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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
If the amount is \(2\frac{1}{4}\) times of the sum after 2 years, then the rate of compound interest must be (a) 60% (b) 40% (c) 64% (d) 50% |
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Answer» (d) 50% Let the sum be Rs x. Then, Amount =Rs \(\frac{9X}{4}\) n = 2, r = ? \(\therefore\) \(\frac{9X}{4}\) = x\(\big(1+\frac{r}{100}\big)^2\) \(\Rightarrow\) \(\frac{9}{4}\) = \(\big(1+\frac{r}{100}\big)^2\) \(\Rightarrow\) \(\big(\frac{3}{2}\big)^2\) = \(\big(1+\frac{r}{100}\big)^2\) \(\Rightarrow\) \(\frac{r}{100}\) = \(\frac{3}{2}\) - 1 = \(\frac{1}{2}\) \(\Rightarrow\) r = 50% p.a. |
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| 52. |
The simple interest on a sum of money for 2 years at 8% per annum is Rs. 2400. What will be the compound interest on that sum at the same rate and for the same period? |
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Answer» Simple interest = Rs.2400 Interest rate = 8% per annum Time = 2 years Simple interest (SI) = PRT/100 [where, P = Present value R = Interest rate ∴ 2400 = (P × 8 × 2)/100 T = Time] ⇒ 2400 = P × 16/100 ⇒ 2400 = P × 4/25 ⇒ P = 2400 × 25/4 ⇒ P = 600 × 25 ⇒ P = 15000 ∴ Sum = Rs.15000 Now, Amount (A) = P (1 + R/100)n [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 15000 [1 + 8/100]2 ⇒ A = 15000 [1 + 2/25]2 ⇒ A = 15000 [27/25]2 ⇒ A = 15000 × 27/25 × 27/25 ⇒ A = 600 × 27 × 27/25 ⇒ A = 24 × 27 × 27 ⇒ A = 17496 ∴ Amount = Rs.17496 ∴ Compound interest = Rs.(17496 – 15000) = Rs.2496 |
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| 53. |
A sum of money invested at compound interest amounts in 3 years to Rs 2400 and in 4 years to Rs 2520. The interest rate per annum is (a) 5% (b) 6% (c) 10% (d) 12% |
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Answer» (a) 5% Let the sum of money invested be Rs x and interest rate per annum = r% Then, x\(\big(1+\frac{r}{100}\big)^3\) = Rs 2400 .....(i) and x\(\big(1+\frac{r}{100}\big)^4\) = Rs 2520 ........(ii) Dividing equation (ii) by (i), we get \(\big(1+\frac{r}{100}\big)\) = \(\frac{2520}{2400}\) \(\Rightarrow\) \(\frac{r}{100}\) = \(\frac{2520-2400}{2400}\) = \(\frac{120}{2400}\) \(\Rightarrow\) r = \(\frac{1}{20}\)x 100 = 5% p.a |
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| 54. |
A financial company claims that it charges only 20% interest on loans. But if a person takes out a loan of Rs 100 he would get only Rs 80, after subtracting the annual interest of rupees 20 at the outset. And he has to pay back Its 100 after 1 year. How much is their real interest? |
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Answer» A borrower gets only Rs 80 when he borrows Rs 100 He has to repay Rs 100. ie. Rs.20 as additional. Real interest \(\frac{20}{80}\times100\) = 25% |
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| 55. |
The population of Kerala increases by 3% every year. The current population is 5 crore. What would he the population after 2 years. |
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Answer» Current population = 50000000 Percentages increase in every year = 3% The population in Kerala after 3 year = 50000000 [1 + \(\frac{3}{10}\)]2 = 50000000 × \(\frac{(103)^2}{10000}\) = 530,45000 |
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| 56. |
Rajan borrowed Rs. 15000 from a cooperative bank for business purpose. The bank computes 9% interest. How much money she has to repay after 5 months. |
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Answer» Amount borrowed = Rs. 15000 Rate of interest = 9% Time duration = 5 month months Interest . = 15000 × \(\frac{9}{100}\) × \(\frac{5}{12}\) = 562.50 Amount he has to repay = 15000 + 562.50 = Rs 15562.50 |
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| 57. |
Balu and Ramu decided to borrow Rs 15000 each for a joint business. Balu borrowed Rs 15000 from a financier Who imputes Rs 5 per month for Rs 100 as interest. Ramu borrowed Rs 15000 from a bank where compound interest of 12 % is computed. How much money both of them have to repay after 2 years? |
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Answer» Interest Balu has to pay after 2 years = 15000 × \(\frac{60}{100}\) × 2 = Rs. 18000 (Interest for Rs 100 per month in Rs 5. Interest for Rs 100 in an year = 5 × 12 = Rs. 60. The rate of interest = 60%) Amount Balu has to repay after 2 years = 15000 + 18000 = Rs 33000 In the case of Ramu interest is computed as compound interest. Principal for first year = Rs 15000 Interest for 1st year = 15000 × \(\frac{12}{100}\) × 1 = Rs 1800 Principal for 2nd year = 15000 + 1800 = Rs 16800 Interest for the second year = 16800 × \(\frac{12}{100}\) × 1 = Rs. 2018 Amount Ramu has to repay = 16800 + 2018 = Rs 18816 |
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| 58. |
Raju has Rs 800 with him. He spent 25 % of it. How much amount left with him? |
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Answer» Total amount = Rs 800 Amount spent = 800 × \(\frac{25}{100}\) = 200 Amount left with Raju = 800 – 200 = Rs 600 |
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| 59. |
The production of a company has ups and downs every year. The production increases for two consecutive years consistently by 15% and in the third year it decreases by 10%. Again in the next two years it increases by 15% each year and decreases by 10% in the third year. If we start counting from the year 1998, approximately what will be the effect on production of the company in 2002 ? (a) 37% increase (b) 32% increase (c) 42% increase (d) 52% increase |
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Answer» (a) 37% increase Let the production in 1998 be 100 units. Then, Production in 2002 = 100\(\big(1+\frac{15}{100}\big)^2\)\(\big(1-\frac{10}{100}\big)\)\(\big(1+\frac{15}{100}\big)\) = 100 x \(\frac{23}{20}\) x \(\frac{23}{20}\) x \(\frac{9}{10}\)x \(\frac{23}{20}\) = 136.88 \(\therefore\) Increase in production = (136.88 – 100)% = 36.88% = 37% |
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| 60. |
Ramu borrowed Rs 50,000 from a bank at the interest rate of 10% for agricultural purpose. The interest rate will be reduced by 5% if he repays the amount properly within 2 years. If he fails to repay in time there will be fine as 1%. Ramu could not repay the amount in time. How much amount Ramu repair? |
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Answer» Amount borrowed = Rs 50,000 Rate of interest = 10% Interest rate including fine = 10 + 1 = 11% Amount to be repaid = 50000 + [50000 × \(\frac{11}{100}\) × 2] = 50000 + 11000 = 61000 = Rs 61000 |
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| 61. |
Find the amount of Rs. 12500 for 2 years compounded annually, the rate of interest being 15% for the first year and 16% for the second year |
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Answer» Given, Principal = Rs.12500 Time = 2 years = R1 = 15% = R2 =16% Hence, Amount\(={p}[({1}+\frac{R_1}{100})]\) \([({1}+\frac{R_2}{100})]\) \(={12500}[{1}+\frac{15}{100}]\) \([{1}+\frac{16}{100}]\) \(={12500}\times\frac{23}{20}\times\frac{29}{25}\) = Rs. 16675 |
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| 62. |
A company which manufactures computers increases its production by 10% every year. In 2009 the company produced 80,000 computers. How many computers would it produced in 2011? |
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Answer» Here the number of computers produced every year is 10% more than the number produced the year before. So starting from 80,000. We have to find the number of computers produced every year after that for two years. Number of computers produced in 2009 = 80,000 Number of computers produced in 2010 = 80000 + 80000 × \(\frac{10}{100}\) = 80000 + 8000 = 88000 Number of computers produced in 2011 = 88000 + 88000 × \(\frac{10}{100}\) = 88000 + 8800 = 96800 |
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| 63. |
The price of a car is rupees 5 lakh and it depreciates by 6% every year. What would be the price after 2 year ? |
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Answer» Here the price every year is 6% less than the previous years price. First year’s price = Rs 500000 First year’s depreciation = 500000 × \(\frac{6}{100}\) = Rs. 300000 Second year’s price = Rs 4,70000 Second year’s depreciation = 470000 × \(\frac{6}{100}\) = 28200 The price of the car after 2 years = Rs 4,70,000 – Rs 28200 = Rs 441800 |
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| 64. |
The simple interest of an amount is Rs 50 for two years and the compound interest is Rs 55. The rate of interest is same in both the cases. Find the rate? Find the amount? |
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Answer» The simple interest of the amount for two years = Rs 50 The simple interest of the amount for 1 year = Rs 25 The compound interest of the amount for 2 years = Rs 55 The interest in the 2nd year = 55 – 25 = Rs 30 Interest for Rs 25 = Rs 5 |
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| 65. |
In a factory the production of scooters rose to 46305 from 40000 in 3 years. Find the annual rate of growth of the production of scooters. |
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Answer» Given, Initial production of scooters = 40000 Final production of scooters = 46305 Time duration = 3 years Let annual growth rate = R% So \(= 40000({1}+\frac{R}{100})({1}+\frac{R}{100})({1}+\frac{R}{100})\) = 46305 \(=({1}+\frac{R}{100})^3\) \(=\frac{46305}{40000}\) \(=\frac{9261}{8000}\) = \((\frac{21}{20})^3\) \(=1+\frac{R}{100}\) \(=\frac{21}{20}\) \(=\frac{R}{100}\) \(=\frac{21}{20}-1\) \(=\frac{1}{20}\) \(=R = \frac{1}{20}\times100\) = 5% Hence, Annual growth rate of production of scooters = 5 % |
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| 66. |
In the month of January, the railway police caught 4000 ticketless travellers. In February, the number rose by 5%. However, due to constant vigil by the police and the railway staff, the number reduced by 5% and in April it further reduced by 10% . The total number of ticketless travellers caught in the month of April was (a) 3125 (b) 3255 (c) 3575 (d) 3591 |
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Answer» (d) 3591 Total number of ticketless travellers in April = 4000 x \(\big(1+\frac{5}{100}\big)\)\(\big(1-\frac{5}{100}\big)\)\(\big(1-\frac{10}{100}\big)\) = 4000 x \(\frac{21}{20}\) x \(\frac{19}{20}\) x \(\frac{9}{10}\) = 3591. |
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| 67. |
Ramnath deposited Rs 30,000 in a financial institution where compound interest is computed in every four months. The annual interest rate in 9%. How much amount Ramnath gets after 1 year? |
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Answer» Amount deposited = Rs 30,000 Rate of interest = 9 % Amount gets after 1 year = 30000 (1 × \(\frac{3}{100}\))3 = 30000 × \(\frac{103}{100}\) × \(\frac{103}{100}\) × \(\frac{103}{100}\) = Rs. 32781.81 |
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| 68. |
A T.V manufacturer reduces the price of a particular model by 5% every year. The current price of this model is Rs. 8000. What would be the price after 3 years? |
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Answer» The present price of the T.V = Rs. 8000 rate of reduction = 5% , Price after 2 years = 8000 (1 – \(\frac{5}{100}\))2 = 8000 × (\(\frac{95}{100}\))2 = 8000 × \(\frac{95}{100}\) × \(\frac{95}{100}\) = Rs.7220 |
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| 69. |
The population of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will its production after 3 years? |
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Answer» Given, Population of mixi company in 1996 was = 8000 mixies Production growth rate in next 2 years is = 15 % Decrease rate in 3rd year is = 5% By using the formula, A = P (1 + R/100) x = 8000 (1 + 15/100) (1 + 15/100) (1 – 5/100) = 8000 (115/100) (115/100) (95/100) = 8000 (1.15) (1.15) (0.95) = 10051 ∴ Production after three years will be 10051 mixies. |
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| 70. |
Jitendra set up a factory by investing Rs. 2500000. During the first two successive years his profits were 5% and 10% respectively. If each year the profit was on previous year’s capital, compute his total profit. |
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Answer» Given details are, Initial investment by Jitendra was = Rs 2500000 Profit in first 2 successive years were = 5% and 10% Final investment after two successive profits = 2500000 (1+ 5/100) (1 + 10/100) = 2500000 (105/100) (110/100) = 2500000 (1.05) (1.1) = 2887500 ∴ Jitendra total profit is = 2887500 – 2500000 = Rs 387500. |
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| 71. |
The population of a city increases at the rate of 4% per annum. There is additional annual increase of 1% due to influx of job seekers. The % increase in the population after 2 years is (a) 10 (b) 10.25 (c) 10.5 (d) 10.75 |
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Answer» (b) 10.25 Let the population of the city be 100. Then Population of city after 2 years = 100\(\big(1+\frac{5}{100}\big)^2\) = 100 x \(\frac{21}{20}\) x \(\frac{21}{20}\) = 110.25 \(\therefore\) Increase per hundred in 2 years = 110.25 – 100 = 10.25% |
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| 72. |
The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. If its present value is Rs 8748, its purchase price was (a) Rs 10000 (b) Rs 11372 (c) Rs 12000 (d) Rs 12500 |
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Answer» (c) Rs 12000 Let the purchase price of the machine be Rs x. Then, 8748 = x\(\big(1-\frac{10}{100}\big)^3\) = x x \(\frac{9}{10}\) x \(\frac{9}{10}\) x \(\frac{9}{10}\) \(\Rightarrow\)x = \(\frac{8748\times1000}{729}\) = Rs 12000. |
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| 73. |
Pritam bought a plot of land for Rs. 640000. Its value is increasing by 5% of its previous value after every six months. What will be the value of the plot after 2 years? |
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Answer» Given details are, Price of land is = Rs 640000 Rate of increase = 5% in every six month By using the formula, A = P (1 + R/100)n = 640000 (1 + 5/100) (1 + 5/100) (1 + 5/100) (1 + 5/100) = 640000 (105/100) (105/100) (105/100) (105/100) = 640000 (1.025) (1.025) (1.025) (1.025) = 706440.25 ∴ The value of the plot after two years will be Rs 706440.25. |
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| 74. |
Ashish started the business with an initial investment of Rs. 500000. In the first year he incurred a loss of 4%. However during the second year he earned a profit of 5% which in third year rose to 10%. Calculate the net profit for the entire period of 3 years. |
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Answer» Given, Initial investment by Ashish = Rs.500000 Loss in first year = 4% Profit in 2nd year = 5 % Profit in 3rd year = 10% Hence, Finally investment becomes \(= 500000({1}-\frac{4}{100})({1}+\frac{5}{100})({1}+\frac{10}{100})\) \(= 500000\times\frac{24}{25}\times\frac{21}{20}\times\frac{11}{10}\) = Rs.5090400 Net profit = Rs.(5090400 – 500000) = Rs.554400 |
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| 75. |
The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (1) 2001 (ii) 1997. |
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Answer» Given, Annually increase rate of population of city = 4% Population in 1999 = 6760000 So , i) Population of city in 2001 (2 years after) \(= 6760000({1}+\frac{4}{100})({1}+\frac{4}{100})\) \(= 6760000\times\frac{26}{25}\times\frac{26}{25}\) = 7311616. ii) Population of city in 1997 (2 years ago) \(= 6760000({1}-\frac{4}{100})({1}-\frac{4}{100})\) \(= 6760000\times\frac{21}{25}\times\frac{21}{25}\) = 6750000 |
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| 76. |
The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 100000? Also, find the total depreciation during this period. |
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Answer» Given details are, Present value of machine is = Rs 100000 Rate of depreciation = 10% per annum By using the formula, A = P (1 + R/100) = 100000 (1 – 10/100) (1 – 10/100) = 100000 (90/100) (90/100) = 100000 (0.9) (0.9) = 81000 Value of machine after two years will be Rs 81000 ∴ Total depreciation during this period is Rs (100000 – 81000) = Rs 19000 |
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| 77. |
Ishita invested a sum of Rs. 12000 at 5% per annum compound interest. She received an amount of Rs. 13230 after n years. Find the value of n. |
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Answer» Given, Principal = Rs.12000 Rate = 5% per annum Amount = Rs.13230 Let time = T years So, A \(={P}[({1}+\frac{R}{100})^T]\) \(={12000}[({1}+\frac{5}{100})^T]\) = 13230 \(=(\frac{21}{20})^T\) \(=\frac{13230}{12000}\) \(=\frac{441}{4000}\) \(=(\frac{21}{20})^2\) = T= 2 years Hence , Time = 2 years |
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| 78. |
The cost of a T.V. set was quoted Rs. 17000 at the beginning of 1999. In the beginning of 2000 the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4% in the beginning of 2001. What was the cost of the T.V. set in 2001? |
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Answer» Given, Cost of T.V at beginning og 1999 = Rs.17000 Hiked in price in 2000 = 5% Depreciation in 2001 = 4% So, Price of T.V in 2001 \(= 17000({1}+\frac{5}{100})({1}-\frac{5}{10})\) \(= 17000\times\frac{21}{20}\times\frac{24}{25}\) = Rs 17136 |
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| 79. |
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs. 43740, find its purchase price |
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Answer» Given, Depreciation rate of machine = 10% p.a Present value of machine = Rs.43740 Let its purchase price 3 years ago = Rs. x So, \(=x({1}+\frac{10}{100})({1}+\frac{10}{100})({1}+\frac{10}{100})\) = 43740 \(x\times\frac{11}{10}\times\frac{11}{10}\times\frac{11}{10}\) = 43740 \(= x =\frac{43740 \times 10 \times 10 \times 10}{11 \times 11 \times 11}\) = 60000 Hence, Purchase price of machine was = Rs.60000 |
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| 80. |
In how much time will a sum of Rs. 1600 amount to Rs. 1852.20 at 5% per annum compound interest? |
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Answer» Given, Principal = Rs.1600 Amount = Rs.1852.20 Rate = 5 % per annum Let time = T years So, A = \({P}[({1}+\frac{R}{100})^T]\) \(={1600}({1}+\frac{5}{100})^T\) = 185.20 \(=(\frac{21}{20})^T\) \(=\frac{1852.20}{1600}\) \(=\frac{9261}{8000}\) \(=(\frac{21}{20})^3\) = T = 3 years Hence, Time = 3 years |
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| 81. |
Three years ago, Beeru purchased a buffalo from Surjeet for Rs. 11000. What payment will discharge his debt now, the rate of interest being 10% per annum, compounded annually? |
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Answer» Present value = Rs.11000 Interest rate = 10% per annum Time = 3 years Amount (A) = P (1 + R/100)n [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 11000 (1 + 10/100)3 ⇒ A = 11000 (1 + 1/10)3 ⇒ A = 11000 (11/10)3 ⇒ A = 11000 × 11/10 × 11/10 × 11/10 ⇒ A = 11000 × 1331/1000 ⇒ A = 11 × 1331 ⇒ A = 14641 ∴ Amount = Rs.14641 ∴ Beeru has to pay Rs.14641 to clear the debt. |
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| 82. |
Three years ago, Beeru purchased a buffalo from Surjeet for ₹ 11000. What paymentWill discharge his debt now, the rate of interest being 10% per annum, compounded annually? |
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Answer» Given: Present value= ₹ 11000 Interest rate= 10% per annum Time=3 years To find the amount we have the formula, Amount (A) = P (1+(R/100))n Where P is present value, r is rate of interest, n is time in years. Now substituting the values in above formula we get, ∴ A = 11000 (1 + 10/100)3 ⇒ A = 11000 (11/10)3 ⇒ A = 11000 (1331/1000) ⇒ A = 11 (1331) ⇒ A = ₹ 14614 ∴ Beeru has to pay ₹ 14614 |
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| 83. |
The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs. 9680, for how much was it purchased? |
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Answer» Given details are, Present value of refrigerator is = Rs 9680 Depreciation rate is = 12% Let the price of refrigerator 2 years ago be = Rs x By using the formula, A = P (1 + R/100)n 9680 = x (1 – 12/100) (1 – 12/100) 9680 = x (88/100) (88/100) 9680 = x (0.88) (0.88) 9680 = 0.7744x x = 9680/0.7744 = 12500 ∴ The refrigerator was purchased for Rs 12500. |
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| 84. |
The population of a village is 10000. If the population increases by 10% in the first year, by 20% in the second year and due to mass exodus it decreases by 5% in the third year, what will be its population after 3 years ? |
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Answer» Population after 3 years = 10000\(\big(1+\frac{10}{100}\big)\)\(\big(1+\frac{20}{100}\big)\)\(\big(1-\frac{5}{100}\big)\) = 10000 x \(\frac{11}{10}\) x \(\frac{6}{5}\) x \(\frac{19}{20}\) = 12540. |
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| 85. |
Anand obtained a loan of Rs. 125000 from the Allahabad Bank for buying computers. The bank charges compound interest at 8% per annum, compounded annually. What amount pwill he have to pay after 3 years to clear the debt? |
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Answer» Present value = Rs.125000 Interest rate = 8% per annum Time = 3 years Amount (A) = P (1 + R/100)n [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 125000 (1 + 8/100)3 ⇒ A = 125000 (108/100)3 ⇒ A = 125000 × 108/100 × 108/100 × 108/100 ⇒ A = 125000 × 1259712/1000000 ⇒ A = 125 × 1259712/1000 ⇒ A = 1259712/8 ⇒ A = 157464 ∴ Amount = Rs.157464 ∴ Anand has to pay Rs.157464 after 3 years to clear the debt. |
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| 86. |
Anand obtained a loan of ₹ 125000 from the Allahabad bank for buying computers. The bank charges compound interest at 8% per annum, compounded annually. What amount will have to pay after 3 years to clear the debt? |
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Answer» Given: Present value= ₹ 125000 Interest rate= 8 % per annum Time= 3 years To find the amount we have the formula, Amount (A) = P (1+(R/100))n Where P is present value, r is rate of interest, n is time in years. Now substituting the values in above formula we get, ∴ A = 125000 (1 + 8/100)3 ⇒ A = 125000 (1+108/100)3 ⇒ A = 125000 (108/100) (108/100) (108/100) ⇒ A = 125000 × 1259712/1000 = 1259712/8 ⇒ A = ₹ 157464 Anand has to pay ₹ 157464 |
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| 87. |
The value of a machine depreciates at the rate of 20% per annum. It was purchased 2 years ago. If its present value is Rs. 40000 for how much was it purchased? A. Rs. 56000 B. Rs. 62500 C. Rs. 65200 D. Rs. 56500 |
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Answer» Let value of a machine 2 years ago, = P Present value of machine = Rs.40000 Time, n = 2 years Rate of depreciates, R = 20% per annum Now, Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest P = Present value R = Annual interest rate n = Time in years] ∴ Value = P (1 - R/100)n [∵ Rate decreases] ⇒ 40000 = P (1 - 20/100)2 ⇒ 40000 = P (1 - 1/5)2 ⇒ 40000 = P (4/5)2 ⇒ 40000 = P × 16/25 ⇒ P = 40000 × 25/16 ⇒ P = 2500 × 25 ⇒ P = 62500 ∴ Value of a machine 2 years ago is Rs.62500. |
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| 88. |
If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 1200 then the compound interest on the same sum for the same period at the same rate will be A. Rs. 1225 B. Rs. 1236 C. Rs. 1248 D. Rs. 1261 |
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Answer» Interest rate, R = 5% per annum Time = 3 years Simple interest = Rs.1200 Simple interest = PRT/100 ⇒ 1200 = (P × 5 × 3)/100 ⇒ 1200 = P × 15/100 ⇒ P = 1200 × 100/15 ⇒ P = 8000 Now, Amount (A) = P (1 + R/100)n = 8000 (1 + 5/100)3 = 8000 (1 + 1/20)3 = 8000 (21/20)3 = 8000 × 9261/8000 = 9261 ∴ Amount = 9261 ∴ Compound interest = Rs.(9261 – 8000) [∵CI = A – P] = Rs.1261 |
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| 89. |
The population of a town increases by 5% annually. If the population in 2009 is 1,38,915 what was it in 2006 ? |
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Answer» Let the population in 2006 be x. Pop. in 2009 = x x \(\big(1+\frac{5}{100}\big)^3\) \(\Rightarrow\) 1,38,915 = x x \(\big(\frac{21}{20}\big)^3\) \(\Rightarrow\) x = \(\frac{138915\times 8000}{9261}\) = 120000 |
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| 90. |
The annual increase in the population of a town is 10%. If the present population of the town is 180000, then what will be its population after two years ? |
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Answer» Population after two years = 180000\(\big(1+\frac{10}{100}\big)^2\) = 180000 x \(\big(\frac{11}{10}\big)^2\) = 180000 x \(\frac{121}{100}\) = 217800. |
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| 91. |
Sum of money amounts to Rs. 453690 in 2 years at 6.5% per annum compounded annually. Find the sum. |
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Answer» Given, Amount = Rs.453690 Time = 2 years Rate = 6.5 % p.a Let principal = Rs. P So, A \(={P}[({1}+\frac{R}{100})^T]\) \(={P}[({1}+\frac{6.5}{100})^2]\) = 453690 \(={P}\times\frac{213}{200}\times\frac{213}{200}\) = 453690 \(={P}=\frac{453690 \times 200 \times 200}{213\times213}\) = Rs. 400000 Hence , Principal = Rs. 400000 |
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| 92. |
The annual rate of growth in population of a town is 10%. If its present population is 33275, what was it 3 years ago? A. Rs. 25000 B. Rs. 27500 C. Rs. 30000 D. Rs. 26000 |
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Answer» Let 3 years ago population = P Present population = 33275 Time, n = 3 years Increases rate, R = 10% per annum Now, Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest P = Present value R = Annual interest rate n = Time in years] ∴ Population = P (1 + R/100)n ⇒ 33275 = P (1 + 10/100)3 ⇒ 33275 = P (1 + 1/10)3 ⇒ 33275 = P (11/10)3 ⇒ 33275 = P × 1331/1000 ⇒ P = 33275 × 1000/1331 ⇒ P = 25 × 1000 ⇒ P = 25000 ∴ 3 years ago population is 25000. |
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| 93. |
The population of a town was 1,60,000 three years ago. If it increased by 3%, 2.5% and 5% respectively in the last three years, then what is the present population ? |
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Answer» Present population = 1,60,000\(\big(1+\frac{3}{100}\big)\big(1+\frac{2.5}{100}\big)\big(1+\frac{5}{100}\big)\) = 160000 x \(\frac{103}{100}\) x \(\frac{102.5}{100}\) x \(\frac{105}{100}\) = 177366. |
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| 94. |
The population of a city was 120000 in the year 2013. During next year it increased by 6% but due to an epidemic it decreased by 5% in the following year. What was its population in the year 2015? |
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Answer» Population of a city in 2013, P = 120000 Time, n = 3 years Increasing rate, R = 6% per annum Now, Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest P = Present value R = Annual interest rate n = Time in years] ∴Population of the city in the year 2014, ∴ Population = P (1 + R/100)n = 120000 (1 + 6/100)1 = 120000 (1 + 3/50) = 120000 (53/50) = 120000 × 53/50 = 2400 × 53 = 127200 ∴ Population of a city in 2014 is 127200. Now, Decreasing rate = 8% ∴ Population of the city in the year 2015, ∴ Population = P (1 - R/100)n = 127200 (1 - 5/100)1 = 127200 (1 - 1/20) = 127200 (19/20) = 127200 × 19/20 = 6360 × 19 = 120840 ∴ Population of a city in 2015 is 120840. |
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| 95. |
The population of a town is 125000. It is increasing at the rate of 2% per annum. What will be its population after 3 years? |
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Answer» Population of a town, P = 125000 Time, n = 3 years Increasing rate, R = 2% per annum Now, Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest P = Present value R = Annual interest rate n = Time in years] ∴ Population = P (1 + R/100)n = 125000 (1 + 2/100)3 = 125000 (1 + 1/50)3 = 125000 (51/50)3 = 125000 × 51/50 × 51/50 × 51/50 = 1 × 51 × 51 × 51 = 132651 ∴ Population of a town after 3 years is 132651. |
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| 96. |
A sum of money deposited at 2% per annum compounded annually becomes Rs. 10404 at the end of 2 years. Find the sum deposited. |
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Answer» Given details are, Time = 2years Amount = Rs 10404 Rate be = 2% per annum Let principal be = Rs P By using the formula, A = P [(1 + R/100)n 10404 = P [(1 + 2/100)2] 10404 = P [1.0404] P = 10404/1.0404 = 10000 ∴ Required sum is Rs 10000. |
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| 97. |
Find the amount of Rs. 2400 after 3 years, when the interest is compounded annually at the rate of 20% per annum. |
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Answer» Given details are, Principal (p) = Rs 2400 Rate (r) = 20% per annum Time (t) = 3 years By using the formula, A = P (1 + R/100) n = 2400 (1 + 20/100)3 = 2400 (120/100)3 = Rs 4147.2 ∴ Amount is Rs 4147.2 |
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| 98. |
Compute the amount and the compound interest by using the formulae when : Principal = Rs 5000, Rate = 10 paise per rupee per annum, Time = 2 years. |
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Answer» Given, P = Rs 5000, rate = 10%, time = 2years A = P (1 + R/100) n = 5000 (1 + 10/100)2 = 5000 (110/100)2 = Rs 6050 Compound interest (CI) = A-P = Rs 6050 – 5000 = Rs 1050 |
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| 99. |
Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 5%, 4% and 3% respectively, what is its present population? |
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Answer» Population of a town, P = 50000 Interest rate for 1st year, p = 5% Interest rate for 2nd year, q = 4% Interest rate for 3rd year, r = 3% Time, n = 3 years Now, Present population = P × (1 + p/100) × (1 + q/100) × (1 + r/100) = 50000 × (1 + 5/100) × (1 + 4/100) × (1 + 3/100) = 50000 × (1 + 1/20) × (1 + 1/25) × (1 + 3/100) = 50000 × 21/20 × 26/25 × 103/100 = 50 × 21/2 × 26/25 × 103 = 1 × 21 × 26 × 103 = 56238 ∴ Present population of a town is 56238. |
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| 100. |
Find the amount of Rs. 406 for 18 months at 12 1/2 % per annum, the interest being compounded semiannually. |
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Answer» Given, Principal = Rs.406 Time = 18 Months \(=\frac{18}{12}\) \(={1}\frac{1}{2}\) \(=\frac{3}{2}\times{2}\) = 3 half year Rate \(={12}\frac{1}{2}\text%\) per annum \(=\frac{25}{4}\text%\) half yearly So, \(={406}[({1}+\frac{25}{4\times100})^3]\) \(={406}(\frac{17}{16})^3\) \(=\frac{406\times4913}{4096}\) = Rs. 486.98 Amount = Rs.486.98 |
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