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EG13Question 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.

Answer 1:

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.

Answer 2:

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.

Answer 3:

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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EG8Question 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.

Answer 4:

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.

Answer 5:

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.

Answer 6:

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]

Answer 7:

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.

Answer 8:

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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EG3Question 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.

Answer 9:

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.

Answer 10:

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.

Answer 11:

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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EG10Question 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.

Answer 12:

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.

Answer 13:

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 

i) Interest paid for 2nd year = Rs. \ 350

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.

Answer 14:

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.

Answer 15: 

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]

Answer 16:

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.

KG4Answer 17:

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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