Question 1: A sum is invested at compound interest compounded yearly. If the interest for two successive years be $Rs. \ 5700$ and $Rs. \ 7410$. calculate the rate of interest.

$Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years$

$= 7410-5700 = Rs. \ 1710$

$\Rightarrow Rs. \ 1710$  $\ is \ the \ interest \ on \ Rs. \ 5700$

$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 1710}{5700 \times 1} \% = 30\%$

$or \ you \ could \ also \ do \ it \ directly \ using \ the \ following \ approach:$

$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$

$= \frac{(7410-5700) \times 100}{5700 \times 1} \% = 30\%$

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Question 2: A certain sum of money is put at compound interest, compounded half-yearly. If the interests for two successive half-years are $Rs. \ 650$ and $Rs. \ 760.50$; find the rate of interest.

$Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years$

$= 760.50-650 = Rs. \ 110.50$

$\Rightarrow Rs. \ 110.50$ is the interest on $Rs. \ 650$

$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times \frac{1}{2}} \% = \frac{100 \times 110.50}{650 \times \frac{1}{2}} \% = 34\%$

$or \ you \ could \ also \ do \ it \ directly \ using \ the \ following \ approach:$

$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$

$= \frac{(760.50-650) \times 100}{650 \times \frac{1}{2}} \% = 34\%$

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Question 3: A certain sum amounts to $Rs. \ 5292$ in two years and $Rs. \ 5556.60$ in three years, interest being compounded annually. Find; The rate of interest. The original sum.

$Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years$

$= 5556.60-5292 = Rs. \ 264.60$

$\Rightarrow Rs. \ 264.60$ is the interest on $Rs. \ 5292$

$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 264.60}{5292 \times 1} \% = 5\%$

$or \ you \ could \ also \ do \ it \ directly \ using \ the \ following \ approach:$

$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$

$= \frac{(5556.60-5292) \times 100}{5292 \times 1} \% = 5\%$

Let the sum of money $= Rs. \ 100$

Therefore Interest on it for 1st Year $= 5\% \ of \ Rs. \ 100 = Rs. \ 5$

$\Rightarrow Amount \ in \ one \ year = 100 + 5 = Rs. \ 105$

$\therefore Interest \ on \ it \ for \ 2^{nd} \ Year = 5\% \ of \ 105 = Rs. \ 5.25$

$\Rightarrow Amount \ in \ 2nd \ year = 105 + 5.25 = Rs. \ 110.25$

When amount in two year $= Rs. \ 110.25, sum \ is \ Rs. \ 100$

$\Rightarrow \ When \ amount \ in \ two \ years= Rs. \ 5292$, then

$sum= \frac{100}{110.25} \times 5292 = Rs. \ 4800$

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Question 4:  The compound interest, calculated yearly, on a certain sum of money for the second year is $Rs. \ 1089$ and for the third year it is $Rs. \ 1197.90$. Calculate the rate of interest and the sum of money.

$Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years$

$= 1197.90-1089 = Rs. \ 108.90$

$\Rightarrow Rs. \ 108.90$ is the interest on $Rs. \ 1089$

$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 108.90}{1089 \times 1} \% = 10\%$

$or \ you \ could \ also \ do \ it \ directly \ using \ the \ following \ approach:$

$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$

$= \frac{(1197.90-1089) \times 100}{1089 \times 1} \% = 10\%$

Let the sum of money $= Rs. \ 100$

Therefore Interest on it for 1st Year $= 10\% \ of \ Rs. \ 100 = Rs. \ 10$

$\Rightarrow Amount \ in \ one \ year = 100 + 10 = Rs. \ 110$

$\therefore Interest \ on \ it \ for \ 2^{nd} \ Year = 10\% \ of \ 110 = Rs. \ 11$

When interest of 2nd year $= Rs. \ 11, sum \ is \ Rs. \ 100$

$\Rightarrow \ When \ interest \ in \ two \ years= Rs. \ 1089$, then

$sum= \frac{100}{11} \times 1089 = Rs. \ 9900$

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Question 5:  A person invests $Rs. \ 8000$ for $3$ years at a certain rate of interest, compounded annually. At the end of one year it amounts to $Rs. \ 9440$. Calculate;

1. The rate of interest per annum.
2. The amount at the end of the second year.
3. The interest accrued in the third year.

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

Therefore $Interest = \frac{8000 \times x \times 1}{100} = Rs. \ 80x$

and, $Amount = 8000+ 80x = Rs. \ 9440 \Rightarrow x=\frac{9440-8000}{80}=18\%$

For 2nd year: $P = Rs. \ 9440; \ R=18\% \ and \ T=1 \ year$

Therefore $Interest = \frac{9440 \times 18 \times 1}{100} = Rs. \ 1699.20$

and, $Amount = 9440+1699.20 = Rs. \ 11139.2$

For 3rd year: $P = Rs. \ 11139.2; \ R=18\% \ and \ T=1 \ year$

Therefore $Interest = \frac{11139.2 \times 18 \times 1}{100} = Rs. \ 2005.05$

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Question 6:  A person borrowed $Rs. \ 15,000$ for $18$ months at a certain rate of interest compounded semi-annually. If at the end of six months it amounted to $Rs. \ 15600$; calculate:

1. The rate of interest per annum;
2. The total amount of money that person must pay at the end of $18$ months in order to clear the account.

For 1st year: $P = Rs. \ 15000; \ R=x\% \ and \ T=\frac{1}{2} \ year$

Therefore $Interest =600 = \frac{15000 \times x \times \frac{1}{2}}{100} \Rightarrow x=8\%$

For 2nd year: $P = Rs. \ 15600; \ R=8\% \ and \ T=\frac{1}{2} \ year$

Therefore $Interest = \frac{15600 \times 8 \times \frac{1}{2}}{100} = Rs. \ 624$

and, $Amount = 15600+624 = Rs. \ 16224$

For 3rd year: $P = Rs. \ 16224; \ R=8\% \ and \ T=\frac{1}{2} \ year$

Therefore $Interest = \frac{16224 \times 8 \times \frac{1}{2}}{100} = Rs. \ 648.96$

and, $Amount = 16224+648.96 = Rs. \ 16872.96$

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Question 7: Ramesh invests $Rs. \ 12800$ for three years at the rate of $10\%$ per annum compound interest. Find;

1. The sum due to that person at the end of the first year.
2. The interest he earns for the second year.
3. The total amount due to him at the end of the third year. [2007]

For 1st year: $P = Rs. \ 12800; \ R=10\% \ and \ T=1 \ year$

Therefore $Interest = \frac{12800 \times 10 \times 1}{100} = Rs. \ 1280$

and, $Amount = 12800+1280 = Rs. \ 14080$

For 2nd year: $P = Rs. \ 14080; \ R=10\% \ and \ T=1 \ year$

Therefore $Interest = \frac{14080 \times 10 \times 1}{100} = Rs. \ 1408$

and, $Amount = 14080+1408 = Rs. \ 15488$

For 3rd year: $P = Rs. \ 15488; \ R=10\% \ and \ T=1 \ year$

Therefore $Interest = \frac{15488 \times 10 \times 1}{100} = Rs. \ 1548.80$

and, $Amount = 15488+1548.80 = Rs. \ 17036.80$

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Question 8: The simple interest on a certain sum computes to $Rs. \ 256 in 2$ years, whereas the compound interest on the same sum at the same rate and for the same time computes to $Rs. \ 276.48$. Find the rate per cent and the sum.

Let the amount be $P$ and the rate be $x\%$

Simple Interest

$P \times \frac{x}{100} \times 2 = 256$

$\Rightarrow \frac{Px}{100}=128$

Compound Interest

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

Therefore $Interest = \frac{P \times x \times 1}{100} = Rs. \ \frac{Px}{100}$

and, $Amount = P+\frac{Px}{100} = Rs. \ P(1+\frac{x}{100})$

For 2nd year: $P = Rs. \ P(1+\frac{x}{100}); \ R=x\% \ and \ T=1 \ year$

Therefore $Interest = P(1+\frac{x}{100})\times \frac{x}{100} \times 1 = 276.48$

$\Rightarrow x=\frac{276.48}{128}\times 100 = 16\%$

Therefore $P \times \frac{16}{100}= 128 \Rightarrow P=Rs. \ 900$

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Question 9: On the certain sum and at a certain rate percent, the simple interest for the first year is $Rs. \ 270$ and the compound interest for the first two years is $Rs. \ 580.50$. Find the sum and the rate per cent.

Let the amount be $P$ and the rate be $x\%$

Simple Interest

$P \times \frac{x}{100} \times 1 = 270$

$\Rightarrow \frac{Px}{100}=270$

Compound Interest

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

Therefore $Interest = \frac{P \times x \times 1}{100} = Rs. \ \frac{Px}{100}$

and, $Amount = P+\frac{Px}{100} = Rs. \ P(1+\frac{x}{100})$

For 2nd year: $P = Rs. \ P(1+\frac{x}{100}); \ R=x\% \ and \ T=1 \ year$

Therefore $Interest = P(1+\frac{x}{100})\times \frac{x}{100} \times 1$

$\frac{Px}{100} + P(1+\frac{x}{100})\times \frac{x}{100} \times 1=580.50$

$\frac{Px}{100}(1+1+\frac{x}{100}) = 580.50$

$\Rightarrow x=15\%$

$\frac{P \times 15}{100}=270$

$\Rightarrow P=Rs. \ 1800$

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Question 10: The interest charged on a certain sum is $Rs. \ 720$ for one year and $Rs. \ 1497.60$ for two years. Find, whether the interest is simple or compound. Also, calculate the rate per cent and the sum.

Interest charged for 1st Year $Rs. \ 720$

Interest Charged for 2 years $Rs. \ 1497.60$

Interest charged for the 2nd year $1497.60-720 = \ Rs. 777.60$

Since the interest for the 2nd year is more than the 1st year, it is not simple interest. It is compound interest.

Difference between the Compound interest of two successive years

$= 777.60-720 = Rs. \ 57.60$

$\Rightarrow Rs. \ 57.60$ is the interest on $Rs. \ 720$

$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 57.60}{720 \times 1} \% = 8\%$

or you could also do it directly using the following approach:

$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$

$= \frac{(777.60-720) \times 100}{720 \times 1} \% = 8\%$

Sum borrowed is $P = \frac{720 \times 100}{8} = 9000$

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Question 11: The compound interest, calculated yearly, on a certain sum of money for the second year is $Rs. \ 864$ and for the third year is $Rs. \ 933.12$. Calculate the rate of interest and the compound interest on the same sum and at the same rate, for the fourth year.

Difference between the Compound interest of two successive years

$= 933.12-864 = Rs. \ 69.12$

$\Rightarrow Rs. \ 69.12$ is the interest on $Rs. \ 864$

$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 69.12}{864 \times 1} \% = 8\%$

or you could also do it directly using the following approach:

$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$

$= \frac{(933.12-864) \times 100}{864 \times 1} \% = 8\%$

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

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

and, $Amount = P+\frac{8P}{100} = Rs. \ P(1+\frac{8}{100})$

For 2nd year: $P = Rs. \ P(1+\frac{8}{100}); \ R=8\% \ and \ T=1 \ year$

Therefore $Interest = P(1+\frac{8}{100})\times \frac{8}{100} \times 1$

Given $P(1+\frac{8}{100}) \times \frac{8}{100} \times 1=864$

$\Rightarrow P (sum for 1st year)=Rs. \ 10000$

For 3rd year: $P = Rs. \ 11664; \ R=8\% \ and \ T=1 \ year$

Therefore $Interest = \frac{11664 \times 8 \times 1}{100} = Rs. \ 933.12$

and, $Amount = 11664+933.12 = Rs. \ 12597.12$

For 4th year: $P = Rs. \ 12597.12; \ R=8\% \ and \ T=1 \ year$

Therefore $Interest = \frac{12597.12 \times 8 \times 1}{100} = Rs. \ 1007.77$

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Question 12: A sum of money to $Rs. \ 20160$ in $3$ years and to $Rs. \ 24192$ in $4$ years. Calculate:

1. The rate of interest.
2. Amount in $2$ years and
3. Amount in $5$ years.

Difference between the Compound interest of two successive years

$= 24192-20160 = Rs. \ 4032$

$\Rightarrow Rs. \ 4032$ is the interest on $Rs. \ 20160$

$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 4032}{20160 \times 1} \% = 20\%$

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

Therefore $Interest = \frac{P \times 20 \times 1}{100} = Rs. \ \frac{20P}{100}$

and, $Amount = P+\frac{20P}{100} = Rs. \ 1.2P$

For 2nd year: $P = Rs. \ 1.2P; \ R=20\% \ and \ T=1 \ year$

Therefore $Interest = 1.2P \times \frac{20}{100} \times 1 =0.24P$

and, $Amount = 1.2P+0.24P = Rs. \ 1.44P$

For 3rd year: $P = Rs. \ 1.44P; \ R=20\% \ and \ T=1 \ year$

Therefore $Interest =1.44P \times \frac{20}{100} \times 1 =0.288P$

and, $Amount = 1.44P+0.288P = Rs. \ 1.728P$

$1.728P=20160 \Rightarrow P= Rs. \ 11666.70$

$Amount \ in \ 2 \ years = 1.44 \ 11666.70 = Rs. 16800.00$

For 4th year: $P = Rs. \ 24192; \ R=20\% \ and \ T=1 \ year$

Therefore $Interest = 24192 \times \frac{20}{100} \times 1 = Rs. \ 4838.40$

and, $Amount = 24192+4838.40 = Rs. \ 29030.4$

For 5th year: $P = Rs. \ 24192; \ R=20\% \ and \ T=1 \ year$

Therefore $Interest = 24192 \times \frac{20}{100} \times 1 = Rs. \ 4838.40$

and, $Amount = 24192+4838.40 = Rs. \ 29030.4$

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Question 13:  $Rs. \ 8000$ is lent out at $7\%$ compound interest for $2$ years. At the end of the first year $Rs. \ 3560$ are returned. Calculate:

1. The interest paid for the second year.
2. The total interest paid in two years.
3. The total amount of money paid in two years to clear the debt.

For 1st year: $P = Rs. \ 8000; \ R=7\% \ and \ T=1 \ year$

Therefore $Interest = \frac{8000 \times 7 \times 1}{100} = Rs. \ 560$

and, $Amount = 8000 + 560 = Rs. \ 8560$

For 2nd year: $P = Rs. \ (8560-3560)=5000; \ R=7\% \ and \ T=1 \ year$

Therefore $Interest = \frac{5000 \times 7 \times 1}{100} = Rs. \ 350$

and, $Amount = 5000 + 350 = Rs. \ 5350$

1. i) Interest paid for 2nd year $= Rs. \ 350$
2. ii) Total interest paid in 2 years $= 560+350 = Rs. \ 910$

iii) Total money paid to clear the dues $= 8000 + 910 = Rs. \ 8910$

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Question 14: A sum of $Rs. \ 24000$ is lent out for $2$ years at compound interest, the rate of interest being $10\%$ per year. The borrower returns some money at the end of the first year and on paying $Rs. \ 12540$ at the end of the second year the total debt is cleared. Calculate the amount of money returned at the end of the first year.

For 1st 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 2nd year: $P = Rs. \ (26400-x); \ R=10\% \ and \ T=1 \ year$

Therefore $Interest = \frac{(26400-x) \times 10 \times 1}{100} = Rs. \ (2640-\frac{x}{10})$

and, $Amount = (26400-x)+(2640-\frac{x}{10})=12540 \Rightarrow x=Rs. \ 15000$

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Question 15: A man invests $Rs. \ 1200$ for two years at compound interest. After one year his money amounts to $Rs. \ 1275$. Find the interest for the second year correct to the nearest rupee.

For 1st year: $P = Rs. \ 1200; \ R=x\% \ and \ T=1 \ year$

Therefore $Interest = \frac{1200 \times x \times 1}{100} = Rs. \ 12x$

and, $Amount = 1200+ 12x = Rs. \ 1275 \Rightarrow x=\frac{75}{12}$

For 2nd year: $P = Rs. \ 1275; \ R=\frac{75}{12} \% \ and \ T=1 \ year$

Therefore $Interest = \frac{1275 \times\frac{75}{12} \times 1}{100} = Rs. \ 79.6875 \ or\ 80$

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Question 16: The compound interest, calculated yearly, on a certain sum of money for the second year is $Rs. \ 880$ and for the third year is $Rs. \ 968$. Calculate the rate of interest and the sum of money. [1995]

Difference between the Compound interest of two successive years

$= 968-880 = Rs. \ 88$

$\Rightarrow Rs. \ 88$ is the interest on $Rs. \ 880$

$\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 88}{880 \times 1} \% = 10\%$

or you could also do it directly using the following approach:

$Rate \ of \ Interest = \frac{Difference \ in \ Interest \ of \ two \ consecutive \ periods \times 100}{Compound \ Interest \ of \ Preceding \ Year \times Time}$

$= \frac{(968-880) \times 100}{880 \times 1} \% = 10\%$

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

Given $0.11x = 880 \Rightarrow x=Rs. \ 8000$

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Question 17: The cost of a machine depreciated by $Rs. \ 4000$ during the first year and by $Rs. \ 3600$ during the second year. Calculate:

1. The rate of depreciation.
2. The original cost of the machine.
3. Its cost at the end of the third year.

Let the value of the machine is Rs. x and the rate of depreciation is r%.

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

Therefore $Depreciation = \frac{x \times r \times 1}{100} = Rs. \ \frac{xr}{100}$

and, $Value = x-\frac{xr}{100} = Rs. x(1-\frac{r}{100})$

For 2nd year: $P = Rs. x(1-\frac{r}{100}); \ R=r \% \ and \ T=1 \ year$

Therefore $Interest = x(1-\frac{r}{100}) \times \frac{r}{100} \times 1$

Given

$\frac{xr}{100} = 4000 ... ... ... ... ... ... (i)$

and

$x(1-\frac{r}{100}) \times \frac{r}{100} = 3600... ... ... ... ... ... (i)$

$Solving \ i) \ and \ ii) \ we \ get \ r=10 \% \ and \ x= Rs. \ 40000$

$Given \ 0.11x = 880 \Rightarrow x=Rs. \ 8000$

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