Q1. Which ratio is greater?
(i) 3 : 5 or 7 : 11
(ii) 8 : 15 or 11 : 20
(iii) $2\dfrac{1}{2} : 3\dfrac{1}{3}$ or 3 .5 : 4.5
Show Answer
(i) $\Rightarrow$ using cross multiplication method
$\Rightarrow \dfrac{3}{5} = \dfrac{7}{11}$
$\Rightarrow 3\times11 = 7\times5$
$\Rightarrow$ 33 = 35
$\therefore 33 < 35$ Answer
(ii) $\Rightarrow$ 8 : 5 or 11 : 20
$\Rightarrow \dfrac{8}{15} = \dfrac{11}{20}$
$\Rightarrow 8\times20 = 11\times15$
$\Rightarrow$ 160 = 165
$\therefore 160 < 165$ Answer
(iii) $\Rightarrow 2\dfrac{1}{2} : 3\dfrac{1}{3}$ or 3.5 : 4.5
$\Rightarrow \dfrac{5}{2} : \dfrac{10}{3} = \dfrac{35}{10} : \dfrac{45}{10}$
$\Rightarrow \dfrac{5\times3} {2\times10} = \dfrac{35\times10}{10\times45}$
$\Rightarrow \dfrac{15}{20} : \dfrac{350}{450}$
$\therefore 3.5 : 4.5 > 2\dfrac{1}{2} : 3\dfrac{1}{3}$ Answer
$\Rightarrow \dfrac{3}{5} = \dfrac{7}{11}$
$\Rightarrow 3\times11 = 7\times5$
$\Rightarrow$ 33 = 35
$\therefore 33 < 35$ Answer
(ii) $\Rightarrow$ 8 : 5 or 11 : 20
$\Rightarrow \dfrac{8}{15} = \dfrac{11}{20}$
$\Rightarrow 8\times20 = 11\times15$
$\Rightarrow$ 160 = 165
$\therefore 160 < 165$ Answer
(iii) $\Rightarrow 2\dfrac{1}{2} : 3\dfrac{1}{3}$ or 3.5 : 4.5
$\Rightarrow \dfrac{5}{2} : \dfrac{10}{3} = \dfrac{35}{10} : \dfrac{45}{10}$
$\Rightarrow \dfrac{5\times3} {2\times10} = \dfrac{35\times10}{10\times45}$
$\Rightarrow \dfrac{15}{20} : \dfrac{350}{450}$
$\therefore 3.5 : 4.5 > 2\dfrac{1}{2} : 3\dfrac{1}{3}$ Answer
Q2. A field is 60m long and 40m wide. Find the ratio between its:
(i) Breadth and Length
(ii) Length and Perimeter
Show Answer
(i) $\Rightarrow$ given, length = 60m
$\Rightarrow$ and, breadth = 40m
Breadth : length = $40 : 60$
$\therefore \dfrac{40}{60} = \dfrac{2}{3}$ Answer
(ii) Perimeter is the length of the boundary of the area:
$\therefore$ field’s perimeter = $(40+40) + (60+60) = 80 + 120 = 200$m
Now, length : perimeter = $60 : 200$
$\Rightarrow \dfrac{60}{200} = \dfrac{6}{20}$
$\Rightarrow \dfrac{3}{10} = 3 : 10$Answer
$
$\Rightarrow$ and, breadth = 40m
Breadth : length = $40 : 60$
$\therefore \dfrac{40}{60} = \dfrac{2}{3}$ Answer
(ii) Perimeter is the length of the boundary of the area:
$\therefore$ field’s perimeter = $(40+40) + (60+60) = 80 + 120 = 200$m
Now, length : perimeter = $60 : 200$
$\Rightarrow \dfrac{60}{200} = \dfrac{6}{20}$
$\Rightarrow \dfrac{3}{10} = 3 : 10$Answer
$
Q3. The monthlyl income and the monthly expenditure of a man are Rs.15,000 and Rs.10,500. Find the ratio between his monthly:
(i) Income and Expenditure
(ii) Expenditure and Savings
(iii) Savings and Income
Show Answer
(i) $\Rightarrow$ given, monthly income = Rs. 15,000
$\Rightarrow$ and, monthly savings = Rs. 10,500
$\therefore$ monthly savings = 15,000 – 10,500
$\Rightarrow$ Rs. 4500
$\Rightarrow$ income : expenditure = 15,000 : 10,500
$\Rightarrow \dfrac{15,000}{10,500} = \dfrac{150}{105}$
$\Rightarrow \dfrac{30}{21} = \dfrac{10}{7} = 10 : 7$Answer
(ii) $\Rightarrow$ expenditure : savings = 10,500 : 4500
$\Rightarrow \dfrac{10,500}{4500} = \dfrac{105}{45}$
$\Rightarrow \dfrac{21}{9} = \dfrac{7}{3} = 7 : 3$Answer
(iii)$\Rightarrow$ savings : income = 4500 : 15,000
$\Rightarrow \dfrac{4500}{15,000} = \dfrac{45}{150}$
$\Rightarrow \dfrac{9}{30} = \dfrac{3}{10} = 3 : 10$ Answer
$\Rightarrow$ and, monthly savings = Rs. 10,500
$\therefore$ monthly savings = 15,000 – 10,500
$\Rightarrow$ Rs. 4500
$\Rightarrow$ income : expenditure = 15,000 : 10,500
$\Rightarrow \dfrac{15,000}{10,500} = \dfrac{150}{105}$
$\Rightarrow \dfrac{30}{21} = \dfrac{10}{7} = 10 : 7$Answer
(ii) $\Rightarrow$ expenditure : savings = 10,500 : 4500
$\Rightarrow \dfrac{10,500}{4500} = \dfrac{105}{45}$
$\Rightarrow \dfrac{21}{9} = \dfrac{7}{3} = 7 : 3$Answer
(iii)$\Rightarrow$ savings : income = 4500 : 15,000
$\Rightarrow \dfrac{4500}{15,000} = \dfrac{45}{150}$
$\Rightarrow \dfrac{9}{30} = \dfrac{3}{10} = 3 : 10$ Answer
Q4. The ratio of the number of boys to the number of girls in a school is 672 students is 5 : 7. When some new boys and girls are admitted, the number of new girls increases by 8 and the ratio of boys to girls changes to 3 : 4. Calculate the number of new boys admitted.
Show Answer
$\Rightarrow$ given ratio = $5 : 7$
$\Rightarrow$ and, total no. of students = 672
$\Rightarrow$ let no. of boys and girls be = $5x$ and $7x$
$\Rightarrow 5x + 7x = 672$
$\Rightarrow 12x = 672$
$\Rightarrow x = \dfrac{672}{12} = 56$
$\Rightarrow$ now, total no. of boys in school = $5\times56 = 280$
$\Rightarrow$ and, total no. of girls in school = $7\times56 = 392$
$\Rightarrow$ given, upon new admission no. of girls increased by 8,
$\therefore$ new no. of girls $= 392 + 8 = 400$
$\Rightarrow$ now, let no. of newly admitted boys be = $x$
$\therefore$ new no. of boys = $280 + x$
$\Rightarrow$ and, given new ratio = $3 : 4$
so, we have the following equation:
$\Rightarrow \dfrac{280 + x}{400}$ = $\dfrac{3}{4}$
$\Rightarrow 4(280 + x) = 3(400)
$\Rightarrow 1120 + 4x = 1200
$\Rightarrow 4x = 1200 – 1120
$\Rightarrow 4x = 80
$\Rightarrow x = \dfrac{80}{4} = 20$
$\therefore$ no. of newly admitted boys = 20 Answer
$\Rightarrow$ and, total no. of students = 672
$\Rightarrow$ let no. of boys and girls be = $5x$ and $7x$
$\Rightarrow 5x + 7x = 672$
$\Rightarrow 12x = 672$
$\Rightarrow x = \dfrac{672}{12} = 56$
$\Rightarrow$ now, total no. of boys in school = $5\times56 = 280$
$\Rightarrow$ and, total no. of girls in school = $7\times56 = 392$
$\Rightarrow$ given, upon new admission no. of girls increased by 8,
$\therefore$ new no. of girls $= 392 + 8 = 400$
$\Rightarrow$ now, let no. of newly admitted boys be = $x$
$\therefore$ new no. of boys = $280 + x$
$\Rightarrow$ and, given new ratio = $3 : 4$
so, we have the following equation:
$\Rightarrow \dfrac{280 + x}{400}$ = $\dfrac{3}{4}$
$\Rightarrow 4(280 + x) = 3(400)
$\Rightarrow 1120 + 4x = 1200
$\Rightarrow 4x = 1200 – 1120
$\Rightarrow 4x = 80
$\Rightarrow x = \dfrac{80}{4} = 20$
$\therefore$ no. of newly admitted boys = 20 Answer
Q5.The weights of Mr.Sharma and Mrs.Sharma are in the ratio 7 : 8 and their combined (total) is 120 kg. after taking a dieting course, the weight of Mr.Gupta reduces by 6 kg. and the ratio between their weights changes to 5 : 6. Find the reduction of weight of Mrs. Gupta due to this dieting course.
Show Answer
$\Rightarrow$ given, weight ratio = $7 : 8$
$\Rightarrow$ and, total combined weight = 120 kg
$\Rightarrow$ let individual weights be = $7x$ and $8x$
$\Rightarrow 7x + 8x = 120$
$\Rightarrow 15x = 120$
$\Rightarrow x = \dfrac{120}{15} = 8$
$\Rightarrow$ Now, weight of Mr. Gupta = $7\times8 = 56$
$\Rightarrow$ and, weight of Mrs. Gupta = $8\times8 = 64$
$\Rightarrow$ given, after dieting Mr. Gupta lost 6 kg of weight,
$\therefore$ new weight of Mr. Gupta $= 56 – 6 = 50$ kg
$\Rightarrow$ now, let weight reduced by Mrs. Gupta be = $x$ kg
$\therefore$ new weight of Mrs. Gupta = $64 – x$
$\Rightarrow$ and, given new ratio = $5 : 6$
so, we have the following equation:
$\Rightarrow \dfrac{50}{646 – x} = \dfrac{5}{6}$
$\Rightarrow 6(50) = 5(64 – x)$
$\Rightarrow 300 = 320 – 5x$
$\Rightarrow 5x = 320 – 300$
$\Rightarrow 5x = 20$
$\Rightarrow x = \dfrac{20}{5} = 4$
$\therefore$ weight reduced by Mrs. Gupta = 4 kg. Answer
$\Rightarrow$ and, total combined weight = 120 kg
$\Rightarrow$ let individual weights be = $7x$ and $8x$
$\Rightarrow 7x + 8x = 120$
$\Rightarrow 15x = 120$
$\Rightarrow x = \dfrac{120}{15} = 8$
$\Rightarrow$ Now, weight of Mr. Gupta = $7\times8 = 56$
$\Rightarrow$ and, weight of Mrs. Gupta = $8\times8 = 64$
$\Rightarrow$ given, after dieting Mr. Gupta lost 6 kg of weight,
$\therefore$ new weight of Mr. Gupta $= 56 – 6 = 50$ kg
$\Rightarrow$ now, let weight reduced by Mrs. Gupta be = $x$ kg
$\therefore$ new weight of Mrs. Gupta = $64 – x$
$\Rightarrow$ and, given new ratio = $5 : 6$
so, we have the following equation:
$\Rightarrow \dfrac{50}{646 – x} = \dfrac{5}{6}$
$\Rightarrow 6(50) = 5(64 – x)$
$\Rightarrow 300 = 320 – 5x$
$\Rightarrow 5x = 320 – 300$
$\Rightarrow 5x = 20$
$\Rightarrow x = \dfrac{20}{5} = 4$
$\therefore$ weight reduced by Mrs. Gupta = 4 kg. Answer
Q6. If A : B = 3 : 4 and B : C = 6 : 7, find:
(i) A : B : C
(ii) A : C
Show Answer
(i) $\Rightarrow \dfrac{A}{B} = \dfrac{3}{4} : 1 and \dfrac{C}{B} = 1 : \dfrac{7}{6}$
$\Rightarrow$ LCM of 4 and 6 = 12
$\Rightarrow$ \dfrac{3}{4} \times 12 : 1 \times 12 : \dfrac{7}{6}$
$\Rightarrow 9 : 12 : 14$
$\therefore$ A : B : C = $9 : 12 : 14$ Answer
(ii) $\Rightarrow \dfrac{A}{B} \times \dfrac{B}{C} = \dfrac{3}{4} \times \dfrac{6}{7}= \dfrac{9}{14}$
$\therefore$ A : C = 9 : 14 Answer
$\Rightarrow$ LCM of 4 and 6 = 12
$\Rightarrow$ \dfrac{3}{4} \times 12 : 1 \times 12 : \dfrac{7}{6}$
$\Rightarrow 9 : 12 : 14$
$\therefore$ A : B : C = $9 : 12 : 14$ Answer
(ii) $\Rightarrow \dfrac{A}{B} \times \dfrac{B}{C} = \dfrac{3}{4} \times \dfrac{6}{7}= \dfrac{9}{14}$
$\therefore$ A : C = 9 : 14 Answer
Q7. If A : B = 4 : 9 and A : C = 2 : 3, find:
(i) B : C
(ii) A : B : C
Show Answer
(i) $\Rightarrow \dfrac{B}{A} = \dfrac{9}{4}$ and $\dfrac{A}{C} = \dfrac{2}{3}$
$\Rightarrow \dfrac{9}{4} \times \dfrac{2}{3} = \dfrac{3}{2}$
$\therefore$ B : C = 3 : 2 Answer
(ii) $\Rightarrow \dfrac{B}{A} = \dfrac{9}{4} : 1$ and 1 : $\dfrac{3}{2}$
$\Rightarrow$ LCM of 4 and 2 = 4
$\Rightarrow$ $\dfrac{9}{4} \times 4 : 1 \times 4 : \dfrac{3}{2} \times 4$
$\Rightarrow$ 9 : 4 : 6
$\therefore$ A : B : C = 9 : 4 : 6 Answer
$\Rightarrow \dfrac{9}{4} \times \dfrac{2}{3} = \dfrac{3}{2}$
$\therefore$ B : C = 3 : 2 Answer
(ii) $\Rightarrow \dfrac{B}{A} = \dfrac{9}{4} : 1$ and 1 : $\dfrac{3}{2}$
$\Rightarrow$ LCM of 4 and 2 = 4
$\Rightarrow$ $\dfrac{9}{4} \times 4 : 1 \times 4 : \dfrac{3}{2} \times 4$
$\Rightarrow$ 9 : 4 : 6
$\therefore$ A : B : C = 9 : 4 : 6 Answer
Q8. If A : C = 5 : 8 and B : C = 5 : 6, find:
(i) A : B
(ii) A : B : C
Show Answer
(i) $\Rightarrow \dfrac{A}{C} = \dfrac{5}{8}$ and $\dfrac{C}{B} = \dfrac{6}{5}$
$\Rightarrow \dfrac{5}{8} \times \dfrac{6}{5} = \dfrac{3}{4}$
$\therefore$ A : B = 3 : 4 Answer
(ii) $\Rightarrow \dfrac{A}{C} = \dfrac{5}{8} : 1$ and $\dfrac{B}{C} = 1 : \dfrac{6}{5}$
$\Rightarrow$ LCM of 8 and 6 = 24
$\Rightarrow \dfrac{5}{8} \times 24 : 1 \times 24 : \dfrac{5}{6} \times 24$
$\Rightarrow 15 : 24 : 20$
$\Rightarrow$ A : C : B = 15 : 24 : 20
$\therefore$ A : B : C = 15 : 20 : 24 Answer
$\Rightarrow \dfrac{5}{8} \times \dfrac{6}{5} = \dfrac{3}{4}$
$\therefore$ A : B = 3 : 4 Answer
(ii) $\Rightarrow \dfrac{A}{C} = \dfrac{5}{8} : 1$ and $\dfrac{B}{C} = 1 : \dfrac{6}{5}$
$\Rightarrow$ LCM of 8 and 6 = 24
$\Rightarrow \dfrac{5}{8} \times 24 : 1 \times 24 : \dfrac{5}{6} \times 24$
$\Rightarrow 15 : 24 : 20$
$\Rightarrow$ A : C : B = 15 : 24 : 20
$\therefore$ A : B : C = 15 : 20 : 24 Answer
Q9. If 3A = 4B = 6C, find A : B : C.:
Show Answer
$\Rightarrow$ Let 3A = 4B = 6C = N
$\therefore$A = $\dfrac{N}{3} ; B = \dfrac{N}{4} ; C = \dfrac{N}{6}$
$\Rightarrow$ LCM = 3, 4, 6 = 12
$\Rightarrow \dfrac{N}{3} \times 12 : \dfrac{N}{4} \times 12 : \dfrac{N}{6} \times 12$
$\Rightarrow$ 4N : 3 N : 2N
$\therefore$ A : B : C = 4 : 3 : 2 Answer
$\therefore$A = $\dfrac{N}{3} ; B = \dfrac{N}{4} ; C = \dfrac{N}{6}$
$\Rightarrow$ LCM = 3, 4, 6 = 12
$\Rightarrow \dfrac{N}{3} \times 12 : \dfrac{N}{4} \times 12 : \dfrac{N}{6} \times 12$
$\Rightarrow$ 4N : 3 N : 2N
$\therefore$ A : B : C = 4 : 3 : 2 Answer
Q10. Divide 282 into two parts such that the eight part of the first and the fifth part of the second are in the ratio 4 : 3.
Show Answer
$\Rightarrow$ Let first part be = x
$\therefore$ Second part = $282 – x$
$\Rightarrow$ Given, eighth part of first part = $\dfrac{x}{8}$
$\Rightarrow$ and, fifth part of second part = $\dfrac{282 - x}{5}$
$\Rightarrow$ Given ratio = $4 : 3$
$\Rightarrow$ So, we have the following equation:
$\Rightarrow \dfrac{\dfrac{x}{8}}{\dfrac{282 - x}{5}} = \dfrac{4}{3}$
$\Rightarrow \dfrac{5x}{8(282 - x)} = \dfrac{4}{3}$
$\Rightarrow 15x = 32(282 - x)$
$\Rightarrow 15x = 9024 - 32x$
$\Rightarrow 15x + 32x = 9024$
$\Rightarrow 47x = 9024$
$\Rightarrow x = \dfrac{9024}{47} = 192$
$\therefore$ First part = 192
$\Rightarrow$ And, second part = $282 - 192 = 90$ Answer
$\therefore$ Second part = $282 – x$
$\Rightarrow$ Given, eighth part of first part = $\dfrac{x}{8}$
$\Rightarrow$ and, fifth part of second part = $\dfrac{282 - x}{5}$
$\Rightarrow$ Given ratio = $4 : 3$
$\Rightarrow$ So, we have the following equation:
$\Rightarrow \dfrac{\dfrac{x}{8}}{\dfrac{282 - x}{5}} = \dfrac{4}{3}$
$\Rightarrow \dfrac{5x}{8(282 - x)} = \dfrac{4}{3}$
$\Rightarrow 15x = 32(282 - x)$
$\Rightarrow 15x = 9024 - 32x$
$\Rightarrow 15x + 32x = 9024$
$\Rightarrow 47x = 9024$
$\Rightarrow x = \dfrac{9024}{47} = 192$
$\therefore$ First part = 192
$\Rightarrow$ And, second part = $282 - 192 = 90$ Answer
Q11. A school has 1260 students. The ratio of the number of girls to the number of boys is 2 : 3. If 180 new students join the school, the ratio changes to 5 : 7. How many new boys joined the school?
Show Answer
$\Rightarrow$ Given no. of total students = 1260
$\Rightarrow$ and ratio = $2 : 3$
$\Rightarrow$ sum of ratio $= 2 + 3 = 5$
$\therefore$ no. of girls = $\dfrac{2}{5} \times 1260 = 2 \times 252 = 504$
$\Rightarrow$ and, no. of boys = $\dfrac{3}{5} \times 1260 = 3 \times 252 = 756$
$\Rightarrow$ given, new no. of students that joined the school = 180
$\therefore$ New total no. of students $= 1260 + 180 = 1440$
$\Rightarrow$ given, new ratio = $5 : 7$
$\Rightarrow$ sum of ratio $= 5 + 7 = 12$
$\Rightarrow$ new total no. of girl students = $\dfrac{5}{12} \times 1440 = 5\times120 = 600$
$\Rightarrow$ and, new no. of total boy students $= \dfrac{7}{12} \times 1440$ = 7\times120 = 840$
$\therefore$ No. of new girl students $= 600 – 504 = 96$
$\Rightarrow$ No. of new boy students $= 840 – 756 = 84$ Answer
$\Rightarrow$ and ratio = $2 : 3$
$\Rightarrow$ sum of ratio $= 2 + 3 = 5$
$\therefore$ no. of girls = $\dfrac{2}{5} \times 1260 = 2 \times 252 = 504$
$\Rightarrow$ and, no. of boys = $\dfrac{3}{5} \times 1260 = 3 \times 252 = 756$
$\Rightarrow$ given, new no. of students that joined the school = 180
$\therefore$ New total no. of students $= 1260 + 180 = 1440$
$\Rightarrow$ given, new ratio = $5 : 7$
$\Rightarrow$ sum of ratio $= 5 + 7 = 12$
$\Rightarrow$ new total no. of girl students = $\dfrac{5}{12} \times 1440 = 5\times120 = 600$
$\Rightarrow$ and, new no. of total boy students $= \dfrac{7}{12} \times 1440$ = 7\times120 = 840$
$\therefore$ No. of new girl students $= 600 – 504 = 96$
$\Rightarrow$ No. of new boy students $= 840 – 756 = 84$ Answer
Q12. Divide Rs.4,864 into three parts such that the second part is five times the first and the ratio of the second part to the third part is 3 : 4.
Show Answer
$\Rightarrow$ Let 3 parts be = A : B : C
$\Rightarrow$ Given, second part is five times the first part:
$\therefore$ if A = 1 then B = 5
$\Rightarrow$ Given, the ratio of second part to third part is = 3 : 4
$\therefore$ B : C = 3 : 4
$\Rightarrow$ So, we have A : B = 1 : 5 and B : C = 3 : 4
$\Rightarrow$ By making B common, LCM of 3 and 5 = 15
$\therefore \dfrac{A}{B} = \dfrac{1}{5} : 1$ and $\dfrac{C}{B} = 1 : \dfrac{4}{3}$
$\Rightarrow$ $\dfrac{1}{5} \times 15 : 1\times15 : \dfrac{4}{3} \times 15$
$\Rightarrow$ A : B : C = 3 : 15 : 20
$\Rightarrow$ sum of ratio $= 3 + 15 + 20 = 38$
$\therefore$ A’s share = $\dfrac{3}{38} \times 4864 = 3\times128$ = Rs.384
$\Rightarrow$ B’s share = $\dfrac{15}{38} \times 4864 = 15\times128$ = Rs.1920
$\Rightarrow$ C’s share = $\dfrac{20}{38} \times 4864 = 20\times128$ = Rs. 2560 Answer
$\Rightarrow$ Given, second part is five times the first part:
$\therefore$ if A = 1 then B = 5
$\Rightarrow$ Given, the ratio of second part to third part is = 3 : 4
$\therefore$ B : C = 3 : 4
$\Rightarrow$ So, we have A : B = 1 : 5 and B : C = 3 : 4
$\Rightarrow$ By making B common, LCM of 3 and 5 = 15
$\therefore \dfrac{A}{B} = \dfrac{1}{5} : 1$ and $\dfrac{C}{B} = 1 : \dfrac{4}{3}$
$\Rightarrow$ $\dfrac{1}{5} \times 15 : 1\times15 : \dfrac{4}{3} \times 15$
$\Rightarrow$ A : B : C = 3 : 15 : 20
$\Rightarrow$ sum of ratio $= 3 + 15 + 20 = 38$
$\therefore$ A’s share = $\dfrac{3}{38} \times 4864 = 3\times128$ = Rs.384
$\Rightarrow$ B’s share = $\dfrac{15}{38} \times 4864 = 15\times128$ = Rs.1920
$\Rightarrow$ C’s share = $\dfrac{20}{38} \times 4864 = 20\times128$ = Rs. 2560 Answer
Q13. Find he value of $x$, if:
(i) $3.5 : 5.6 :: 1.25 : x$
(ii) $x : 3.5 :: 1.8 : 2.8$
Show Answer
(i) $\Rightarrow \dfrac{3.5}{5.6} = \dfrac{1.25}{x}$
$\Rightarrow 3.5x = 5.6 \times 1.25$
$\Rightarrow x = \dfrac{5.6 \times 1.25}{3.5}$
$\Rightarrow x = \dfrac{8 \times 1.25}{5} = 8 \times 0.25 = 2.00$
$\therefore x = 2$ Answer
(ii) $\Rightarrow \dfrac{x}{3.5} = \dfrac{1.8}{2.8}$
$\Rightarrow 2.8x = 1.8 \times 3.5$
$\Rightarrow x = \dfrac{1.8 \times 3.5}{2.8}$
$\Rightarrow x = \dfrac{18 \times 35 \times 10}{28 \times 10 \times 10}$
$\Rightarrow x = \dfrac{9 \times 7}{14 \times 2}$
$\Rightarrow x = \dfrac{9}{2 \times 2}$
$\therefore x = \dfrac{9}{4} = 2.252.25$ Answer
$\Rightarrow 3.5x = 5.6 \times 1.25$
$\Rightarrow x = \dfrac{5.6 \times 1.25}{3.5}$
$\Rightarrow x = \dfrac{8 \times 1.25}{5} = 8 \times 0.25 = 2.00$
$\therefore x = 2$ Answer
(ii) $\Rightarrow \dfrac{x}{3.5} = \dfrac{1.8}{2.8}$
$\Rightarrow 2.8x = 1.8 \times 3.5$
$\Rightarrow x = \dfrac{1.8 \times 3.5}{2.8}$
$\Rightarrow x = \dfrac{18 \times 35 \times 10}{28 \times 10 \times 10}$
$\Rightarrow x = \dfrac{9 \times 7}{14 \times 2}$
$\Rightarrow x = \dfrac{9}{2 \times 2}$
$\therefore x = \dfrac{9}{4} = 2.252.25$ Answer
Q14. Find the fourth proportional to:
(i) 5, 7 and 8
(ii) 1.2, 3.8 and 9
(iii) $2\dfrac{1}{2}, 3\dfrac{1}{4}$ and $3\dfrac{1}{3}$
Show Answer
(i) $\Rightarrow 5 : 7 :: 8 : x$
$\Rightarrow \dfrac{5}{7} = \dfrac{8}{x}$
$\Rightarrow 5x = 7 \times 8$
$\Rightarrow x = \dfrac{56}{5}$
$\therefore x = 11.2$ Answer
(ii) $\Rightarrow 1.2 : 3.8 :: 9 : x$
$\Rightarrow \dfrac{1.2}{3.8} = \dfrac{9}{x}$
$\Rightarrow 1.2x = 3.8 \times 9$
$\Rightarrow x = \dfrac{3.8 \times 9}{1.2}$
$\Rightarrow x = \dfrac{38 \times 9 \times 10}{12 \times 10}$
$\Rightarrow x = \dfrac{38 \times 3}{4}$
$\Rightarrow x = \dfrac{19 \times 3}{2}$
$\Rightarrow x = \dfrac{57}{2}$
$\therefore x = 28.5$ Answer
(iii) $\Rightarrow \dfrac{5}{2} : \dfrac{5}{4} :: \dfrac{10}{3} : x$
$\Rightarrow \dfrac{\dfrac{5}{2}}{\dfrac{5}{4}} :: \dfrac{\dfrac{10}{3}}{x}$
$\Rightarrow \dfrac{5 \times 4}{5 \times 2} :: \dfrac{10}{3x}$
$\Rightarrow 2 = \dfrac{10}{3x}$
$\Rightarrow 2 \times 3x = 10$
$\Rightarrow 6x = 10$
$\Rightarrow x = \dfrac{10}{6}$
$\Rightarrow x = \dfrac{5}{3}$
$\therefore x = 1\dfrac{2}{3}$ Answer
$\Rightarrow \dfrac{5}{7} = \dfrac{8}{x}$
$\Rightarrow 5x = 7 \times 8$
$\Rightarrow x = \dfrac{56}{5}$
$\therefore x = 11.2$ Answer
(ii) $\Rightarrow 1.2 : 3.8 :: 9 : x$
$\Rightarrow \dfrac{1.2}{3.8} = \dfrac{9}{x}$
$\Rightarrow 1.2x = 3.8 \times 9$
$\Rightarrow x = \dfrac{3.8 \times 9}{1.2}$
$\Rightarrow x = \dfrac{38 \times 9 \times 10}{12 \times 10}$
$\Rightarrow x = \dfrac{38 \times 3}{4}$
$\Rightarrow x = \dfrac{19 \times 3}{2}$
$\Rightarrow x = \dfrac{57}{2}$
$\therefore x = 28.5$ Answer
(iii) $\Rightarrow \dfrac{5}{2} : \dfrac{5}{4} :: \dfrac{10}{3} : x$
$\Rightarrow \dfrac{\dfrac{5}{2}}{\dfrac{5}{4}} :: \dfrac{\dfrac{10}{3}}{x}$
$\Rightarrow \dfrac{5 \times 4}{5 \times 2} :: \dfrac{10}{3x}$
$\Rightarrow 2 = \dfrac{10}{3x}$
$\Rightarrow 2 \times 3x = 10$
$\Rightarrow 6x = 10$
$\Rightarrow x = \dfrac{10}{6}$
$\Rightarrow x = \dfrac{5}{3}$
$\therefore x = 1\dfrac{2}{3}$ Answer
Q15. Find the mean proportiono between:
(i) 2 and 8
(ii) 12 and 192
(iii) 0.2 and 0.8
Show Answer
(i) $\Rightarrow$ 2 and 8
$\Rightarrow 2 : x :: x : 8$
$\Rightarrow \dfrac{2}{x} = \dfrac{x}{8}$
$\Rightarrow 2 \times 8 = x^2$
$\Rightarrow x^2 = 16$
$\Rightarrow x = \sqrt{16}$
$\therefore x = 4$ Answer
(ii) $\Rightarrow$ 12 and 192
$\Rightarrow 12 : x :: x : 192$
$\Rightarrow \dfrac{12}{x} = \dfrac{x}{192}$
$\Rightarrow 12 \times 192 = x^2$
$\Rightarrow x^2 = 2304$
$\Rightarrow x = \sqrt{2304}$
$\therefore x = 48$ Answer
(iii) $\Rightarrow$ 0.2 and 0.8
$\Rightarrow 0.2 : x :: x : 0.8$
$\Rightarrow \dfrac{0.2}{x} = \dfrac{x}{0.8}$
$\Rightarrow 0.2 \times 0.8 = x^2$
$\Rightarrow x^2 = 0.16$
$\Rightarrow x = \sqrt{0.16}$
$\therefore x = 0.4$ Answer
$\Rightarrow 2 : x :: x : 8$
$\Rightarrow \dfrac{2}{x} = \dfrac{x}{8}$
$\Rightarrow 2 \times 8 = x^2$
$\Rightarrow x^2 = 16$
$\Rightarrow x = \sqrt{16}$
$\therefore x = 4$ Answer
(ii) $\Rightarrow$ 12 and 192
$\Rightarrow 12 : x :: x : 192$
$\Rightarrow \dfrac{12}{x} = \dfrac{x}{192}$
$\Rightarrow 12 \times 192 = x^2$
$\Rightarrow x^2 = 2304$
$\Rightarrow x = \sqrt{2304}$
$\therefore x = 48$ Answer
(iii) $\Rightarrow$ 0.2 and 0.8
$\Rightarrow 0.2 : x :: x : 0.8$
$\Rightarrow \dfrac{0.2}{x} = \dfrac{x}{0.8}$
$\Rightarrow 0.2 \times 0.8 = x^2$
$\Rightarrow x^2 = 0.16$
$\Rightarrow x = \sqrt{0.16}$
$\therefore x = 0.4$ Answer