Question 1: Express the following ratios in simplest form:

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

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

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

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

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

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

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

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

$\displaystyle \$

Question 2: Express the following ratios in simplest form:

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

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

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

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

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

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

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

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

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

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

$\displaystyle \$

Question 3: Which ratio is greater:

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

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

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

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

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

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

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

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

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

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

$\displaystyle \$

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

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

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

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

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

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

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

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

$\displaystyle \$

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

$\displaystyle A : B = 5:4$

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

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

$\displaystyle \$

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

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

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

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

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

$\displaystyle \$

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

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

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

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

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

$\displaystyle \$

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

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

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

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

$\displaystyle \$

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

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

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

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

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

$\displaystyle \$

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.

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 }$\displaystyle 105:126:168$

$\displaystyle \$

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

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

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

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

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

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

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

$\displaystyle A : B = 15 : 10$

$\displaystyle B : C = 10 : 08$

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

$\displaystyle \$

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

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

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

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

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

$\displaystyle \$

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

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

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

$\displaystyle A : C = 15 : 24$

$\displaystyle A : B = 15 : 20$

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

$\displaystyle \$

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.

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

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

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

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

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

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

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

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

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

$\displaystyle \$

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.

Let the total money to be divided be $\displaystyle x$

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

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

$\displaystyle \$

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

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

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

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

$\displaystyle \$

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 $\displaystyle B \text{ and } C$.

Let the total money to be divided be $\displaystyle x$

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

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

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

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

$\displaystyle \$

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 $\displaystyle 4:3$.

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 $\displaystyle 4 : 3$

$\displaystyle \$

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.

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 $\displaystyle 4 : 5$

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.

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.

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?

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

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

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 $\displaystyle 1 : 2$?

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