Percent and Percentage

Q1: Multiple Choice Type

i. A number, whose 4% is 6, is:

Step 1: Let the number be \( x \) \[ \frac{4}{100} \times x = 6 \Rightarrow x = \frac{6 \times 100}{4} = 150 \] Answer: c. 150

ii. What percent of 50 is 10?

\[ \frac{10}{50} \times 100 = 20\% \] Answer: a. 20%

iii. 18 hours as a percentage of 3 days is:

3 days = \( 3 \times 24 = 72 \) hours \[ \frac{18}{72} \times 100 = 25\% \\ \Rightarrow \frac{18}{3 \times 24} \times 100\% \] Answer: c. \( \frac{18}{3 \times 24} \times 100\% \)

iv. An alloy contains 30% of 30% of Zinc and rest nickel. The amount of nickel in 400 gm of is:

\[ \text{Nickel} = 100\% – (30\% + 30\%) = 40\% \Rightarrow \frac{40}{100} \times 400 = 160 \text{ gm} \] Answer: a. 40% of 400 gm

v. A number is first decreased by 40% and then increased by 40%. The equivalent change is:

Let number = 100
After 40% decrease = 60
Then 40% of 60 = 24, so final = 60 + 24 = 84 \[ \text{Net change} = 100 – 84 = 16\% \text{ decrease} \] Answer: c. 16% decrease

vi. Out of 700 eggs, 20% are rotten. The number of good eggs is:

\[ 20\% \text{ of } 700 = \frac{20}{100} \times 700 = 140\\ \text{Good eggs} = 700 – 140 = 560 \] Answer: b. 560

vii. 80% of 200 – 50 is equal to:

\[ \frac{80}{100} \times 200 = 160\\ 160 – 50 = 110 \] Only option b. 110 matches.
Answer: b. 110

viii. A number 80 is wrongly taken 100. The percentage error is:

\[ \text{Error} = \frac{100 – 80}{80} \times 100 = \frac{20}{80} \times 100 = 25\% \] Answer: b. 25%

ix. The price of an article was ₹ 680 last year. This year its price is ₹816. The percentage change in the price is:

\[ \text{Increase} = 816 – 680 = 136\\ \frac{136}{680} \times 100 = 20\% \\ \Rightarrow \frac{816 – 680}{680} \times 100\% \text{ increase} \] Answer: c. \( \frac{816 – 680}{680} \times 100\% \) increases

x. Statement 1 : To change a number in percentage to a ratio. write it as fraction with denominator 100 and then reduce it to the lowest terms if possible.
Statement 2 : A percentage can be converted to a fraction by removing sign of % dividing by 100.
Which of the following options is correct?

✅ Statement 1: TRUE → Write % as fraction over 100, reduce.
✅ Statement 2: TRUE → Remove % and divide by 100.
Answer: a. Both the statements are true

xi. Assertion(A): 9% of \(\sqrt{0.0169}\) is 0.0117.
Reason (R) : To a percentage of a quantity, we change the percentage to a fraction or a decimal and multiply by the quantity.

Step 1: Find \(\sqrt{0.0169} = 0.13\)
9% of 0.13 = \(\frac{9}{100} \times 0.13 = 0.0117\)
✔️ Assertion is Correct
Step 2: ✔️ Reason is also Correct and gives the correct method used in A
Answer: a. Both A and R are correct, and R is the correct explanation for A.

xii. Assertion (A): If we decrease ₹120 by \(12\frac{1}{2}\)%, then decreased amount = ₹105.
Reason (R) : Percentage change = \((\frac{Actual\ charge\ (increase/decrease)}{Original\ quantity}\times100\)%

Step 1: ₹120 decrease by \(12\frac{1}{2}\)% = \(\frac{25}{2}\)% = \(\frac{25}{200} \times 120 = 15\)
So, new amount = ₹120 – ₹15 = ₹105 ✔️
✔️ Assertion is Correct
Step 2: ✔️ Reason is Correct as it gives the formula to calculate % change
Step 3: ❌ But R is not the reason why A is true (we’re not calculating % change in A, just applying %)
Answer: b. Both A and R are correct, and R is not the correct explanation for A.

xiii. Assertion (A): By increasing ₹ 320 by 20%, we obtain the increased amount = ₹348.
Reason (R): To increase a quantity by a percentage. we first find the percentage of the quantity and then add it to the original quantity.

Step 1: 20% of ₹320 = \(\frac{20}{100} \times 320 = ₹64\)
₹320 + ₹64 = ₹384 ❌
But Assertion claims increased amount is ₹348 → FALSE
Step 2: ✔️ Reason is Correct, method is valid
Answer: d. A is false, but R is true.

xiv. Assertion (A): The sum of two numbers is \(\frac{28}{25}\) of the first number. Then the second number is 12% of the first number.
Reason (R): To express one quantity as a percentage of the other we write the other quantity as the fraction of the one and then multiply by 100.

Step 1: Let first number = \(x\)
Sum = \(\frac{28}{25}x\) → So second number = \(\frac{28}{25}x – x = \frac{3}{25}x\)
Now \(\frac{3}{25} \times 100 = 12\%\) ✔️
✔️ Assertion is Correct
✔️ Reason is Correct, and explains the method used
Answer: a. Both A and R are correct, and R is the correct explanation for A.

Q2: A family spends 30% of its income on house rent and 60% of the rest on house hold expenses. If the total savings of the family is ₹12,600 per month, find the total monthly income of the family.

Step 1: Let the total monthly income = ₹x
Step 2: House rent = 30% of income = \[ \frac{30}{100} \times x = \frac{3x}{10} \]Step 3: Remaining income after rent = \[ x – \frac{3x}{10} = \frac{7x}{10} \]Step 4: Household expenses = 60% of the remaining = \[ \frac{60}{100} \times \frac{7x}{10} = \frac{42x}{100} \]Step 5: Savings = Total income − (Rent + Household expenses) \[ \text{Savings} = x – \left(\frac{3x}{10} + \frac{42x}{100}\right) \]Step 6: Simplify the expenses: \[ \frac{3x}{10} = \frac{30x}{100},\quad \frac{42x}{100} = \frac{42x}{100} \\ \text{Total expenses} = \frac{30x + 42x}{100} = \frac{72x}{100} \]Step 7: Savings = \[ x – \frac{72x}{100} = \frac{28x}{100} \]Step 8: Given savings = ₹12,600 \[ \frac{28x}{100} = 12600 \\ \Rightarrow x = \frac{12600 \times 100}{28} = 45000 \]Answer: ₹45,000 is the total monthly income of the family.

Q3: Geeta saves 20% of her monthly salary and saves ₹43.500 per month. Find her monthly expenditure.

Step 1: Let Geeta’s monthly salary be ₹x
Step 2: Savings = 20% of salary \[ \frac{20}{100} \times x = \frac{x}{5} \]Step 3: Given savings = ₹43,500 \[ \frac{x}{5} = 43500 \Rightarrow x = 43500 \times 5 = ₹2,17,500 \]Step 4: Expenditure = Salary − Savings \[ 217500 – 43500 = ₹1,74,000 \]Answer: ₹1,74,000 is her monthly expenditure.

Q4: In an examination, 92% of the candidates passed and 96 failed. Find the number of candidates who appeared for this exam.

Step 1: Let the total number of candidates who appeared = x
Step 2: Passed candidates = 92% of x = \(\frac{92}{100} \times x = \frac{92x}{100}\)
Failed candidates = Total − Passed = \(x – \frac{92x}{100} = \frac{8x}{100}\)
Step 3: Given failed candidates = 96 \[ \frac{8x}{100} = 96 \\ \Rightarrow x = \frac{96 \times 100}{8} = 1200 \]Answer: 1,200 candidates appeared for the exam.

Q5: A number is increased by 30% and then this increased number is decreased by 30%. Find the net change.

Step 1: Let the original number be ₹100 (assumed for simplicity).
Step 2: Increase by 30%: \[ 100 + \frac{30}{100} \times 100 = 100 + 30 = ₹130 \]Step 3: Now decrease ₹130 by 30%: \[ 130 – \frac{30}{100} \times 130 = 130 – 39 = ₹91 \]Step 4: Compare final result with original ₹100: \[ \text{Net change} = 100 – 91 = ₹9 \text{ decrease} \]Step 5: Percentage change = \[ \frac{9}{100} \times 100 = 9\% \text{ decrease} \]Answer: 9% decrease

Q6: A number is decreased by 30% and then this decreased number is increased by 30%. Find the net change as percent.

Step 1: Let the original number be ₹100 (assumed for easy calculation).
Step 2: Decrease by 30%: \[ 100 – \frac{30}{100} \times 100 = 100 – 30 = ₹70 \]Step 3: Now increase ₹70 by 30%: \[ 70 + \frac{30}{100} \times 70 = 70 + 21 = ₹91 \]Step 4: Compare final result with original ₹100: \[ \text{Net change} = 100 – 91 = ₹9 \text{ decrease} \]Step 5: Percentage decrease = \[ \frac{9}{100} \times 100 = 9\% \]Answer: 9% decrease

Q7: The population of a village increases by 10% per year. If the present population of the village is 24,000; find it at the end of 2 years.

Step 1: Let present population = 24,000
Step 2: Population growth rate = 10% per year
Step 3: Use compound growth formula: \[ \text{Future Population} = P \left(1 + \frac{r}{100}\right)^t \] where P = 24,000, r = 10, t = 2
Step 4: Substitute the values: \[ \text{Population after 2 years} = 24000 \left(1 + \frac{10}{100}\right)^2 = 24000 \left(\frac{11}{10}\right)^2 = 24000 \times \frac{121}{100} \]Step 5: Simplify: \[ \frac{24000 \times 121}{100} = \frac{2904000}{100} = 29040 \]Answer: Population after 2 years will be 29,040

Q8: The cost of a machine decreases by 10% per year. If its present cost is ₹24,000; find its value at the beginning of 3rd year.

Step 1: Present cost of the machine = ₹24,000
Depreciation rate = 10% per year
Step 2: Use depreciation formula: \[ \text{Value after } t \text{ years} = P \left(1 – \frac{r}{100}\right)^t \] Here, P = 24000, r = 10, t = 2 (as we want value at beginning of 3rd year)
Step 3: Substitute the values: \[ \text{Value} = 24000 \left(1 – \frac{10}{100}\right)^2 = 24000 \left(\frac{9}{10}\right)^2 = 24000 \times \frac{81}{100} \]Step 4: Simplify: \[ 24000 \times \frac{81}{100} = \frac{1944000}{100} = ₹19,440 \]Answer: ₹19,440 is the value of the machine at the beginning of the 3rd year.

Q9: The price of sugar has been increased by 50%. By how much percent can the consumption of the sugar be decreased in order to keep the expenditure on sugar the same?

Step 1: Let the original price of sugar be ₹100 per kg.
Let the original consumption be 1 kg.
So, original expenditure = ₹100 × 1 = ₹100
Step 2: Price increased by 50%, so new price = ₹100 + 50% of 100 = ₹150
Step 3: Let new required quantity (to keep expenditure same) = x kg
New expenditure = ₹150 × x
We want new expenditure = old expenditure = ₹100
Step 4: Equating both expenditures: \[ 150x = 100 \Rightarrow x = \frac{100}{150} = \frac{2}{3} \text{ kg} \]Step 5: So, consumption decreased from 1 kg to \(\frac{2}{3}\) kg
Decrease in consumption = \(1 – \frac{2}{3} = \frac{1}{3}\) kg
Step 6: Percentage decrease in consumption = \[ \frac{1/3}{1} \times 100 = \frac{100}{3} = 33 \frac{1}{3}\% \]Answer: Consumption must be decreased by \(33 \frac{1}{3}\)% to keep the expenditure same.