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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