Class 10: Compound Interest – Exercise 1(a) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Calculate the amount and the compound interest on:

$\displaystyle \text{ i) Rs. } 12000 \text{ for 2 years at} 5\% \text{ per annum compounded annually. }$

$\displaystyle \text{ ii) Rs. } 8000 \text{ for years at} 10\% \text{ per annum compounded yearly.}$

$\displaystyle \text{ iii) Rs. } 8000 \text{ for years at} 10\% \text{ per annum compounded half-yearly.}$

Answer:

i) $\displaystyle \text{For 1st year: } P = \text{ Rs. } 12000; R=5\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{12000 \times 5 \times 1}{100} = \text{ Rs. } 600$

$\displaystyle \text{and, Amount } = 12000 + 600 = \text{ Rs. } 12600$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 12600; R=5\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{12600 \times 5 \times 1}{100} = \text{ Rs. } 630$

$\displaystyle \text{and, Amount } = 12600 + 630 = \text{ Rs. } 13230$

$\displaystyle \text{and Compound Interest } = 600+630 = \text{ Rs. } 1230$

$\displaystyle \\$

ii) $\displaystyle \text{For 1st year: } P = \text{ Rs. } 8000; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{8000 \times 10 \times 1}{100} = \text{ Rs. } 800$

$\displaystyle \text{and, Amount } = 8000 + 800 = \text{ Rs. } 8800$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 8800; R=10\% \text{ and } T=0.5 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{8800 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 440$

$\displaystyle \text{and, Amount } = 8800 + 440 = \text{ Rs. } 9240$

$\displaystyle \text{and Compound Interest } = 800+440 = \text{ Rs. } 1240$

$\displaystyle \\$

iii) $\displaystyle \text{For 1st half-year: } P = \text{ Rs. } 8000; R=10\% \text{ and } T= \frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{8000 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 400$

$\displaystyle \text{and, Amount } = 8000 + 400 = \text{ Rs. } 8400$

$\displaystyle \text{For 2nd half-year: } P = \text{ Rs. } 8400; R=10\% \text{ and } T= \frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{8400 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 420$

$\displaystyle \text{and, Amount } = 8400 + 420 = \text{ Rs. } 8820$

$\displaystyle \text{For 3rd half-year: } P = \text{ Rs. } 8820; R=10\% \text{ and } T= \frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{8820 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 441$

$\displaystyle \text{and, Amount } = 8820 + 441 = \text{ Rs. } 9261$

$\displaystyle \text{and Compound Interest } = 400+420+441 = \text{ Rs. } 1261$

$\displaystyle \\$

Question 2: Calculate the amount and the compound interest on $\displaystyle \text{ Rs. } 12,500 in 3$ years when the rates of interest for successive years are $\displaystyle 8\%, 10\% \text{ and } 10\%$ respectively:

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 12500; R=8\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{12500 \times 8 \times 1}{100} = \text{ Rs. } 1000$

$\displaystyle \text{and, Amount } = 12500 + 1000 = \text{ Rs. } 13500$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 13500; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{13500 \times 10 \times 1}{100} = \text{ Rs. } 1350$

$\displaystyle \text{and, Amount } = 13500 + 1350 = \text{ Rs. } 14850$

$\displaystyle \text{For 3rd year: } P = \text{ Rs. } 14850; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{14850 \times 10 \times 1}{100} = \text{ Rs. } 1485$

$\displaystyle \text{and, Amount } = 14850 + 1485 = \text{ Rs. } 16335$

$\displaystyle \text{and Compound Interest } = 1000+1350+1485 = \text{ Rs. } 3835$

$\displaystyle \\$

Question 3: A Man lends $\displaystyle \text{ Rs. } 5500 \text{ at the rate of } 8\%$ per annum. Find the amount if the interest is compounded half-yearly and the duration is one year.

Answer:

$\displaystyle \text{For 1st half-year: } P = \text{ Rs. } 5500; R=8\% \text{ and } T= \frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{5500 \times 8 \times 1}{100 \times 2} = \text{ Rs. } 220$

$\displaystyle \text{and, Amount } = 5500 + 220 = \text{ Rs. } 5720$

$\displaystyle \text{For 2nd half-year: } P = \text{ Rs. } 5720; R=8\% \text{ and } T= \frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{5720 \times 8 \times 1}{100 \times 2} = \text{ Rs. } 228.80$

$\displaystyle \text{and, Amount } = 5720 + 228.80 = \text{ Rs. } 5948.80$

$\displaystyle \text{and Compound Interest } = 220+228.80 = \text{ Rs. } 448.80$

$\displaystyle \\$

Question 4: A man borrows $\displaystyle \text{ Rs. } 8500 \ at \ 10\%$ compound interest. If he repays $\displaystyle \text{ Rs. } 2700$ at the end of each year, find the amount of the loan outstanding at the beginning of the third year.

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 8500; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{8500 \times 10 \times 1}{100} = \text{ Rs. } 850$

$\displaystyle \text{and, Amount } = 8500 + 850 = \text{ Rs. } 9350$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } (9350-2700)=6650; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{6650 \times 10 \times 1}{100} = \text{ Rs. } 665$

$\displaystyle \text{and, Amount } = 6650 + 665 = \text{ Rs. } 7315$

and Amount left at the beginning of 3rd year $\displaystyle = 7315-2700=\text{ Rs. } 4615$

$\displaystyle \\$

Question 5: A man borrows $\displaystyle \text{ Rs. } 10,000 \ at \ 5\%$ per annum compound interest. He repays $\displaystyle 35\%$ of the sum borrowed at the end of the first year and $\displaystyle 42\%$ of the sum borrowed at the end of the second year. How much must he pay at the end of the third year in order the debt?

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 10000; R=5\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{10000 \times 5 \times 1}{100} = \text{ Rs. } 500$

$\displaystyle \text{and, Amount } = 10000 + 500 = \text{ Rs. } 10500$

He repays $\displaystyle 35\% of 10000 = \text{ Rs. } 3500$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } (10500-3500)=7000; R=5\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{7000 \times 5 \times 1}{100} = \text{ Rs. } 350$

$\displaystyle \text{and, Amount } = 7000 + 350 = \text{ Rs. } 7350$

He repays $\displaystyle 42\% of 10000 = \text{ Rs. } 4200$

and Amount left at the beginning of 3rd year $\displaystyle = 7350-4200=\text{ Rs. } 3150$

$\displaystyle \text{For 3rd year: } P = \text{ Rs. } 3150; R=5\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{3150 \times 5 \times 1}{100} = \text{ Rs. } 157.50$

$\displaystyle \text{and, Amount } = 3150+157.50 = \text{ Rs. } 3307.50$

$\displaystyle \\$

Question 6: Rachana borrows $\displaystyle \text{ Rs. } 12,000 \ at \ 10\%$ per annum interest compounded half-yearly. She repays $\displaystyle \text{ Rs. } 4000$ at the end of every six months. Calculate the third payment she has to make at the end of $\displaystyle 18$ months in order to clear the entire loan.

Answer:

$\displaystyle \text{For 1st half-year: } P = \text{ Rs. } 12000; R=10\% \text{ and } T= \frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{12000 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 600$

$\displaystyle \text{and, Amount } = 12000 + 600 = \text{ Rs. } 12600$

She repays $\displaystyle \text{ Rs. } 4000$

$\displaystyle \text{For 2nd half year: } P = \text{ Rs. } (12600-4000)=8600; R=10\% \text{ and } T= \frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{8600 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 430$

$\displaystyle \text{and, Amount } = 8600 + 430 = \text{ Rs. } 9030$

She repays $\displaystyle \text{ Rs. } 4000$

$\displaystyle \text{For 3rd half-year: } P = \text{ Rs. } (9030-4000)=5030; R=10\% \text{ and } T= \frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{5030 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 251.50$

$\displaystyle \text{and, Amount } = 5030+251.50= \text{ Rs. } 5281.50$

$\displaystyle \\$

Question 7: On a certain sum of money, invested at the rate of $\displaystyle 10\%$ per annum compounded annually, the interest for the first year plus the interest for the third year is $\displaystyle \text{ Rs. } 2652$ . Find the sum.

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } x; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{x \times 10 \times 1}{100} = \text{ Rs. } 0.1x$

$\displaystyle \text{and, Amount } = x + 0.1x = \text{ Rs. } 1.1x$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 1.1x; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{1.1x \times 10 \times 1}{100} = \text{ Rs. } 0.11x$

$\displaystyle \text{and, Amount } = 1.1x + 0.11x = \text{ Rs. } 1.21x$

$\displaystyle \text{For 3rd year: } P = \text{ Rs. } 1.21x; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{1.21x \times 10 \times 1}{100} = \text{ Rs. } 0.121x$

and, $\displaystyle \therefore 0.1x+0.121x= \text{ Rs. } 2652 \Rightarrow x=\text{ Rs. } 12000$

$\displaystyle \\$

Question 8: A sum of money is lent at $\displaystyle 8\%$ per annum compound interest. If the interest for the second year exceeds that for the first year by $\displaystyle \text{ Rs. } 96$ , find the sum of money.

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } x; R=8\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{x \times 8 \times 1}{100} = \text{ Rs. } 0.0.08x$

$\displaystyle \text{and, Amount } = x + 0.08x = \text{ Rs. } 1.08x$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 1.08x; R=8\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{1.08x \times 8 \times 1}{100} = \text{ Rs. } 0.0864x$

$\displaystyle \text{and, Amount } = 1.08x + 0.0864x = \text{ Rs. } 1.8864x$

$\displaystyle \therefore 0.0864x - 0.08x= \text{ Rs. } 96 \Rightarrow x=\text{ Rs. } 15000$

$\displaystyle \\$

Question 9: A person invested $\displaystyle \text{ Rs. } 8000$ every year at the beginning of the year, at $\displaystyle 10\%$ per annum compounded interest. Calculate his total savings at the beginning of the third year.

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 8000; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{8000 \times 10 \times 1}{100} = \text{ Rs. } 800$

$\displaystyle \text{and, Amount } = 8000+800 = \text{ Rs. } 8800$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 8800+8000= \text{ Rs. } 16800; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{16800 \times 10 \times 1}{100} = \text{ Rs. } 1680$

$\displaystyle \text{and, Amount } = 16800 + 1680 = \text{ Rs. } 18480$

Therefore the amount at the start of third year $\displaystyle 18480+8000 = \text{ Rs. } 26480$

$\displaystyle \\$

Question 10: A person saves $\displaystyle \text{ Rs. } 8000$ every year and invests it at the end of the year at $\displaystyle 10\%$ per annum compound interest. Calculate her total amount of savings at the end of the third year.

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 8000; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{8000 \times 10 \times 1}{100} = \text{ Rs. } 800$

$\displaystyle \text{and, Amount } = 8000+800 = \text{ Rs. } 8800$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 8800+8000= \text{ Rs. } 16800; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{16800 \times 10 \times 1}{100} = \text{ Rs. } 1680$

$\displaystyle \text{and, Amount } = 16800 + 1680 = \text{ Rs. } 18480$

$\displaystyle \text{For 3rd year: } P = \text{ Rs. } 18480+8000= \text{ Rs. } 26480; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{26480 \times 10 \times 1}{100} = \text{ Rs. } 2648$

and, Therefore the amount at the start of third year $\displaystyle Amount = 26480 + 2648 = \text{ Rs. } 29128$

$\displaystyle \\$

Question 11: During every financial year, the value of the machine depreciates by $\displaystyle 12\%$ . Find the original cost of the machine which depreciates by $\displaystyle \text{ Rs. } 2640$ during the second financial year of its purchase.

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } x; R=12\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Depreciation } = \frac{x \times 12 \times 1}{100} = \text{ Rs. } 0.12x$

$\displaystyle \text{and, Amount } = x-0.12x = \text{ Rs. } 0.88x$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 0.88x; R=12\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Depreciation } = \frac{0.88x \times 12 \times 1}{100} = \text{ Rs. } 0.1056x$

$\displaystyle \text{and, Original Cost } = \frac{2640}{0.1056} = \text{ Rs. } 25000$

$\displaystyle \\$

Question 12: Find the sum on which the difference between the simple interest and the compound interest at a rate of $\displaystyle 8\%$ per annum compounded annually be $\displaystyle \text{ Rs. } 64$ in $\displaystyle 2$ years.

Answer:

Let the sum be $\displaystyle x$

Simple Interest

$\displaystyle \frac{x \times 8 \times 1}{100} = \text{ Rs. } 0.16x$

Compound Interest

$\displaystyle \text{For 1st year: } P = \text{ Rs. } x; R=8\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{x \times 8 \times 1}{100} = \text{ Rs. } 0.08x$

$\displaystyle \text{and, Amount } = x + 0.08x = \text{ Rs. } 1.08x$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 1.08x; R=8\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{1.08x \times 8 \times 1}{100} = \text{ Rs. } 0.0864x$

Total Compound Interest $\displaystyle 0.08x + 0.0864x = \text{ Rs. } 0.1664$

Given $\displaystyle 0.1664x-0.16x = 64 \Rightarrow x=\text{ Rs. } 10000$

$\displaystyle \\$

Question 13: A person borrows $\displaystyle \text{ Rs. } 18000$ at $\displaystyle 10\%$ simple interest. He immediately invests the money borrowed at $\displaystyle 10\%$ compound interest compounded half yearly. How much money does he gain in one year.

Answer:

Sum borrowed $\displaystyle 18000$

Simple Interest

$\displaystyle \frac{18000 \times 10 \times 1}{100} = \text{ Rs. } 1800$

Compound Interest

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 18000; R=10\% \text{ and } T= \frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{18000 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 900$

$\displaystyle \text{and, Amount } = 18000+900 = \text{ Rs. } 18900$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 18900; R=10\% \text{ and } T= \frac{1}{2} \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{18900 \times 10 \times 1}{100 \times 2} = \text{ Rs. } 945$

Total Compound Interest earned $\displaystyle 900 + 945 = \text{ Rs. } 1845$

Gain $\displaystyle 1845-1800 = \text{ Rs. } 45$

$\displaystyle \\$

Question 14: A sum of $\displaystyle \text{ Rs. } 13500$ is invested at $\displaystyle 16\%$ per annum compound interest for $\displaystyle 5$ yea\text{ Rs. } Calculate i) interest for the first year, ii) the amount at the end of the first year, iii) interest for the second year.

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 13500; R=16\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{13500 \times 16 \times 1}{100} = \text{ Rs. } 2160$

$\displaystyle \text{and, Amount } = 13500 + 2160 = \text{ Rs. } 15660$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 15660; R=16\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{15660 \times 16 \times 1}{100} = \text{ Rs. } 2505.6$

$\displaystyle \text{and, Amount } = 15660 + 2505.60 = \text{ Rs. } 18165.60$

$\displaystyle \text{For 3rd year: } P = \text{ Rs. } 18165.60; R=16\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{18165.60 \times 16 \times 1}{100} = \text{ Rs. } 2906.50$

$\displaystyle \text{and, Amount } = 18165.60 + 2906.50 = \text{ Rs. } 21072.10$

For 4th year: $\displaystyle P = \text{ Rs. } 21072.10; R=16\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{21072.10 \times 16 \times 1}{100} = \text{ Rs. } 3371.54$

$\displaystyle \text{and, Amount } = 21072.10 + 3371.54 = \text{ Rs. } 24443.64$

For 5th year: $\displaystyle P = \text{ Rs. } 24443.64; R=16\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{24443.64 \times 16 \times 1}{100} = \text{ Rs. } 3911.00$

$\displaystyle \text{and, Amount } = 24443.64 + 3911.00 = \text{ Rs. } 28354.62$

Hence,

i) the interest for 1st year $\displaystyle = \text{ Rs. } 2160$

ii) Amount at the end of 1st year $\displaystyle = \text{ Rs. } 15660$

iii) the interest for 2nd year $\displaystyle = \text{ Rs. } 2505.6$

$\displaystyle \\$

Question 15: A person invests $\displaystyle \text{ Rs. } 48000$ for $\displaystyle 7$ years at $\displaystyle 10\%$ per annum compound interest. Calculate i) the interest for the first year, ii) the amount at the end of the 2nd year, iii) interest for the third year,

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 48000; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{48000 \times 10 \times 1}{100} = \text{ Rs. } 4800$

$\displaystyle \text{and, Amount } = 48000 + 4800 = \text{ Rs. } 52800$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 52800; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{52800 \times 10 \times 1}{100} = \text{ Rs. } 5280$

$\displaystyle \text{and, Amount } = 52800 + 5280 = \text{ Rs. } 58080$

$\displaystyle \text{For 3rd year: } P = \text{ Rs. } 58080; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{58080 \times 10 \times 1}{100} = \text{ Rs. } 5808$

$\displaystyle \text{and, Amount } = 58080 + 5808 = \text{ Rs. } 63888$

Hence,

i) the interest for 1st year $\displaystyle = \text{ Rs. } 4800$

ii) Amount at the end of 2nd year $\displaystyle = \text{ Rs. } 58080$

iii) the interest for 2nd year $\displaystyle = \text{ Rs. } 5808$

$\displaystyle \\$

Question 16: A person borrowed $\displaystyle \text{ Rs. } 7500$ from another person at $\displaystyle 8\%$ per annum compound interest. After $\displaystyle 2$ years he gave $\displaystyle \text{ Rs. } 6248$ back and a TV set to clear the debt. Find the value of the TV set.

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 7500; R=8\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{7500 \times 8 \times 1}{100} = \text{ Rs. } 600$

$\displaystyle \text{and, Amount } = 7500 + 600 = \text{ Rs. } 8100$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 8100; R=8\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{8100 \times 10 \times 1}{100} = \text{ Rs. } 648$

$\displaystyle \text{and, Amount } = 8100 + 648 = \text{ Rs. } 8748$

$\displaystyle \text{Amount Paid } = \text{ Rs. } 6248 \therefore Cost of TV = (8748-6248)=\text{ Rs. } 2500$ .

$\displaystyle \\$

Question 17: It is estimated that every year, the value of the asset depreciates at $\displaystyle 20\%$ of its value at the beginning of the year. Calculate the original value of the asset if its value after two years is $\displaystyle \text{ Rs. } 10240$ .

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } x; R=20\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Depreciation } = \frac{x \times 20 \times 1}{100} = \text{ Rs. } 0.2x$

$\displaystyle \text{ and, Value } = x-0.2x = \text{ Rs. } 0.8x$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 0.8x; R=20\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Depreciation } = \frac{0.8x \times 20 \times 1}{100} = \text{ Rs. } 0.16x$

$\displaystyle \text{ and, Value } = 0.8x-0.16x = \text{ Rs. } 0.64x$

$\displaystyle \text{ and, Original Value } = \frac{10240}{0.64} = \text{ Rs. } 16000$

$\displaystyle \\$

Question 18: Find the sum that will amount to $\displaystyle \text{ Rs. } 4928$ in $\displaystyle 2$ years at compound interest, if the rates for the successive year are at $\displaystyle 10\% \text{ and } 12\%$ respectively.

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } x; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{x \times 10 \times 1}{100} = \text{ Rs. } 0.1x$

$\displaystyle \text{and, Amount } = x+0.1x = \text{ Rs. } 1.1x$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 1.1; R=12\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{1.1 \times 12 \times 1}{100} = \text{ Rs. } 0.132x$

$\displaystyle \text{and, Amount } = 1.1x+0.132x = \text{ Rs. } 1.232x$

$\displaystyle \text{ Given Amount } = 4928 = 1.232x \Rightarrow x = \text{ Rs. } 4000$

$\displaystyle \\$

Question 19: A person opens up a bank account on 1st jan 2010 with $\displaystyle \text{ Rs. } 24000$ . If the bank pays $\displaystyle 10\%$ per annum and the person deposits $\displaystyle \text{ Rs. } 4000$ at the end of each year, find the sum in the account on 1st Jan 2012.

Answer:

For 2010 year: $\displaystyle P = \text{ Rs. } 24000; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{24000 \times 10 \times 1}{100} = \text{ Rs. } 2400$

$\displaystyle \text{and, Amount } = 24000+2400 = \text{ Rs. } 26400$

For 2011 year: $\displaystyle P = \text{ Rs. } 26400+4000= \text{ Rs. } 30400; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{30400 \times 10 \times 1}{100} = \text{ Rs. } 3040$

$\displaystyle \text{and, Amount } = 30400 + 3040 = \text{ Rs. } 33440$

For 2012 year: $\displaystyle P = \text{ Rs. } 33400+4000= \text{ Rs. } 37400$

$\displaystyle \\$

Question 20: A person borrows $\displaystyle \text{ Rs. } 12000$ at some rate per cent compound interest. After a year, the person paid back $\displaystyle \text{ Rs. } 4000$ . If the compound interest for the second year is $\displaystyle \text{ Rs. } 920$ , find: i) the rate of interest charged ii) amount of debt at the end of the second year.