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

1. $Rs. \ 12000 \ for \ 2 \ years \ at \ 5\%$ per annum compounded annually.
2. $Rs. \ 8000 \ for \ years \ at \ 10\%$ per annum compounded yearly.
3. $Rs . \ 8000 \ for \ years \ at \ 10\%$ per annum compounded half-yearly.

i)

For 1st year: $P = Rs. \ 12000; \ R=5\% \ and \ T=1 \ year$

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

and, $Amount = 12000 + 600 = Rs. \ 12600$

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

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

and, $Amount = 12600 + 630 = Rs. \ 13230$

and $Compound \ Interest = 600+630 = Rs. \ 1230$

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ii)

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

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

and, $Amount = 8000 + 800 = Rs. \ 8800$

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

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

and, $Amount = 8800 + 440 = Rs. \ 9240$

and $Compound \ Interest = 800+440 = Rs. \ 1240$

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iii)

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

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

and, $Amount = 8000 + 400 = Rs. \ 8400$

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

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

and, $Amount = 8400 + 420 = Rs. \ 8820$

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

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

and, $Amount = 8820 + 441 = Rs. \ 9261$

and $Compound \ Interest = 400+420+441 = Rs. \ 1261$

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

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

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

and, $Amount = 12500 + 1000 = Rs. \ 13500$

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

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

and, $Amount = 13500 + 1350 = Rs. \ 14850$

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

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

and, $Amount = 14850 + 1485 = Rs. \ 16335$

and $Compound \ Interest = 1000+1350+1485 = Rs. \ 3835$

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

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

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

and, $Amount = 5500 + 220 = Rs. \ 5720$

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

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

and, $Amount = 5720 + 228.80 = Rs. \ 5948.80$

and $Compound \ Interest = 220+228.80 = Rs. \ 448.80$

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

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

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

and, $Amount = 8500 + 850 = Rs. \ 9350$

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

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

and, $Amount = 6650 + 665 = Rs. \ 7315$

and Amount left at the beginning of  3rd Year $= 7315-2700=Rs.\ 4615$

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Question 5: A man borrows $Rs.\ 10,000 \ at \ 5\%$ per annum compound interest. He repays $35\%$ of the sum borrowed at the end of the first year and $42\%$ 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?

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

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

and, $Amount = 10000 + 500 = Rs. \ 10500$

He repays $35\% \ of \ 10000 = Rs.\ 3500$

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

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

and, $Amount = 7000 + 350 = Rs. \ 7350$

He repays $42\% \ of \ 10000 = Rs.\ 4200$

and Amount left at the beginning of  3rd Year $= 7350-4200=Rs.\ 3150$

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

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

and, $Amount = 3150+157.50 = Rs. \ 3307.50$

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

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

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

and, $Amount = 12000 + 600 = Rs. \ 12600$

She repays $Rs.\ 4000$

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

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

and, $Amount = 8600 + 430 = Rs. \ 9030$

She repays $Rs.\ 4000$

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

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

and, $Amount = 5030+251.50= Rs. \ 5281.50$

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

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

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

and, $Amount = x + 0.1x = Rs. \ 1.1x$

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

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

and, $Amount = 1.1x + 0.11x = Rs. \ 1.21x$

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

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

and, $\therefore 0.1x+0.121x= Rs. \ 2652 \Rightarrow x=Rs. \ 12000$

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

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

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

and, $Amount = x + 0.08x = Rs. \ 1.08x$

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

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

and, $Amount = 1.08x + 0.0864x = Rs. \ 1.8864x$

$\therefore 0.0864x - 0.08x= Rs. \ 96 \Rightarrow x=Rs. \ 15000$

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

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

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

and, $Amount = 8000+800 = Rs. \ 8800$

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

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

and, $Amount = 16800 + 1680 = Rs. \ 18480$

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

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

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

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

and, $Amount = 8000+800 = Rs. \ 8800$

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

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

and, $Amount = 16800 + 1680 = Rs. \ 18480$

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

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

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

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

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

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

and, $Amount = x-0.12x = Rs. \ 0.88x$

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

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

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

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

Let the sum be $x$

Simple Interest

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

Compound Interest

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

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

and, $Amount = x + 0.08x = Rs. \ 1.08x$

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

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

Total Compound Interest $0.08x + 0.0864x = Rs. \ 0.1664$

Given $0.1664x-0.16x = 64 \Rightarrow x=Rs. \ 10000$

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

Sum borrowed $18000$

Simple Interest

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

Compound Interest

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

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

and, $Amount = 18000+900 = Rs. \ 18900$

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

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

Total Compound Interest earned $900 + 945 = Rs. \ 1845$

Gain $1845-1800 = Rs. \ 45$

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

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

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

and, $Amount = 13500 + 2160 = Rs. \ 15660$

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

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

and, $Amount = 15660 + 2505.60 = Rs. \ 18165.60$

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

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

and, $Amount = 18165.60 + 2906.50 = Rs. \ 21072.10$

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

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

and, $Amount = 21072.10 + 3371.54 = Rs. \ 24443.64$

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

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

and, $Amount = 24443.64 + 3911.00 = Rs. \ 28354.62$

Hence,

1. i) the interest for 1st Year $= Rs. \ 2160$
2. ii) Amount at the end of 1st year $= Rs. \ 15660$

iii) the interest for 2nd Year $= Rs. \ 2505.6$

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Question 15: A person invests  $Rs. \ 48000$ for  $7$ years at  $10\%$  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,

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

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

and, $Amount = 48000 + 4800 = Rs. \ 52800$

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

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

and, $Amount = 52800 + 5280 = Rs. \ 58080$

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

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

and, $Amount = 58080 + 5808 = Rs. \ 63888$

Hence,

1. i) the interest for 1st Year $= Rs. \ 4800$
2. ii) Amount at the end of 2nd year $= Rs. \ 58080$

iii) the interest for 2nd Year $= Rs. \ 5808$

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

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

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

and, $Amount = 7500 + 600 = Rs. \ 8100$

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

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

and, $Amount = 8100 + 648 = Rs. \ 8748$

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

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

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

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

and, $Value = x-0.2x = Rs. \ 0.8x$

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

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

and, $Value = 0.8x-0.16x = Rs. \ 0.64x$

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

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

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

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

and, $Amount = x+0.1x = Rs. \ 1.1x$

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

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

and, $Amount = 1.1x+0.132x = Rs. \ 1.232x$

Given $Amount = 4928 = 1.232x \Rightarrow x = Rs. \ 4000$

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

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

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

and, $Amount = 24000+2400 = Rs. \ 26400$

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

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

and, $Amount = 30400 + 3040 = Rs. \ 33440$

For 2012 year: $P = Rs. \ 33400+4000= Rs. \ 37400$

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

For 1st year: $P = Rs. \ 12000; \ R=x\% \ and \ T=1 \ year$

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

and, $Amount = Rs. \ (12000 + 120x)$

For 2nd year:

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

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

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

$\Rightarrow x=10\%$

and, $Debt = Rs. \ (8000 + 120 \times 10) + 920 = Rs. 10120$

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