Question 1: Express the following ratios in simplest form:

\displaystyle \text{ i)  } 18:30       \displaystyle \text{ ii)  } 7.5 \colon 9       \displaystyle \text{ iii)  } 6 \frac{2}{3} \colon 7 \frac{1}{2}       \displaystyle \text{ iv)  } \frac{1}{6} \colon \frac{ 1}{9} \colon \frac{ 1}{12}

\displaystyle \text{v)  } 7 \colon 5 \colon \frac{ 9}{2}       \displaystyle \text{ vi ) } 3 \frac{1}{5} \colon 5 \frac{1}{3} \colon 6 \frac{2}{3}  

Answer:

\displaystyle \text{ i ) } 18:30 = \frac{18}{30} = \frac{3}{5}  

\displaystyle \text{ ii ) } 7.5 \colon 9 = \frac{7.5}{ 9} = \frac{75}{ 90} = \frac{5}{6}  

\displaystyle \text{ iii ) } 6 \frac{2}{3} \colon 7 \frac{1}{2} = \frac{20}{3} \times \frac{2}{15} = \frac{8}{9}  

\displaystyle \text{        iv)       } \frac{1}{6} \colon \frac{ 1}{9} \colon \frac{ 1}{12} = \frac{1}{6} \times 36 \colon \frac{ 1}{9} \times 36 \colon \frac{ 1}{12} \times 36=6 \colon 4 \colon 3

\displaystyle \text{        v)       } 7 \colon 5 \colon \frac{ 9}{2} = 7 \times 2 \colon 5 \times 2 \colon \frac{ 9}{2} \times 2=14 \colon 10 \colon 9

\displaystyle \text{ vi) } 3 \frac{1}{5} \colon 5 \frac{1}{3} \colon 6 \frac{2}{3} = \frac{16}{5} \colon \frac{16}{3} \colon \frac{20}{3} = \frac{16}{5} \times 15 \colon \frac{16}{3} \times 15 \colon \frac{20}{3} \times 15 =48 \colon 80 \colon 100=12 \colon 20 \colon 25

\displaystyle \

Question 2: Express the following ratios in simplest form:

\displaystyle \text{ i)       } \text{ 75 paisa : 4 Rupees}

\displaystyle \text{ ii } \text{ 1 m 8 cm : 72 cm}  

\displaystyle \text{ iii ) } \text{ 1 hour 15 minutes : 45 minutes }

\displaystyle \text{        iv)       } \text{ 2 kg 750 g : 3 kg}  

\displaystyle \text{        v)       } \text{ 1 year 9 months : 2 years 4 months}

Answer:

\displaystyle \text{ i ) } \text{ 75 paisa : 4 Rupees} = \frac{75}{400} = \frac{3}{16}  

\displaystyle \text{        ii)       } \text{ 1 m 8 cm : 72 cm } = \frac{108}{72} = \frac{3}{2}  

\displaystyle \text{        iii)       } \text{ 1 hour 15 minutes : 45 minutes } = \frac{75}{45} = \frac{5}{3}  

\displaystyle \text{        iv)       } \text{ 2 kg 750 g : 3 kg } = \frac{2750}{3000} = \frac{11}{12}  

\displaystyle \text{ v ) } \text{ 1 year 9 months : 2 years 4 months } = \frac{21}{28} = \frac{3}{4}  

\displaystyle \

Question 3: Which ratio is greater:

\displaystyle \text{ i ) } (4 : 9) \text{ or } (3:7) \displaystyle \text{ ii ) } \Big( 2 \frac{1}{3} \colon 3 \frac{1}{3} \Big) \text{ or } (3.6\colon 4.8)

\displaystyle \text{ iii ) } \Big( \frac{1}{2} \colon \frac{1}{3} \Big) \text{ or } \Big( \frac{1}{6} \colon \frac{1}{4} \Big) \displaystyle \text{        iv)       } \Big( 3 \frac{1}{3} \colon 4 \frac{1}{6} \Big) \text{ or } (0.9 \colon 1)

Answer:

\displaystyle \text{ i)       } (4 : 9) \text{ or } (3:7) = \frac{4}{9} \text{ or } \frac{3}{7} = \frac{28}{63} \text{ or } \frac{27}{63} .

\displaystyle \text{   Therefore   }  (4:9)  \text{ is greater than } (3:7)

\displaystyle \text{        ii)       } \Big( 2 \frac{1}{3} \colon 3 \frac{1}{3} \Big) \text{ or } (3.6\colon 4.8) = \Big( \frac{7}{3} \times \frac{3}{10} \Big) \text{ or } \frac{3.6}{4.8} = \frac{7}{10} \text{ or } \frac{3}{4} = \frac{14}{20} \text{ or } \frac{15}{20}  

\displaystyle \text{   Therefore   }  (3.6\colon 4.8)  \text{ is greater than } \Big( 2 \frac{1}{3} \colon 3 \frac{1}{3} \Big)

\displaystyle \text{ iii ) } \Big( \frac{1}{2} \colon \frac{1}{3} \Big) \text{ or } \Big( \frac{1}{6} \colon \frac{1}{4} \Big) = \frac{3}{2} \text{ or } \frac{2}{3}  

\displaystyle \text{   Therefore   }  \Big( \frac{1}{2} \colon \frac{1}{3} \Big)  \text{ is greater than } \Big( \frac{1}{6} \colon \frac{1}{4} \Big)

\displaystyle \text{        iv)       } \Big( 3 \frac{1}{3} \colon 4 \frac{1}{6} \Big) \text{ or } (0.9 \colon 1) = \frac{4}{5} \text{ or } \frac{9}{10} = \frac{8}{10} \text{ or } \frac{9}{10}  

\displaystyle \text{   Therefore   }  (0.9 \colon 1)  \text{ is greater than } \Big( 3 \frac{1}{3} \colon 4 \frac{1}{6} \Big)

\displaystyle \

Question 4: Arrange the following ratios in ascending order of magnitude:

\displaystyle \text{ i ) } (2:3) , (5:9) , (11:15) \displaystyle \text{        ii)       } (5:7) , (9:14) , (20:21)  \text{ and } (3:8)

Answer:

\displaystyle \text{ i ) } (2:3) , (5:9) , (11:15)

\displaystyle \frac{2}{3} \colon \frac{5}{9} \colon \frac{11}{15} = \frac{30}{45} \colon \frac{25}{45} \colon \frac{33}{45}

\displaystyle \text{   Therefore   }  \frac{5}{9} < \frac{2}{3} < \frac{11}{15}  

\displaystyle \text{ ii ) } (5:7) , (9:14) , (20:21)  \text{ and } (3:8)

\displaystyle \frac{5}{7} \colon \frac{9}{14} \colon \frac{20}{21} \colon \frac{3}{8} = \frac{120}{168} \colon \frac{108}{168} \colon \frac{160}{168} \colon \frac{63}{168}

\displaystyle \text{   Therefore   }  \frac{3}{8} < \frac{9}{14} < \frac{5}{7} < \frac{20}{21}  

 \displaystyle \

Question 5: Divide Rs. \displaystyle 142.20 between \displaystyle A  \text{ and } B in the ratio \displaystyle \frac{1}{4} \colon \frac{1}{5}  

Answer:

\displaystyle A : B = 5:4

\displaystyle \text{ A's Share }  = 142.20 \times \frac{5}{9} = 79 \text{ Rs. }

\displaystyle \text{ B's Share }  = 142.20 \times \frac{4}{9} = 63.2 \text{ Rs. }

 \displaystyle \

Question 6: Divide Rs. \displaystyle 3726 among A, B, C in the ratio of \displaystyle \frac{1}{3} \colon \frac{1}{4} \colon \frac{1}{6}  

Answer:

\displaystyle A : B: C = \frac{1}{3} \colon \frac{1}{4} \colon \frac{1}{6} = 4: 3: 2

\displaystyle \text{ A's Share }  = \frac{4}{9} \times 3726 = 1656 \text{ Rs. }

\displaystyle \text{ B's Share }  = \frac{3}{9} \times 3726 = 1242 \text{ Rs. }

\displaystyle \text{ C's Share }  = \frac{2}{9} \times 3726 = 828 \text{ Rs. }

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Question 7: Divide Rs. \displaystyle 810 among A, B, C in the ration of \displaystyle \frac{1}{4} : \frac{2}{5} : 1 \frac{3}{8}  

Answer:

\displaystyle A : B: C = \frac{1}{4} \colon \frac{2}{5} \colon \frac{11}{8} = 10: 16: 55

\displaystyle \text{ A's Share }  = \frac{10}{81} \times 810 = 100 \text{ Rs. }

\displaystyle \text{ B's Share }  = \frac{16}{81} \times 810 = 160 \text{ Rs. }

\displaystyle \text{ C's Share }  = \frac{55}{81} \times 810 = 550 \text{ Rs. }

\displaystyle \

Question 8: Divide Rs. \displaystyle 1050 between \displaystyle A  \text{ and } B in the ratio \displaystyle 2 \frac{2}{3} : 6 \frac{2}{3}  

Answer:

\displaystyle A : B = \frac{8}{3} : \frac{20}{3} \text{ or } 8:20

\displaystyle \text{ A's Share }  = \frac{8}{28} \times 1050 = 300 \text{ Rs. }

\displaystyle \text{ B's Share }  = \frac{20}{28} \times 1050 = 750 \text{ Rs. }

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Question 9: Divide Rs. \displaystyle 747 among \displaystyle A, B, C such that \displaystyle 4A = 5B = 7C

Answer:

\displaystyle A : B : C = \frac{a}{4} : \frac{a}{5} : \frac{a}{7} = 35 : 28 : 20

\displaystyle \text{ A's Share }  = \frac{35}{83} \times 747 = 315 \text{ Rs. }

\displaystyle \text{ B's Share }  = \frac{28}{83} \times 747 = 252 \text{ Rs. }

\displaystyle \text{ C's Share }  = \frac{20}{83} \times 747 = 180 \text{ Rs. }

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Question 10: A big contains one rupee, \displaystyle 50 p and \displaystyle 25 p coins in the ration of \displaystyle 5 : 6 : 8 amounting to Rs \displaystyle 210 . Find the number of coins in each type.

Answer:

Coins: \displaystyle x One Rupee \displaystyle : 50 paisa \displaystyle : 25 paisa \displaystyle = 5x : 6x : 8x

Amount: One Rupee \displaystyle : 50 paisa \displaystyle : 25 paisa \displaystyle = 500x : 300x : 200x

\displaystyle \text{   Therefore   }  500x+300x+200x=21000 \text{ or } x=21

Therefore   the coins are in ratio of \displaystyle 5x:6x:8x \text{ or }\displaystyle 105:126:168

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Question 11: Find \displaystyle A : B : C when:

\displaystyle \text{ i } A : B = 2 : 5  \text{ and } B : C = 7 : 9

\displaystyle \text{ ii } A : B = 3 : 4  \text{ and } B : C = 6 : 11

\displaystyle \text{ iii } A : B = \frac{1}{2} : \frac{1}{3} and B : C = \frac{1}{4} : \frac{1}{5}  

Answer:

\displaystyle \text{ i } A : B = 2 : 5  \text{ and } B : C = 7 : 9

\displaystyle A : B = 2 : 5  \text{ and } B : C = 7 :9

LCM of \displaystyle 5  \text{ and } 7 = 35 .   Therefore   we can say

\displaystyle A: B = 14 : 35

\displaystyle B: C = 35 : 45

\displaystyle \text{   Therefore   }  A : B : C = 14 : 35 : 45

\displaystyle \text{ ii } A : B = 3 : 4  \text{ and } B : C = 6 : 11

\displaystyle A : B = 3 : 4  \text{ and } B : C = 6 : 11

LCM of \displaystyle 4  \text{ and } 6 = 12 .   Therefore   we can say

\displaystyle A : B = 9 : 12

\displaystyle B : C = 12 : 22

\displaystyle \text{   Therefore   }  A : B : C = 9 : 12 : 22

\displaystyle \text{ iii } A : B = \frac{1}{2} : \frac{1}{3} and B : C = \frac{1}{4} : \frac{1}{5}  

\displaystyle A : B = 3 : 2  \text{ and } B : C = 5 : 4

LCM of \displaystyle 2  \text{ and } 5 = 10 .   Therefore   we can say

\displaystyle A : B = 15 : 10

\displaystyle B : C = 10 : 08

\displaystyle \text{   Therefore   }  A : B : C = 15 : 10 : 08

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Question 12: If \displaystyle A : B = 4 : 9  \text{ and } A : C = 2 : 3 , find \displaystyle B : C  \text{ and } A : B : C .

Answer:

\displaystyle \frac{A}{B} = \frac{4}{9} \displaystyle \frac{A}{C} = \frac{2}{3}  

Therefore  , \displaystyle \frac{B}{C} = \frac{A}{C} \times \frac{B}{A} = \frac{2}{3} \times \frac{9}{4} = \frac{3}{2} \displaystyle \text{ or } B : C = 3 : 2

Now, \displaystyle A : B = 4 : 9  \text{ and } B : C = 3 : 2

LCM of \displaystyle 9  \text{ and } 3 = 9 ,   Therefore   we can say

\displaystyle A : B = 4 : 9 \displaystyle B : C = 9 : 6

\displaystyle \text{   Therefore   }  A : B : C = 4 : 9 : 6

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Question 13: If \displaystyle A : C = 5 : 8  \text{ and } B : C = 5 : 6 , find \displaystyle A : B  \text{ and } A : B : C

Answer:

\displaystyle \frac{A}{C} = \frac{5}{8} \displaystyle \frac{B}{C} = \frac{5}{6}  

Therefore  , \displaystyle \frac{A}{B} = \frac{A}{C} \times \frac{C}{B} = \frac{5}{8} \times \frac{6}{5} = \frac{3}{4} \text{ or } A:B = 3: 4

Now, \displaystyle A : C = 5 : 8  \text{ and } A : B = 3 : 4

LCM of \displaystyle 5  \text{ and } 3 = 15 ,   Therefore   we can say

\displaystyle A : C = 15 : 24

\displaystyle A : B = 15 : 20

\displaystyle \text{   Therefore   }  A : B : C = 15 : 20 : 24

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Question 14: Two numbers are in the ration of \displaystyle 6 : 11 . On adding \displaystyle 2 to the first and \displaystyle 7 to the second, their rations become \displaystyle 8 : 15 . Find the numbers.

Answer:

Let the two numbers be \displaystyle a  \text{ and } b .   Therefore  

\displaystyle \frac{a}{b} = \frac{6}{11} \text{ or }\displaystyle b= \frac{11}{6} a

\displaystyle \Rightarrow \frac{a+2}{b+7} = \frac{8}{15}  

\displaystyle \Rightarrow   a+2= \frac{8}{15} (b+7)

\displaystyle \Rightarrow   a+2= \frac{8}{15} ( \frac{11}{6} a + 7)

\displaystyle \Rightarrow   a+2= \frac{8}{15} \frac{11}{6} a + \frac{8}{15} \times 7

\displaystyle \Rightarrow   a- \frac{88}{90} a= \frac{56}{15} -2

\displaystyle \Rightarrow   a= \frac{90}{2} \times \frac{26}{15} = 78

\displaystyle \Rightarrow   b= \frac{11}{6} \times 78 = 143

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Question 15: A sum of money is divided between \displaystyle A  \text{ and } B in the ratio \displaystyle 5 : 7 . If B’s share is Rs. \displaystyle 665 , find the total amount.

Answer:

Let the total money to be divided be \displaystyle x

\displaystyle \frac{7}{12} \times x = 665

\displaystyle x= \frac{12}{7} \times 665=1140 \text{ Rs. } .

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Question 16: Two numbers are in the ratio of \displaystyle 7:4 . If their difference is \displaystyle 72 , find the numbers.

Answer:

Let the two numbers be \displaystyle a  \text{ and } b

\displaystyle \text{ Given } \frac{a}{b} = \frac{7}{4} \text{ or } b= \frac{4}{7} a  \text{ and } (a - b) = 72

Substituting \displaystyle (a- \frac{4}{7} a) =72

\displaystyle a= \frac{7}{3} \times 72=168

\displaystyle b= \frac{4}{7} \times 168=96

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Question 17: A certain sum of money is divided between \displaystyle A, B  \text{ and } C in the ration of \displaystyle 5 : 6 : 7 . If A’s share is Rs. \displaystyle 175 , find the total amount and also the shares of each one of \displaystyle B  \text{ and } C .

Answer:

Let the total money to be divided be \displaystyle x

\displaystyle \frac{5}{18} \times x=175

\displaystyle x= \frac{18}{5} \times 175 = 630 \text{ Rs. } .

\displaystyle \text{ B's share } = \frac{6}{18} \times 630 = 210 \text{ Rs. } .

\displaystyle \text{ C's share } = \frac{7}{18} \times 630 = 245 \text{ Rs. } .

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Question 18: The ratio of the number of boys and number of girls in a school of \displaystyle 1440 students is \displaystyle 7:5 . If \displaystyle 40 new boys are admitted, find how many new girls can be admitted to make the ratio \displaystyle 4:3 .

Answer:

Let the number of boys and girls be \displaystyle a  \text{ and } b respectively.

\displaystyle a+b=1440

\displaystyle \frac{a}{b} = \frac{7}{5}  \text{ or } b= \frac{5}{7} a

\displaystyle a+ \frac{5}{7} a=1440  \text{ hence }  a (boys) =840

\displaystyle b (girls) =1440-840=600

Let the number of girls added \displaystyle = x

\displaystyle \frac{880}{600+x} = \frac{4}{3}  

Solve for \displaystyle x = 60 .   Therefore   we need to add \displaystyle 60 girls to make the new ratio to \displaystyle 4 : 3

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Question 19: The sum of three numbers is \displaystyle 212 . If the ratio of the first to the second is \displaystyle 13 : 16 and that of the second to the third is \displaystyle 2 : 3 , then find the numbers.

Answer:

Let the three numbers be \displaystyle A, B  \text{ and } C .

\displaystyle A : B = 13 : 16

\displaystyle B : C = 2 : 3

LCM of \displaystyle 16  \text{ and } 2 is \displaystyle 16 .

\displaystyle \text{   Therefore   }  B : C = 16 : 24

\displaystyle \text{   Therefore   }  A : B : C = 13 : 16 : 24

Hence:

\displaystyle A = \frac{13}{53} \times 212=52 \displaystyle B = \frac{16}{53} \times 212=64 \displaystyle C = \frac{24}{53} \times 212=96

\displaystyle \

Question 20: Find the number which when added to each term of the ratio \displaystyle 27 : 35 , changes the ratio to \displaystyle 4 : 5

Answer:

Let \displaystyle x be added to both the terms

\displaystyle \frac{27+x}{35+x} = \frac{4}{5}  

\displaystyle x=5

Question 21: The present ages of Mr. X and his Son are in the ratio of \displaystyle 17 : 19 . If the ratio of their ages \displaystyle 9 years ago was \displaystyle 7 : 3 , then find their present ages.

Answer:

Let the present age of Mr. X be \displaystyle a and his son be \displaystyle b .   Therefore  

 \displaystyle \frac{a}{b} = \frac{17}{9} \text{ or }\displaystyle b= \frac{9}{17}  

\displaystyle \frac{(a-9) }{(b-9) } = \frac{7}{3}  

\displaystyle 3a-27=7b-63

\displaystyle 3a-27=7 \times \frac{9}{17} a-63

Solving for \displaystyle a=51  \text{ and } b=27

Question 22: Salaries of \displaystyle A, B  \text{ and } C are in the ratio of \displaystyle 2 : 3: 5 . If the increment of \displaystyle 15\%, 10\%  \text{ and } 20\% are allowed respectively in their salaries, then what will be the new ratio of their salaries.

Answer:

If A’s salary was \displaystyle 200 , B’s salary was \displaystyle 300 and C’s salary was \displaystyle 500 , then the ratios of their salaries would be \displaystyle 2 : 3 : 5 .

With increments, A’s salary becomes \displaystyle 230 , B’s salary becomes \displaystyle 330 and C’s salary becomes \displaystyle 600 .

Therefore   the new ratio would be \displaystyle A : B : C = 23 : 33 : 60

Question 23: A got \displaystyle 58 marks out of \displaystyle 75 in Physics and \displaystyle 97 marks out of \displaystyle 120 in Math. In which of the two subjects did he perform better?

Answer:

\displaystyle \frac{Physics}{Total} = \frac{58}{75}  

\displaystyle \frac{Math}{Total} = \frac{97}{120}  

LCM of \displaystyle 75  \text{ and } 120 = 600   Therefore  

\displaystyle \frac{Physics}{Total} = \frac{58}{75} \times \frac{8}{8} = \frac{464}{600}  

\displaystyle \frac{Math}{Total} = \frac{97}{120} \times \frac{5}{5} = \frac{485}{600}  

Therefore   A performed better in Math.

Question 24: A mixture consists of only two components \displaystyle A  \text{ and } B . In \displaystyle 60  \text{ liters }  of this mixture, the components \displaystyle A  \text{ and } B are in the ratio of \displaystyle 2 :1 . What quantity of component B $ has to be added in this mixture so that the new ratio is \displaystyle 1 : 2 ?

Answer:

Quantity of \displaystyle A in the mixture \displaystyle = 60 \times \frac{2}{3} =40  \text{ liters } 

Quantity of \displaystyle B in the mixture \displaystyle = 60 \times \frac{1}{3} =20  \text{ liters } 

Let the quantity of \displaystyle B added to the mixture is \displaystyle x

\displaystyle \frac{40}{(20+x) } = \frac{1}{2}  \text{ or } x=60  \text{ liters }