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

Answer:

\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}  

\displaystyle \\

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

Answer:

\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}  

Answer:

\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)

Answer:

\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}  

Answer:

\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}  

Answer:

\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

Answer:

\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}  

Answer:

\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.

Answer:

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.

Answer:

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.

Answer:

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]

Answer:

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 .

Answer:

\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.

Answer:

\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.

Answer:

\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.

Answer:

\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.

Answer:

\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  

Answer:

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  

Answer:

\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

Answer:

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}  

Answer:

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}  

Answer:

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

Answer:

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  

Answer:

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}  

Answer:

\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  

Answer:

\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  

Answer:

\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]

Answer:

\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 .

Answer:

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 .

Answer:

\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}