Percentage And Its Applications — Practice Quiz
NIOS · Class 10 · Maths
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Quick Quiz: Percentage And Its Applications
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What percent of 250 is 75?
A shopkeeper buys an article for ₹800 and sells it for ₹960. What is the profit percent?
A shirt marked at ₹600 is sold at a discount of 15%. Find the selling price.
Find the simple interest on ₹5000 for 3 years at 8% per annum.
Sample Questions
A sum of ₹10000 becomes ₹12100 in 2 years at compound interest. What is the rate of interest?
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10%
Step 1: Given P = ₹10000, A = ₹12100, T = 2 years. We need to find R. Step 2: Use the compound interest formula: A = P(1 + R/100)². Step 3: Substitute values: 12100 = 10000(1 + R/100)². Step 4: Simplify: (1 + R/100)² = 12100/10000 = 1.21 = (1.1)². Step 5: Therefore, 1 + R/100 = 1.1, which gives R = 10%.
The population of a town increases from 50000 to 52000. Find the percent increase.
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4%
Step 1: Initial population = 50000, Final population = 52000. Step 2: Calculate increase: Increase = 52000 - 50000 = 2000. Step 3: Use percent increase formula: Percent increase = (Increase/Original value) × 100. Step 4: Substitute values: Percent increase = (2000/50000) × 100 = 4%. Step 5: Therefore, the population increased by 4%.
A man saves 20% of his income. If his savings are ₹4000, what is his income?
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₹20000
Step 1: Let income = x. Given: 20% of income = ₹4000. Step 2: Write the equation: (20/100) × x = 4000. Step 3: Simplify: 0.20x = 4000. Step 4: Solve for x: x = 4000/0.20 = 20000. Step 5: Therefore, his income is ₹20000. Verification: 20% of 20000 = 4000 ✓
A machine worth ₹100000 depreciates at 10% per annum. What will be its value after 2 years?
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₹81000
Step 1: Initial value = ₹100000, depreciation rate = 10% per annum, time = 2 years. Step 2: Use depreciation formula: Final value = Initial value × (1 - rate/100)ⁿ. Step 3: Substitute values: Final value = 100000 × (1 - 10/100)². Step 4: Calculate: Final value = 100000 × (0.9)² = 100000 × 0.81 = ₹81000. Step 5: Therefore, the value after 2 years is ₹81000.
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