Class 10: Compound Interest – Exercise 1(a) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

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

Answer:

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:

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:

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.

Answer:

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:

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:

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:

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:

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:

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.

Answer:

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:

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:

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:

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:

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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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,

Answer:

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:

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:

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:

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:

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.