myhelper: RS Aggarwal solution class 8 chapter 9 Percentage Exercise 9A

Exercise 9A

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Q1 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 1:

Express each of the following as a fraction:
(i) 48%
(ii) 220%
(iii) 2.5%

Answer 1:

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

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

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

Q2 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 2:

Express each of the following as a decimal:
(i) 6%
(ii) 72%
(iii) 125%

Answer 2:

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

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

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

Q3 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 3:

Express each of the following as a percentage:
(i) $\frac{9}{25}$
(ii) $\frac{3}{125}$
(iii) $\frac{12}{5}$

Answer 3:

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

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

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

Q4 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 4:

Convert the ratio 4 : 5 to percentage.

Answer 4:

$4 : 5$

$= \frac{4}{5}$

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

$= 80 %$

Q5 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 5:

Express 125% as a ratio.

Answer 5:

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

Q6 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 6:

Which is largest in $6 \frac{2}{3} %, \frac{3}{20}$ and 0.14?

Answer 6:

$We have : 6 \frac{2}{3} % = \frac{20}{3} %$

$= (\frac{20}{3} \times \frac{1}{100})$

$= \frac{1}{15}$

$= 0.06$
$Also, \frac{3}{20} = 0.15$
$The third number is 0.14 . Clearly, 0.15 is the largest .$

$Hence, \frac{3}{20} is the largest .$

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Q7 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 7:

(i) What per cent of 150 is 96?
(ii) What per cent of 5 kg is 200 g?
(iii) What per cent of 2 litres is 250 mL?

Answer 7:

$(i) Required percentage$

$= (\frac{96}{150} \times 100) %$

$= 64 %$

$(ii) Required percentage$

$= (\frac{200}{5 \times 1000} \times 100) %$

$= 4 %$

$(iii) Required percentage$

$= (\frac{250}{2 \times 1000} \times 100) %$

$= 12.5 %$

Q8 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 8:

Find $4 \frac{1}{2} %$ of Rs 3600.

Answer 8:

$4 \frac{1}{2} % = \frac{9}{2 \times 100}$

$∴ \frac{9}{200} of Rs 3600$

$= \frac{9}{200} \times 3600 = Rs 162$

Q9 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 9:

If 16% of a number is 72, find the number.

Answer 9:

$Let the number be x . 16 % of x is 72 .$

$\Rightarrow \frac{16}{100} \times x = 72$

$\Rightarrow 16 x = 72 \times 100$

$\Rightarrow 16 x = 7200$

$\Rightarrow x = \frac{7200}{16} = 450$

$∴ The required number is 450 .$

Q10 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 10:

A man saves 18% of his monthly income. If he saves Rs 1890 per month, what is his monthly income?

Answer 10:

$Let Rs x be his monthly income .$

$His savings = 18 % of Rs x$

$= Rs (x \times \frac{18}{100}) = Rs \frac{9 x}{50}$
$Now, \frac{9 x}{50} = 1890$

$\Rightarrow x = Rs (1890 \times \frac{50}{9})$

$\Rightarrow x = Rs 10500$

$∴ His monthly income is Rs . 10500 .$

Q11 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 11:

A football team wins 7 games, which is 35% of the total games played. How many games were played in all?

Answer 11:

$Let x be the total number of games played .$

$Percentage of games won = 35 % of x$
$= (x \times \frac{35}{100}) = \frac{35 x}{100}$
$Now, \frac{35 x}{100} = 7$

$\Rightarrow x = (7 \times \frac{100}{35})$

$\Rightarrow x = 20$

$∴ The total number games played is 20 .$

Q12 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 12:

Amit was given an increment of 20% on his salary. If his new salary is Rs 15300, what was his salary before the increment?

Answer 12:

$Let R s x be Amit' s old salary .$

$His salary after increment will be Rs (x + \frac{20}{100} x)$

$According to the question, we have :$

$\Rightarrow x + \frac{20}{100} x = 15300$

$\Rightarrow \frac{100 x + 20 x}{100} = 15300 (LCM = 100)$

$\Rightarrow \frac{120 x}{100} = 15300$

$\Rightarrow 120 x = 15300 \times 100$

$\Rightarrow x = \frac{15300 \times 100}{120}$

$\Rightarrow x = 12750$

$∴ The old salary is Rs 12, 750 .$

Q13 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 13:

cAnswer 13:

$Let x be the number of days the school was opened .$

$Number of days Sonal attended school = 204 days$

$Percentage of her attendance = 85 % of x$
$= (x \times \frac{85}{100}) = \frac{85 x}{100}$
$Now, \frac{85 x}{100} = 204$

$\Rightarrow x = (204 \times \frac{100}{85})$

$\Rightarrow x = 240$

$∴ The school was opened for 240 day .$

Q14 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 14:

A's income is 20% less than that of B. By what per cent is B's income more than A's?

Answer 14:

$Let B' s income be Rs 100 Then, A' s income = Rs 80$

$Therefore, B' s income is more than A' s income by$

$= \frac{(100 - 80)}{80} \times 100 %$

$= \frac{20}{80} \times 100 % = 25 %$

$= R s 125$
$∴ B' s income is more than that of A' s by$

$(125 - 100) %, i . e ., 25 % .$

Q15 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 15:

The price of petrol goes up by 10%. By how much per cent must a motorist reduce the consumption of petrol so that the expenditure on it remains unchanged?

Answer 15:

$Let the consumption of petrol originally be 1 unit and$

$let its cost be Rs 100 . New cost of 1 unit of petrol = Rs 110$

$Now, Rs 110 will yield 1 unit of petrol . i . e .,$

$Rs 100 will yield (\frac{1}{110} \times 100),$

$i . e ., \frac{10}{11} units of petrol .$

$Now, reduction in consumption$

$= (1 - \frac{10}{11})$ $= \frac{1}{11} unit$
$Percentage of reduction$

$= (\frac{1}{11} \times \frac{1}{1} \times 100) %$ $= 9 \frac{1}{11} %$

$∴ A motorist must reduce the consumption of petrol by 9 \frac{1}{11} % .$

Q16 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 16:

The population of a town increases by 8% annually. If the present population is 54000, what was it a year ago?

Answer 16:

$Let x be the population of the town a year ago .$

$Then, present population = 108 % of x$
$= (x \times \frac{108}{100}) = \frac{27 x}{25}$
$Now, \frac{27 x}{25} = 54000$

$\Rightarrow x = (54000 \times \frac{25}{27})$

$\Rightarrow x = 50000$

Hence, the population of the town a year ago was 50000.

Q17 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 17:

The value of a machine depreciates every year by 20%. If the present value of the machine be Rs 160000, what was its value last year?

Answer 17:

$Let Rs x be the value of the machine last year .$

$Then, present value = 80 % of Rs x$
$= Rs (x \times \frac{80}{100}) = Rs \frac{4 x}{5}$
$Now, \frac{4 x}{5} = 160000$

$\Rightarrow x = (160000 \times \frac{5}{4})$

$\Rightarrow x = 40000 \times 5 = 200000$

Hence, the value of the machine last year was Rs 2,00,000.

Q18 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 18:

An alloy contains 40% copper, 32% nickel and rest zinc. Find the mass of zinc in one kg of the alloy.

Answer 18:

$Mass of the alloy = 1 kg$

$Percentage of copper = 40 % Percentage of nickel = 32 %$

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

Q19 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 19:

Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in his food daily, find in calories the amount of each of these in his daily food intake.

Answer 19:

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

Q20 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

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Question 20:

Gunpowder contains 75% nitre and 10% sulphur. Find the amount of gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.5 kg sulphur?

Answer 20:

$Let x be the amount of gunpowder . Amount of nitre = 75 %$

Let x kg be the amount of gunpowder containing 9 kg of nitre.

i.e., 75% of x=9 kg

$\Rightarrow (x \times \frac{75}{100}) = 9$

$\Rightarrow \frac{75 x}{100} = 9$

$\Rightarrow x = (9 \times \frac{100}{75})$

$\Rightarrow x = 12 kg$
$Hence, 12 kg of gunpowder contains 9 kg of nitre .$

$Now, amount of sulphur = 10 %$

$Let x kg be the amount of gunpowder containing 2.5 kg of sulphur$

$. i . e ., (10 % of x) = 2.5 kg$
$\Rightarrow (x \times \frac{10}{100}) = 2.5$

$\Rightarrow \frac{10 x}{100} = 2.5$

$\Rightarrow \frac{x}{10} = 2.5$

$\Rightarrow x = (2.5 \times 10)$

$\Rightarrow x = 25 kg$
$Hence, 25 kg of gunpowder contains 2.5 kg of sulphur .$

Q21 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

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Question 21:

Divide Rs 7000 among A, B and C such that A gets 50% of what B gets and B gets 50% of what C gets.

Answer 21:

$Let Rs x be the amount of money received by C .$

$Then, amount of money B gets = (50 % of Rs x)$

$Amount of money A gets = (50 % of B) = (25 % of Rs x)$

$Now, x + (50 % of Rs x) + (25 % of Rs x) = Rs 7000$

$\Rightarrow x + (x \times \frac{50}{100}) + (x \times \frac{25}{100}) = Rs 7000$

$\Rightarrow x + \frac{50 x}{100} + \frac{25 x}{100} = Rs 7000$

$\Rightarrow (x + \frac{50 x}{100} + \frac{25 x}{100}) = Rs 7000$

$\Rightarrow \frac{175 x}{100} = Rs 7000$

$\Rightarrow x = Rs (7000 \times \frac{100}{175})$

$\Rightarrow x = Rs 4000$
$∴ C gets Rs 4000 . Amount of money B gets = (50 % of Rs x)$
$= (50 % of Rs 4000) = Rs (4000 \times \frac{50}{100}) = Rs 2000$

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

Q22 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

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Question 22:

Find the percentage of pure gold in 22-carat gold, if 24-carat gold is 100% pure.

Answer 22:

$22 carat gold contains 22 parts pure gold out of 24 parts .$

$Also, 24 carat gold is given to be 100 % pure .$

$∴ Percentage of pure gold in 22 carat gold = (\frac{22}{24} \times 100) %$
$= 91 \frac{2}{3} %$
$Hence, 22 carat gold contains 91 \frac{2}{3} % of pure gold .$

Q23 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

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Question 23:

The salary of an officer is increased by 25%. By what per cent should the new salary be decreased to restore the original salary?

Answer 23:

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.

RS Aggarwal solution class 8 chapter 9 Percentage Exercise 9A

Exercise 9A

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Q1 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

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Answer 1:

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