RS Aggarwal Class 7 Mathematics Solutions Chapter 10 Percentage

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Exercise 10A

Question 1. Convert the following to percentages:
(i) \( \frac{47}{100} \)
(ii) \( \frac{9}{20} \)
(iii) \( \frac{3}{8} \)
(iv) \( \frac{6}{25} \)
(v) \( \frac{19}{500} \)
(vi) \( \frac{1}{6} \)
(vii) \( \frac{2}{3} \)
(viii) \( 1\frac{3}{8} \)
Answer:
(i) \( \frac{47}{100} = \left(\frac{47}{100} \times 100\right)\% = 47\% \)
(ii) \( \frac{9}{20} = \left(\frac{9}{20} \times 100\right)\% = (9 \times 5)\% = 45\% \)
(iii) \( \frac{3}{8} = \left(\frac{3}{8} \times 100\right)\% = \left(\frac{300}{8}\right)\% = \left(\frac{75}{2}\right)\% = 37\frac{1}{2}\% \)
(iv) \( \frac{6}{25} = \left(\frac{6}{25} \times 100\right)\% = \left(\frac{24}{4}\right)\% = \left(\frac{6}{1}\right)\% = 6.4\% \)
(v) \( \frac{19}{500} = \left(\frac{19}{500} \times 100\right)\% = \left(\frac{19}{5}\right)\% = 3.8\% \)
(vi) \( \frac{1}{6} = \left(\frac{1}{6} \times 100\right)\% = \left(\frac{4 \times 20}{3}\right)\% = \left(\frac{50}{3}\right)\% = 20\frac{2}{3}\% \)
(vii) \( \frac{2}{3} = \left(\frac{2}{3} \times 100\right)\% = \left(\frac{200}{3}\right)\% = 66\frac{2}{3}\% \)
(viii) \( 1\frac{3}{8} = \frac{11}{8} = \left(\frac{11}{8} \times 100\right)\% = (8 \times 20)\% = 160\% \)
In simple words: To change any fraction to a percentage, multiply it by 100 and add the % sign.

Exam Tip: Always simplify fractions first before multiplying by 100 - this makes cancellation easier and reduces errors.

Question 2. Find the value of each of the following:
(i) 32% of \( \frac{32}{100} \)
(ii) \( 6\frac{1}{4}\% \) of a number
(iii) \( 26\frac{2}{5}\% \) of a measurement
(iv) 120% of a quantity
(v) 6.25% of an amount
(vi) 0.8% of a value
(vii) 0.06% of another value
(viii) 22.75% of a figure
Answer:
(i) \( 32\% = \left(\frac{32}{100}\right) = \frac{8}{25} \)
(ii) \( 6\frac{1}{4}\% = \left(\frac{25}{4}\right)\% = \left(\frac{25}{4} \times \frac{1}{100}\right) = \frac{1}{16} \)
(iii) \( 26\frac{2}{5}\% = \left(\frac{132}{5}\right)\% = \left(\frac{132}{5} \times \frac{1}{100}\right) = \left(\frac{264}{1000}\right) = \frac{4}{15} \)
(iv) \( 120\% = \left(\frac{120}{100}\right) = \frac{6}{5} = 1\frac{1}{5} \)
(v) \( 6.25\% = \left(\frac{6.25}{100}\right) = \left(\frac{625}{10000}\right) = \left(\frac{625}{10000}\right) = \frac{1}{16} \)
(vi) \( 0.8\% = \left(\frac{0.8}{100}\right) = \left(\frac{8}{1000}\right) = \left(\frac{1}{125}\right) = \frac{1}{125} \)
(vii) \( 0.06\% = \left(\frac{0.06}{100}\right) = \left(\frac{6}{10000}\right) = \left(\frac{3}{5000}\right) = \frac{3}{5000} \)
(viii) \( 22.75\% = \left(\frac{22.75}{100}\right) = \left(\frac{2275}{10000}\right) = \frac{91}{400} \)
In simple words: To turn a percentage into a fraction, write the number over 100 and simplify by cancelling common factors.

Exam Tip: For percentages with decimals, multiply numerator and denominator by 100 to remove the decimal, then reduce to lowest terms.

Question 3. Express each of the following percentages as a ratio:
(i) 43%
(ii) 36%
(iii) 7.5%
(iv) 125%
Answer:
(i) \( 43\% = \frac{43}{100} = 43 : 100 \)
(ii) \( 36\% = \frac{36}{100} = \frac{9}{25} = 9 : 25 \)
(iii) \( 7.5\% = \left(\frac{7.5}{100}\right) = \left(\frac{75}{10 \times 100}\right) = \frac{3}{40} = 3 : 40 \)
(iv) \( 125\% = \frac{125}{100} = \frac{5}{4} = 5 : 4 \)
In simple words: Write the percentage as a fraction with 100 in the denominator, then simplify. The numerator and denominator form the ratio.

Exam Tip: Always simplify the fraction completely before writing the ratio in its lowest terms.

Question 4. Express each of the following as a percentage:
(i) 37 : 100
(ii) 16 : 25
(iii) 3 : 5
(iv) 5 : 4
Answer:
(i) \( 37 : 100 = \frac{37}{100} = \left(\frac{37}{100} \times 100\right)\% = 37\% \)
(ii) \( 16 : 25 = \frac{16}{25} = \left(\frac{16}{25} \times 100\right)\% = (16 \times 4)\% = 64\% \)
(iii) \( 3 : 5 = \frac{3}{5} = \left(\frac{3}{5} \times 100\right)\% = (3 \times 20)\% = 60\% \)
(iv) \( 5 : 4 = \frac{5}{4} = \left(\frac{5}{4} \times 100\right)\% = (5 \times 25)\% = 125\% \)
In simple words: Change the ratio to a fraction, then multiply by 100 and add the % symbol.

Exam Tip: Remember that a ratio greater than 1 (like 5:4) will give a percentage greater than 100%.

Question 5. Find:
(i) 45% of 200
(ii) 127% of 400
(iii) 3.6% of 500
(iv) 0.23% of 1200
Answer:
(i) \( 45\% = \left(\frac{45}{100}\right) = 0.45 \)
(ii) \( 127\% = \left(\frac{127}{100}\right) = 1.27 \)
(iii) \( 3.6\% = \left(\frac{3.6}{100}\right) = \left(\frac{36}{10 \times 100}\right) = \frac{36}{1000} = 0.036 \)
(iv) \( 0.23\% = \left(\frac{0.23}{100}\right) = \left(\frac{23}{100 \times 100}\right) = \frac{23}{10000} = 0.0023 \)
In simple words: To find a percentage of a number, divide the percentage by 100 to get a decimal, then multiply by the number.

Exam Tip: For small percentages (less than 1%), count your zeros carefully when converting to decimal form.

Question 6. Find the percentage increase or decrease:
(i) Let x% of Rs 120 be Rs 15
(ii) Let x% of 2 h be 36 min
(iii) Let x% of 2 days be 8 h
(iv) Let x% of 4 km be 160 m
(v) Let x% of 1 L be 175 mL
(vi) Let x% of Rs 4 be 25 paise
Answer:
(i) Let x% of Rs 120 be Rs 15
Then, \( \text{Rs}\left(\frac{x}{100} \times 120\right) = \text{Rs } 15 \)
\( = \left(\frac{6x}{5}\right) = 15 \)
\( \therefore x = \left(\frac{15 \times 5}{6}\right) = 12.5\% \)
Hence, 12.5% of Rs 120 is Rs 15.

(ii) Let x% of 2 h be 36 min
Then, \( \left(\frac{x}{100} \times 2 \times 60\right) \text{min} = 36 \text{ min} \)
\( = \left(\frac{120x}{100}\right) = 36 \)
\( \therefore x = \left(\frac{36 \times 100}{120}\right)\% = 30\% \)
Hence, 30% of 2 h is 36 min.

(iii) Let x% of 2 days be 8 h
Then, \( \left(\frac{x}{100} \times 2 \times 24\right) \text{h} = 8 \text{ h} \)
\( = \left(\frac{48x}{100}\right) = 8 \)
\( \therefore x = \left(\frac{8 \times 100}{48}\right)\% = 16\frac{2}{3}\% \)
Hence, 16 2/3 % of 2 days is 8 h.

(iv) Let x% of 4 km be 160 m
Then, \( \left(\frac{x}{100} \times 4 \times 1000\right) \text{m} = 160 \text{ m} \)
\( \Rightarrow 40x = 160 \)
\( \therefore x = \left(\frac{160}{40}\right)\% = 4\% \)
Hence, 4% of 4 km is 160 m.

(v) Let x% of 1 L be 175 mL
Then, \( \left(\frac{x}{100} \times 1 \times 1000\right) \text{mL} = 175 \text{ mL} \)
\( = 10x = 175 \)
\( \therefore x = \left(\frac{175}{10}\right)\% = 17.5\% \)
Hence, 17.5% of 1 L is 175 mL.

(vi) Let x% of Rs 4 be 25 paise
Then, \( \left(\frac{x}{100} \times 4 \times 100\right) \text{paise} = 25 \text{paise} \)
\( = 4x = 25 \)
\( \therefore x = \left(\frac{25}{4}\right)\% = 6\frac{1}{4}\% \)
Hence, 6 1/4 % of Rs 4 is 25 paise.
In simple words: Set up an equation where x% of the larger amount equals the smaller amount. Then solve for x by rearranging the equation.

Exam Tip: Always convert all units to the same measurement (hours to minutes, km to metres, rupees to paise) before setting up your equation.

Question 7. Calculate the required value in each case:
(i) 32% of 425
(ii) 16 2/3 % of 16
(iii) 6.5% of 400
(iv) 136% of 70
(v) 2.8% of 35
(vi) 0.6% of 45
Answer:
(i) \( 32\% \text{ of } 425 = \left(\frac{32}{100} \times 425\right) = \left(\frac{32 \times 17}{4}\right) = (8 \times 17) = 136 \)
(ii) \( 16\frac{2}{3}\% \text{ of } 16 = \frac{50}{3}\% \text{ of } 16 = \left(\frac{50}{3 \times 100} \times 16\right) = \left(\frac{1}{3} \times 16\right) = \frac{8}{3} = 2\frac{2}{3} \)
(iii) \( 6.5\% \text{ of } 400 = \left(\frac{6.5}{100} \times 400\right) = \left(\frac{65}{10 \times 100} \times 400\right) = \left(\frac{65 \times 4}{10}\right) = \frac{260}{10} = 26 \)
(iv) \( 136\% \text{ of } 70 = \left(\frac{136}{100} \times 70\right) = \left(\frac{136 \times 7}{10}\right) = \left(\frac{952}{10}\right) = 95.2 \)
(v) \( 2.8\% \text{ of } 35 = \left(\frac{2.8}{100} \times 35\right) = \left(\frac{28}{10 \times 100} \times 35\right) = \left(\frac{14 \times 7}{100}\right) = \frac{98}{100} = 0.98 \)
(vi) \( 0.6\% \text{ of } 45 = \left(\frac{0.6}{100} \times 45\right) = \left(\frac{6}{10 \times 100} \times 45\right) = \left(\frac{3 \times 45}{5 \times 100}\right) = \left(\frac{3 \times 9}{100}\right) = \frac{27}{100} = 0.27 \)
In simple words: Turn the percentage into a fraction, then multiply it by the number. Simplify and cancel whenever you can to make the arithmetic easier.

Exam Tip: Look for common factors between the percentage numerator and the given number before multiplying - this cuts down on your calculation work.

Question 8. Solve:
(i) 25% of Rs 76
(ii) 20% of Rs 132
(iii) 7.5% of 600 m
(iv) 3 3/4 % of 90 km
(v) 8.5% of 5 kg
(vi) 20% of 12 L
Answer:
(i) \( 25\% \text{ of Rs } 76 = \text{Rs}\left(76 \times \frac{25}{100}\right) = \text{Rs}\left(76 \times \frac{1}{4}\right) = \text{Rs } 19 \)
(ii) \( 20\% \text{ of Rs } 132 = \text{Rs}\left(132 \times \frac{20}{100}\right) = \text{Rs}\left(132 \times \frac{1}{5}\right) = \text{Rs } 26.4 \)
(iii) \( 7.5\% \text{ of } 600 \text{ m} = \left(600 \times \frac{7.5}{100}\right) \text{m} = (6 \times 7.5) \text{ m} = 45 \text{ m} \)
(iv) \( 3\frac{3}{4}\% \text{ of } 90 \text{ km} = \frac{15}{4}\% \text{ of } 90 \text{ km} = \left(90 \times \frac{15}{4 \times 100}\right) \text{ km} = \left(90 \times \frac{1}{30}\right) \text{ km} = 3 \text{ km} \)
(v) \( 8.5\% \text{ of } 5 \text{ kg} = \left(5 \times \frac{8.5}{100}\right) \text{ kg} = \left(5 \times \frac{85}{1000}\right) \text{ kg} = 0.425 \text{ kg} = 425 \text{ g } [1 \text{ kg} = 1000 \text{ g}] \)
(vi) \( 20\% \text{ of } 12 \text{ L} = \left(12 \times \frac{20}{100}\right) \text{ L} = \left(12 \times \frac{1}{5}\right) \text{ L} = 2.4 \text{ L} \)
In simple words: Multiply the amount by the percentage (as a decimal or fraction), remembering to include the unit in your final answer.

Exam Tip: When the answer involves unit conversion (like grams from kilograms), always show the conversion factor and complete the conversion in your final statement.

Question 9. Find the number, if 13% of it is 65.
Answer: Let y be the required number.
Then, 13% of y = 65
\( \Rightarrow \left(\frac{13}{100} \times y\right) = 65 \)
\( \Rightarrow y = \left(65 \times \frac{100}{13}\right) = 500 \)
Hence, the required number is 500.
In simple words: When you know the percentage and its value, divide the value by the percentage (as a decimal) to find the whole amount.

Exam Tip: Always check your answer by computing the percentage of your result - it should equal the given value.

Question 10. Find the number, if 6 1/4 % of it is 2.
Answer: Let x be the required number.
Then, 6 1/4 % of x = 2
\( \Rightarrow \left(6\frac{1}{4}\% \times x\right) = 2 \)
\( \Rightarrow \left(\frac{25}{400} \times x\right) = 2 \)
\( \Rightarrow x = \left(2 \times \frac{400}{25}\right) = 32 \)
Hence, the required number is 32.
In simple words: Convert the mixed number percentage to an improper fraction, then divide the given value by this fraction to find the original number.

Exam Tip: For mixed number percentages, convert to an improper fraction first - this avoids rounding errors and keeps all values exact.

Question 11. Find 10% more than Rs 90.
Answer: 10% of Rs 90 = Rs \(\left(\frac{10}{100} \times 90\right) = \text{Rs } 9\)
\( \therefore \text{Amount that is } 10\% \text{ more than Rs } 90 = \text{Rs}(90 + 9) = \text{Rs } 99\)
Hence, the required amount is Rs 99.
In simple words: Find 10% of the original amount, then add it to the original amount to get the new total.

Exam Tip: "More than" means you add the percentage increase to the original value; "less than" means you subtract the percentage decrease.

Question 12. Find 20% less than Rs 60.
Answer: 20% of Rs 60 = Rs \(\left(60 \times \frac{20}{100}\right) = \text{Rs } 12\)
\( \therefore \text{Amount that is } 20\% \text{ less than Rs } 60 = \text{Rs}(60 - 12) = \text{Rs } 48\)
Hence, the required amount is Rs 48.
In simple words: Calculate 20% of the starting value, then subtract it from the starting value.

Exam Tip: Always check reasonableness: 20% less should be a noticeably smaller number, and 20% more should be noticeably larger.

Question 13. If 3% of x is 9, find the value of x.
Answer: 3% of x = 9
\( \Rightarrow \left(\frac{3}{100} \times x\right) = 9 \)
\( \Rightarrow x = \left(9 \times \frac{100}{3}\right) = 300 \)
Hence, the value of x is 300.
In simple words: Set the percentage expression equal to 9, then solve for x by multiplying both sides by the reciprocal of the fraction.

Exam Tip: To solve percentage equations, isolate x by multiplying both sides by 100 and dividing by the percentage number.

Question 14. If 12.5% of x is 6, find the value of x.
Answer: 12.5% of x = 6
\( \Rightarrow \left(\frac{12.5}{100} \times x\right) = 6 \)
\( \Rightarrow x = \left(6 \times \frac{100}{12.5}\right) = (6 \times 8) = 48 \)
Hence, the value of x is 48.
In simple words: Divide the given value by the percentage (converted to decimal form) to find x.

Exam Tip: When working with decimal percentages, multiply numerator and denominator by 10 to clear the decimal and simplify your calculation.

Question 15. What percentage of 84 is 14?
Answer: Let r% of 84 be 14.
Then, \( \left(\frac{r}{100} \times 84\right) = 14 \)
\( \Rightarrow \frac{21r}{25} = 14 \)
\( \Rightarrow x = \left(14 \times \frac{25}{21}\right) = \left(\frac{2 \times 25}{3}\right) = \frac{50}{3} = 16\frac{2}{3}\% \)
Hence, 16 2/3 % of 84 is 14.
In simple words: Set up an equation with the percentage as unknown, place the smaller number over the larger one, and solve for the percentage.

Exam Tip: When asked "what percentage of A is B," always set it up as (percentage/100) × A = B, then solve for the percentage.

Question 16.
(i) Let x% of Rs 120 be Rs 15.
Then, Rs \(\left(\frac{x}{100} \times 120\right) = \text{Rs } 15\)
\( = \left(\frac{6x}{5}\right) = 15\)
\( \therefore x = \left(2 \times \frac{400}{25}\right) = 12.5\%\)
Hence, 12.5% of Rs 120 is Rs 15.

(ii) Let x% of 2 h be 36 min
Then, \( \left(\frac{x}{100} \times 2 \times 60\right) \text{min} = 36 \text{ min}\)
\( = \left(\frac{120x}{100}\right) = 36\)
\( \therefore x = \left(\frac{36 \times 100}{120}\right)\% = 30\%\)
Hence, 30% of 2 h is 36 min.

(iii) Let x% of 2 days be 8 h
Then, \( \left(\frac{x}{100} \times 2 \times 24\right) \text{h} = 8 \text{ h}\)
\( = \left(\frac{48x}{100}\right) = 8\)
\( \therefore x = \left(\frac{8 \times 100}{48}\right)\% = 16\frac{2}{3}\%\)
Hence, 16 2/3 % of 2 days is 8 h.

(iv) Let x% of 4 km be 160 m
Then, \( \left(\frac{x}{100} \times 4 \times 1000\right) \text{m} = 160 \text{ m}\)
\( \Rightarrow 40x = 160\)
\( \therefore x = \left(\frac{160}{40}\right)\% = 4\%\)
Hence, 4% of 4 km is 160 m.

(v) Let x% of 1 L be 175 mL
Then, \( \left(\frac{x}{100} \times 1 \times 1000\right) \text{mL} = 175 \text{ mL}\)
\( = 10x = 175\)
\( \therefore x = \left(\frac{175}{10}\right)\% = 17.5\%\)
Hence, 17.5% of 1 L is 175 mL.

(vi) Let x% of Rs 4 be 25 paise
Then, \( \left(\frac{x}{100} \times 4 \times 100\right) \text{paise} = 25 \text{paise}\)
\( = 4x = 25\)
\( \therefore x = \left(\frac{25}{4}\right)\% = 6\frac{1}{4}\%\)
Hence, 6 1/4 % of Rs 4 is 25 paise.
Answer:
In simple words: Write the percentage equation, convert units if needed, solve for x, and always check that your final answer makes sense with the original problem.

Exam Tip: Unit conversion is essential - convert everything to the smallest unit (minutes, grams, millilitres, paise) before setting up the equation.

Exercise 10B

Definition

Percent is used to describe a quantity 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

Key Formulas

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

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

Conversion table: Percent, Decimal, Fraction
50% = 0.50 = \( \frac{50}{100} \)

Example: 60% = \( \frac{60}{100} \) = 0.6
(Note: "Percent" means "per one hundred," so to change a percent to a fraction, divide by 100.)

Question 1. Rupesh scored 495 marks in an examination with a maximum of 750 marks. What percentage did he score?
Answer: Maximum marks in the examination = 750
Marks obtained by Rupesh = 495
Percentage of marks obtained = \( \left(\frac{495}{750} \times 100\right)\% = 66\% \)
Hence, Rupesh scored 66% in the examination.
In simple words: Divide the marks he got by the total marks, then multiply by 100 to find what percentage he scored.

Exam Tip: Always state the final answer as a percentage - include the % sign to show you understand what the question is asking.

Question 2. A typist's monthly salary is Rs 15,625. His salary is increased by 12%. What is his new salary?
Answer: Total monthly salary = Rs 15,625
Increase percentage = 12%
\( \therefore \text{Amount increase} = 12\% \text{ of Rs } 15,625 = \text{Rs}\left(15,625 \times \frac{12}{100}\right) = \text{Rs } 1,875 \)
\( \therefore \text{New salary} = \text{Rs } 15,625 + \text{Rs } 1,875 = \text{Rs } 17,500 \)
Hence, the new salary of the typist is Rs 17,500.
In simple words: Find 12% of the original salary, then add this amount to the original salary to get the new total.

Exam Tip: For salary, price, or value increases, always add the percentage increase to the original amount.

Question 3. The original excise duty on an item is Rs 950. If the excise duty is reduced to Rs 760, by what percentage is the duty reduced?
Answer: Original excise duty on the item = Rs 950
Amount reduced on excise duty = Rs (950 - 760) = Rs 190
\( \therefore \text{Reduction percent} = \left(\frac{\text{Amount reduced}}{\text{Original value}} \times 100\right) = \left(\frac{190}{950} \times 100\right) = 20 \)
Hence, the excise duty on that item is reduced by 20%.
In simple words: Find how much the duty dropped by subtracting the new amount from the original. Then divide this decrease by the original amount and multiply by 100.

Exam Tip: For percentage reduction, always use the original value as the denominator, not the new value.

Question 4. If 96% of the total cost of a TV set is Rs 10,464, find the total cost of the TV set.
Answer: Let Rs x be the total cost of the TV set.
Now, 96% of the total cost of TV = Rs 10,464
\( \Rightarrow 96\% \text{ of Rs } x = \text{Rs } 10,464 \)
\( \Rightarrow \left(\frac{96}{100} \times x\right) = 10,464 \)
\( \therefore x = \left(\frac{10,464 \times 100}{96}\right) = 10,900 \)
Hence, the total cost of the TV set is Rs 10,900.
In simple words: If a percentage of a total equals a known amount, divide that amount by the percentage to find the full total.

Exam Tip: Always set up the equation carefully: (percentage/100) × total = known value, then solve for the total.

Question 5. In a school, the number of boys is 70% of the total number of students. If there are 30 girls, how many boys are there in the school?
Answer: Let the total number of students be 100.
Then, number of boys = 70
\( \therefore \text{Number of girls} = (100 - 70) = 30 \)
Now, total number of students when the number of girls is 30 = 100
Then, total number of students when the number of girls is 504 = \( \left(\frac{100}{30} \times 504\right) = 1,680 \)
\( \therefore \text{Number of boys} = (1,680 - 504) = 1,176 \)
Hence, there are 1,176 boys in the school.
In simple words: If 70% are boys, then 30% are girls. Use this ratio to find how many boys correspond to 30 girls.

Exam Tip: Use a simple base (like 100 total students) to find the ratio of boys to girls, then scale up to the actual numbers.

Question 6. If 12% of an ore contains 69 kg of pure copper, find the total quantity of ore required to get 69 kg of copper.
Answer: Let x kg be the amount of the required ore.
Then, 12% of x kg = 69 kg
\( \Rightarrow \left(\frac{12}{100} \times x\right) \text{kg} = 69 \text{ kg} \)
\( \Rightarrow x = \left(\frac{69 \times 100}{12}\right) \text{kg} = 575 \text{ kg} \)
Hence, 575 kg of ore is required to get 69 kg of copper.
In simple words: Divide the amount of pure copper by the percentage it represents to find the total ore needed.

Exam Tip: When a percentage of one substance yields a known quantity, divide the known quantity by the percentage to find the whole amount.

Question 7. In an examination, the pass marks are 36% of the maximum marks. If a student gets 162 marks and just passes the examination, find the maximum marks.
Answer: Let x be the maximum marks.
Pass marks = (123 + 39) = 162
Then, 36% of x = 162
\( \Rightarrow \left(\frac{36}{100} \times x\right) = 162 \)
\( \Rightarrow x = \left(\frac{162 \times 100}{36}\right) = 450 \)
\( \therefore \text{Maximum marks} = 450 \)
In simple words: If the pass percentage is 36% and the pass marks are 162, divide 162 by 0.36 to get the maximum marks.

Exam Tip: For "just passes" questions, the marks scored equal the pass marks, so use this value directly in your percentage equation.

Question 8. Suppose a fruit seller initially had 100 apples. He sold 40 apples. How many apples remain? If 60 of the remaining apples account for 100% of the final count, what percentage of 420 apples still remain?
Answer: Suppose that the fruit seller initially had 100 apples.
Apples sold = 40
\( \therefore \text{Remaining apples} = (100 - 40) = 60 \)
Initial amount of apples if 60 of them are remaining = 100
Initial amount of apples if 1 of them is remaining = \( \left(\frac{100}{60}\right) \)
Initial amount of apples if 420 of them are remaining = \( \left(\frac{100}{60} \times 420\right) = 700 \)
Hence, the fruit seller originally had 700 apples.
In simple words: Set up a proportion: if 60 apples remain out of 100 initial, then 420 remaining apples came from a larger starting amount. Use cross-multiplication to find the original total.

Exam Tip: For word problems involving percentages and quantities, always identify what percentage is being asked and set up a clear proportion.

Question 9. Suppose 100 candidates took an examination. If 72 candidates passed, how many total candidates would there be if 392 of them failed?
Answer: Suppose that 100 candidates took the examination.
Number of passed candidates = 72
Number of failed candidates = (100 - 72) = 28
Total number of candidates if 28 of them failed = 100
Total number of candidates if 392 of them failed = \( \left(\frac{100}{28} \times 392\right) = 1,400 \)
Hence, the total number of examinees is 1,400.
In simple words: If 28 out of 100 failed, then set up a ratio to find how many total candidates there were when 392 failed: (28/100) = (392/total).

Exam Tip: Always verify your answer: 392 failed ÷ 1,400 total should equal 28/100.

Question 10.
Answer: Suppose that the gross value of the moped is Rs x.
Commission on the moped = 5%
Price of moped after deducting the commission = Rs \( (x - 5\% \text{ of } x) = \text{Rs}\left(x - \frac{5x}{100}\right) = \text{Rs}\left(\frac{95x}{100}\right) \)
Now, price of the moped after deducting the commission = Rs 15,200
Then, Rs \( \left(\frac{95x}{100}\right) = \text{Rs } 15,200 \)
\( \therefore x = \text{Rs}\left(\frac{15,200 \times 100}{95}\right) = \text{Rs}(160 \times 100) = \text{Rs } 16,000 \)
Hence, the gross value of the moped is Rs 16,000.
In simple words: If the price after a 5% deduction is Rs 15,200, work backwards by dividing by 0.95 to find the original price.

Exam Tip: For "price after deduction" problems, remember that the final price is 95% of the original (100% minus the 5% deduction).

Question 11. The total quantity of gunpowder in 8 kg is 8,000 g. If the quantity of nitre in it is 75% of 8,000 g, and the quantity of sulphur is 10% of 8,000 g, find the quantity of charcoal in the gunpowder.
Answer: Total quantity of gunpowder = 8 kg = 8,000 g (1 kg = 1,000 g)
Quantity of nitre in it = 75% of 8,000 g = \( \left(\frac{75}{100} \times 8,000\right) \text{ g} = 6,000 \text{ g} = 6 \text{ kg} \)
Quantity of sulphur in it = 10% of 8,000 g = \( \left(\frac{10}{100} \times 8,000\right) \text{ g} = 800 \text{ g} = 0.8 \text{ kg} \)
\( \therefore \text{Quantity of charcoal in it} = (8,000 - (6,000 + 800)) \text{ g} = (8,000 - 6,800) \text{ g} = 1,200 \text{ g} = 1.2 \text{ kg} \)
Hence, the amount of charcoal in 8 kg of gunpowder is 1.2 kg.
In simple words: Find the amount of each ingredient by taking its percentage of the total. Charcoal is whatever remains after subtracting nitre and sulphur from the total.

Exam Tip: For mixture problems, always verify that all ingredients add up to 100% of the total weight.

Question 12. The total quantity of chalk in 1 kg is 1,000 g. Find the quantity of each ingredient given the following percentages: carbon (3%), calcium (10%), and oxygen (12%).
Answer: Total quantity of chalk = 1 kg = 1,000 g
We have the following:
Quantity of carbon in it = 3% of 1,000 g = \( \left(\frac{3}{100} \times 1,000\right) \text{ g} = 30 \text{ g} \)
Quantity of calcium in it = 10% of 1,000 g = \( \left(\frac{10}{100} \times 1,000\right) \text{ g} = 100 \text{ g} \)
Quantity of oxygen in it = 12% of 1,000 g = \( \left(\frac{12}{100} \times 1,000\right) \text{ g} = 120 \text{ g} \)
In simple words: Multiply the total amount by each percentage (converted to decimal) to get the quantity of each ingredient in the mixture.

Exam Tip: For composition problems, always show the percentage calculation step clearly so the examiner can follow your reasoning.

Question 13. Let x be the total number of days on which the school was open. Sonai went to school for 219 days. The percentage of attendance is 75%. What is the total number of days the school was open?
Answer: Let x be the total number of days on which the school was open.
Number of days when Sonai went to school = 219
Percentage of attendance = 75%
Thus, 75% of x = 219
\( \Rightarrow \left(\frac{75}{100} \times x\right) = 219 \)
\( \therefore x = \left(\frac{219 \times 100}{75}\right) = 292 \text{ days} \)
Hence, the school was open for a total of 292 days.
In simple words: If 75% attendance means 219 days attended, divide 219 by 0.75 to find the total number of school days.

Exam Tip: For attendance problems, remember that attendance percentage = (days attended / total days) × 100.

Question 14. Let the total value of the property be Rs x. The percentage of commission is 3%. The amount of commission is Rs 42,660. Find the total value of the property.
Answer: Let the total value of the property be Rs x.
Percentage of commission = 3
Amount of commission = Rs 42,660
Thus, 3% of Rs x = Rs 42,660
\( \Rightarrow \left(\frac{3}{100} \times x\right) = 42,660 \)
\( \therefore x = \left(\frac{42,660 \times 100}{3}\right) = 1,422,000 \)
Hence, the total value of the property is Rs 14,22,000.
In simple words: If 3% commission equals Rs 42,660, divide this amount by 0.03 to find the original property value.

Exam Tip: For commission problems, set up the equation (percentage/100) × total value = commission amount, then solve for the total value.

Question 15. A candidate received 60% of the votes cast and won by 19,200 votes. How many votes were cast in total?
Answer: Let the total number of eligible voters be 60,000. Voters who gave their votes make up 80% of 60,000, which equals 48,000. The number of votes in favor of candidate A is 60% of 48,000, which is \( \left( \frac{60}{100} \times 48000 \right) = 28,800 \). The number of votes received by candidate B is \( (48,000 - 28,800) = 19,200 \). Therefore, candidate B received 19,200 votes.
In simple words: Candidate A got 60% of all the votes cast. The difference between A's votes and B's votes was 19,200. You can work backwards from this difference to find the total votes cast.

Exam Tip: Always set up the relationship between percentages and the vote difference - it's the key to solving election problems quickly.

Question 16. A shirt costs Rs x. A discount of 12% is given on it. After the discount, the price is Rs 1,188. Find the original price of the shirt.
Answer: Let the original price of the shirt be Rs x. The discount rate is 12%. The discount amount on the shirt is 12% of Rs x, which equals \( Rs \left( \frac{12}{100} \times x \right) = Rs \left( \frac{12x}{100} \right) \). The price after applying the discount is \( Rs \left( x - \frac{12x}{100} \right) = Rs \left( \frac{88x}{100} \right) \). Given that the price after discount is Rs 1,188, we have \( Rs \left( \frac{88x}{100} \right) = Rs 1,188 \). Solving, \( 88x = 1,188 \times 100 \), which gives \( 88x = 118,800 \). Therefore, \( x = \left( \frac{118,800}{88} \right) = 1,350 \). The original price of the shirt is Rs 1,350.
In simple words: If you reduce the original price by 12%, you get Rs 1,188. Work backwards by dividing 1,188 by the fraction that remains (which is 88/100) to find what the starting price must have been.

Exam Tip: Remember that after a discount, you pay the percentage that remains (100% - 12% = 88%), not the discount percentage itself.

Question 17. A sweater costs Rs x. Its price increased by 8%. The new price is Rs 1,566. Find the original price of the sweater.
Answer: Let the original price of the sweater be Rs x. The increase in price is 8%. The value of the price increase on the sweater is 8% of Rs x, which equals \( Rs \left( \frac{8}{100} \times x \right) = Rs \left( \frac{8x}{100} \right) \). The new price of the sweater is \( Rs \left( x + \frac{8x}{100} \right) = Rs \left( \frac{108x}{100} \right) \). Given that the new price is Rs 1,566, we have \( Rs \left( \frac{108x}{100} \right) = Rs 1,566 \). Solving, \( x = \left( \frac{1,566 \times 100}{108} \right) = 1,450 \). The original price of the sweater is Rs 1,450.
In simple words: The price went up by 8%, so the new price is 108% of the original. To find the starting price, divide the new price by 1.08 (or multiply by 100/108).

Exam Tip: After an increase, the new amount is 100% plus the increase percentage. Always use (100 + increase%) / 100 as your multiplier.

Question 18. A man's income is Rs x. He spends 80% of his income and gives 10% of what remains to a charity. He is left with Rs 46,260. Find his income.
Answer: Let the man's income be Rs x. He spends 80% of Rs x, which equals \( Rs \left( \frac{80}{100} \times x \right) = Rs \left( \frac{4x}{5} \right) \). The amount remaining after spending is \( Rs \left( x - \frac{4x}{5} \right) = Rs \left( \frac{x}{5} \right) \). He donates 10% of the remaining amount to charity, which is 10% of Rs \( \left( \frac{x}{5} \right) \), equaling \( Rs \left( \frac{10}{100} \times \frac{x}{5} \right) = Rs \left( \frac{x}{50} \right) \). The amount left after the charity gift is \( Rs \left( \frac{x}{5} - \frac{x}{50} \right) = Rs \left( \frac{10x - x}{50} \right) = Rs \left( \frac{9x}{50} \right) \). According to the problem, this final amount is Rs 46,260. Therefore, \( Rs \left( \frac{9x}{50} \right) = Rs 46,260 \). Solving, \( x = Rs \left( \frac{46,260 \times 50}{9} \right) = Rs 257,000 \). The income of the man is Rs 2,57,000.
In simple words: He keeps 20% after spending 80%. Then he gives away 10% of that 20%. The final remaining 9% equals Rs 46,260. Divide backwards to get the original income.

Exam Tip: Track each operation carefully - spending comes first, then the charity gift is from what's left, not from the original amount.

Question 19. A number is increased by 20%. Then it is decreased by 20% of the new value. The net decrease is 4%. What is the number?
Answer: Let the number be 100. An increase of 20% in the number gives \( (100 + 20) = 120 \). Now, a decrease of 20% is applied to the new number, which is 20% of 120. The decrease equals \( \left( \frac{20}{100} \times 120 \right) = 24 \). The final number is \( (120 - 24) = 96 \). The net decrease from the original is \( (100 - 96) = 4 \). The net decrease percentage is \( \left( \frac{4}{100} \times 100 \right) = 4 \). Therefore, the net decrease is 4%.
In simple words: Start with any number. Increase it by 20%, then decrease the result by 20%. You'll find that you end up 4% below where you started. This is because the 20% decrease is applied to a larger amount (the increased value), not the original.

Exam Tip: Always apply percentage changes to the current value at each step, not the original. The order and base matter in successive percentage problems.

Question 20. An original salary is Rs 100. It is increased by 20%. To get back to the original salary, by what percentage should the new salary be reduced?
Answer: Let the original salary be Rs 100. An increase of 20% raises it to \( Rs (100 + 20) = Rs 120 \). To restore the original salary, a reduction of \( Rs (120 - 100) = Rs 20 \) is needed. The reduction percentage is calculated as \( \left( \frac{20}{120} \times 100 \right) = \left( \frac{100}{6} \right) = 16 \frac{2}{3} \% \). Therefore, the required reduction on the new salary is \( 16 \frac{2}{3} \% \).
In simple words: After raising a salary by 20%, you need to cut it back down. But the cut must be calculated as a percentage of the new (higher) salary, not the original one. That's why it's slightly less than 20%.

Exam Tip: The percentage needed to reverse an increase is always less than the original percentage because it applies to a larger base.

Question 21. A property costs Rs 5,40,000. Commission is charged on the first Rs 2,00,000 at 2%, on the next Rs 2,00,000 at 1%, and on the remaining amount at 0.5%. Calculate the total commission.
Answer: The total cost of the property is Rs 5,40,000. The commission on the first Rs 2,00,000 at 2% is \( \left( \frac{2}{100} \times 2,00,000 \right) = Rs 4,000 \). The commission on the next Rs 2,00,000 at 1% is \( \left( \frac{1}{100} \times 2,00,000 \right) = Rs 2,000 \). The remaining amount is \( Rs (5,40,000 - 4,00,000) = Rs 1,40,000 \). The commission on Rs 1,40,000 at 0.5% is \( \left( \frac{0.5}{100} \times 1,40,000 \right) = Rs 700 \). The total commission on the property worth Rs 5,40,000 is \( Rs (4,000 + 2,000 + 700) = Rs 6,700 \). Therefore, the commission of the property dealer on the property that has been sold for Rs 5,40,000 is Rs 6,700.
In simple words: Break the property price into three chunks and calculate the commission for each chunk at its own rate. Add all three commission amounts together to get the total.

Exam Tip: Slab-based commission problems require you to identify each bracket carefully and not apply one rate to the whole amount.

Question 22. Akhil's income, when Nikhil's income is Rs 80, equals Rs 100. What percentage more is Akhil's income than Nikhil's income?
Answer: Let Akhil's income be Rs 100. When Nikhil's income is Rs 80, Akhil's income equals Rs 100. This means when Nikhil's income is Rs 80, Akhil's income when Nikhil's income is Rs 100 becomes \( Rs \left( \frac{100}{80} \times 100 \right) = Rs 125 \). In other words, if Nikhil's income is Rs 100, then Akhil's income is Rs 125. Therefore, Akhil's income is more than Nikhil's by 25%.
In simple words: Set Nikhil's income as the base reference (Rs 100). Calculate what Akhil's income would be at that same level. The difference tells you the percentage by which Akhil earns more.

Exam Tip: When comparing two quantities, always set one as your baseline (100) and express the other relative to it. The difference directly gives your percentage.

Question 23. Mr. Thomas's income, when John's income is Rs 100, is Rs 120. How much less is Mr. Thomas's income than John's, expressed as a percentage?
Answer: Let Rs 100 be Mr. Thomas's income. John's income is Rs 120. Mr. Thomas's income when John's income is Rs 120 equals Rs 100. Mr. Thomas's income when John's income is Rs 100 becomes \( Rs \left( \frac{100}{120} \times 100 \right) = Rs 83 \frac{1}{3} \). Therefore, Mr. Thomas's income is less than John's income by \( 16 \frac{2}{3} \% \).
In simple words: If John makes Rs 120 when Thomas makes Rs 100, then when John makes Rs 100, Thomas would make only about Rs 83.33. The gap represents how much less Thomas earns.

Exam Tip: Always compute percentages using the higher earner's income as the denominator to find "how much less" the other person earns.

Question 24. A machine is worth Rs 4,30,000 today. Its value was 90% of this amount one year ago. What was the value of the machine a year ago?
Answer: Let Rs x be the value of the machine one year ago. The current value is 90% of Rs x, which means \( Rs \left( \frac{90}{100} \times x \right) = Rs \left( \frac{9x}{10} \right) \). It is given that the current value is Rs 4,30,000. Therefore, \( x = Rs \left( \frac{4,30,000 \times 10}{9} \right) = Rs (43,000 \times 10) = Rs 4,30,000 \). The value of the machine a year ago was Rs 4,30,000.
In simple words: The present value is 90% of the past value. To work backwards, divide the present amount by 0.9 (or multiply by 10/9) to find what it was worth before.

Exam Tip: Depreciation problems require you to use the percentage formula correctly - if the current value is 90% of the original, solve accordingly using reverse operations.

Question 25. A car costs Rs 4,50,000. It depreciates by 20% in the first year and by 20% in the second year. Find its value after two years.
Answer: The current value of the car is Rs 4,50,000. A depreciation of 20% in the first year reduces it by \( Rs \left( \frac{20}{100} \times 4,50,000 \right) = Rs 90,000 \). The depreciated value after the first year is \( Rs (4,50,000 - 90,000) = Rs 3,60,000 \). A depreciation of 20% in the second year on this amount is \( Rs \left( \frac{20}{100} \times 3,60,000 \right) = Rs 72,000 \). The depreciated value after the second year is \( Rs (3,60,000 - 72,000) = Rs 2,88,000 \). Therefore, the value of the car after two years will be Rs 2,88,000.
In simple words: Multiply the starting value by 0.8 (which represents 80% remaining). Do this twice - once for year one, then again for year two - to get the final value after depreciation both times.

Exam Tip: Each year's depreciation is calculated on that year's starting value, not the original price. This is compound depreciation.

Question 26. A town's population is 60,000 today. It increases by 10% in the first year and by 10% in the second year. What will be the population after two years?
Answer: The current population of the town is 60,000. An increase of 10% in the population after the first year equals \( \left( \frac{10}{100} \times 60,000 \right) = 6,000 \). Thus, the population after one year becomes \( 60,000 + 6,000 = 66,000 \). An increase of 10% in the population after two years is \( \left( \frac{10}{100} \times 66,000 \right) = 6,600 \). Thus, the population after the second year becomes \( 66,000 + 6,600 = 72,600 \). Therefore, the population of the town after two years will be 72,600.
In simple words: Each year, the population grows by 10% of that year's starting population. Calculate year one's increase, add it to get the new base, then calculate year two's increase from that new base.

Exam Tip: Population growth is compound growth - each year's increase is based on the previous year's total, not the original population.

Question 27. Sugar's cost was Rs 100 per unit. The cost increased to Rs 125 per unit. By what percentage did the consumption have to decrease so that the total expenditure on sugar remained the same?
Answer: Let the consumption of sugar originally be 1 unit at a cost of Rs 100. The new cost of 1 unit of sugar is Rs 125. When Rs 125 produces 1 unit of sugar, Rs 100 will produce \( \left( \frac{125}{100} \right) \text{ unit} = \left( \frac{5}{4} \right) \text{ unit} \). The reduction in consumption is \( \left( 1 - \frac{4}{5} \right) = \left( \frac{1}{5} \right) \text{ unit} \). The reduction percentage in consumption is \( \left( \frac{1}{5} \times \frac{1}{1} \times 100 \right) \% = \left( \frac{100}{5} \right) \% = 20 \% \).
In simple words: When the price goes up by 25%, you have to buy less to spend the same amount. The percentage you need to reduce consumption is found by comparing the old and new quantities you can afford with the same money.

Exam Tip: Use the inverse relationship: if price increases by a factor, consumption must decrease proportionally to keep total spending constant. A 25% price rise requires a 20% consumption cut.

Exercise 10C

Question 1. What is 75% as a fraction?
Answer: \( \frac{3}{4} = \left( \frac{3}{4} \times 100 \right) \% = 75 \% \)
In simple words: 75% equals 3 out of 4 parts, which is \( \frac{3}{4} \).

Exam Tip: Always convert percentages to fractions by dividing by 100 and simplifying - it makes further calculations easier.

Question 2. What is 2.5 ÷ 5 as a percentage?
Answer: \( 2.5 \div 5 = \frac{2.5}{5} = \left( \frac{2.5}{5} \times 100 \right) \% = 40 \% \)
In simple words: Divide 2.5 by 5 to get 0.5, then multiply by 100 to convert to a percentage, giving 40%.

Exam Tip: To express any fraction or decimal as a percentage, simply multiply by 100 and add the % symbol.

Question 3. Express 8\( \frac{1}{3} \)% as a fraction.
Answer: \( 8 \frac{1}{3} \% = \frac{25}{3} \% = \left( \frac{25}{3} \times \frac{1}{100} \right) = \left( \frac{1 \times 1}{3 \times 4} \right) = \frac{1}{12} \)
In simple words: Convert the mixed percentage to an improper fraction, then divide by 100. Simplify to get \( \frac{1}{12} \).

Exam Tip: Mixed number percentages require careful conversion - turn the mixed number into an improper fraction first, then divide by 100.

Question 4. If x% of 75 = 9, find the value of x.
Answer: We know that x% of 75 = 9. This gives \( \left( \frac{x}{100} \times 75 \right) = 9 \). Solving, \( x = \left( \frac{9 \times 100}{75} \right) = 12 \). Therefore, the value of x is 12.
In simple words: Set up the equation x% × 75 = 9, then isolate x by multiplying 9 by 100 and dividing by 75.

Exam Tip: Always convert percentage to decimal form (divide by 100) when setting up equations, then solve using basic algebra.

Question 6. If x% of \( \frac{2}{5} \) = \( \frac{1}{35} \), find x.
Answer: Let x be the required percentage. Then, x% of \( \frac{2}{5} \) equals \( \frac{1}{35} \). This gives \( \left( \frac{x}{100} \times \frac{2}{5} \right) = \frac{1}{35} \). Rearranging, \( x = \left( \frac{100 \times \frac{1}{35}}{\frac{2}{5}} \right) = \left( \frac{100 \times 1 \times 5}{35 \times 2} \right) = 10 \). Therefore, 10% of \( \frac{2}{5} \) is \( \frac{1}{35} \).
In simple words: Set the percentage equation equal to the given result, then solve by cross-multiplying and isolating x. You get x = 10.

Exam Tip: When fractions are involved, clear all denominators by cross-multiplication before solving for the variable.

Question 7. Find the number such that 20% of it equals 42.
Answer: Let the required number be x. Then, 20% of x = 42. This gives \( \left( x + \frac{30x}{100} \right) = 42 \). Simplifying, \( \left( x + \frac{x}{5} \right) = 42 \). Combining, \( \left( \frac{5x + x}{5} \right) = 42 \). So, \( \left( \frac{6x}{5} \right) = 42 \). Solving, \( x = \left( \frac{42 \times 5}{6} \right) = 35 \). Therefore, the required number is 35.
In simple words: If 20% of a number equals 42, divide 42 by 0.2 (or multiply by 5) to get the full number.

Exam Tip: To find the whole when given a percentage part, always divide the part by the percentage (in decimal form).

Question 8. Find the number such that 8% of it minus 8% of it equals 69.
Answer: Let the required number be x. Then, x - 8% of x = 69. This gives \( \left( x - \frac{8x}{100} \right) = 69 \). Simplifying, \( \left( x - \frac{2x}{25} \right) = 69 \). Combining, \( \left( \frac{25x - 2x}{25} \right) = 69 \). So, \( \left( \frac{23x}{25} \right) = 69 \). Solving, \( x = \left( \frac{69 \times 25}{23} \right) = 75 \). Therefore, the required number is 75.
In simple words: A number minus 8% of itself leaves 92% of the number. Divide 69 by 0.92 to find the original number.

Exam Tip: When a percentage is subtracted from a whole, you are left with (100% - that percentage) of the original. Use this shortcut to speed up solving.

Question 9. How much ore is needed to get 400 g of copper if 5% of the ore is copper?
Answer: Let x kg be the required amount of ore. Then, 5% of x kg = 400 g = 0.4 kg. This gives \( \left( \frac{5}{100} \times x \right) = 0.4 \). Solving, \( x = \left( \frac{0.4 \times 100}{5} \right) = 8 \). Therefore, 8 kg of ore is required to obtain 400 g of copper.
In simple words: If 5% of ore yields copper, then to get 400 g of copper, divide 400 by 0.05 to find the total ore needed.

Exam Tip: Always match units before calculating - convert grams to kilograms or vice versa so the percentage applies to compatible quantities.

Question 11. The gross value of a TV is Rs x. The commission is 10%. After deducting the commission, the price is Rs 18,000. Find the gross value of the TV.
Answer: Assume the gross value of the TV is Rs x. The commission charged is 10%. The price of the TV after deducting the commission becomes \( Rs \left( x - \frac{10x}{100} \right) = Rs \left( \frac{100x - 10x}{100} \right) = Rs \left( \frac{90x}{100} \right) \). We are told the price after deducting the commission is Rs 18,000. Therefore, \( Rs \left( \frac{90x}{100} \right) = Rs 18,000 \). Solving, \( x = \left( \frac{18,000 \times 100}{90} \right) = Rs (2,000 \times 10) = Rs 20,000 \). Therefore, the gross value of the TV is Rs 20,000.
In simple words: After removing a 10% commission, Rs 18,000 remains. This means Rs 18,000 is 90% of the original. Divide by 0.9 to get the starting value.

Exam Tip: When a commission or discount is subtracted, the remaining price represents a specific percentage of the gross value - use this to work backwards.

Question 12. A man's original salary is Rs x. It increases by 25%. His new salary is Rs 20,000. Find the original salary.
Answer: Assume the original salary of the man is Rs x. The increase rate is 25%. The value increased in the salary is 25% of Rs x, which equals \( Rs \left( \frac{25}{100} \times x \right) = Rs \left( \frac{x}{4} \right) \). The salary after increment is \( Rs \left( x + \frac{x}{4} \right) = Rs \left( \frac{5x}{4} \right) \). We are told that the increased salary equals Rs 20,000. Therefore, \( Rs \left( \frac{5x}{4} \right) = Rs 20,000 \). Solving, \( x = Rs \left( \frac{20,000 \times 4}{5} \right) = Rs 16,000 \). Therefore, the original salary of the man is Rs 16,000.
In simple words: When a 25% raise is added, the new salary becomes 125% of the old. Divide the new salary by 1.25 to recover the original amount.

Exam Tip: After an increase of 25%, the new value is always 125% of the original - use division to reverse the operation.

Question 13. Suppose that originally there are 100 apples. 40 apples were sold. The remaining apples would be 100 - 40 = 60. If 60 apples represent 420 apples originally available, how many apples did the seller originally have?
Answer: Assume the fruit seller initially had 100 apples. Apples sold total 40. The remaining apples equal \( (100 - 40) = 60 \). If 60 apples represent 420 apples still in stock originally, the original number of apples must be calculated. If 60 apples remaining means 420 were available initially, then the original amount is \( \left( \frac{100}{60} \times 420 \right) = 700 \). Therefore, the fruit seller originally had 700 apples with him.
In simple words: After selling 40 apples, 60 remain. This 60 represents 420 of the starting quantity. Use proportions to find: if 60 left means 420 started, then 100 left would mean 700 started.

Exam Tip: Work with ratios and proportions when asked about original quantities after some are removed - keep fractions consistent.

Question 14. A machine currently worth Rs 25,000 loses 10% of its value annually. What will be the value of the machine after 1 year?
Answer: The current value of the machine is Rs 25,000. The annual depreciation rate is 10% of Rs 25,000. Depreciation equals \( Rs \left( \frac{10}{100} \times 25,000 \right) = Rs 2,500 \). The depreciated value after one year is \( Rs (25,000 - 2,500) = Rs 22,500 \). Therefore, the value of the machine after 1 year will be Rs 22,500.
In simple words: Calculate 10% of the current value and subtract it. Alternatively, multiply the value by 0.9 (which represents 90% remaining).

Exam Tip: For depreciation, multiply by (100% - depreciation rate) as a decimal - it's faster than calculating the loss separately.

Question 15. A number x is such that 8% of x equals 6. Find x.
Answer: Let the required number be x. Then, 8% of x equals 6. This gives \( \left( \frac{8}{100} \times x \right) = 6 \). Solving, \( x = \left( \frac{6 \times 100}{8} \right) = 75 \). Therefore, the required number is 75.
In simple words: If 8% of a number is 6, then the whole number is 6 divided by 0.08, which gives 75.

Exam Tip: To find the whole from a percentage, divide the given part by the percentage in decimal form.

Question 16. Find 60% of 450.
Answer: 60% of 450 equals \( \left( \frac{60}{100} \times 450 \right) = \left( \frac{3}{5} \times 450 \right) = (3 \times 90) = 270 \)
In simple words: Multiply 450 by the decimal 0.6 to get 270.

Exam Tip: Convert the percentage to a simplified fraction first (60% = 3/5) to make mental calculation faster.

Question 17. Assume a chair originally costs Rs x. A reduction of 6% is applied to its price. The reduced price is Rs 658. Find the original cost of the chair.
Answer: Let the original price of the chair be Rs x. A 6% reduction in price means the discount is \( Rs \left( \frac{6}{100} \times x \right) = Rs \left( \frac{3x}{50} \right) \). The reduced price of the chair is \( Rs \left( x - \frac{3x}{50} \right) = Rs \left( \frac{50x - 3x}{50} \right) = Rs \left( \frac{47x}{50} \right) \). We are told the present price of the chair is Rs 658. Therefore, \( Rs \left( \frac{47x}{50} \right) = Rs 658 \). Solving, \( Rs x = Rs \left( \frac{658 \times 50}{47} \right) = Rs (14 \times 50) = Rs 700 \). Therefore, the original price of the chair is Rs 700.
In simple words: After a 6% discount, you pay 94% of the original. Divide the discounted price by 0.94 to find the starting price.

Exam Tip: Always multiply (100% - discount%) to find what percentage you actually pay, then use this to work backwards.

Question 18. There are 100 total students in a school. 70% are boys. If there are 240 girls instead, how many boys would there be?
Answer: Let the total number of students be 100. Then, the number of boys is 70% of 100 = 70. The number of girls is \( (100 - 70) = 30 \). Now, if there are 30 girls instead, the total would be 100. If there are 240 girls instead, the total number of students is \( \left( \frac{100}{30} \times 240 \right) = 800 \). The number of boys is \( (800 - 240) = 560 \). Therefore, there are 560 boys in the school.
In simple words: Find the ratio of girls to total students (30/100). Use this ratio: if 30 girls represents 1/3 of the school, then 240 girls represents a school of 800. Subtract girls from total to get boys.

Exam Tip: Maintain the original ratio structure when changing the quantities - scale up or down proportionally.

Question 19. A number x is increased by the difference between 11% and 7% of x, giving a total of 18. Find x.
Answer: Let x be the number. The difference between 11% and 7% of x equals \( \left( \frac{11x}{100} - \frac{7x}{100} \right) = \frac{4x}{100} = \frac{x}{25} \). When x is increased by this difference, we get \( x + \frac{x}{25} = 18 \). Simplifying, \( \frac{25x + x}{25} = 18 \). So, \( \frac{26x}{25} = 18 \). Solving, \( x = \left( \frac{18 \times 25}{26} \right) = (18 \times 25) = 450 \). Therefore, the required number is 450.
In simple words: Calculate 11% of x minus 7% of x (which is 4% of x). Add this 4% back to x to get 18. Solve for x.

Exam Tip: Break down composite operations step by step - find the difference first, then apply it to the original number.

Question 21. A number when increased by 35% plus 39 equals itself. Find the number.
Answer: Let x be the number. When increased by 35%, the value becomes \( (35\% \text{ of } x) + 39 = x \). This gives \( \left( \frac{35}{100} \times x \right) + 39 = x \). Rearranging, \( 39 = x - \frac{35x}{100} = \frac{65x}{100} = \frac{13x}{20} \). Solving, \( x = \left( \frac{39 \times 20}{13} \right) = 60 \). Therefore, the required number is 60.
In simple words: If 35% of a number plus 39 brings you back to the original number, then 39 must equal 65% of that number. Divide 39 by 0.65 to find it.

Exam Tip: When a percentage increase is offset by a fixed amount to return to the original, set up the equation carefully and isolate the variable.

Question 22. If 36% of a certain number is 180, what are the maximum marks for that test?
Answer: Let x be the maximum marks. Then, 36% of x = 180. This gives \( \left( \frac{36}{100} \times x \right) = 180 \). Solving, \( x = \left( \frac{180 \times 100}{36} \right) = 500 \). Therefore, the maximum marks for the test are 500.
In simple words: If 36% of the total equals 180 marks, divide 180 by 0.36 to find the full maximum marks.

Exam Tip: To find the maximum or total from a percentage portion, always divide the portion by its percentage in decimal form.

Question 23. When a number is reduced by 40%, the result is 135. Find the number.
Answer: Let x be the number. When reduced by 40%, the result is 135. This gives \( x - \frac{40x}{100} = 135 \). Simplifying, \( \frac{60x}{100} = 135 \). So, \( x = \left( \frac{135 \times 100}{60} \right) = 225 \). Therefore, the required number is 225.
In simple words: If reducing a number by 40% leaves 135, then 135 represents 60% of the original. Divide 135 by 0.6 to recover the starting number.

Exam Tip: A reduction of 40% means you keep 60% - use this multiplier (0.6) to set up and solve the equation quickly.