Rs Aggarwal 2020 2021 for Class 8 Maths Chapter 9 - Percentage

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Rs Aggarwal 2020 2021 Solutions for Class 8 Maths Chapter 9 Percentage are provided here with simple step-by-step explanations. These solutions for Percentage are extremely popular among Class 8 students for Maths Percentage Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 8 Maths Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

Page No 121:

Answer:

$(i) 48 % = \frac{48}{100} = \frac{12}{25}$

$(i i) 220 % = \frac{220}{100} = \frac{11}{5}$

$(i i i) 2.5 % = \frac{2.5}{100} = \frac{25}{1000} = \frac{1}{40}$

Page No 121:

Answer:

$(i) 6 % = \frac{6}{100} = 0.06$

$(i i) 72 % = \frac{72}{100} = 0.72$

$(i i i) 125 % = \frac{125}{100} = 1.25$

Page No 121:

Answer:

$(i) \frac{9}{25} = (\frac{9}{25} \times 100) % = (9 \times 4) % = 36 %$

$(ii) \frac{3}{125} = (\frac{3}{125} \times 100) % = 2.4 %$

$(iii) \frac{12}{5} = (\frac{12}{5} \times 100) % = 240 %$

Page No 121:

Answer:

$4 : 5 = \frac{4}{5} = (\frac{4}{5} \times 100) % = 80 %$

Page No 121:

Answer:

$125 % = \frac{125}{100} = \frac{5}{4} = 5 : 4$

Page No 121:

Answer:

   $We have : 6 \frac{2}{3} % = \frac{20}{3} % = (\frac{20}{3} \times \frac{1}{100}) = \frac{1}{15} = 0.06$
   $Also, \frac{3}{20} = 0.15$
   $The third number is 0.14 . Clearly, 0.15 is the largest . Hence, \frac{3}{20} is the largest .$

Page No 122:

Answer:

$(i) Required percentage = (\frac{96}{150} \times 100) % = 64 %$

$(ii) Required percentage = (\frac{200}{5 \times 1000} \times 100) % = 4 %$

$(iii) Required percentage = (\frac{250}{2 \times 1000} \times 100) % = 12.5 %$

Page No 122:

Answer:

$4 \frac{1}{2} % = \frac{9}{2 \times 100} ∴ \frac{9}{200} of Rs 3600 = \frac{9}{200} \times 3600 = Rs 162$

Page No 122:

Answer:

$Let the number be x . 16 % of x is 72 . \Rightarrow \frac{16}{100} \times x = 72 \Rightarrow 16 x = 72 \times 100 \Rightarrow 16 x = 7200 \Rightarrow x = \frac{7200}{16} = 450 ∴ The required number is 450 .$

Page No 122:

Answer:

$Let Rs x be his monthly income . His savings = 18 % of Rs x$

$= Rs (x \times \frac{18}{100}) = Rs \frac{9 x}{50}$
$Now, \frac{9 x}{50} = 3780 \Rightarrow x = Rs (3780 \times \frac{50}{9}) \Rightarrow x = Rs 21000 ∴ His monthly income is Rs . 21000 .$

Page No 122:

Answer:

$Let x be the total number of games played . Percentage of games won = 35 % of x$
$= (x \times \frac{35}{100}) = \frac{35 x}{100}$
$Now, \frac{35 x}{100} = 7 \Rightarrow x = (7 \times \frac{100}{35}) \Rightarrow x = 20 ∴ The total number games played is 20 .$

Page No 122:

Answer:

$Let R s x be Amit' s old salary . His salary after increment will be Rs (x + \frac{20}{100} x) According to the question, we have : \Rightarrow x + \frac{20}{100} x = 30600 \Rightarrow \frac{100 x + 20 x}{100} = 30600 (LCM = 100) \Rightarrow \frac{120 x}{100} = 30600 \Rightarrow 120 x = 30600 \times 100 \Rightarrow x = \frac{30600 \times 100}{120} \Rightarrow x = 25500 ∴ The old salary is Rs 25500 .$

Page No 122:

Answer:

$Let x be the number of days the school was opened . Number of days Sonal attended school = 204 days Percentage of her attendance = 85 % of x$
$= (x \times \frac{85}{100}) = \frac{85 x}{100}$
$Now, \frac{85 x}{100} = 204 \Rightarrow x = (204 \times \frac{100}{85}) \Rightarrow x = 240 ∴ The school was opened for 240 day .$

Page No 122:

Answer:

$Let B' s income be Rs 100 Then, A' s income = Rs 80 Therefore, B' s income is more than A' s income by = \frac{(100 - 80)}{80} \times 100 % = \frac{20}{80} \times 100 % = 25 %$
$= R s 125$
$∴ B' s income is more than that of A' s by (125 - 100) %, i . e ., 25 % .$

Page No 122:

Answer:

$Let the consumption of petrol originally be 1 unit and let its cost be Rs 100 . New cost of 1 unit of petrol = Rs 110 Now, Rs 110 will yield 1 unit of petrol . i . e ., Rs 100 will yield (\frac{1}{110} \times 100), i . e ., \frac{10}{11} units of petrol .$

$Now, reduction in consumption = (1 - \frac{10}{11})$ $= \frac{1}{11} unit$
$Percentage of reduction = (\frac{1}{11} \times \frac{1}{1} \times 100) %$ $= 9 \frac{1}{11} %$
$∴ A motorist must reduce the consumption of petrol by 9 \frac{1}{11} % .$

Page No 122:

Answer:

$Let x be the population of the town a year ago . Then, present population = 108 % of x$
$= (x \times \frac{108}{100}) = \frac{27 x}{25}$
$Now, \frac{27 x}{25} = 54000 \Rightarrow x = (54000 \times \frac{25}{27}) \Rightarrow x = 50000$
$Hence, the population of the town a year ago was 50000 .$

Page No 122:

Answer:

$Let Rs x be the value of the machine last year . Then, present value = 80 % of Rs x$
$= Rs (x \times \frac{80}{100}) = Rs \frac{4 x}{5}$
$Now, \frac{4 x}{5} = 160000 \Rightarrow x = (160000 \times \frac{5}{4}) \Rightarrow x = 40000 \times 5 = 200000$
$Hence, the value of the machine last year was Rs 2, 00, 000 .$

Page No 122:

Answer:

$Mass of the alloy = 1 kg Percentage of copper = 40 % Percentage of nickel = 32 % Percentage of zinc = \{100 - (40 + 32)\} %$
$= 28 %$
$∴ Mass of zinc in 1 kg of alloy = (\frac{28}{100} \times 1) kg$
$= 0.28 kg = 0 . 28 \times 1000 g = 280 g$

Page No 122:

Answer:

$Amount of protein = 12 % of 2600$
   $= (2600 \times \frac{12}{100}) = 312 cal$
$Amount of fat = 25 % of 2600$
   $= (2600 \times \frac{25}{100}) = 650 cal$
$Amount of carbohydrate = 63 % of 2600$
   $= (2600 \times \frac{63}{100}) = 1638 cal$

Page No 122:

Answer:

$Let x be the amount of gunpowder . Amount of nitre = 75 % Let x kg be the amount of gunpowder containing 9 kg of nitre . i . e ., (75 % of x) = 9 kg$
$\Rightarrow (x \times \frac{75}{100}) = 9 \Rightarrow \frac{75 x}{100} = 9 \Rightarrow x = (9 \times \frac{100}{75}) \Rightarrow x = 12 kg$
$Hence, 12 kg of gunpowder contains 9 kg of nitre . Now, amount of sulphur = 10 % Let x kg be the amount of gunpowder containing 2.5 kg of sulphur . i . e ., (10 % of x) = 2.5 kg$
$\Rightarrow (x \times \frac{10}{100}) = 2.5 \Rightarrow \frac{10 x}{100} = 2.5 \Rightarrow \frac{x}{10} = 2.5 \Rightarrow x = (2.5 \times 10) \Rightarrow x = 25 kg$
$Hence, 25 kg of gunpowder contains 2.5 kg of sulphur .$

Page No 122:

Answer:

$Let Rs x be the amount of money recieved by C . Then, amount of money B gets = (50 % of Rs x) Amount of money A gets = (50 % of B) = (25 % of Rs x) Now, x + (50 % of Rs x) + (25 % of Rs x) = Rs 7000 \Rightarrow x + (x \times \frac{50}{100}) + (x \times \frac{25}{100}) = Rs 7000 \Rightarrow x + \frac{50 x}{100} + \frac{25 x}{100} = Rs 7000 \Rightarrow (x + \frac{50 x}{100} + \frac{25 x}{100}) = Rs 7000 \Rightarrow \frac{175 x}{100} = Rs 7000 \Rightarrow x = Rs (7000 \times \frac{100}{175}) \Rightarrow x = Rs 4000$
$∴ C gets Rs 4000 . Amount of money B gets = (50 % of Rs x)$
$= (50 % of Rs 4000) = Rs (4000 \times \frac{50}{100}) = Rs 2000$

$Amount of money A gets = (25 % of Rs x)$
$= (25 % of Rs 4000) = Rs (4000 \times \frac{25}{100}) = Rs 1000$

Page No 122:

Answer:

$22 carat gold contains 22 parts pure gold out of 24 parts . Also, 24 carat gold is given to be 100 % pure . ∴ Percentage of pure gold in 22 carat gold = (\frac{22}{24} \times 100) %$
$= 91 \frac{2}{3} %$
$Hence, 22 carat gold contains 91 \frac{2}{3} % of pure gold .$

Page No 122:

Answer:

Let the original salary be Rs 100
Then, after increment of 25% the salary becomes
= $100 (1 + \frac{25}{100}) = 100 (\frac{125}{100}) = R s 125$

To restore the original salary, let the new salary be decreased by x%.
Thus, we get

$125 (1 - \frac{x}{100}) = 100 \Rightarrow (1 - \frac{x}{100}) = \frac{100}{125} = \frac{4}{5} \Rightarrow \frac{x}{100} = \frac{1}{5} \Rightarrow x = \frac{100}{5} = 20 %$
Therefore, the new salary must be reduced by 20% to restore the original salary.

Page No 123:

Answer:

(d) 60%

$\frac{3}{5} = (\frac{3}{5} \times 100) % = 60 %$

Page No 123:

Answer:

(b) 0.008
$0.8 % = \frac{0.8}{100} = 0.008$

Page No 123:

Answer:

Page No 123:

Answer:

(d) 180

$Let x be the required number . Then, we have : 5 % o f x = 9 \Rightarrow (x \times \frac{5}{100}) = 9 \Rightarrow \frac{5 x}{100} = 9 \Rightarrow x = (9 \times \frac{100}{5}) \Rightarrow x = 180$

Page No 123:

Answer:

Page No 123:

Answer:

(d) 2.5%

$Required percentage = (\frac{250}{(10 \times 1000)} \times 100) % = 2.5 %$

Page No 123:

Answer:

(b) 600

$Let the required number be x . Then, we have : 40 % o f x = 240 \Rightarrow (x \times \frac{40}{100}) = 240 \Rightarrow \frac{40 x}{100} = 240 \Rightarrow x = (240 \times \frac{100}{40}) \Rightarrow x = 600$

Page No 123:

Answer:

(c) 15

$Let the required number be x . Then, we have : x % o f 400 = 60 \Rightarrow (400 \times \frac{x}{100}) = 60 \Rightarrow \frac{400 x}{100} = 60 \Rightarrow 4 x = 60 \Rightarrow x = \frac{60}{4} \Rightarrow x = 15$

Page No 123:

Answer:

(d) 560

$Let the required number be x . Then, we have : (180 % o f x) \div 2 = 504 \Rightarrow (x \times \frac{180}{100}) \div 2 = 504 \Rightarrow (\frac{180 x}{100}) \div 2 = 504 \Rightarrow (\frac{180 x}{100} \times \frac{1}{2}) = 504 \Rightarrow \frac{9 x}{10} = 504 \Rightarrow x = (504 \times \frac{10}{9}) \Rightarrow x = 560$

Page No 123:

Answer:

(a) Rs 160
$20 % o f R s 800 = R s (800 \times \frac{20}{100})$
$= R s 160$

Page No 123:

Answer:

(c) 175

$Let the maximum marks be x . Then, we have : 56 % o f x = (x \times \frac{56}{100})$
$= \frac{56 x}{100}$
$Now, \frac{56 x}{100} = 98 \Rightarrow x = (98 \times \frac{100}{56}) \Rightarrow x = 175$

Page No 123:

Answer:

(b) decrease by 1 %

$Let x be the number . A 10 % increase will give a new number, \frac{110}{100} x = \frac{11}{10} x The number is then reduced by 10 % . The new number will be \frac{90}{100} (\frac{11}{10} x) = \frac{990}{1000} x = \frac{99}{100} x Difference = x - \frac{99}{100} x = \frac{1}{100} x Percentage of decrease = \frac{1}{100} x \times \frac{1}{x} \times 100 = 1 %$

Page No 123:

Answer:

(a) $18 \frac{3}{4} %$

$4 h 30 m i n = (4 \times 60 \times 60) + (30 \times 60)$
$= 16200 s e c$
$24 h = (24 \times 60 \times 60)$
$= 86400 s e c$
$Now, (\frac{16200}{86400} \times 100) % = 18 \frac{3}{4} %$

Page No 123:

Answer:

(c) 1200

$Let x be the total number of examinees . Percentage of the examinees passed = 65 % Percentage of the examinees failed = 35 % Number of the examinees failed = (35 % of x)$
$= (x \times \frac{35}{100}) = \frac{35 x}{100}$
$Now, \frac{35 x}{100} = 420 \Rightarrow x = (420 \times \frac{100}{35}) \Rightarrow x = 1200$

Page No 124:

Answer:

(a) 50

$Let x be the required number . Then, we have : 20 % o f x + 40 = x \Rightarrow (x \times \frac{20}{100}) + 40 = x \Rightarrow \frac{20 x}{100} + 40 = x \Rightarrow (\frac{20 x}{100} - x) = - 40 \Rightarrow \frac{- 80 x}{100} = - 40 \Rightarrow x = (40 \times \frac{100}{80}) \Rightarrow x = 50$

Page No 124:

Answer:

(c) 120

$Let the required number be x . Then, we have : x - (27 \frac{1}{2} % of x) = 87 \Rightarrow x - (\frac{55}{2} % of x) = 87 \Rightarrow x - (x \times \frac{55}{2} \times \frac{1}{100}) = 87 \Rightarrow x - \frac{11 x}{40} = 87 \Rightarrow \frac{29 x}{40} = 87 \Rightarrow x = (87 \times \frac{40}{29}) \Rightarrow x = 120$

Page No 124:

Answer:

Page No 124:

Answer:

(d) 300%

$Required percentage = (\frac{1206}{3} \times \frac{1}{134} \times 100) % = 300 %$

Page No 124:

Answer:

(a) x

$Let the required number be z . Then, we have : x % o f y = y % o f z \Rightarrow (y \times \frac{x}{100}) = (z \times \frac{y}{100}) \Rightarrow \frac{y x}{100} = \frac{z y}{100} \Rightarrow z = (\frac{y x}{100} \times \frac{100}{y}) \Rightarrow z = x$

Page No 124:

Answer:

(a) x
$Required percentage = (\frac{1}{35} \times \frac{7}{2} \times 100) % = 10 %$

Page No 125:

Answer:

$(i) 24 % = \frac{24}{100} = \frac{6}{25}$

$(i i) 105 % = \frac{105}{100} = 1.05 (i i i) 4 : 5 = \frac{4}{5} = (\frac{4}{5} \times 100) % = 80 % (i v) 56 % = \frac{56}{100} = \frac{14}{25} = 14 : 25$

Page No 125:

Answer:

$Let the required number be x . Then, we have : (34 % of x) = 85 \Rightarrow (x \times \frac{34}{100}) = 85 \Rightarrow \frac{34 x}{100} = 85 \Rightarrow x = (85 \times \frac{100}{34}) \Rightarrow x = 250 Hence, the required number is 250 .$

Page No 125:

Answer:

$Let the value of the machine last year be Rs x . Then, its present value = 90 % of Rs x$
$= R s (x \times \frac{90}{100}) = R s \frac{90 x}{100}$
$Now, \frac{90 x}{100} = 54000 \Rightarrow x = (54000 \times \frac{100}{90}) \Rightarrow x = Rs 60000$
$Hence, the value of the machine last year was Rs 60, 000 .$

Page No 125:

Answer:

$Percentage of copper = 30 % Percentage of nickel = 42 % Percentage of zinc = \{100 - (30 + 42)\} %$
      $= 28 %$

$∴ Mass of zinc in 1 kg of the alloy = (\frac{28}{100} \times 1) kg = 0.28 kg = 280 g$

Page No 125:

Answer:

$Let the total number of students be x . Then, we have : Percentage of boys = 60 % Percentage of girls = 40 % ∴ Number of girls = 40 % of x$
$= (x \times \frac{40}{100}) = \frac{40 x}{100}$
$Now, \frac{40 x}{100} = 14 \Rightarrow x = (14 \times \frac{100}{40}) \Rightarrow x = 35$

$∴ Total number of students = 35$

Page No 125:

Answer:

$We have : 8 \frac{1}{3} % = \frac{25}{3} % = (\frac{25}{3} \times \frac{1}{100}) = \frac{1}{12} = 0.083$
$Also, \frac{4}{25} = 0.16 The third number is 0.15 . Clearly, 0.16 is the largest . i . e ., \frac{4}{25} is the largest .$

Page No 125:

Answer:

(d) 10%

$Required percentage = (\frac{1}{45} \times \frac{9}{2} \times 100) % = 10 %$

Page No 125:

Answer:

(c) 120

$Let the required number be x$
$x - (30 % o f x) = 84 \Rightarrow \{x - (x \times \frac{30}{100})\} = 84 \Rightarrow (x - \frac{30 x}{100}) = 84 \Rightarrow \frac{70 x}{100} = 84 \Rightarrow x = (84 \times \frac{100}{70}) \Rightarrow x = 120$

Page No 125:

Answer:

(b) 15%

$Let the required number be x . Then, we have : (x % o f 320) = 48 \Rightarrow (320 \times \frac{x}{100}) = 48 \Rightarrow \frac{320 x}{100} = 48 \Rightarrow x = (48 \times \frac{100}{320}) \Rightarrow x = 15 %$

Page No 125:

Answer:

(d) 120%
$Required percentage = (\frac{54}{45} \times 100) % = 120 %$

Page No 125:

Answer:

(c) 80

$Let the required number be x . Then, we have : (25 % o f x) + 60 = x \Rightarrow (x \times \frac{25}{100}) + 60 = x \Rightarrow \frac{25 x}{100} + 60 = x \Rightarrow (\frac{25 x}{100} - x) = - 60 \Rightarrow \frac{- 75 x}{100} = - 60 \Rightarrow x = (60 \times \frac{100}{75}) \Rightarrow x = 80$

Page No 125:

Answer:

(c) 240

$Let the required number be x . Then, we have : (5 % of x) = 12 \Rightarrow (x \times \frac{5}{100}) = 12 \Rightarrow \frac{5 x}{100} = 12 \Rightarrow x = (12 \times \frac{100}{5}) \Rightarrow x = 240$

Page No 125:

Answer:

$(i) 7 \frac{1}{2} % of Rs 1200 = (\frac{15}{2} % of Rs 1200)$
$= Rs (\frac{15}{2} \times \frac{1}{100} \times 1200) = Rs 90$
$Hence, 7 \frac{1}{2} % of Rs 1200 = Rs 90$

$(ii) Required percentage = (\frac{240}{3 \times 1000} \times 100) % = 8 % Hence, 240 ml is 8 % of 3 L .$

$(iii) (x % of 35) = 42 \Rightarrow (35 \times \frac{x}{100}) = 42 \Rightarrow \frac{35 x}{100} = 42 \Rightarrow x = (42 \times \frac{100}{35}) \Rightarrow x = 120 % ∴ If x % of 35 is 42, then x = 120 % .$

$(iv) (\frac{12}{5} \times 100) % = 240 % Hence, \frac{12}{5} = 240 %$

$(v) Let the required number be x . Then, we have : 120 = x % o f 80 \Rightarrow (80 \times \frac{x}{100}) = 120 \Rightarrow \frac{80 x}{100} = 120 \Rightarrow x = (120 \times \frac{100}{80}) \Rightarrow x = 150 % ∴ 120 = 150 % o f 80$

Page No 126:

Answer:

$(i) 6 % of 8 = (8 \times \frac{6}{100}) = 0.48 Hence, it is false .$

$(ii) 6 : 5 = \frac{6}{5} = (\frac{6}{5} \times 100) % = 120 % Hence, it is false .$

$(iii) \frac{3}{5} = (\frac{3}{5} \times 100) % = 60 % Hence, it is true .$

$(iv) 6 h o u r s = (\frac{6}{24} \times 100) % = 25 %$
Hence, it is true.

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