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

Answer:

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

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

Answer:

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

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

Answer:

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

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

Answer:

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

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

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

Answer:

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

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

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

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

Answer:

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

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

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

Answer:

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

Dividing by $y^2$

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

Let $\frac{x}{y} = a$

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

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

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

Answer:

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

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

$14x=27y$

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

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

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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 $x \ and \ y$. Therefore

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

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

$7x+35= 5y+25$

$7x-5y+10=0$

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

$14y-15y+30=0$

or $y = 30.$

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

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

Answer:

Let the number be $x \ and \ y$

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

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

Given $y^2-x^2= 400$

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

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

or $y = 25$

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 $x$ be subtracted. Therefore

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

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

or $x = 5$

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Question 12: The monthly pocket money of Ravi and Sanjeev are 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 $x and y$ respectively.

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

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

Substituting

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

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

Substituting

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

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

Answer:

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

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

$2x=7$

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

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

Answer:

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

i) $9 \times 245 = 7 \times Increased \ Fare$

$\Rightarrow Increase \ Fare = 315$

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

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

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

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

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

Answer:

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

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

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

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

Answer:

$\frac{Apples}{Orracnges} =\frac{7}{13}$

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

$2 Oranges -16 = Apples - 11$

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

Substituting

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

$13 Apples + 65 = 14 Apples$

Therefore  $Apples = 65.$

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

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

Answer:

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

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

$Boys + 20 = 2Girls -24$

or  $Boys = 2Girls - 44$

Substituting in i)

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

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

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

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Question 18: a) If $A:B=3:4$ and $B:C=6:7$ , find $A:B:C$ and $A:C$

b)  If $A:B=2:5$ and $A:C=3:4$ , find $A:B:C$

Answer:

a)

$A:B=3:4 ... ... ... ... ... ... ... i)$

$B:C=6:7 ... ... ... ... ... ... ... ii)$

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

$A:B=18:24 ... ... ... ... ... ... ... iii)$

$B:C=24:28 ... ... ... ... ... ... ... iv)$

Therefore $A:B:C = 18:24:28 or A:B:C = 9:12:14$

Hence $A:C = 9:14$

b)

$A:B=2:5 ... ... ... ... ... ... ... i)$

$A:C=3:4 ... ... ... ... ... ... ... ii)$

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

$A:B=6:15 ... ... ... ... ... ... ... iii)$

$A:C=6:8 ... ... ... ... ... ... ... iv)$

Therefore $A:B:C = 6:15:8$

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

Answer:

$A:B=4:3 ... ... ... ... ... ... ... i)$

$B:C=6:4 ... ... ... ... ... ... ... ii)$

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

$A:B=24:18 ... ... ... ... ... ... ... iii)$

$B:C=18:12 ... ... ... ... ... ... ... iv)$

Therefore $A:B:C = 24:18:12 or A:B:C = 4:3:2$

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

Question 20: Find the compound ratio of

i) $3:5 \ and \ 8:15$

ii) $2:3, 9:14 \ and \ 14:27$

iii) $2a:3b, mn:x^2 \ and \ x:n$

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

Answer:

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

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

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

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

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

Answer:

i) Duplicate ratio of $3:4 = 9:14$

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

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

Answer:

i) Triplicate ratio of $1:3 = 1:27$

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) $9:16$  ii) $(x-y)^4:(x+y)^6$

Answer:

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

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

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

Answer:

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

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

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

Answer:

i) The reciprocal ratio of  $5:8 = 8:5$

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 $3x+4:x+5$ is the duplicate ratio of $8:5$ , find $x$

Answer:

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

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

Answer:

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

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

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

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

Hence $x^2=mn$

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

Answer:

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

$81x-729=48x+96$

$x=25$

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

Answer:

Reciprocal Ratio $15:28 = 28:15$

Sub Duplicate Ratio $36:49 = 6:7$

Triplicate Ratio $5:4 = 125:16$

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

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

Answer:

$\frac{a+b}{am+bn}=\frac{b+c}{mb+nc}=\frac{c+a}{mc+na} = \frac{a+b+b+c+c+a}{ am+bn + mb+nc + mc+na} = \frac{2}{m+n}$

Note: if you have  $\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b} =\frac{Sum \ of \ antecedents}{sum \ of\ consequents} = \frac{a+b+c}{2(a+b+c)} = \frac{1}{2}$

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