Exercise : Percentage - General Questions
✔ Percentage -
General Questions
11.
The population of a town increases by 5% annually. If the current population is 10,000, what will it be after 2 years?
View Answer
Answer: Option B
Explanation:
Explanation:
This is a compound growth problem:
1. Population after 1 year: \(10,000 \times 1.05 = 10,500\).
2. Population after 2 years: \(10,500 \times 1.05 = 11,025\).
3. Alternatively, use the formula: \(P(1 + r)^n = 10,000(1.05)^2 = 10,000 \times 1.1025 = 11,025\).
1. Population after 1 year: \(10,000 \times 1.05 = 10,500\).
2. Population after 2 years: \(10,500 \times 1.05 = 11,025\).
3. Alternatively, use the formula: \(P(1 + r)^n = 10,000(1.05)^2 = 10,000 \times 1.1025 = 11,025\).
12.
In an exam, a student scored 35% and failed by 40 marks. If the passing percentage is 40%, find the maximum marks.
View Answer
Answer: Option C
Explanation:
Explanation:
1. Find the percentage gap between the student's score and the passing score: \(40\% - 35\% = 5\%\).
2. This 5% corresponds to the marks the student failed by: \(5\% \text{ of Max Marks} = 40\).
3. Solve for Max Marks: \(0.05 \times M = 40 \implies M = \frac{40}{0.05}\).
4. \(M = 800\).
2. This 5% corresponds to the marks the student failed by: \(5\% \text{ of Max Marks} = 40\).
3. Solve for Max Marks: \(0.05 \times M = 40 \implies M = \frac{40}{0.05}\).
4. \(M = 800\).
13.
A man spends 75% of his income. If his income increases by 20% and expenditure increases by 10%, find the % increase in savings.
View Answer
Answer: Option B
Explanation:
Explanation:
1. Assume original Income = 100, Expenditure = 75, Savings = 25.
2. New Income: \(100 + 20 = 120\).
3. New Expenditure: \(75 + 10\% \text{ of } 75 = 75 + 7.5 = 82.5\).
4. New Savings: \(120 - 82.5 = 37.5\).
5. Increase in savings: \(37.5 - 25 = 12.5\).
6. Percentage increase in savings: \(\left( \frac{12.5}{25} \right) \times 100 = 50\%\).
2. New Income: \(100 + 20 = 120\).
3. New Expenditure: \(75 + 10\% \text{ of } 75 = 75 + 7.5 = 82.5\).
4. New Savings: \(120 - 82.5 = 37.5\).
5. Increase in savings: \(37.5 - 25 = 12.5\).
6. Percentage increase in savings: \(\left( \frac{12.5}{25} \right) \times 100 = 50\%\).
14.
2 liters of water are evaporated from 10 liters of a solution containing 4% sugar. Find the % of sugar in the remaining solution.
View Answer
Answer: Option C
Explanation:
Explanation:
1. Calculate initial amount of sugar: \(4\% \text{ of } 10 = 0.4\) liters.
2. After evaporation, the amount of sugar remains the same, but total volume changes: \(10 - 2 = 8\) liters.
3. Calculate new percentage: \(\left( \frac{0.4}{8} \right) \times 100\).
4. \(0.05 \times 100 = 5\%\).
2. After evaporation, the amount of sugar remains the same, but total volume changes: \(10 - 2 = 8\) liters.
3. Calculate new percentage: \(\left( \frac{0.4}{8} \right) \times 100\).
4. \(0.05 \times 100 = 5\%\).
15.
If the side of a square is increased by 20%, find the percentage increase in its area.
View Answer
Answer: Option B
Explanation:
Explanation:
Area of a square is \(s^2\).
1. Let the original side be 10. Area = 100.
2. New side = \(10 + 20\% = 12\). New Area = \(12^2 = 144\).
3. Increase = \(144 - 100 = 44\).
4. Percentage increase = 44%.
5. Alternatively, use the successive change formula: \(x + y + \frac{xy}{100} = 20 + 20 + \frac{400}{100} = 44\%\).
1. Let the original side be 10. Area = 100.
2. New side = \(10 + 20\% = 12\). New Area = \(12^2 = 144\).
3. Increase = \(144 - 100 = 44\).
4. Percentage increase = 44%.
5. Alternatively, use the successive change formula: \(x + y + \frac{xy}{100} = 20 + 20 + \frac{400}{100} = 44\%\).
16.
The price of sugar is increased by 25%. By how much percent should a family reduce consumption to keep the expenditure same?
View Answer
Answer: Option A
Explanation:
Explanation:
Use the formula for constant expenditure when price changes: \(\text{Reduction} \% = \left[ \frac{r}{100 + r} \right] \times 100\).
1. Here \(r = 25\).
2. Reduction = \(\left[ \frac{25}{100 + 25} \right] \times 100\).
3. Reduction = \(\frac{25}{125} \times 100 = \frac{1}{5} \times 100 = 20\%\).
1. Here \(r = 25\).
2. Reduction = \(\left[ \frac{25}{100 + 25} \right] \times 100\).
3. Reduction = \(\frac{25}{125} \times 100 = \frac{1}{5} \times 100 = 20\%\).
17.
A number is first increased by 10% and then decreased by 10%. Find the net change.
View Answer
Answer: Option B
Explanation:
Explanation:
1. Let the number be 100.
2. Increase by 10%: \(100 \times 1.10 = 110\).
3. Decrease the result by 10%: \(110 \times 0.90 = 99\).
4. Net change: \(100 - 99 = 1\).
5. Since the value decreased, it is a 1% decrease.
6. Formula: \(\frac{x^2}{100}\) decrease always applies when increase and decrease percentages are same: \(\frac{10^2}{100} = 1\%\) decrease.
2. Increase by 10%: \(100 \times 1.10 = 110\).
3. Decrease the result by 10%: \(110 \times 0.90 = 99\).
4. Net change: \(100 - 99 = 1\).
5. Since the value decreased, it is a 1% decrease.
6. Formula: \(\frac{x^2}{100}\) decrease always applies when increase and decrease percentages are same: \(\frac{10^2}{100} = 1\%\) decrease.
18.
In an election between two candidates, one got 55% of total valid votes. 20% of votes were invalid. If total votes were 7500, find valid votes for the other candidate.
View Answer
Answer: Option B
Explanation:
Explanation:
1. Find total valid votes: \(7500 - 20\% \text{ of } 7500 = 7500 \times 0.80 = 6000\).
2. If one candidate got 55% of valid votes, the other candidate got \(100\% - 55\% = 45\%\) of valid votes.
3. Calculate 45% of 6000: \(0.45 \times 6000 = 2700\).
2. If one candidate got 55% of valid votes, the other candidate got \(100\% - 55\% = 45\%\) of valid votes.
3. Calculate 45% of 6000: \(0.45 \times 6000 = 2700\).
19.
If the radius of a circle is decreased by 50%, find the % decrease in area.
View Answer
Answer: Option C
Explanation:
Explanation:
Area of a circle is \(\pi r^2\).
1. Let the original radius be 10. Area = \(100\pi\).
2. New radius = 5 (50% decrease). New Area = \(25\pi\).
3. Decrease = \(100\pi - 25\pi = 75\pi\).
4. Percentage decrease = \(\left( \frac{75\pi}{100\pi} \right) \times 100 = 75\%\).
5. Successive formula: \(-50 - 50 + \frac{(-50 \times -50)}{100} = -100 + 25 = -75\%\).
1. Let the original radius be 10. Area = \(100\pi\).
2. New radius = 5 (50% decrease). New Area = \(25\pi\).
3. Decrease = \(100\pi - 25\pi = 75\pi\).
4. Percentage decrease = \(\left( \frac{75\pi}{100\pi} \right) \times 100 = 75\%\).
5. Successive formula: \(-50 - 50 + \frac{(-50 \times -50)}{100} = -100 + 25 = -75\%\).
20.
A fruit seller had some apples. He sells 40% and still has 420 apples. How many did he have originally?
View Answer
Answer: Option B
Explanation:
Explanation:
1. If he sells 40%, he has \(100\% - 40\% = 60\%\) of apples left.
2. Let total apples be \(x\). Then \(60\% \text{ of } x = 420\).
3. \(0.60x = 420\).
4. \(x = \frac{420}{0.60} = 700\).
2. Let total apples be \(x\). Then \(60\% \text{ of } x = 420\).
3. \(0.60x = 420\).
4. \(x = \frac{420}{0.60} = 700\).