RS Aggarwal Class 8 Mathematics Solutions Chapter 9 Percentage

Definition

Percent means "per one hundred." It refers to a ratio expressed out of 100.

Percentage Formula

\( \frac{x}{n} \times 100 = p \)
where:
x = given quantity
n = total amount
p = percentage of the quantity compared to the total

Percentage Increase

\( \text{Percentage increase} = \frac{\text{actual increase}}{\text{original amount}} \times 100\% \)

Percentage Decrease

\( \text{Percentage decrease} = \frac{\text{actual decrease}}{\text{original amount}} \times 100\% \)

Percent - Decimal - Fraction Conversions

50% = 0.50 = \( \frac{50}{100} \)

Question 1. Convert to a fraction in lowest terms.
(i) 48%
(ii) 220%
(iii) 2.5%
Answer:
(i) 48% = \( \frac{48}{100} = \frac{12}{25} \)
(ii) 220% = \( \frac{220}{100} = \frac{11}{5} \)
(iii) 2.5% = \( \frac{2.5}{100} = \frac{25}{1000} = \frac{1}{40} \)

Exam Tip: Always simplify fractions to their lowest form by dividing both numerator and denominator by their greatest common factor.

Question 2. Convert to decimal form.
(i) 6%
(ii) 72%
(iii) 125%
Answer:
(i) 6% = \( \frac{6}{100} = 0.06 \)
(ii) 72% = \( \frac{72}{100} = 0.72 \)
(iii) 125% = \( \frac{125}{100} = 1.25 \)

Exam Tip: To convert a percentage to a decimal, divide by 100 - this is equivalent to moving the decimal point two places to the left.

Question 3. Convert to a percentage.
(i) \( \frac{9}{25} \)
(ii) \( \frac{3}{125} \)
(iii) \( \frac{12}{5} \)
Answer:
(i) \( \frac{9}{25} = \left(\frac{9}{25} \times 100\right)\% = (9 \times 4)\% = 36\% \)
(ii) \( \frac{3}{125} = \left(\frac{3}{125} \times 100\right)\% = 2.4\% \)
(iii) \( \frac{12}{5} = \left(\frac{12}{5} \times 100\right)\% = 240\% \)

Exam Tip: To convert any fraction to a percentage, multiply it by 100 and add the % symbol - this shows how many parts out of every 100 the fraction represents.

Question 4. Express 4:5 as a percentage.
Answer: 4:5 = \( \frac{4}{5} = \left(\frac{4}{5} \times 100\right)\% = 80\% \)

Exam Tip: A ratio can be converted to a percentage by writing it as a fraction, then multiplying by 100.

Question 5. Express 125% as a ratio.
Answer: 125% = \( \frac{125}{100} = \frac{5}{4} \) = 5:4

Exam Tip: Convert the percentage to a fraction first, simplify it, then express it as a ratio using the numerator and denominator.

Question 6. Which is the largest: 6\(\frac{2}{3}\)%, \(\frac{3}{20}\), or 0.14?
Answer: We have: 6\(\frac{2}{3}\)% = \(\frac{20}{3}\)% = \( \left(\frac{20}{3} \times \frac{1}{100}\right) = \frac{1}{15} = 0.067 \)
Also, \( \frac{3}{20} = 0.15 \)
The third number is 0.14. Clearly, 0.15 is the largest, so \( \frac{3}{20} \) is the largest.

Exam Tip: When comparing different forms - percentages, fractions, and decimals - convert all to the same form (usually decimal) to compare easily.

Question 7. Find the required percentage.
(i) 64 out of 150
(ii) 4 out of 5 × 1000
(iii) 12.5 out of 2 × 1000
Answer:
(i) Required percentage = \( \left(\frac{64}{150} \times 100\right)\% = 64\% \)
(ii) Required percentage = \( \left(\frac{300}{5 \times 1000} \times 100\right)\% = 4\% \)
(iii) Required percentage = \( \left(\frac{250}{2 \times 1000} \times 100\right)\% = 12.5\% \)

Exam Tip: Always arrange the numbers as (part/whole) × 100 to find the percentage - this shows what fraction of the whole the part represents.

Question 8. If 4\(\frac{1}{4}\)% of a sum equals Rs 162, find the sum.
Answer: 4\(\frac{1}{4}\)% = \( \frac{9}{200} \)
\( \therefore \frac{9}{200} \) of Rs 3600 = \( \frac{9}{200} \times 3600 = Rs 162 \)

Exam Tip: When a percentage of an unknown amount is given, set up an equation where (percentage/100) × (unknown amount) = (given value), then solve.

Question 9. If 16% of a number is 72, what is the number?
Answer: Let the number be x.
16% of x is 72.
\( \Rightarrow \frac{16}{100} \times x = 72 \)
\( \Rightarrow 16x = 7200 \)
\( \Rightarrow 16x = 7200 \)
\( \Rightarrow x = \frac{7200}{16} = 450 \)
\( \therefore \) The required number is 450.

Exam Tip: To find a number when its percentage is given, divide the given value by the percentage (expressed as a decimal or fraction).

Question 11. If a man's monthly savings are 18% of his monthly income and his savings equal Rs 1890, find his monthly income.
Answer: Let Rs x be his monthly income.
His savings = 18% of Rs x
\( = Rs \left(x \times \frac{18}{100}\right) = Rs \frac{9x}{50} \)
Now, \( \frac{9x}{50} = 1890 \)
\( \Rightarrow x = Rs \left(1890 \times \frac{50}{9}\right) \)
\( \Rightarrow x = Rs 10500 \)
\( \therefore \) His monthly income is Rs 10,500.

Exam Tip: Set up an equation where the percentage of the unknown equals the known amount, then solve for the unknown by multiplying by the reciprocal of the percentage.

Question 12. In a series of games, a team won 35% of all matches played. If the team won 7 games, how many total games were played?
Answer: Let z be the total number of games played.
Percentage of games won = 35% of z
\( = \left(z \times \frac{35}{100}\right) = \frac{35z}{100} \)
Now, \( \frac{35z}{100} = 7 \)
\( \Rightarrow z = \left(7 \times \frac{100}{35}\right) \)
\( \Rightarrow z = 20 \)
\( \therefore \) The total number of games played is 20.

Exam Tip: When you know the percentage and the actual count, divide the count by the percentage (as a decimal) to find the total.

Question 13. Amit's salary increased by a certain percentage and is now Rs 15,300. If his salary was Rs x before the increase, and the percentage increase was 25%, find his previous salary.
Answer: Let Rs x be Amit's old salary.
His salary after increment will be Rs \( \left(x + \frac{25}{100}x\right) \)
According to the question, we have:
\( \Rightarrow x + \frac{25}{100}x = 15300 \)
\( \Rightarrow \frac{100x + 25x}{100} = 15300 \) (LCM = 100)
\( \Rightarrow \frac{125x}{100} = 15300 \)
\( \Rightarrow 125x = 1530000 \)
\( \Rightarrow x = \frac{15300 \times 100}{125} = 12750 \)
\( \therefore \) The old salary is Rs 12,750.

Exam Tip: For percentage increase problems, express the new amount as (original + percentage increase of original), then solve the resulting equation.

Question 14. B's income is how much more than A's income in percentage terms, if A's income is Rs 80 and B's income is Rs 100?
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 = \( \frac{(100 - 80)}{80} \times 100\% = \frac{20}{80} \times 100\% = 25\% \)
\( = Rs 125 \)
\( \therefore \) B's income is more than that of A's by (125 - 100)%, i.e., 25%.

Exam Tip: To compare two quantities as a percentage, find their difference, divide by the reference quantity (usually the smaller or original), and multiply by 100.

Question 16. If the cost of 1 unit of petrol was originally Rs 100 and is now Rs 110, by how much should consumption be reduced so that the expense remains the same?
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 \( \left(\frac{1}{110} \times 100\right) \), i.e., \( \frac{10}{11} \) units of petrol.
Now, reduction in consumption = \( \left(1 - \frac{10}{11}\right) = \frac{1}{11} \) unit
Percentage of reduction = \( \left(\frac{1}{11} \times \frac{1}{1} \times 100\right)\% = 9\frac{1}{11}\% \)
\( \therefore \) A motorist must reduce the consumption of petrol by \( 9\frac{1}{11}\% \).

Exam Tip: For such problems, assume a fixed total amount and show how changing prices affects quantity when spending stays constant.

Question 17. The population of a town increased by 8% in one year. If the current population is 54,000, what was the population one year ago?
Answer: Let x be the population of the town a year ago. Then, present population = 108% of x
\( = \left(x \times \frac{108}{100}\right) = \frac{27x}{25} \)
Now, \( \frac{27x}{25} = 54000 \)
\( \Rightarrow x = \left(54000 \times \frac{25}{27}\right) \)
\( \Rightarrow x = 50000 \)
Hence, the population of the town a year ago was 50,000.

Exam Tip: When a percentage increase is involved, multiply the original amount by (100 + increase percentage)/100 to get the new amount.

Question 18. A machine was worth Rs x last year. Now its value is 80% of what it was. If the current value is Rs 160,000, find its value last year.
Answer: Let Rs x be the value of the machine last year.
Then, present value = 80% of Rs x
\( = Rs \left(x \times \frac{80}{100}\right) = Rs \frac{4x}{5} \)
Now, \( \frac{4x}{5} = 160000 \)
\( \Rightarrow x = \left(160000 \times \frac{5}{4}\right) \)
\( = x = 40000 \times 5 = 200000 \)
Hence, the value of the machine last year was Rs 2,00,000.

Exam Tip: For depreciation problems, multiply the original value by (100 - decrease percentage)/100 to find the reduced value.

Question 19. An alloy contains 40% copper, 32% nickel, and the rest is zinc. Find the mass of zinc in 1 kg of alloy.
Answer: Mass of the alloy = 1 kg
Percentage of copper = 40%
Percentage of nickel = 32%
Percentage of zinc = (100 - (40 + 32))% = 28%
\( \therefore \) Mass of zinc in 1 kg of alloy = \( \left(\frac{28}{100} \times 1\right) \) kg = 0.28 kg = 0.28 × 1000 g = 280 g

Exam Tip: When several percentages make up a total, find the missing percentage by subtracting all given percentages from 100.

Question 20. A food item contains 12% protein, 25% fat, and 63% carbohydrate. If the total mass is 2600 grams, find the mass of each component.
Answer: Amount of protein = 12% of 2600
\( = \left(2600 \times \frac{12}{100}\right) = 312 \) cal
Amount of fat = 25% of 2600
\( = \left(2600 \times \frac{25}{100}\right) = 650 \) cal
Amount of carbohydrate = 63% of 2600
\( = \left(2600 \times \frac{63}{100}\right) = 1638 \) cal

Exam Tip: When finding percentages of a total, multiply the percentage (as a fraction or decimal) by the total amount.

Question 21. Gunpowder contains 75% nitre and 10% sulphur. Find the quantities of these materials in 12 kg of gunpowder.
Answer: Let x be the amount of gunpowder.
Amount of nitre = 75%
Let z kg be the amount of gunpowder containing 9 kg of nitre.
i.e., (75% of x) = 9 kg
\( \Rightarrow \left(x \times \frac{75}{100}\right) = 9 \)
\( \Rightarrow \frac{75x}{100} = 9 \)
\( \Rightarrow x = \left(9 \times \frac{100}{75}\right) \)
\( \Rightarrow x = 12 \) kg
Hence, 12 kg of gunpowder contains 9 kg of nitre.
Now, amount of sulphur = 10%
Let z kg be the amount of gunpowder containing 2.5 kg of sulphur.
i.e., (10% of x) = 2.5 kg
\( \Rightarrow \left(x \times \frac{10}{100}\right) = 2.5 \)
\( \Rightarrow \frac{10x}{100} = 2.5 \)
\( \Rightarrow \frac{x}{10} = 2.5 \)
\( \Rightarrow x = (2.5 \times 10) \)
\( \Rightarrow x = 25 \) kg
Hence, 25 kg of gunpowder contains 2.5 kg of sulphur.

Exam Tip: Use the percentage composition to set up equations where the percentage of a component equals a known mass.

Question 22. Money was distributed among A, B, and C such that C receives Rs x. B gets 50% of x, and A gets 25% of what B receives. If the total is Rs 7000, find each person's share.
Answer: Let Rs x be the amount 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, z + (50% of Rs x) + (25% of Rs x) = Rs 7000
\( \Rightarrow x + \left(x \times \frac{50}{100}\right) + \left(x \times \frac{25}{100}\right) = Rs 7000 \)
\( \Rightarrow x + \frac{50x}{100} + \frac{25x}{100} = Rs 7000 \)
\( \Rightarrow \left(x + \frac{50x + 25x}{100}\right) = Rs 7000 \)
\( \Rightarrow \frac{175x}{100} = Rs 7000 \)
\( \Rightarrow x = Rs \left(7000 \times \frac{100}{175}\right) \)
\( \Rightarrow x = Rs 4000 \)
\( \therefore \) C gets Rs 4000.
Amount of money B gets = (50% of Rs 4000) = (50% of Rs 4000)
\( = Rs \left(4000 \times \frac{50}{100}\right) = Rs 2000 \)
Amount of money A gets = (25% of Rs x) = (25% of Rs 4000)
\( = Rs \left(4000 \times \frac{25}{100}\right) = Rs 1000 \)

Exam Tip: When money or items are divided in percentages, express each person's share as a percentage of a variable, then add them to equal the total.

Question 23. 22 carat gold contains 22 parts pure gold out of 24 parts. Also, 24 carat gold is given to be 100% pure. Find the percentage of pure gold in 22 carat gold.
Answer: 22 carat gold contains 22 parts pure gold out of 24 parts.
Also, 24 carat gold is given to be 100% pure.
\( \therefore \) Percentage of pure gold in 22 carat gold = \( \left(\frac{22}{24} \times 100\right)\% = 91\frac{2}{3}\% \)
Hence, 22 carat gold contains \( 91\frac{2}{3}\% \) of pure gold.

Exam Tip: To find the percentage purity or composition, express the desired component as a fraction of the total, then multiply by 100.

Question 24. A salary was increased by 25%. To restore the original salary, by what percentage must the new salary be reduced?
Answer: Let the original salary be Rs 100
Then, after increment of 25% the salary becomes = \( 100\left(1 + \frac{25}{100}\right) = 100\left(\frac{125}{100}\right) = Rs 125 \)
To restore the original salary, let the new salary be decreased by x%.
Thus, we get
\( 125\left(1 - \frac{x}{100}\right) = 100 \)
\( \Rightarrow \left(1 - \frac{x}{100}\right) = \frac{100}{125} = \frac{4}{5} \)
\( \Rightarrow \frac{x}{100} = \frac{1}{5} \)
\( \Rightarrow x = 20\% \)
Therefore, the new salary must be reduced by 20% to restore the original salary

Exam Tip: When restoring to an original amount after a percentage change, the required percentage decrease is not the same as the initial increase - it must be calculated based on the new (larger) amount.

Exercise 9B

Question 1. Convert \( \frac{3}{5} \) to a percentage.
Answer: \( \frac{3}{5} = \left(\frac{3}{5} \times 100\right)\% = 60\% \)

Exam Tip: Multiply the fraction by 100 to convert it directly to a percentage.

Question 3. Convert 6:5 to a percentage.
Answer: (c) 120%
\( 6:5 = \frac{6}{5} = \left(\frac{6}{5} \times 100\right)\% = 120\% \)

Exam Tip: Express the ratio as a fraction first, then multiply by 100 to get the percentage.

Question 4. If 5% of a number is 9, find the number.
Answer: (d) 180
Let x be the required number. Then, we have:
5% of x = 9
\( \Rightarrow \left(x \times \frac{5}{100}\right) = 9 \)
\( \Rightarrow \frac{5x}{100} = 9 \)
\( \Rightarrow x = \left(9 \times \frac{100}{5}\right) \)
\( \Rightarrow x = 180 \)

Exam Tip: To find a number when its percentage is given, divide the given value by the percentage (in decimal form).

Question 6. What is 4 \( \frac{1}{6} \) as a percentage?
Answer: (c) \( 133\frac{1}{3}\% \)
Required percentage = \( \left(\frac{25}{30} \times 100\right)\% = 133\frac{1}{3}\% \)

Exam Tip: Convert mixed numbers to improper fractions, then apply the percentage formula.

Question 7. If 40% of a number equals 240, what is the number?
Answer: (b) 600
Let the required number be x. Then, we have:
40% of x = 240
\( \Rightarrow \left(x \times \frac{40}{100}\right) = 240 \)
\( \Rightarrow \frac{40x}{100} = 240 \)
\( \Rightarrow x = \left(240 \times \frac{100}{40}\right) \)
\( \Rightarrow x = 600 \)

Exam Tip: Always isolate the variable by dividing both sides by the percentage coefficient.

Question 8. If x% of 400 equals 60, find x.
Answer: (c) 15
Let the required number be x. Then, we have:
x% of 400 = 60
\( \Rightarrow \left(400 \times \frac{x}{100}\right) = 60 \)
\( \Rightarrow \frac{400x}{100} = 60 \)
\( \Rightarrow 4x = 60 \)
\( \Rightarrow x = \frac{60}{4} = 15 \)

Exam Tip: When finding the percentage itself, set up the equation and solve for the percentage variable directly.

Question 9. (180% of x):2 = 504. Find x.
Answer: (d) 560
Let the required number be x. Then, we have:
(180% of x):2 = 504
\( \Rightarrow \left(x \times \frac{180}{100}\right) \div 2 = 504 \)
\( \Rightarrow \left(\frac{180x}{100}\right) \div 2 = 504 \)
\( \Rightarrow \left(\frac{180x}{100} \times \frac{1}{2}\right) = 504 \)
\( \Rightarrow \frac{90x}{100} = 504 \)
\( \Rightarrow x = \left(504 \times \frac{10}{9}\right) \)
\( \Rightarrow x = 560 \)

Exam Tip: Break down compound expressions step by step, applying operations in the correct order.

Question 10. Find 20% of Rs 800.
Answer: (a) Rs 160
20% of Rs 800 = \( Rs \left(800 \times \frac{20}{100}\right) = Rs 160 \)

Exam Tip: To find a percentage of any amount, multiply the amount by the percentage (in fraction or decimal form).

Question 12. If 56% of a number is 98, what is the number?
Answer: (c) 175
Let the maximum marks be x. Then, we have:
56% of x = \( \left(x \times \frac{56}{100}\right) = \frac{56x}{100} \)
Now, \( \frac{56x}{100} = 98 \)
\( \Rightarrow x = \left(98 \times \frac{100}{56}\right) \)
\( \Rightarrow x = 175 \)

Exam Tip: When a percentage of a number is known, divide the known value by the percentage to find the number.

Question 13. A number increased by 10% and then reduced by 10%. What is the net change?
Answer: (b) decrease by 1%
Let z 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 \( \frac{90}{100}\left(\frac{11}{10}x\right) = \frac{990}{1000} = \frac{99}{100}x \)
Difference = \( x - \frac{99}{100}x = \frac{1}{100}x \)
Percentage of decrease = \( \frac{\frac{1}{100}x}{x} \times 100 = 1\% \)

Exam Tip: When a quantity undergoes successive percentage changes, apply each change sequentially to the result of the previous change.

Question 14. If 65% of the examinees passed, and 35% failed, how many total examinees were there if 420 failed?
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)
\( = \left(x \times \frac{35}{100}\right) = \frac{35x}{100} \)
Now, \( \frac{35x}{100} = 420 \)
\( \Rightarrow x = \left(420 \times \frac{100}{35}\right) \)
\( \Rightarrow x = 1200 \)

Exam Tip: When complementary percentages are given (such as pass/fail), use either one to set up the equation - both will yield the same total.

Question 15. If 20% of x + 40 = x, find x.
Answer: (a) 50
Let x be the required number. Then, we have:
20% of x + 40 = x
\( \Rightarrow \left(x \times \frac{20}{100}\right) + 40 = x \)
\( \Rightarrow \frac{20x}{100} + 40 = x \)
\( \Rightarrow \left(\frac{20x}{100} - x\right) = -40 \)
\( \Rightarrow \frac{-80x}{100} = -40 \)
\( \Rightarrow x = \left(40 \times \frac{100}{80}\right) \)
\( \Rightarrow x = 50 \)

Exam Tip: Rearrange such equations to collect all terms with the variable on one side, then solve.

Question 17. If x - (27\(\frac{1}{9}\)% of x) = 87, find x.
Answer: (c) 120
Let the required number be x. Then, we have:
\( x - \left(27\frac{1}{9}\% \text{ of } x\right) = 87 \)
\( \Rightarrow x - \left(\frac{35}{9}\% \text{ of } x\right) = 87 \)
\( \Rightarrow x - \left(x \times \frac{35}{9} \times \frac{1}{100}\right) = 87 \)
\( \Rightarrow x - \frac{11x}{40} = 87 \)
\( \Rightarrow \frac{29x}{40} = 87 \)
\( \Rightarrow x = \left(87 \times \frac{40}{29}\right) \)
\( \Rightarrow x = 120 \)

Exam Tip: Convert fractional percentages to improper fractions before performing calculations.

Question 18. Find the required percentage: \( \left(\frac{0.08}{30} \times 100\right)\% \)
Answer: (d) 300%
Required percentage = \( \left(\frac{1296}{5} \times \frac{1}{13} \times 100\right)\% = 300\% \)

Exam Tip: Simplify fractions and decimals before multiplying by 100 to find the percentage.

Question 19. If x% of y = y% of z, find z.
Answer: (a) x
Let the required number be z. Then, we have:
x% of y = y% of z
\( \Rightarrow \left(y \times \frac{x}{100}\right) = \left(z \times \frac{y}{100}\right) \)
\( \Rightarrow \frac{xy}{100} = \frac{zy}{100} \)
\( \Rightarrow z = \left(\frac{xy}{100} \times \frac{100}{y}\right) \)
\( \Rightarrow z = x \)

Exam Tip: When dealing with percentage equations involving variables, simplify by expressing each percentage algebraically, then solve.

Question 20. Find the required percentage: \( \left(\frac{1}{35} \times \frac{5}{2} \times 100\right)\% \)
Answer: (a) x
Required percentage = \( \left(\frac{1}{35} \times \frac{5}{2} \times 100\right)\% = 10\% \)

Exam Tip: Multiply the fractions first, then multiply by 100 to obtain the final percentage.

Exercise 9C

Question 1. Convert to a fraction or decimal as indicated.
(i) 24% to fraction
(ii) 105% to decimal
(iii) Express 4:5 as a percentage
(iv) 56% to ratio
Answer:
(i) 24% = \( \frac{24}{100} = \frac{6}{25} \)
(ii) 105% = \( \frac{105}{100} = 1.05 \)
(iii) 4:5 = \( \left(\frac{4}{5} \times 100\right)\% = 80\% \)
(iv) 56% = \( \frac{56}{100} = \frac{14}{25} \) = 14:25

Exam Tip: Always simplify fractions to their lowest terms and verify that conversions between forms are mathematically equivalent.

Question 2. If 34% of a number is 85, find the number.
Answer: Let the required number be x. Then, we have:
(34% of x) = 85
\( \Rightarrow \left(x \times \frac{34}{100}\right) = 85 \)
\( \Rightarrow \frac{34x}{100} = 85 \)
\( \Rightarrow x = \left(85 \times \frac{100}{34}\right) \)
\( \Rightarrow x = 250 \) Hence, the required number is 250.

Exam Tip: To find a number when its percentage is known, divide the known value by the percentage expressed as a decimal.

Question 3. The value of a machine last year was Rs x. Its present value is 90% of Rs x. If the present value is Rs 54,000, find its value last year.
Answer: Let the value of the machine last year be Rs x.
Then, its present value = 90% of Rs x
\( = Rs \left(x \times \frac{90}{100}\right) = Rs \frac{9x}{10} \)
Now, \( \frac{9x}{10} = 54000 \)
\( \Rightarrow x = \left(54000 \times \frac{10}{9}\right) \)
\( \Rightarrow x = Rs 60000 \)
Hence, the value of the machine last year was Rs 60,000.

Exam Tip: For depreciation problems, use the formula: new value = original value × (100 - decrease %)/100, then solve for the original value.

Question 4. An alloy consists of 30% copper, 42% nickel, and the rest is zinc. Find the mass of zinc in 1 kg of alloy.
Answer: Percentage of copper = 30%
Percentage of nickel = 42%
Percentage of zinc = (100 - (30 + 42))% = 28%
\( \therefore \) Mass of zinc in 1 kg of the alloy = \( \left(\frac{28}{100} \times 1\right) \) kg = 0.28 kg = 280 g

Exam Tip: When percentages of a whole are given, find the missing percentage by subtracting all known percentages from 100.

Question 5. Of the total students in a school, 60% are boys. If there are 14 girls, how many students are in the school?
Answer: Let the total number of students be x. Then, we have:
Percentage of boys = 60%
Percentage of girls = 40%
\( \therefore \) Number of girls = 40% of x
\( = \left(x \times \frac{40}{100}\right) = \frac{40x}{100} \)
Now, \( \frac{40x}{100} = 14 \)
\( \Rightarrow x = \left(14 \times \frac{100}{40}\right) \)
\( \Rightarrow x = 35 \)
\( \therefore \) Total number of students = 35

Exam Tip: Use complementary percentages (e.g., boys and girls must sum to 100%) to solve problems where one group's percentage is known.

Question 6.
Answer: We have: 8\(\frac{1}{3}\)% = \( \frac{25}{3}\)% = \( \left(\frac{25}{3} \times \frac{1}{100}\right) = \frac{1}{12} = 0.083 \)
Also, \( \frac{3}{20} = 0.15 \)
The third number is 0.15. Clearly, 0.16 is the largest.
i.e., \( \frac{3}{20} \) is the largest.

Exam Tip: Convert all values to the same form (usually decimals) before comparing different types of numbers.

Question 7. Find the required percentage: \( \left(\frac{1}{6} \times \frac{3}{2} \times 100\right)\% \)
Answer: (d) 10%
Required percentage = \( \left(\frac{1}{6} \times \frac{3}{2} \times 100\right)\% = 10\% \)

Exam Tip: Simplify products of fractions before multiplying by 100.

Question 8. If x - (30% of x) = 84, find x.
Answer: (c) 120
Let the required number be x
x - (30% of x) = 84
\( \Rightarrow x - \left(x \times \frac{30}{100}\right) = 84 \)
\( \Rightarrow x - \frac{30x}{100} = 84 \)
\( \Rightarrow \frac{70x}{100} = 84 \)
\( \Rightarrow x = \left(84 \times \frac{100}{70}\right) \)
\( \Rightarrow x = 120 \)

Exam Tip: When subtracting a percentage from a number, express it as (number - percentage of number) = given value, then solve.

Question 9. If x% of 320 = 48, find x.
Answer: (b) 15%
Let the required number be x. Then, we have:
(x% of 320) = 48
\( \Rightarrow \left(320 \times \frac{x}{100}\right) = 48 \)
\( \Rightarrow \frac{320x}{100} = 48 \)
\( \Rightarrow x = \left(48 \times \frac{100}{320}\right) \)
\( \Rightarrow x = 15\% \)

Exam Tip: To find the percentage value, set up the equation and isolate the percentage variable by cross-multiplication if needed.

Question 10. The required percentage is \( \left(\frac{54}{45} \times 100\right)\% \)
Answer: (d) 120%
Required percentage = \( \left(\frac{54}{45} \times 100\right)\% = 120\% \)

Exam Tip: Simplify fractions to their lowest terms before converting to percentages.

Question 11. If (25% of z) + 60 = x, find x.
Answer: (c) 80
Let the required number be x. Then, we have:
(25% of x) + 60 = x
\( \Rightarrow \left(x \times \frac{25}{100}\right) + 60 = x \)
\( \Rightarrow \frac{25x}{100} + 60 = x \)
\( \Rightarrow \left(\frac{25x}{100} - x\right) = -60 \)
\( \Rightarrow \frac{-75x}{100} = -60 \)
\( \Rightarrow x = \left(60 \times \frac{100}{75}\right) \)
\( \Rightarrow x = 80 \)

Exam Tip: Rearrange the equation to collect the variable on one side, simplify, and solve by isolating the variable.

Question 12. If (5% of x) = 12, find x.
Answer: (c) 240
Let the required number be x. Then, we have:
(5% of x) = 12
\( \Rightarrow \left(x \times \frac{5}{100}\right) = 12 \)
\( \Rightarrow \frac{5x}{100} = 12 \)
\( \Rightarrow x = \left(12 \times \frac{100}{5}\right) \)

Exam Tip: Solve percentage equations by expressing the percentage as a fraction and solving the resulting linear equation.

Question 13.
Answer:
(i) 7\(\frac{1}{2}\)% of Rs 1200 = \( \left(\frac{15}{2}\% \text{ of } Rs 1200\right) = Rs \left(\frac{15}{2} \times \frac{1}{100} \times 1200\right) = Rs 90 \)
Hence, 7\(\frac{1}{2}\)% of Rs 1200 = Rs 90
(ii) Required percentage = \( \left(\frac{240}{3 \times 1000} \times 100\right)\% = 8\% \)
Hence, 240 ml is 8% of 3 L.
(iii) (x% of 35) = 42
\( \Rightarrow \left(35 \times \frac{x}{100}\right) = 42 \)
\( \Rightarrow \frac{35x}{100} = 42 \)
\( \Rightarrow x = \left(42 \times \frac{100}{35}\right) \)
\( \Rightarrow x = 120\% \)
\( \therefore \) If x% of 35 is 42, then x = 120%.
(iv) \( \left(\frac{12}{5} \times 100\right)\% = 240\% \)
Hence, \( \frac{12}{5} = 240\% \)
(v) Let the required number be x. Then, we have:
120 = x% of 80
\( \Rightarrow \left(80 \times \frac{x}{100}\right) = 120 \)
\( \Rightarrow \frac{80x}{100} = 120 \)
\( \Rightarrow x = \left(120 \times \frac{100}{80}\right) \)
\( \Rightarrow x = 150\% \)
\( \therefore \) 120 = 150% of 80

Exam Tip: Always set up equations with clear identification of what is being asked - whether finding a percentage of a number, finding the percentage itself, or finding the original number.

Question 14.
Answer:
(i) 6% of 8 = \( \left(8 \times \frac{6}{100}\right) = 0.48 \)
Hence, it is false.
(ii) 6:5 = \( \frac{6}{5} = \left(\frac{6}{5} \times 100\right)\% = 120\% \)
Hence, it is false.
(iii) \( \frac{3}{5} = \left(\frac{3}{5} \times 100\right)\% = 60\% \)
Hence, it is true.
(iv) 6 hours = \( \left(\frac{6}{24} \times 100\right)\% = 25\% \)
Hence, it is true.

Exam Tip: Verify each statement by converting to a common form and comparing - this prevents calculation errors in true/false problems.