Practice Questions
Concept Explanation
Growth (or appreciation) increases a quantity each year at a fixed rate, using the compound interest formula with an increase.
Key Formula
Value after T years = P(1 + R/100)^T
Example
Population 1,00,000 grows at 10% p.a. After 2 years: 100000 × 1.1^2 = 1,21,000.
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Q1 EasyThe population of a town is 1,00,000. It grows at 10% per annum. Population after 2 years is:ExplanationFormula: Value after T years = P(1 + R/100)^T. Population = 100000 × (1.1)^2 = 100000 × 1.21 = 1,21,000. Hence the answer is 1,21,000.
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Q2 EasyA property worth Rs. 50,000 appreciates at 20% per annum. Its value after 2 years is:ExplanationFormula: Value = P(1 + R/100)^T. Value = 50000 × (1.2)^2 = 50000 × 1.44 = Rs. 72,000. Hence the answer is Rs. 72,000.
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Q3 EasyRs. 40,000 appreciates at 10% per annum for 2 years. The appreciated value is:ExplanationFormula: Value = P(1 + R/100)^T. Value = 40000 × (1.1)^2 = 40000 × 1.21 = Rs. 48,400. Hence the answer is Rs. 48,400.
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Q4 MediumPopulation of a town increases by 10% annually. If present population is 22000, population 2 years ago:ExplanationFormula: Past population = Present / (1 + R/100)^T. Population 2 years ago = 22000 / (1.1)^2 = 22000 / 1.21 = 18181.81 (approx.). Hence the answer is 18,181.81.
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Q5 MediumTown population is 80,000 and grows at 5% per annum. Population after 2 years is:ExplanationFormula: Population = P(1 + R/100)^T. Population = 80000 × (1.05)^2 = 80000 × 1.1025 = 88,200. Hence the answer is 88,200.
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Q6 MediumAn investment of Rs. 1,00,000 appreciates at 8% per annum. Value after 2 years is:ExplanationFormula: Value = P(1 + R/100)^T. Value = 100000 × (1.08)^2 = 100000 × 1.1664 = Rs. 1,16,640. Hence the answer is Rs. 1,16,640.
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Q7 MediumPresent population of a city is 1,21,000. It grew at 10% annually for the last 2 years. Population 2 years ago was:ExplanationFormula: Past = Present / (1 + R/100)^T. Past = 121000 / (1.1)^2 = 121000 / 1.21 = 1,00,000. Hence the answer is 1,00,000.
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Q8 MediumPopulation of a town increases by 8% annually. If present population is 1,16,640, population 2 years ago was:ExplanationFormula: Past = Present / (1 + R/100)^T. Past = 116640 / (1.08)^2 = 116640 / 1.1664 = 1,00,000. Hence the answer is 1,00,000.
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Q9 MediumRs. 60,000 grows at 10% per annum for 3 years. The amount becomes:ExplanationFormula: Value = P(1 + R/100)^T. Value = 60000 × (1.1)^3 = 60000 × 1.331 = Rs. 79,860. Hence the answer is Rs. 79,860.
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Q10 HardThe population of a city increases at 5% per annum. Present population 185220. Population 3 years ago:ExplanationFormula: Past = Present / (1 + R/100)^T. Population 3 years ago = 185220 / (1.05)^3 = 185220 / 1.157625 = 1,60,000. Check: 160000 × 1.05^3 = 185220. Hence the answer is 1,60,000.
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Q11 HardA tree increases annually by 1/8th of its height. Present height 2187 cm. Height 3 years ago:ExplanationEach year height becomes 9/8 of previous height. Height 3 years ago = 2187 × (8/9)^3 = 2187 × 512 / 729. Since 2187 = 3 × 729, height = 3 × 512 = 1,536 cm. Hence the answer is 1,536 cm.
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Q12 HardPopulation of a village is 50,000. It increases by 10% every year. Population after 3 years is:ExplanationFormula: Population = P(1 + R/100)^T. Population = 50000 × (1.1)^3 = 50000 × 1.331 = 66,550. Hence the answer is 66,550.
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Q13 HardPopulation 1,20,000 grows at 5% per annum for 3 years. The population becomes:ExplanationFormula: Population = P(1 + R/100)^T. Population = 120000 × (1.05)^3 = 120000 × 1.157625 = 1,38,915. Hence the answer is 1,38,915.
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Q14 HardA plant grows by 20% every year. If its present height is 1,728 cm, height 3 years ago was:ExplanationFormula: Past = Present / (1 + R/100)^T. Past height = 1728 / (1.2)^3 = 1728 / 1.728 = 1,000 cm. Hence the answer is 1,000 cm.
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DS1 Easy
What is the population after 2 years?
Statement IPresent population is 1,00,000 growing at 10% per annum for 2 years.
Statement IIThe annual growth rate is 10%.
ExplanationFormula: Population = P(1 + R/100)^T. Statement I alone: 100000 × (1.1)^2 = 1,21,000. Statement II alone is incomplete. Statement I alone is sufficient. Answer: A. -
DS2 Medium
A company worksheet asks for the appreciated value. What is the value after 2 years?
Statement IThe initial value is Rs. 50,000.
Statement IIIt appreciates at 20% per annum for 2 years.
ExplanationNeed initial value, rate and time. Together: Value = 50000 × (1.2)^2 = Rs. 72,000. Both are needed. Answer: C.
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DS3 Medium
Is the value after appreciation greater than Rs. 1,00,000?
Statement IRs. 80,000 appreciates at 8% per annum for 2 years.
Statement IIThe appreciation rate is 8% per annum.
ExplanationStatement I alone: Value = 80000 × (1.08)^2 = Rs. 93,312, which is not greater than Rs. 1,00,000, so No. Statement II alone cannot decide. Statement I alone is sufficient. Answer: A. -
DS4 Hard
Find the population 3 years ago. What was the population?
Statement IPresent population is 1,85,220 and growth rate is 5% per annum.
Statement IIThe growth has been continuous for exactly 3 years.
ExplanationFormula: Past = Present / (1 + R/100)^T. Statement I alone lacks time. Statement II alone lacks present population and rate. Together: Past = 185220 / (1.05)^3 = 1,60,000. Both are needed. Answer: C.