$\displaystyle \text{Question 1: If } a : b = 5 : 3 \text{, find } \frac{5a-3b}{5a+3b}$

$\displaystyle \frac{5a-3b}{5a+3b} = \frac{5(\frac{a}{b})-3}{5(\frac{a}{b})+3} = \frac{25-9}{25+9} = \frac{8}{17}$

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$\displaystyle \text{Question 2: If } x : y = 4 : 7 \text{, find the value of } (3x+2y):(5x+y)$

$\displaystyle \frac{3x+2y}{5x+y} = \frac{3(\frac{x}{y})+2}{5(\frac{x}{y})+1} = \frac{12+14}{20+7} = \frac{26}{27}$

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$\displaystyle \text{Question 3: If } a : b = 3 : 8 \text{, find the value of } \frac{4a+3b}{6a-b}$

$\displaystyle \frac{4a+3b}{6a-b} = \frac{4(\frac{a}{b})+3}{6(\frac{a}{b})-1} = \frac{12+24}{18-8} = \frac{18}{5}$

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$\displaystyle \text{Question 4: If } (a-b) : (a+b) = 1 : 11 \text{, find the ratio} (5a+4b+15):(5a-4b+3)$

$\displaystyle \frac{a-b}{a+b} = \frac{1}{11} \Rightarrow 11a-11b=a+b \Rightarrow \frac{a}{b} = \frac{6}{5}$

$\displaystyle \frac{5a+4b+15}{5a-4b+3} = \frac{5(\frac{a}{b})+4+\frac{15}{b}}{5(\frac{a}{b})-4+\frac{3}{b}} = \frac{10b+15}{2b+3} = \frac{5(2b+3)}{(2b+3)} = 5$

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$\displaystyle \text{Question 5: If } \frac{y-x}{x} = \frac{3}{8} \text{, find the ratio} \frac{y}{x}$

$\displaystyle \frac{y-x}{x} = \frac{3}{8}$

$\displaystyle \frac{y}{x} -1 = \frac{3}{8}$

$\displaystyle \frac{y}{x} = \frac{11}{8}$

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$\displaystyle \text{Question 6: If } \frac{m+n}{m+3n} = \frac{2}{3} \text{, find the ratio} \frac{2n^2}{3m^2+mn}$

$\displaystyle \frac{m+n}{m+3n} = \frac{2}{3} \Rightarrow \frac{m}{n} = 3$

$\displaystyle \frac{2n^2}{3m^2+mn} = \frac{2}{3(\frac{m}{n})^2+\frac{m}{n}} = \frac{2}{3.3^2+3} = \frac{1}{15}$

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$\displaystyle \text{Question 7: Find } \frac{x}{y} \text{ ; , when } x^2+6y^2=5xy$

$\displaystyle x^2+6y^2=5xy$

Dividing by $\displaystyle y^2$

$\displaystyle ( \frac{x}{y} )^2+6=5( \frac{x}{y} )$

$\displaystyle \text{Let } \frac{x}{y} = a$

$\displaystyle a^2-5a+6=0$

$\displaystyle (a-3)(a-2)=0 \text{ or } a=3, 2 \text{ or } \frac{x}{y} = 3, 2$

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Question 8: If the ratio between 8 and 11 is the same as the ratio of $\displaystyle 2x-y \text{ to } x+2y$ , find the value of $\displaystyle \frac{7x}{9y}$

$\displaystyle \text{Given } \frac{2x-y}{x+2y} = \frac{8}{11}$

$\displaystyle 22x-11y=8x+16y$

$\displaystyle 14x=27y$

$\displaystyle \frac{x}{y} = \frac{27}{14}$

$\displaystyle \text{Therefore } \frac{7x}{9y} = \frac{7}{9} \times \frac{27}{14} = \frac{3}{2}$

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Question 9: Two numbers are in the ratio 2:3. If 5 is added to each number, the ratio becomes 5:7. Find the numbers.

Let the number be $\displaystyle x \text{ and } y$. Therefore

$\displaystyle \frac{x}{y} = \frac{2}{3} \Rightarrow x = \frac{2}{3} y$

$\displaystyle \frac{x+5}{y+5} = \frac{5}{7}$

$\displaystyle 7x+35= 5y+25$

$\displaystyle 7x-5y+10=0$

$\displaystyle \text{Substituting} 7( \frac{2}{3} y) -5y+10=0$

$\displaystyle 14y-15y+30=0$

$\displaystyle \text{Or } y = 30.$

$\displaystyle \text{Therefore } x= \frac{2}{3} \times 30 = 20$

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Question 10: Two positive numbers are in the ratio of 3:5 and the difference between their squares is 400. Find the numbers.

Let the number be $\displaystyle x \text{ and } y$

$\displaystyle \frac{x}{y} = \frac{3}{5}$

$\displaystyle \Rightarrow x = \frac{3}{5} y$

$\displaystyle \text{Given } y^2-x^2= 400$

$\displaystyle (y-x)(y+x)=400$

$\displaystyle (y- \frac{3}{5} y)(y+ \frac{3}{5} y)=400$

$\displaystyle \text{Or } y = 25$

$\displaystyle \text{Therefore } x = \frac{3}{5} y = 15$

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Question 11: What quantity must be subtracted from each term of the ratio 9:17 to make it equal to 1:3.

Let $\displaystyle x$ be subtracted. Therefore

$\displaystyle \frac{9-x}{17-x} = \frac{1}{3}$

$\displaystyle \Rightarrow 27-3x=17-x$

$\displaystyle \text{Or } x = 5$

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Question 12: The monthly pocket money of Ravi and Sanjeev is in the ratio of 5:7 Their expenditures are in the ratio of 3:5. If each saves Rs. 80 per month, find their monthly pocket money. [2012]

Let monthly pocket of Rave and Sanjeev by $\displaystyle x \text{ and } y$ respectively.

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

$\displaystyle \frac{x-80}{y-80} = \frac{3}{5}$

Substituting

$\displaystyle \frac{ \frac{5}{7} y-80}{y-80} = \frac{3}{5}$

$\displaystyle \frac{25}{7} x-400=3x-240 \Rightarrow x=280$

Substituting

$\displaystyle y = \frac{5}{7} \times 280 = 200$

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Question 13: The work done by $\displaystyle (x-2)$ men in $\displaystyle (4x+1)$ days and the work done by $\displaystyle (4x+1)$ men in $\displaystyle (2x-3)$ days are in the ratio $\displaystyle 3:8$ . Find the value of $\displaystyle x$ .

$\displaystyle \frac{(x-2)(4x+1)}{(4x+1)(2x-3)} = \frac{3}{8}$

$\displaystyle 8(x-2) = 3(2x-3)$

$\displaystyle 2x=7$

$\displaystyle x= \frac{7}{2}$

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Question 14: The bus fare between two cities is increased in the ratio of $\displaystyle 7:9$. Find the increase in the fare if: i) the original fare is Rs. 245 and ii) the increased fare is Rs. 207.

$\displaystyle \frac{ \text{ Original Fare } }{ \text{ Increased Fare } } = \frac{7}{9}$

i) $\displaystyle 9 \times 245 = 7 \times \text{ Increased Fare }$

$\displaystyle \Rightarrow \text{ Increased Fare } = 315$

Therefore Increase in the fare $\displaystyle = 315-245 = 60$

ii) $\displaystyle \frac{ \text{ Original Fare } }{ \text{ Increased Fare } } = \frac{7}{9}$

$\displaystyle \text{ Original Fare } = \frac{7}{9} \times 207 = 161$

Therefore Increase in the fare $\displaystyle = 207-161 = 46$

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Question 15: By increasing the cost of the entry ticket to a fair in the ratio of $\displaystyle 10:13$ , the number of visitors to the fair has decreased in the ratio of $\displaystyle 6:5$ . In what ratio has the total collection increased or decreased.

$\displaystyle \frac{ \text{ Original Ticket } }{ \text{ Increased Ticket } } = \frac{10}{13}$

$\displaystyle \frac{ \text{ Original Visitors } }{ \text{ Final Visitors } } = \frac{6}{5}$

$\displaystyle \text{Collections Ratio } = \frac{ \text{ Original Ticket } \times \text{ Original Visitors } }{ \text{ Increased Ticket } \times \text{ Final Visitors } } = \frac{10}{13} \times \frac{6}{5} = \frac{12}{13}$

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Question 16: In a basket, the ratio of the number of oranges and the number of apples is $\displaystyle 7:13$. If 8 oranges and 11 apples are eaten the ratio between the number of oranges and the number of apples becomes $\displaystyle 1:2$. Find the original number of oranges and apples in the basket.

$\displaystyle \frac{ \text{ Apples } }{ \text{ Oranges } } = \frac{7}{13}$

$\displaystyle \frac{ \text{ Oranges } -8}{ \text{ Apples } - 11} = \frac{1}{2}$

$\displaystyle 2 \text{ Oranges } -16 = \text{ Apples } - 11$

$\displaystyle \text{ Oranges } = \frac{1}{2} ( \text{ Apples } + 5)$

Substituting

$\displaystyle \frac{\frac{1}{2}( \text{ Apples } + 5)}{ \text{ Apples } } = \frac{7}{13}$

$\displaystyle 13 \text{ Apples } + 65 = 14 \text{ Apples }$

$\displaystyle \text{Therefore } \text{ Apples } = 65.$

$\displaystyle \text{ Oranges } = \frac{1}{2} ( \text{ Apples } + 5) = \frac{1}{2} (65 + 5) = 35$

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Question 17: The ratio between the number of boys and the number of girls in the class is $\displaystyle 4:3$. If there were 20 more boys and 12 less girls, the ratio would have been $\displaystyle 1:2$. Find the total number of students in the class.

$\displaystyle \frac{Boys}{Girls} = \frac{4}{3}$ … … … … … … … i)

$\displaystyle \frac{Boys+20}{Girls-12} = \frac{2}{1}$ … … … … … … … ii)

$\displaystyle Boys + 20 = 2 Girls -24$

$\displaystyle \text{Or } Boys = 2 Girls - 44$

Substituting in i)

$\displaystyle Girls = \frac{3}{4} (2 Girls -44) \Rightarrow Girls = 66$

$\displaystyle \text{Therefore } Boys = \frac{4}{3} \times 66 = 88$

Hence the total number of students in the class $\displaystyle = 66+88 = 154$

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Question 18: a) If $\displaystyle A:B=3:4 \text{ and } B:C=6:7$ , find $\displaystyle A:B:C \text{ and } A:C$ b) If $\displaystyle A:B=2:5 \text{ and } A:C=3:4$ , find $\displaystyle A:B:C$

a) $\displaystyle A:B=3:4$ … … … … … … … i)

$\displaystyle B:C=6:7$ … … … … … … … ii)

Multiplying i) by 6 and ii) by 4 we get

$\displaystyle A:B=18:24$ … … … … … … … iii)

$\displaystyle B:C=24:28$ … … … … … … … iv)

$\displaystyle \text{Therefore } A:B:C = 18:24:28 \text{ or } A:B:C = 9:12:14$

Hence $\displaystyle A:C = 9:14$

b) $\displaystyle A:B=2:5$ … … … … … … … i)

$\displaystyle A:C=3:4$ … … … … … … … ii)

Multiplying i) by 3 and ii) by 2 we get

$\displaystyle A:B=6:15$ … … … … … … … iii)

$\displaystyle A:C=6:8$ … … … … … … … iv)

$\displaystyle \text{Therefore } A:B:C = 6:15:8$

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Question 19: If $\displaystyle 3A=4B=6C$ , find $\displaystyle A:B:C$

$\displaystyle A:B=4:3$ … … … … … … … i)

$\displaystyle B:C=6:4$ … … … … … … … ii)

Multiplying i) by 6 and ii) by 3 we get

$\displaystyle A:B=24:18$ … … … … … … … iii)

$\displaystyle B:C=18:12$ … … … … … … … iv)

$\displaystyle \text{Therefore } A:B:C = 24:18:12 \text{ or } A:B:C = 4:3:2$

$\displaystyle \\$

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

Question 20: Find the compound ratio of

i) $\displaystyle 3:5\text{ and }8:15$

ii) $\displaystyle 2:3, 9:14\text{ and }14:27$

iii) $\displaystyle 2a:3b, mn:x^2\text{ and }x:n$

iv) $\displaystyle \sqrt{2}:1, 3:\sqrt{5}\text{ and }\sqrt{20}:9$

i) $\displaystyle \text{Compound Ratio of } 3:5\text{ and }8:15 = (3 \times 8):(5 \times 15) = 24:75 or 8:25$

ii) $\displaystyle \text{Compound Ratio of } 2:3, 9:14\text{ and }14:27 = (2 \times 9 \times 14) :(3 \times 14 \times 27) = 2:9$

iii) $\displaystyle \text{Compound Ratio of } 2a:3b, mn:x^2\text{ and }x:n = \frac{2a \times mx \times x}{3b \times x^2 \times n} = \frac{2am}{3bx}$

iv) $\displaystyle \text{Compound Ratio of } 2a:3b, mn:x^2\text{ and }x:n = \frac{\sqrt{2} \times 3 \times \sqrt{20}}{1 \times \sqrt{5} \times 9 } = 2\sqrt{2}:3$

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Question 21: Find duplicate ratio of i) $\displaystyle 3:4$ ii) $\displaystyle 3\sqrt{3}:2\sqrt{5}$

i) $\displaystyle \text{Duplicate ratio of } 3:4 = 9:14$

ii) $\displaystyle \text{Duplicate ratio of } 3\sqrt{3}:2\sqrt{5} = 27:20$

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Question 22: Find triplicate ratio of i) $\displaystyle 1:3$ ii) $\displaystyle \frac{m}{2} : \frac{n}{3}$

i) Triplicate ratio of $\displaystyle 1:3 = 1:27$

$\displaystyle \text{ii) Triplicate Ratio of } \frac{m}{2} : \frac{n}{3} = \frac{m^3n^3}{216}$

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Question 23: Find sub-duplicate ratio of i) $\displaystyle 9:16$ ii) $\displaystyle (x-y)^4:(x+y)^6$

i) The sub-duplicate ratio of $\displaystyle 9:16 = 3: 4$

i) The sub-duplicate ratio of $\displaystyle (x-y)^4:(x+y)^6 = (x+y)^2 : (x+y)^3$

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Question 24: Find sub triplicate ratio of i) $\displaystyle 64:27$ ii) $\displaystyle x^3:125y^3$

i) The sub-triplicate ratio of $\displaystyle 64:27 = \sqrt{3}{64}:\sqrt{3}{27} = 4:3$

ii) The sub triplicate ratio of $\displaystyle x^3:125y^3 = x:5y$

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Question 25: Find the reciprocal ratio of i) $\displaystyle 5:8$ ii) $\displaystyle \frac{x}{3} : \frac{y}{7}$

$\displaystyle \text{i) The reciprocal ratio of } 5:8 = 8:5$

$\displaystyle \text{ii) The reciprocal ratio of } \frac{x}{3} : \frac{y}{7} = \frac{y}{7} : \frac{x}{3} = 3y:7x$

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Question 26: If $\displaystyle 3x+4:x+5$ is the duplicate ratio of $\displaystyle 8:5$ , find $\displaystyle x$

$\displaystyle \frac{64}{25} = \frac{3x+4}{x+5} \Rightarrow x = 20$

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Question 27: If $\displaystyle m:n$ is the duplicate ratio of $\displaystyle (m+x):(n+x)$ ; show that $\displaystyle x^2=mn$

$\displaystyle \frac{(m+x)^2}{(n+x)^2} = \frac{m}{n}$

$\displaystyle \frac{m^2+x^2+2mx}{n^2+x^2+2nx} = \frac{m}{n}$

$\displaystyle m^2n+nx^2+2mnx = n^2m+mx^2+2mnx$

$\displaystyle mn(m-n) = (m-n)x^2$

Hence $\displaystyle x^2=mn$

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Question 28: If $\displaystyle (x-9):(3x+6)$ is the triplicate ratio of $\displaystyle 4:9$ , find $\displaystyle x$ . [2014]

$\displaystyle \frac{x-9}{3x+6} = \frac{4^2}{9^2} = \frac{16}{81}$

$\displaystyle 81x-729=48x+96$

$\displaystyle x=25$

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Question 29: Find the ratio compounded of the reciprocal ratio of $\displaystyle 15:28$ , the sub – duplicate ration of $\displaystyle 36:49$ and the triplicate ratio of $\displaystyle 5:4$ .

Reciprocal Ratio $\displaystyle 15:28 = 28:15$

Sub Duplicate Ratio $\displaystyle 36:49 = 6:7$

Triplicate Ratio $\displaystyle 5:4 = 125:16$

$\displaystyle \text{Compound ratio of the above three } = \frac{28 \times 6 \times 125}{15 \times 7 \times 16} = \frac{25}{2}$

$\displaystyle \\$

$\displaystyle \text{Question 30: If } \frac{a+b}{am+bn} = \frac{b+c}{mb+nc} = \frac{c+a}{mc+na}, \\ \\ \text{ prove that each of these ratio is equal to } \frac{2}{m+n} \text{ provided } a+b+c \neq 0$ .

$\displaystyle \frac{a+b}{am+bn} = \frac{b+c}{mb+nc} = \frac{c+a}{mc+na} = \frac{a+b+b+c+c+a}{ am+bn + mb+nc + mc+na} = \frac{2}{m+n}$
$\displaystyle \text{Note: } \frac{a}{b+c} = \frac{b}{a+c} = \frac{c}{a+b} = \frac{\text{Sum of antecedents}}{\text{sum of consequents}} = \frac{a+b+c}{2(a+b+c)} = \frac{1}{2}$