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KG2Question 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.

Answer 1:

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:

Answer 2:

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.

Answer 3:

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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EG8Question 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.

Answer 4:

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?

Answer 5:

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.

Answer 6:

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.

Answer 7:

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.

Answer 8:

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.

Answer 9:

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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EG1Question 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.

Answer 10:

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.

Answer 11:

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.

Answer 12:

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.

Answer 13:

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.

Answer 14:

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,

i) the interest for 1st Year = Rs. \ 2160 

ii) Amount at the end of 1st year = Rs. \ 15660 

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

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EG12Question 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,

Answer 15:

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,

i) the interest for 1st Year = Rs. \ 4800 

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.

Answer 16:

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  .

Answer 17:

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.

Answer 18:

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.

Answer 19:

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.

Answer 20:

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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