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.

Answer:

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

\displaystyle \text{Therefore Interest } = \frac{12000 \times x \times 1}{100} = \text{ Rs. } 120x  

\displaystyle \text{and, Amount } = \text{ Rs. } (12000 + 120x)  

For 2nd year:

\displaystyle P = \text{ Rs. } (12000 + 120x - 4000) = (8000+120x); R=x\% \text{ and }  T=1 \text{ year }  

\displaystyle \text{ Therefore }  920 = \frac{(8000+120x) \times x \times 1}{100}  

\displaystyle \Rightarrow 92000 = 8000x + 120x^2  

\displaystyle \Rightarrow x=10\%  

\displaystyle \text{and, Debt } = \text{ Rs. } (8000 + 120 \times 10) + 920 = \text{ Rs. } 10120  

\displaystyle \\