Class 8: Ratio and Proportion – Exercise 9A – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Express the following ratios in simplest form:

i) $18:30$ ii) $7.5 \colon 9$ iii) $6$ $\frac{2}{3}$ $\colon 7$ $\frac{1}{2}$ iv) $\frac{1}{6}$ $\colon$ $\frac{ 1}{9}$ $\colon$ $\frac{ 1}{12}$ v) $7 \colon 5 \colon$ $\frac{ 9}{2}$ vi) $3$ $\frac{1}{5}$ $\colon 5$ $\frac{1}{3}$ $\colon 6$ $\frac{2}{3}$

Answer:

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

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

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

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

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

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

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Question 2: Express the following ratios in simplest form:

i) $\ 75 \ paisa : 4 \ Rupees$ ii) $1 \ m \ 8 \ cm : 72 \ cm$

iii) $1 \ hour 15 \ minutes : 45 \ minutes$ iv) $2 \ kg \ 750 \ g : 3 \ kg$

v) $1 \ year \ 9 \ months : 2 \ years \ 4 \ months$

Answer:

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

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

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

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

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

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Question 3: Which ratio is greater:

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

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

Answer:

i) $(4 : 9) \ or \ (3:7) =$ $\frac{4}{9}$ $\ \ or \ \$ $\frac{3}{7}$ $=$ $\frac{28}{63}$ $\ \ or \ \$ $\frac{27}{63}$ . Therefore $(4:9)$ is greater than $(3:7)$

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

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

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

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

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

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

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

i) $(2:3), (5:9), (11:15)$ ii) $(5:7), (9:14), (20:21)$ and $(3:8)$

Answer:

i) $(2:3), (5:9), (11:15)$

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

ii) $(5:7), (9:14), (20:21)$ and $(3:8)$

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

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

Answer:

$A : B = 5:4$

A’s Share $= 142.20 \times$ $\frac{5}{9}$ $= 79$ Rs

B’s share $= 142.20 \times$ $\frac{4}{9}$ $= 63.2$ Rs

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

Answer:

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

A’s share $=$ $\frac{4}{9}$ $\times 3726 = 1656$ Rs

B’s share $=$ $\frac{3}{9}$ $\times 3726 = 1242$ Rs

C’s share $=$ $\frac{2}{9}$ $\times 3726 = 828$ Rs

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

Answer:

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

A’s share $=$ $\frac{10}{81}$ $\times 810 = 100$ Rs

B’s share $=$ $\frac{16}{81}$ $\times 810 = 160$ Rs

C’s share $=$ $\frac{55}{81}$ $\times 810 = 550$ Rs

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

Answer:

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

A’s share $=$ $\frac{8}{28}$ $\times 1050 = 300$ Rs

B’s share $=$ $\frac{20}{28}$ $\times 1050 = 750$ Rs

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

Answer:

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

A’s share $=$ $\frac{35}{83}$ $\times 747 = 315$ Rs

B’s share $=$ $\frac{28}{83}$ $\times 747 = 252$ Rs

C’s share $=$ $\frac{20}{83}$ $\times 747 = 180$ Rs

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

Answer:

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

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

Therefore $500x+300x+200x=21000 \ or \ x=21$

Therefore the coins are in ratio of $5x:6x:8x$ or $105:126:168$

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

i) $A : B = 2 : 5$ and $B : C = 7 : 9$

ii) $A : B = 3 : 4$ and $B : C = 6 : 11$

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

Answer:

i) $A : B = 2 : 5$ and $B : C = 7 : 9$

$A : B = 2 : 5$ and $B : C = 7 :9$

LCM of $5$ and $7 = 35$ . Therefore we can say

$A: B = 14 : 35$

$B: C = 35 : 45$

Therefore $A : B : C = 14 : 35 : 45$

ii) $A : B = 3 : 4$ and $B : C = 6 : 11$

$A : B = 3 : 4$ and $B : C = 6 : 11$

LCM of $4$ and $6 = 12$ . Therefore we can say

$A : B = 9 : 12$

$B : C = 12 : 22$

Therefore $A : B : C = 9 : 12 : 22$

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

$A : B = 3 : 2$ and $B : C = 5 : 4$

LCM of $2$ and $5 = 10$ . Therefore we can say

$A : B = 15 : 10$

$B : C = 10 : 08$

Therefore $A : B : C = 15 : 10 : 08$

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Question 12: If $A : B = 4 : 9$ and $A : C = 2 : 3$ , find $B : C$ and $A : B : C$ .

Answer:

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

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

Now, $A : B = 4 : 9$ and $B : C = 3 : 2$

LCM of $9$ and $3 = 9$ , Therefore we can say

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

Therefore $A : B : C = 4 : 9 : 6$

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Question 13: If $A : C = 5 : 8$ and $B : C = 5 : 6$ , find $A : B$ and $A : B : C$

Answer:

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

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

Now, $A : C = 5 : 8$ and $A : B = 3 : 4$

LCM of $5$ and $3 = 15$ , Therefore we can say

$A : C = 15 : 24$

$A : B = 15 : 20$

Therefore $A : B : C = 15 : 20 : 24$

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

Answer:

Let the two numbers be $a$ and $b$ . Therefore

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

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

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

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

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

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

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

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

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

Answer:

Let the total money to be divided be $x$

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

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

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

Answer:

Let the two numbers be $a$ and $b$

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

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

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

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

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

Answer:

Let the total money to be divided be $x$

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

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

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

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

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

Answer:

Let the number of boys and girls be $a$ and $b$ respectively.

$a+b=1440$

$\frac{a}{b}$ $=$ $\frac{7}{5}$ or $b=$ $\frac{5}{7}$ $a$

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

$b (girls)=1440-840=600$

Let the number of girls added $= x$

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

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

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

Answer:

Let the three numbers be $A, B$ and $C$ .

$A : B = 13 : 16$

$B : C = 2 : 3$

LCM of $16$ and $2$ is $16$ .

Therefore $B : C = 16 : 24$

Therefore $A : B : C = 13 : 16 : 24$

Hence:

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

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

Answer:

Let $x$ be added to both the terms

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

$x=5$

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

Answer:

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

$\frac{a}{b}$ $=$ $\frac{17}{9}$ or $b=$ $\frac{9}{17}$

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

$3a-27=7b-63$

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

Solving for $a=51$ and $b=27$

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

Answer:

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

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

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

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

Answer:

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

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

LCM of $75$ and $120 = 600$ Therefore

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

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

Therefore A performed better in Math.

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