\right)-4+\frac{3}{b}}
\displaystyle =\frac{10b+15}{2b+3}=\frac{5(2b+3)}{2b+3}=5
\displaystyle \therefore (5a+4b+15):(5a-4b+3)=5:1
\\

\displaystyle \textbf{Question 5: } \text{If }\frac{y-x}{x}=\frac{3}{8},\text{ find the ratio }\frac{y}{x}.
\displaystyle \text{Answer:}
\displaystyle \frac{y-x}{x}=\frac{3}{8}
\displaystyle \Rightarrow \frac{y}{x}-1=\frac{3}{8}
\displaystyle \Rightarrow \frac{y}{x}=\frac{11}{8}
\displaystyle \therefore y:x=11:8
\\

\displaystyle \textbf{Question 6: } \text{If }\frac{m+n}{m+3n}=\frac{2}{3},\text{ find }\frac{2n^2}{3m^2+mn}.
\displaystyle \text{Answer:}
\displaystyle \frac{m+n}{m+3n}=\frac{2}{3}
\displaystyle \Rightarrow 3m+3n=2m+6n
\displaystyle \Rightarrow m=3n
\displaystyle \Rightarrow \frac{m}{n}=3
\displaystyle \frac{2n^2}{3m^2+mn}=\frac{2}{3\left(\frac{m}{n}\right)^2+\frac{m}{n}}
\displaystyle =\frac{2}{3(3)^2+3}=\frac{2}{30}=\frac{1}{15}
\\

\displaystyle \textbf{Question 7: } \text{Find }\frac{x}{y},\text{ when }x^2+6y^2=5xy.
\displaystyle \text{Answer:}
\displaystyle x^2+6y^2=5xy
\displaystyle \text{Dividing by }y^2,
\displaystyle \left(\frac{x}{y}\right)^2+6=5\left(\frac{x}{y}\right)
\displaystyle \text{Let }\frac{x}{y}=a
\displaystyle \Rightarrow a^2-5a+6=0
\displaystyle \Rightarrow (a-3)(a-2)=0
\displaystyle \Rightarrow a=3\text{ or }a=2
\displaystyle \therefore \frac{x}{y}=3\text{ or }2
\\

\displaystyle \textbf{Question 8: } \text{If }8:11\text{ is the same as }(2x-y):(x+2y),\text{ find }\frac{7x}{9y}.
\displaystyle \text{Answer:}
\displaystyle \frac{2x-y}{x+2y}=\frac{8}{11}
\displaystyle \Rightarrow 22x-11y=8x+16y
\displaystyle \Rightarrow 14x=27y
\displaystyle \Rightarrow \frac{x}{y}=\frac{27}{14}
\displaystyle \therefore \frac{7x}{9y}=\frac{7}{9}\times\frac{27}{14}=\frac{3}{2}
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\displaystyle \textbf{Question 9: } \text{Two numbers are in the ratio }2:3.\text{ If }5\text{ is added to each number, the ratio becomes }5:7.
\displaystyle \text{Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the numbers be }x\text{ and }y.
\displaystyle \therefore \frac{x}{y}=\frac{2}{3}
\displaystyle \Rightarrow x=\frac{2}{3}y
\displaystyle \text{Also, }\frac{x+5}{y+5}=\frac{5}{7}
\displaystyle \Rightarrow 7x+35=5y+25
\displaystyle \Rightarrow 7x-5y+10=0
\displaystyle \text{Substituting }x=\frac{2}{3}y,
\displaystyle 7\left(\frac{2}{3}y\right)-5y+10=0
\displaystyle \Rightarrow 14y-15y+30=0
\displaystyle \Rightarrow y=30
\displaystyle \therefore x=\frac{2}{3}\times30=20
\displaystyle \therefore \text{The numbers are }20\text{ and }30.
\\

\displaystyle \textbf{Question 10: } \text{Two positive numbers are in the ratio }3:5\text{ and the difference between their squares is }400.
\displaystyle \text{Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the numbers be }x\text{ and }y.
\displaystyle \therefore \frac{x}{y}=\frac{3}{5}
\displaystyle \Rightarrow x=\frac{3}{5}y
\displaystyle \text{Given, }y^2-x^2=400
\displaystyle \Rightarrow \left(y-\frac{3}{5}y\right)\left(y+\frac{3}{5}y\right)=400
\displaystyle \Rightarrow \frac{2y}{5}\times\frac{8y}{5}=400
\displaystyle \Rightarrow \frac{16y^2}{25}=400
\displaystyle \Rightarrow y^2=625
\displaystyle \Rightarrow y=25
\displaystyle \therefore x=\frac{3}{5}\times25=15
\displaystyle \therefore \text{The numbers are }15\text{ and }25.
\\

\displaystyle \textbf{Question 11: } \text{What quantity must be subtracted from each term of the ratio }9:17\text{ to make it }1:3?
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ be subtracted from each term.}
\displaystyle \therefore \frac{9-x}{17-x}=\frac{1}{3}
\displaystyle \Rightarrow 27-3x=17-x
\displaystyle \Rightarrow 10=2x
\displaystyle \Rightarrow x=5
\\

\displaystyle \textbf{Question 12: } \text{The monthly pocket money of Ravi and Sanjeev is in the ratio }
\displaystyle 5:7. \ \text{Their expenditures are in the ratio }3:5.\text{ If each saves Rs. }80 \\ \text{ per month, find their monthly pocket money.}\hfill\text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let Ravi's and Sanjeev's monthly pocket money be Rs. }x\text{ and Rs. }y\text{ respectively.}
\displaystyle \therefore \frac{x}{y}=\frac{5}{7}
\displaystyle \Rightarrow x=\frac{5}{7}y
\displaystyle \text{Since each saves Rs. }80,\text{ their expenditures are }x-80\text{ and }y-80.
\displaystyle \therefore \frac{x-80}{y-80}=\frac{3}{5}
\displaystyle \text{Substituting }x=\frac{5}{7}y,
\displaystyle \frac{\frac{5}{7}y-80}{y-80}=\frac{3}{5}
\displaystyle \Rightarrow 5\left(\frac{5}{7}y-80\right)=3(y-80)
\displaystyle \Rightarrow \frac{25y}{7}-400=3y-240
\displaystyle \Rightarrow 25y-2800=21y-1680
\displaystyle \Rightarrow 4y=1120
\displaystyle \Rightarrow y=280
\displaystyle \Rightarrow x=\frac{5}{7}\times280=200
\displaystyle \therefore \text{Ravi's monthly pocket money is Rs. }200\text{ and Sanjeev's is Rs. }280.
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\displaystyle \textbf{Question 13: } \text{The work done by }(x-2)\text{ men in }(4x+1)\text{ days and by }
\displaystyle (4x+1)\text{ men} \ \text{in }(2x-3)\text{ days are in the ratio }3:8.\text{ Find }x.
\displaystyle \text{Answer:}
\displaystyle \frac{(x-2)(4x+1)}{(4x+1)(2x-3)}=\frac{3}{8}
\displaystyle \Rightarrow \frac{x-2}{2x-3}=\frac{3}{8}
\displaystyle \Rightarrow 8(x-2)=3(2x-3)
\displaystyle \Rightarrow 8x-16=6x-9
\displaystyle \Rightarrow 2x=7
\displaystyle \Rightarrow x=\frac{7}{2}
\\

\displaystyle \textbf{Question 14: } \text{The bus fare between two cities is increased in the ratio }7:9.
\displaystyle \text{Find the increase in the fare if: (i) the original fare is Rs. }245\text{ and (ii) the increased fare is Rs. }207.
\displaystyle \text{Answer:}
\displaystyle \frac{\text{Original Fare}}{\text{Increased Fare}}=\frac{7}{9}
\displaystyle \text{(i) }9\times245=7\times\text{Increased Fare}
\displaystyle \Rightarrow \text{Increased Fare}=\frac{9\times245}{7}=315
\displaystyle \therefore \text{Increase in fare}=315-245=\text{Rs. }60
\displaystyle \text{(ii) Original Fare}=\frac{7}{9}\times207=161
\displaystyle \therefore \text{Increase in fare}=207-161=\text{Rs. }46
\\

\displaystyle \textbf{Question 15: } \text{By increasing the cost of the entry ticket to a fair in the ratio }
\displaystyle 10:13,\text{ the number of visitors} \ \text{decreased in the ratio }6:5.\text{ In what ratio} \\ \text{has the total collection increased or decreased?}
\displaystyle \text{Answer:}
\displaystyle \frac{\text{Original Ticket}}{\text{Increased Ticket}}=\frac{10}{13}
\displaystyle \frac{\text{Original Visitors}}{\text{Final Visitors}}=\frac{6}{5}
\displaystyle \frac{\text{Original Collection}}{\text{New Collection}}=\frac{10}{13}\times\frac{6}{5}=\frac{12}{13}
\displaystyle \therefore \text{Original Collection : New Collection}=12:13
\displaystyle \therefore \text{The total collection has increased in the ratio }12:13.
\\

\displaystyle \textbf{Question 16: } \text{In a basket, the ratio of oranges to apples is }7:13.
\displaystyle \text{If }8\text{ oranges and }11\text{ apples are eaten, the ratio becomes }1:2.\text{ Find the original numbers.}
\displaystyle \text{Answer:}
\displaystyle \frac{\text{Oranges}}{\text{Apples}}=\frac{7}{13}
\displaystyle \frac{\text{Oranges}-8}{\text{Apples}-11}=\frac{1}{2}
\displaystyle \Rightarrow 2(\text{Oranges}-8)=\text{Apples}-11
\displaystyle \Rightarrow 2\text{ Oranges}-16=\text{Apples}-11
\displaystyle \Rightarrow \text{Oranges}=\frac{\text{Apples}+5}{2}
\displaystyle \text{Substituting in }\frac{\text{Oranges}}{\text{Apples}}=\frac{7}{13},
\displaystyle \frac{\frac{\text{Apples}+5}{2}}{\text{Apples}}=\frac{7}{13}
\displaystyle \Rightarrow 13(\text{Apples}+5)=14\text{ Apples}
\displaystyle \Rightarrow \text{Apples}=65
\displaystyle \Rightarrow \text{Oranges}=\frac{65+5}{2}=35
\displaystyle \therefore \text{Original number of oranges }=35\text{ and apples }=65.
\\

\displaystyle \textbf{Question 17: } \text{The ratio between boys and girls in a class is }4:3.
\displaystyle \text{If there were }20\text{ more boys and }12\text{ less girls, the ratio would be }2:1.\text{ Find the total number.}
\displaystyle \text{Answer:}
\displaystyle \frac{\text{Boys}}{\text{Girls}}=\frac{4}{3}
\displaystyle \frac{\text{Boys}+20}{\text{Girls}-12}=\frac{2}{1}
\displaystyle \Rightarrow \text{Boys}+20=2(\text{Girls}-12)
\displaystyle \Rightarrow \text{Boys}=2\text{ Girls}-44
\displaystyle \text{Substituting in }\frac{\text{Boys}}{\text{Girls}}=\frac{4}{3},
\displaystyle \frac{2\text{ Girls}-44}{\text{Girls}}=\frac{4}{3}
\displaystyle \Rightarrow 6\text{ Girls}-132=4\text{ Girls}
\displaystyle \Rightarrow 2\text{ Girls}=132
\displaystyle \Rightarrow \text{Girls}=66
\displaystyle \Rightarrow \text{Boys}=\frac{4}{3}\times66=88
\displaystyle \therefore \text{Total number of students }=88+66=154.
\\

\displaystyle \textbf{Question 18: } \text{(a) If }A:B=3:4\text{ and }B:C=6:7,\text{ find }
\displaystyle A:B:C\text{ and }A:C.  \ \text{(b) If }A:B=2:5\text{ and }A:C=3:4,\text{ find }A:B:C.
\displaystyle \text{Answer:}
\displaystyle \text{(a) }A:B=3:4 \qquad \text{... (i)}
\displaystyle B:C=6:7 \qquad \text{... (ii)}
\displaystyle \text{Multiplying (i) by }6\text{ and (ii) by }4,
\displaystyle A:B=18:24 \qquad \text{... (iii)}
\displaystyle B:C=24:28 \qquad \text{... (iv)}
\displaystyle \therefore A:B:C=18:24:28=9:12:14
\displaystyle \therefore A:C=9:14
\displaystyle \text{(b) }A:B=2:5 \qquad \text{... (i)}
\displaystyle A:C=3:4 \qquad \text{... (ii)}
\displaystyle \text{Multiplying (i) by }3\text{ and (ii) by }2,
\displaystyle A:B=6:15 \qquad \text{... (iii)}
\displaystyle A:C=6:8 \qquad \text{... (iv)}
\displaystyle \therefore A:B:C=6:15:8
\\

\displaystyle \textbf{Question 19: } \text{If }3A=4B=6C,\text{ find }A:B:C.
\displaystyle \text{Answer:}
\displaystyle A:B=4:3 \qquad \text{... (i)}
\displaystyle B:C=6:4=3:2 \qquad \text{... (ii)}
\displaystyle \text{Multiplying (i) by }3\text{ and (ii) by }3,
\displaystyle A:B=12:9 \qquad \text{... (iii)}
\displaystyle B:C=9:6 \qquad \text{... (iv)}
\displaystyle \therefore A:B:C=12:9:6=4:3:2
\\

For question 20 to 29, please refer to the lecture notes on Ratios and Proportions.

\displaystyle \textbf{Question 20: } \text{Find the compound ratio of:}
\displaystyle \text{(i) }3:5\text{ and }8:15 \qquad \text{(ii) }2:3,\ 9:14\text{ and }14:27
\displaystyle \text{(iii) }2a:3b,\ mn:x^2\text{ and }x:n \qquad \text{(iv) }\sqrt{2}:1,\ 3:\sqrt{5}\text{ and }\sqrt{20}:9
\displaystyle \text{Answer:}
\displaystyle \text{(i) Compound ratio }=(3\times8):(5\times15)=24:75=8:25
\displaystyle \text{(ii) Compound ratio }=(2\times9\times14):(3\times14\times27)=252:1134=2:9
\displaystyle \text{(iii) Compound ratio }=(2a\times mn\times x):(3b\times x^2\times n)
\displaystyle \Rightarrow 2amnx:3bx^2n
\displaystyle \Rightarrow 2am:3bx
\displaystyle \text{(iv) Compound ratio }=(\sqrt{2}\times3\times\sqrt{20}):(1\times\sqrt{5}\times9)
\displaystyle \Rightarrow 6\sqrt{10}:9\sqrt{5}
\displaystyle \Rightarrow 2\sqrt{2}:3
\\

\displaystyle \textbf{Question 21: } \text{Find the duplicate ratio of:} \quad \text{(i) }3:4 \qquad \text{(ii) }3\sqrt{3}:2\sqrt{5}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Duplicate ratio of }3:4=3^2:4^2=9:16
\displaystyle \text{(ii) Duplicate ratio of }3\sqrt{3}:2\sqrt{5}=(3\sqrt{3})^2:(2\sqrt{5})^2=27:20
\\

\displaystyle \textbf{Question 22: } \text{Find the triplicate ratio of:} \quad \text{(i) }1:3 \qquad \text{(ii) }\frac{m}{2}:\frac{n}{3}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Triplicate ratio of }1:3=1^3:3^3=1:27
\displaystyle \text{(ii) Triplicate ratio of }\frac{m}{2}:\frac{n}{3}=\left(\frac{m}{2}\right)^3:\left(\frac{n}{3}\right)^3
\displaystyle =\frac{m^3}{8}:\frac{n^3}{27}=27m^3:8n^3
\\

\displaystyle \textbf{Question 23: } \text{Find the sub-duplicate ratio of:} \quad \text{(i) }9:16  \qquad  \\  \text{(ii) }(x-y)^4:(x+y)^6
\displaystyle \text{Answer:}
\displaystyle \text{(i) Sub-duplicate ratio of }9:16=\sqrt{9}:\sqrt{16}=3:4
\displaystyle \text{(ii) Sub-duplicate ratio of }(x-y)^4:(x+y)^6
\displaystyle =\sqrt{(x-y)^4}:\sqrt{(x+y)^6}=(x-y)^2:(x+y)^3
\\

\displaystyle \textbf{Question 24: } \text{Find the sub-triplicate ratio of:} \quad \text{(i) }64:27 \qquad  \text{(ii) }x^3:125y^3
\displaystyle \text{Answer:}
\displaystyle \text{(i) Sub-triplicate ratio of }64:27=\sqrt[3]{64}:\sqrt[3]{27}=4:3
\displaystyle \text{(ii) Sub-triplicate ratio of }x^3:125y^3=\sqrt[3]{x^3}:\sqrt[3]{125y^3}=x:5y
\\

\displaystyle \textbf{Question 25: } \text{Find the reciprocal ratio of:} \quad \text{(i) }5:8 \qquad \text{(ii) }\frac{x}{3}:\frac{y}{7}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Reciprocal ratio of }5:8=8:5
\displaystyle \text{(ii) Reciprocal ratio of }\frac{x}{3}:\frac{y}{7}=\frac{y}{7}:\frac{x}{3}=3y:7x
\\

\displaystyle \textbf{Question 26: } \text{If }(3x+4):(x+5)\text{ is the duplicate ratio of }8:5,\text{ find }x.
\displaystyle \text{Answer:}
\displaystyle \text{Duplicate ratio of }8:5=8^2:5^2=64:25
\displaystyle \therefore \frac{3x+4}{x+5}=\frac{64}{25}
\displaystyle \Rightarrow 25(3x+4)=64(x+5)
\displaystyle \Rightarrow 75x+100=64x+320
\displaystyle \Rightarrow 11x=220
\displaystyle \Rightarrow x=20
\\

\displaystyle \textbf{Question 27: } \text{If }m:n\text{ is the duplicate ratio of }(m+x):(n+x),\text{ show that } \\ x^2=mn.
\displaystyle \text{Answer:}
\displaystyle \frac{(m+x)^2}{(n+x)^2}=\frac{m}{n}
\displaystyle \Rightarrow \frac{m^2+x^2+2mx}{n^2+x^2+2nx}=\frac{m}{n}
\displaystyle \Rightarrow n(m^2+x^2+2mx)=m(n^2+x^2+2nx)
\displaystyle \Rightarrow m^2n+nx^2+2mnx=n^2m+mx^2+2mnx
\displaystyle \Rightarrow mn(m-n)=(m-n)x^2
\displaystyle \Rightarrow x^2=mn
\\

\displaystyle \textbf{Question 28: } \text{If }(x-9):(3x+6)\text{ is the triplicate ratio of }4:9,\text{ find }x.\hfill\text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Triplicate ratio of }4:9=4^3:9^3=64:729
\displaystyle \therefore \frac{x-9}{3x+6}=\frac{64}{729}
\displaystyle \Rightarrow 729(x-9)=64(3x+6)
\displaystyle \Rightarrow 729x-6561=192x+384
\displaystyle \Rightarrow 537x=6945
\displaystyle \Rightarrow x=\frac{6945}{537}=\frac{2315}{179}
\\

\displaystyle \textbf{Question 29: } \text{Find the ratio compounded of the reciprocal ratio of }15:28,\text{ the sub-duplicate}
\displaystyle \text{ratio of }36:49\text{ and the triplicate ratio of }5:4.
\displaystyle \text{Answer:}
\displaystyle \text{Reciprocal ratio of }15:28=28:15
\displaystyle \text{Sub-duplicate ratio of }36:49=6:7
\displaystyle \text{Triplicate ratio of }5:4=125:64
\displaystyle \therefore \text{Compound ratio}=(28\times6\times125):(15\times7\times64)
\displaystyle \Rightarrow 21000:6720=25:8
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\displaystyle \textbf{Question 30: } \text{If }\frac{a+b}{am+bn}=\frac{b+c}{mb+nc}=\frac{c+a}{mc+na},
\displaystyle \text{prove that each ratio is }\frac{2}{m+n}\text{ provided }a+b+c\neq0.
\displaystyle \text{Answer:}
\displaystyle \frac{a+b}{am+bn}=\frac{b+c}{mb+nc}=\frac{c+a}{mc+na}
\displaystyle =\frac{(a+b)+(b+c)+(c+a)}{(am+bn)+(mb+nc)+(mc+na)}
\displaystyle =\frac{2(a+b+c)}{(m+n)(a+b+c)}
\displaystyle =\frac{2}{m+n}\qquad(\because a+b+c\neq0)
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