Compound Interest without Formula

Question 1: Find the amount and the compound interest on $Rs. \ 3000$, at $5\%$ per annum for $2$ years, compounded annually.

For the $1^{st} \ Year$: We have,

Principal $= Rs. 3000$, Rate of Interest $= 5\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{3000 \times 5 \times 1}{100}$ $= 150 \ Rs.$

Therefore, the Amount at the end of $1^{st} year = 3000 + 150 = 3150 \ Rs.$

For the $2^{nd} \ Year$: We have,

Principal $= Rs. 3150$, Rate of Interest $= 5\%$ per annum, $T = 1$

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

Therefore, the Amount at the end of $2^{nd}$ year $= 3150 + 157.50 = 3307.50 \ Rs.$

Compound Interest for 2 years = Amount at the end of $2^{nd}$ Year – Principal $= 3307.50 - 3000 = 307.50 \ Rs.$

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Question 2: Find the amount and the compound interest on $Rs. \ 2000$ at $10\%$ per annum for $2\frac{1}{2}$ years.

For the $1^{st} \ Year$: We have,

Principal $= Rs. 2000$, Rate of Interest $= 10\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{2000 \times 10 \times 1}{100}$ $= 200 \ Rs.$

Therefore, the Amount at the end of $1^{st} year = 2000 + 200 = 2200 \ Rs.$

For the $2^{nd} \ Year$: We have,

Principal $= Rs. 2200$, Rate of Interest $= 10\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{2200 \times 10 \times 1}{100}$ $=220 \ Rs.$

Therefore, the Amount at the end of $2^{nd}$ year $= 2200 + 220 = 2420 \ Rs.$

For the the next six months: We have,

Principal $= Rs. 2420$, Rate of Interest $= 10\%$ per annum, $T =$ $\frac{1}{2}$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{2420 \times 10 \times \frac{1}{2}}{100}$ $=121 \ Rs.$

Therefore, the Amount at the end of $2\frac{1}{2}$ year $= 2420 + 121 = 2541 \ Rs.$

Compound Interest for $2\frac{1}{2}$ years = Amount at the end of $2\frac{1}{2}$ Year – Principal $= 2541 - 2000 = 541 \ Rs.$

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Question 3: Find the compound interest on $Rs. \ 160000$ for one year at the rate of $20\%$, per annum, if the interest is compounded quarterly.

Rate of Interest $=$ $\frac{20}{4}$ $\% = 5\%$ per quarter

For the $1^{st} Quarter$: We have,

Principal $= Rs. 160000$, Rate of Interest $= 5\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{160000 \times 5 \times 1}{100}$ $= 8000 \ Rs.$

Therefore, the Amount at the end of $1^{st} quarter = 160000 + 8000 = 168000 \ Rs.$

For the $2^{nd} Quarter$: We have,

Principal $= Rs. 168000$, Rate of Interest $= 5\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{168000 \times 5 \times 1}{100}$ $= 8400 \ Rs.$

Therefore, the Amount at the end of $2^{nd} quarter = 168000 + 8400 = 176400 \ Rs.$

For the $3^{rd} Quarter$: We have,

Principal $= Rs. 176400$, Rate of Interest $= 5\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{176400 \times 5 \times 1}{100}$ $= 8820 \ Rs.$

Therefore, the Amount at the end of $3^{rd} quarter = 176400 + 8820 = 185220 \ Rs.$

For the $4^{th} Quarter$: We have,

Principal $= Rs. 185220$, Rate of Interest $= 5\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{185220 \times 5 \times 1}{100}$ $= 9261 \ Rs.$

Therefore, the Amount at the end of $4^{th} quarter = 185220 + 9261 = 194481 \ Rs.$

Compound Interest for the year = Amount at the end of $1^{st}$ Year – Principal $= 194481 - 160000 = 34481 \ Rs.$

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Question 4: Calculate the amount and the compound interest on $Rs. \ 6000$ for $2$ years when the rates of interest for successive years are $5\%$ and $6\%$ respectively.

For the $1^{st} \ Year$: We have,

Principal $= Rs. 6000$, Rate of Interest $= 5\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{6000 \times 5 \times 1}{100}$ $= 300 \ Rs.$

Therefore, the Amount at the end of $1^{st} year = 6000 + 300 = 6300 \ Rs.$

For the $2^{nd} \ Year$: We have,

Principal $= Rs. 6300$, Rate of Interest $= 6\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{6300 \times 6 \times 1}{100}$ $=378 \ Rs.$

Therefore, the Amount at the end of $2^{nd}$ year $= 6300 + 378 = 6678 \ Rs.$

Compound Interest for 2 years = Amount at the end of $2^{nd}$ Year – Principal $= 6678 - 6000 = 678 \ Rs.$

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Question 5: The simple interest on a certain sum of money for $2$ years at $6\%$ per annum is $Rs. \ 1680$. Find the amount and the compound interest on the same sum, at the same rate and for the same time, compounded annually.

Simple interest on $Rs. P$ for $2$ years at $6\%$ per annum is $Rs. \ 1680$

$\Rightarrow$ $\frac{P \times R \times T}{100}$ $= 1680$

$\Rightarrow P =$ $\frac{1680 \times 100}{6 \times 2}$ $= 14000 \ Rs.$

Hence the Principal $= 14000 \ Rs.$

For the $1^{st} \ Year$: We have,

Principal $= Rs. 14000$, Rate of Interest $= 6\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{14000 \times 6 \times 1}{100}$ $= 840 \ Rs.$

Therefore, the Amount at the end of $1^{st} year = 14000 + 840 = 14840 \ Rs.$

For the $2^{nd} \ Year$: We have,

Principal $= Rs. 14840$, Rate of Interest $= 6\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{14840 \times 6 \times 1}{100}$ $=890.40 \ Rs.$

Therefore, the Amount at the end of $2^{nd}$ year $= 14840 + 890.40 = 15730.40 \ Rs.$

Compound Interest for 2 years = Amount at the end of $2^{nd}$ Year – Principal $= 15730.40 - 14000 = 1730.40 \ Rs.$

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Question 6: A man invested $Rs. \ 8000$ for $2$ years at $10\%$ per annum, compounded annually. Compute: (i) the amount at the end of first year. (ii) the compound interest for the second year. (iii) the compound interest for $2$ years.

(i)   For the $1^{st} \ Year$: We have,

Principal $= Rs. 8000$, Rate of Interest $= 10\%$ per annum, $T = 1$

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

Therefore, the Amount at the end of $1^{st} year = 8000 + 800 = 8800 \ Rs.$

(ii)  For the $2^{nd} \ Year$: We have,

Principal $= Rs. 8800$, Rate of Interest $= 10\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{8800 \times 10 \times 1}{100}$ $=880 \ Rs.$

Therefore, the Amount at the end of $2^{nd}$ year $= 8800 + 880 = 9680 \ Rs.$

(iii)  Compound Interest for 2 years = Amount at the end of $2^{nd}$ Year – Principal $= 9680 - 8000 = 1680 \ Rs.$

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Question 7: A person invests $Rs. \ 240000$ for $2$ years at $10\%$ per annum compounded annually. If the income tax at $20\%$ is deducted at the end of each year on interest accrued, find the amount she received at the end of $2$ years.

For the $1^{st} \ Year$: We have,

Principal $= Rs. 240000$, Rate of Interest $= 10\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{240000 \times 10 \times 1}{100}$ $= 24000 \ Rs.$

Income tax deducted $=$ $\frac{20}{100}$ $\times 24000 = 4800 \ Rs.$

Therefore, the Amount at the end of $1^{st} year = 240000 + 24000 - 4800 = 259200 \ Rs.$

For the $2^{nd} \ Year$: We have,

Principal $= Rs. 259200$, Rate of Interest $= 10\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{259200 \times 10 \times 1}{100}$ $=25920 \ Rs.$

Income tax deducted $=$ $\frac{20}{100}$ $\times 25920 = 5184 \ Rs.$

Therefore, the Amount at the end of $2^{nd}$ year $= 259200 + 25920 - 5184 = 279936 \ Rs.$

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Question 8: A man borrows $Rs. \ 15000$ at $14\%$ per annum compounded annually. If he repays $Rs. \ 4100$ at the end of first year and $Rs. \ 5220$ at the end of second year, find the amount of the loan outstanding at the beginning of the third year.

For the $1^{st} \ Year$: We have,

Principal $= Rs. 15000$, Rate of Interest $= 14\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{15000 \times 14 \times 1}{100}$ $= 2100 \ Rs.$

Therefore, the Amount at the end of $1^{st} year = 15000 + 2100 = 17100 \ Rs.$

Amount repaid after $1^{st} \ Year = 4100 \ Rs.$

For the $2^{nd} \ Year$ : We have,

Principal $= Rs. 17100 - 4100 = 13000$, Rate of Interest $= 14\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{13000 \times 14 \times 1}{100}$ $= 1820 \ Rs.$

Therefore, the Amount at the end of $2^{nd} year = 13000 + 1820 = 14820 \ Rs.$

Amount repaid after $2^{nd} \ Year = 5220 \ Rs.$

For the $3^{rd} \ Year$ : We have,

Outstanding Principal $= Rs. 14820 - 5220 = 9600 \ Rs.$

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Question 9: A person invested $Rs. \ 5000$ at a certain rate of interest compounded annually for two years. At the end of first year it amounts to $Rs. \ 5325$. Calculate: (i) The rate of interest (ii) The amount at the end of second year

For the $1^{st} \ Year$: We have,

Principal $= Rs. 5000$, Rate of Interest $= r\%$ per annum, $T = 1$

Therefore $Inerest =$ $\frac{P \times R \times T}{100}$

$\Rightarrow 325 =$ $\frac{5000 \times r \times 1}{100}$

$\Rightarrow r = 6.5\%$

For the $2^{nd} \ Year$ : We have,

Principal $= Rs. 5325$, Rate of Interest $= 6.5\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{5325 \times 6.5 \times 1}{100}$ $= 346.125 \ Rs.$

Therefore, the Amount at the end of $2^{nd} year = 5325 + 346.125 = 5671.125 \ Rs.$

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Compound Interest Formula

Question 10: Find the amount and compound interest on $Rs. \ 16000$ for $2$ years at $15\%$, interest being payable annually.

Principal $= Rs. 16000$, Rate of Interest $= 15\%$ per annum, $n = 2$

$A = P \Big( 1+$ $\frac{R}{100}$ $\Big)^n$

$= 16000 \Big( 1+$ $\frac{15}{100}$ $\Big)^2$

$= 21160 \ Rs.$

Therefore compound interest $= A - P = 21160 - 16000 = 5160 \ Rs.$

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Question 11: Find the amount and compound interest on $Rs. \ 25000$ for $3$ years compounded annually and the rate of interest being $8\%, 10\%$ and $12\%$ for three successive years respectively.

Principal $= Rs. 25000$, Rate of Interest $: R_1=8\%, R_2=10\% , R_3=12\%$ per annum,

$A = P \Big(1+$ $\frac{R_1}{100}$ $\Big).\Big(1+$ $\frac{R_2}{100}$ $\Big).\Big(1+$ $\frac{R_3}{100} )$

$= 25000 \Big(1+$ $\frac{8}{100}$ $\Big).\Big(1+$ $\frac{10}{100}$ $\Big).\Big(1+$ $\frac{12}{100})$

$= 25000 \times 1.08 \times 1.10 \times 1.12$

$= 33264 \ Rs.$

Therefore compound interest $= A - P = 33264 - 25000 = 8264 \ Rs.$

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Question 12: Compute the interest earned and amount due if a sum of $Rs. \ 15000$ is invested for $1\frac{1}{2}$ years at $8\%$ per annum compound interest, interest being compounded semi-annually.

Principal $= Rs. 15000$, Rate of Interest $: R=8\%,$ semi-annually, $T= 1\frac{1}{2}$ years

$A = P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$= 15000 \Big( 1+$ $\frac{8}{100 \times 2}$ $\Big)^{1.5 \times 2}$

$= 15000 \times (1.04)^3 = 16872.96 \ Rs.$

Therefore compound interest $= A - P = 16872.96 - 15000 = 1872.96 \ Rs.$

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Question 13: A man borrows $Rs. \ 1000$ at $10\%$ per annum simple interest for $3$ years. He immediately lends this money out at compound interest at the same rate and for the same time. What is his gain at the end of $3$ years?

Simple Interest:

Principal $= Rs. 1000$, Rate of Interest $: R=10\%,$ annually, $T= 3$ years

Interest $=$ $\frac{P \times R \times T}{100}$ $=$ $\frac{1000 \times 10 \times 3}{100}$ $= 300 \ Rs.$

Compound Interest:

Principal $= Rs. 1000$, Rate of Interest $= 10\%$ per annum, $n = 3$

$A = P \Big( 1+$ $\frac{R}{100}$ $\Big)^n$

$= 1000 \Big( 1+$ $\frac{10}{100}$ $\Big)^3$

$= 1331 \ Rs.$

Therefore compound interest $= A - P = 1331 - 1000 = 331 \ Rs.$

Therefore gain $= 331 - 300 = 31 \ Rs.$

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Question 14: What sum of money will amount to $Rs. \ 4374$ in $3$ years at $12\frac{1}{2}\%$ per annum, compounded annually?

Principal $= P$, Rate of Interest $= 10\%$ per annum, $n = 3$ $A = 4374 \ Rs$

$A = P \Big( 1+$ $\frac{R}{100}$ $\Big)^n$

$4374 = P \Big( 1+$ $\frac{12.5}{100}$ $\Big)^3$

$\Rightarrow P =$ $\frac{4734}{1.125^3}$ $= 3072 \ Rs.$

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Question 15: What sum will become $Rs. \ 9826$ in $18$ months if the rate of interest is $2\frac{1}{2}\%$ per is compounded half-yearly?

Principal $= P$, Rate of Interest $: R=2\frac{1}{2}\%,$ semi-annually, $T= 1\frac{1}{2}$ years  $A = Rs. \ 9826$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$9826 = P \Big( 1+$ $\frac{2.5}{100 \times 2}$ $\Big)^{1.5 \times 2}$

$\Rightarrow P =$ $\frac{9826}{1.0125^3}$ $= 9466.54 \ Rs.$

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Question 16: The difference between the compound interest and the simple interest on a certain sum at $10\%$ per annum for $3$ years is $Rs. \ 93$. Find the sum.

Simple Interest:

Principal $= P$, Rate of Interest $: R=10\%,$ annually, $T= 3$ years

Interest $=$ $\frac{P \times R \times T}{100}$ $=$ $\frac{P \times 10 \times 3}{100}$ $=$ $\frac{3}{10}$ $P \ Rs. = 0.3P$

Compound Interest:

Principal $= P$, Rate of Interest $= 10\%$ per annum, $n = 3$

$A = P \Big( 1+$ $\frac{R}{100}$ $\Big)^n$

$A = P \Big( 1+$ $\frac{10}{100}$ $\Big)^3$

$A = 1.331P$

Compound Interest $= 1.331P-P = 0.331 P$

Given $93 = 0.331P - 0.3P$

$\Rightarrow P =$ $\frac{93}{0.031}$ $= 3000 \ Rs.$

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Question 17: The difference between the compound interest for a year payable half-yearly and the simple interest on a certain sum of money lent out at $10\%$ for a year is $Rs. \ 15$. Find the sum of money lent out.

Simple Interest:

Principal $= P$, Rate of Interest $: R=10\%,$ annually, $T= 1$ years

Interest $=$ $\frac{P \times R \times T}{100}$ $=$ $\frac{P \times 10 \times 1}{100}$ $=$ $\frac{1}{10}$ $P \ Rs. = 0.1P$

Compound Interest:

Principal $= P$, Rate of Interest $: R=10\%,$ semi-annually, $T= 1$ year  $Amount = A$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$A = P \Big( 1+$ $\frac{10}{100 \times 2}$ $\Big)^{1 \times 2}$

$A = 1.1025P$

Compound Interest $= 1.1025P-P = 0.1025 P$

Given $15 = 0.1025P - 0.1P$

$\Rightarrow P =$ $\frac{15}{0.0025}$ $= 6000 \ Rs.$

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Question 18: On a certain sum, lent out at $20\%$ per annum for $1\frac{1}{2}$, the difference between the compound interest reckoned yearly and the reckoned yearly half- is $Rs. \ 178.75$. Find the sum.

Yearly:

Principal $= P$, Rate of Interest $: R=20\%,$ annually, $T= 1\frac{1}{2}$ year  $Amount = A$

$A = P \Big(1+$ $\frac{R}{100}$ $\Big)$ $\Big(1 +$ $\frac{\frac{R}{2}}{100}$ $\Big)$

$A = P \Big(1+$ $\frac{20}{100}$ $\Big)$ $\Big(1 +$ $\frac{10}{100}$ $\Big)$

$A = 1.20 \times 1.10 P = 1.32 P$

Therefore Compound Interest $= 1.32P - P = 0.32 P$

Half Yearly:

Principal $= P$, Rate of Interest $: R=20\%,$ semi-annually, $T= 1\frac{1}{2}$ year  $Amount = A$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$A = P \Big( 1+$ $\frac{20}{100 \times 2}$ $\Big)^{1.5 \times 2}$

$A = 1.331P$

Therefore Compound Interest $= 1.331P - P = 0.331 P$

Given $178.50 = 0.331P - 0.332P = 0.011$

$\Rightarrow P =$ $\frac{178.75}{0.011}$ $= 16250 \ Rs.$

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Question 19: The compound interest on a certain sum for $2$ years at $12\%$ per annum is $Rs. \ 795$. Find the simple interest on the same sum for the the same period and at the same rate.

Compound Interest:

Principal $= P$, Rate of Interest $= 12\%$ per annum, $n = 2$

$A = P \Big( 1+$ $\frac{R}{100}$ $\Big)^n$

$A = P \Big( 1+$ $\frac{12}{100}$ $\Big)^2$

$A = 1.2544P$

Compound Interest $795 = 1.2544P-P = 0.2544 P \Rightarrow P = 3125$

Simple Interest:

Principal $= 3125$, Rate of Interest $: R=12\%,$ annually, $T= 2$ years

Interest $=$ $\frac{P \times R \times T}{100}$ $=$ $\frac{3125 \times 12 \times 2}{100}$ $= 750 \ Rs.$

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Question 20: The simple interest on a certain sum for $2$ years at $14\%$ per annum is $Rs. \ 3500$. Find the corresponding compound interest.

Simple Interest:

Principal $= P$, Rate of Interest $: R=14\%,$ annually, $T= 2$ years

Interest $=$ $\frac{P \times R \times T}{100}$

$3500 =$ $\frac{P \times 14 \times 2}{100}$ $\Rightarrow P = 12500 \ Rs.$

Compound Interest:

Principal $= 12500$, Rate of Interest $= 14\%$ per annum, $n = 2$

$A = P \Big( 1+$ $\frac{R}{100}$ $\Big)^n$

$A = 12500 \Big( 1+$ $\frac{14}{100}$ $\Big)^2$

$A = 16245$

Compound Interest $= 16245 - 12500 = 3745 \ Rs.$

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Question 21: The simple interest on a sum of money for $2$ years at $8\%$ per annum is $Rs. \ 900$. Find the compound interest on the sum at the same rate for one year, compounded half -yearly.

Simple Interest:

Principal $= P$, Rate of Interest $: R=8\%,$ annually, $T= 2$ years

Interest $=$ $\frac{P \times R \times T}{100}$

$900 =$ $\frac{P \times 8 \times 2}{100}$ $\Rightarrow P = 5625 \ Rs.$

Compound Interest:

Principal $= 5625$, Rate of Interest $: R=8\%,$ semi-annually, $T= 1$ year  $Amount = A$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$A = 5625 \Big( 1+$ $\frac{8}{100 \times 2}$ $\Big)^{1 \times 2}$

$A = 6084 \ Rs.$

Compound Interest $= 6084 - 5625 = 459 \ Rs.$

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Question 22: A sum of money is lent out at compound interest for $2 \ years$ years at $20\%$ per annum interest being reckoned yearly. If the same sum of money is lent out at compound interest at the same rate per per annum, compound interest being reckoned half-yearly it will fetch $Rs. \ 482$ more by way of interest. Calculate the sum of money-lent out.

Compound Interest (yearly):

Principal $= P$, Rate of Interest $= 20\%$ per annum, $n = 2$

$A = P \Big( 1+$ $\frac{R}{100}$ $\Big)^n$

$A = P \Big( 1+$ $\frac{20}{100}$ $\Big)^2$

$A = 1.44P$

Compound Interest $= 1.44P - P = 0.44P$

Compound Interest (Half yearly):

Principal $= P$, Rate of Interest $: R=20\%,$ semi-annually, $T= 2$ year  $Amount = A$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$A = P \Big( 1+$ $\frac{20}{100 \times 2}$ $\Big)^{2 \times 2}$

$A = 1.4641$

Compound Interest $= 1.4641P - P = 0.4641P$

Given $482 = 0.4641P - 0.44P \Rightarrow P = 20000 \ Rs.$

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Question 23: What sum will amount to $Rs. \ 2782.50$ in $2$ years at compound interest, if the rates are $5\%$ and $6\%$ for the successive years?

Principal $= P$, Rate of Interest $: R_1=5\%, R_2 = 6\%$ per annum, $Amount = 2782.50 \ Rs.$

$A = P \Big(1+$ $\frac{R_1}{100}$ $\Big).\Big(1+$ $\frac{R_2}{100}$ $\Big)$

$2782.50 = P \Big(1+$ $\frac{5}{100}$ $\Big).\Big(1+$ $\frac{6}{100}$ $\Big)$

$\Rightarrow P =$ $\frac{2782.50}{1.05 \times 1.06}$ $= 2500 \ Rs.$

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Question 24: A certain sum of money lent out at compound interest amounts to $Rs. \ 9200$ in one year and to $Rs. \ 12167$ in $3$ years. Find ii) the rate of interest (ii) the original sum.

Compound Interest (yearly):

Principal $= P$, Rate of Interest $= r\%$ per annum, $n = 1$ $Amount = 9200 \ Rs.$

$9200 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^1$ … … … … (i)

Principal $= P$, Rate of Interest $= r\%$ per annum, $n = 3$ $Amount = 12167 \ Rs.$

$12167 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^3$ … … … … (ii)

Dividing (ii) by (i) we get

$\frac{12167}{9200}$ $= \Big( 1+$ $\frac{r}{100}$ $\Big)^2$

$\Rightarrow 1.15 = \Big( 1+$ $\frac{r}{100}$ $\Big)$

$\Rightarrow r = 1.15 - 1 = 15\%$

Now substituting in (i)

$P =$ $\frac{9200}{1.15}$ $= 8000 \ Rs.$

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Question 25: 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 original money.

$1^{st} \ Year$

Principal $= P$, Rate of Interest $= r\%$ per annum, $n = 1$ $Amount = A_1$

$A_1 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^1$ … … … … (i)

$2^{nd} \ Year$

Principal $= P$, Rate of Interest $= r\%$ per annum, $n = 1$ $Amount = A_2$

$A_2 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^2$ … … … … (ii)

$3^{rd} \ Year$

Principal $= P$, Rate of Interest $= r\%$ per annum, $n = 1$ $Amount = A_3$

$A_3 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^3$ … … … … (iii)

Given: $A_2 - A_1 = 880$  and $A_3 - A_2 = 968$

Therefore

$P \Big( 1+$ $\frac{r}{100}$ $\Big)^2$ $- P \Big( 1+$ $\frac{r}{100}$ $\Big)^1 = 880$

$\Rightarrow P (1+$ $\frac{r}{100}$ $)($ $\frac{r}{100}$ $) = 880$ … … … … (iv)

$P \Big( 1+$ $\frac{r}{100}$ $\Big)^3$ $- P \Big( 1+$ $\frac{r}{100}$ $\Big)^1 = 968$

$\Rightarrow P (1+$ $\frac{r}{100}$ $)^2($ $\frac{r}{100}$ $) = 968$ … … … … (v)

Dividing (v) by (iv) we get

$\frac{968}{880}$ $= 1 +$ $\frac{r}{100}$

$\Rightarrow r = 1.1 - 1 = 10\%$

Substituting in (iv) $P =$ $\frac{880 \times 100}{10 \times 1.1}$ $= 8000 \ Rs.$

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Question 26: The compound interest on a sum of money for $2$ years is $Rs. \ 1050$ and the simple interest on the same sum for the same period and at the same rate is $Rs. \ 1000$. Find: (i) the rate of interest (ii) the sum.

Compound Interest:

Principal $= P$, Rate of Interest $= r\%$ per annum, $T = 2 \ years$

$A = P \Big( 1+$ $\frac{r}{100}$ $\Big)^2$

Compound Interest $1050 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^2 - P$

$\Rightarrow 1050 = P \Big[ (1+$ $\frac{r}{100})^2$ $- 1 \Big]$

$\Rightarrow 1050 = P (2+$ $\frac{r}{100}$ $)($ $\frac{r}{100}$ $)$ … … … … … (i)

Simple Interest:

Principal $= P$, Rate of Interest $: R=r\%,$ annually, $T= 2$ years

Interest $=$ $\frac{P \times R \times T}{100}$

$1000 =$ $\frac{P \times r \times 2}{100}$ $\Rightarrow 50000 = Pr$

$\Rightarrow P =$ $\frac{50000}{r}$ … … … … … (ii)

Now solving (i) and (ii) we get

$1050 =$ $\frac{50000}{r}$ $(2+$ $\frac{r}{100}$ $)($ $\frac{r}{100}$ \$latex )

$\Rightarrow 1050 = 1000 + 5r$

$\Rightarrow r = 10\%$

$\Rightarrow P =$ $\frac{50000}{10}$ $= 5000 \ Rs.$

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Question 27: Find the rate percent per annum if $Rs. \ 2000$ amounts to $Rs. \ 2662$ in $1\frac{1}{2}$ years, interest being compounded half-yearly.

Principal $= 2000 \ Rs.$, Rate of Interest $: R=r\%,$ semi-annually, $T= 1.5 \ years$ year  $Amount = 2662 \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$26620 = 2000 \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^{1.5 \times 2}$

$1.331 = \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^{1.5 \times 2}$

$1.10 = 1 +$ $\frac{r}{200}$ $\Rightarrow r = 20\%$ per annum

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Question 28: Find the rate at which a sum of money will double itself in $3$ years if the interest is compounded annually.

Principal $= x \ Rs.$, Rate of Interest $: R=r\%,$ annually, $n= 3$ year  $Amount = 2x \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100}$ $\Big)^{n}$

$2x = x \Big( 1+$ $\frac{r}{100 }$ $\Big)^{3}$

$2 = \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^3$

$1.2599 = 1 +$ $\frac{r}{100}$ $\Rightarrow r = 25.99\%$ per annum

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Question 29: Find the rate at which a sum of money will become four amount times the original in $2$ years if the interest is compounded half yearly.

Principal $= x \ Rs.$, Rate of Interest $: R=r\%,$ annually, $n= 2$ year  $Amount = 4x \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$4x = x \Big( 1+$ $\frac{r}{100 \times 2 }$ $\Big)^{2 \times 2}$

$4 = \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^4$

$1.4142 = 1 +$ $\frac{r}{200}$ $\Rightarrow r = 82.84\%$ per annum

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Question 30: A sum compounded annually becomes $\frac{25}{16}$ times of itself in $2$ years. Determine the rate of interest.

Principal $= x \ Rs.$, Rate of Interest $: R=r\%,$ annually, $n= 2$ year  $Amount = \frac{25}{16}x \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100}$ $\Big)^{n}$

$\frac{25}{16}$ $x = x \Big( 1+$ $\frac{r}{100}$ $\Big)^{2}$

$\frac{25}{16}$ $= \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^2$

$\frac{5}{4}$ $= 1 +$ $\frac{r}{100}$ $\Rightarrow r = 25\%$ per annum

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Question 31: Rishi invested $Rs. \ 30000$ in a finance company and received $Rs. \ 39930$ after $1\frac{1}{2}$ years. Find the rate of interest per annum compounded semi-annually.

Principal $= 30000 \ Rs.$, Rate of Interest $: R=r\%,$ semi-annually, $T= 1.5 \ years$ year  $Amount = 39930 \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$39930 = 30000 \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^{1.5 \times 2}$

$1.331 = \Big( 1+$ $\frac{r}{100 \times 2}$ $\Big)^{1.5 \times 2}$

$1.10 = 1 +$ $\frac{r}{200}$ $\Rightarrow r = 20\%$ per annum

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Question 32: In how much time would $Rs. \ 5000$ amounts to $Rs. \ 6655$ at $10\%$ per annum compound interest?

Principal $= 5000 \ Rs.$, Rate of Interest $: R=10\%,$ annually, $T= n \ years$ year  $Amount = 6655 \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100}$ $\Big)^{n}$

$6655 = 5000 \Big( 1+$ $\frac{10}{100}$ $\Big)^{n}$

$1.331 = \Big( 1+$ $\frac{10}{100}$ $\Big)^{n}$

$1.1^3 = \Big( 1+$ $\frac{10}{100}$ $\Big)^{n}$

$\Rightarrow n = 3 \ years$

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Question 33: In what time will $Rs. \ 4400$ become $Rs. \ 4576$ at $8\%$ per annum interest compounded half-yearly?

Principal $= 4400 \ Rs.$, Rate of Interest $: R=8\%,$ annually, $T= n \ years$ year  $Amount = 4576 \ Rs.$

$A= P \Big( 1+$ $\frac{R}{100k}$ $\Big)^{nk}$

$4576 = 4400 \Big(1+$ $\frac{8}{100 \times 2}$ $\Big)^{n \times 2}$

$1.04 = \Big( 1+$ $\frac{4}{100}$ $\Big)^{2n}$

$1.04 = ( 1.04)^{2n}$

$\Rightarrow 2n = 1 \ year \Rightarrow n =$ $\frac{1}{2}$ $\ year$

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Question 34: The compound interest, calculated yearly, on a certain sum of money for the second year is $Rs. \ 1320$ and for the third year is $Rs. \ 1452$. Calculate the rate of interest and the original sum of money.

$1^{st} \ Year$

Principal $= P$, Rate of Interest $= r\%$ per annum, $n = 1$ $Amount = A_1$

$A_1 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^1$ … … … … (i)

$2^{nd} \ Year$

Principal $= P$, Rate of Interest $= r\%$ per annum, $n = 1$ $Amount = A_2$

$A_2 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^2$ … … … … (ii)

$3^{rd} \ Year$

Principal $= P$, Rate of Interest $= r\%$ per annum, $n = 1$ $Amount = A_3$

$A_3 = P \Big( 1+$ $\frac{r}{100}$ $\Big)^3$ … … … … (iii)

Given: $A_2 - A_1 = 1320$  and $A_3 - A_2 = 1452$

Therefore

$P \Big( 1+$ $\frac{r}{100}$ $\Big)^2$ $- P \Big( 1+$ $\frac{r}{100}$ $\Big)^1 = 1320$

$\Rightarrow P (1+$ $\frac{r}{100}$ $)($ $\frac{r}{100}$ $) = 1320$ … … … … (iv)

$P \Big( 1+$ $\frac{r}{100}$ $\Big)^3$ $- P \Big( 1+$ $\frac{r}{100}$ $\Big)^1 = 1452$

$\Rightarrow P (1+$ $\frac{r}{100}$ $)^2($ $\frac{r}{100}$ $) = 1452$ … … … … (v)

Dividing (v) by (iv) we get

$\frac{1452}{1320}$ $= 1 +$ $\frac{r}{100}$

$\Rightarrow r = 1.1 - 1 = 10\%$

Substituting in (iv) $P =$ $\frac{1320 \times 100}{10 \times 1.1}$ $= 12000 \ Rs.$

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Question 35: On what sum of money will the difference between the compound interest and simple interest for $2$ years be equal to $Rs. \ 25$, if the rate of interest charged for both is $5\%$ per annum?

Compound Interest:

Principal $= P$, Rate of Interest $= 5\%$ per annum, $T = 2 \ years$

$A = P \Big( 1+$ $\frac{5}{100}$ $\Big)^2$

$A = P \Big( 1+$ $\frac{5}{100}$ $\Big)^2 = 1.1025 P$

Compound Interest $= 1.1025 P - P = 0.1025P$

Simple Interest:

Principal $= P$, Rate of Interest $: R=5\%,$ annually, $T= 2$ years

Interest $=$ $\frac{P \times R \times T}{100}$

$=$ $\frac{P \times 5 \times 2}{100}$ $\Rightarrow = 0.1P$

Given: $25 = 0.1025 P - 0.1P \Rightarrow P = 10000 \ Rs.$

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Question 36: Mr Kumar borrowed $Rs. \ 15000$ for two years. The rate of interest for the two successive years are $8\%$ and $10\%$ respectively. If he repays $Rs. \ 6200$ at the end of the first year, find the outstanding amount at the end of second year.

For the $1^{st} \ Year$: We have,

Principal $= Rs. 15000$, Rate of Interest $= 8\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{15000 \times 8 \times 1}{100}$ $= 1200 \ Rs.$

Therefore, the Amount at the end of $1^{st} year =15000 + 1200 = 16200 \ Rs.$

For the $2^{nd} \ Year$: We have,

Principal $= Rs. 16200 - 6200 = 10000$, Rate of Interest $= 10\%$ per annum, $T = 1$

Therefore $Interest =$ $\frac{P \times R \times T}{100}$ $=$ $\frac{10000 \times 10 \times 1}{100}$ $=1000 \ Rs.$

Therefore, the Amount at the end of $2^{nd}$ year $= 10000 + 1000 = 11000 \ Rs.$

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Depreciation Problems

Question 37: The value of a machine depreciates at the rate of $10\%$ per annum. what will be its value $2$ years hence if the present value is $Rs. \ 100000$? Also, find the total depreciation during this period.

Present Value $V_0 = 100000 \ Rs.$  Rate of Depreciation $= 10\%$ No of Years $= 2$

$V_2 = V_0 \Big( 1-$ $\frac{R}{100}$ $\Big)^n$

$V_2 = 100000 \Big( 1-$ $\frac{10}{100}$ $\Big)^2$

$V_2= 81000 \ Rs.$

Total depreciation $= 100000 - 81000 = 19000 \ Rs.$

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Question 38: Pritam bought a plot of land for $Rs. \ 640000$. Its value is increasing by $5\%$ of its previous value after every six months. What will be the value of the plot after $2$ years?

Present Value $= 640000 \ Rs.$, Rate of Interest $= 5\%$ half yearly , $T = 2 \ years$

$V_2 = P \Big( 1+$ $\frac{r}{100k}$ $\Big)^{nk}$

$V_2 = 640000 \Big( 1+$ $\frac{10}{100 \times 2}$ $\Big)^{2 \times 2}$

$V_2 = 777924 \ Rs.$

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Question 39: The value of a machine depreciates at the rate of $10\%$ per annum. It was purchased $3$ years ago. If its present value is $Rs. \ 43740$, find its purchase price.

Present Value $V_0 = 43740 \ Rs.$  Rate of Depreciation $= 10\%$ No of Years $= 3$ Value 3 years back  $V_{-3} = x$

$V_0 = V_{-3} \Big( 1-$ $\frac{R}{100}$ $\Big)^3$

$43740 = V_{-3} \Big( 1-$ $\frac{10}{100}$ $\Big)^3$

$V_{-3} = 60000 \ Rs.$

Total depreciation $= 60000 - 43740 = 16260 \ Rs.$

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Question 40: The cost of a T.V. set was quoted $Rs. \ 17000$ at the beginning of 1999. In the beginning of 2000 the price was hiked by $5\%$. Because of decrease in demand the cost was reduced by $4\%$ in the beginning of 2001. What was the cost of the T.V. set in 2001?

Cost of TV at the beginning of 1999 $= 17000 \ Rs.$

Cost of TV at the beginning of 2000 $= 17000 \times 1.05 = 17850 \ Rs.$

Cost of TV at the beginning of 2001 $= 17850 \times 0.96 = 17136 \ Rs.$

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Question 41: Ashish staffed the business with an initial investment of $Rs. \ 500000$. In the first he incurred a loss of $4\%$. However, during the second year he earned a profit of $5\%$ which in third year rose to $10\%$. Calculate the net profit for the entire period of $3$ years.

Initial Investment at the beginning of 1st Year $= 500000 \ Rs.$

Capital at the beginning of 2nd Year $= 500000 \times 0.96 = 480000 \ Rs.$

Capital at the beginning of 3rd Year $= 480000 \times 1.05 = 504000 \ Rs.$

Capital at the beginning of 4th Year $= 504000 \times 1.10 = 554400 \ Rs.$

Net profit $= 554400 - 500000 = 54400 \ Rs.$

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Population Questions

Question 42: The present population of a town is $28000$. If it increases at the rate of $5\%$ per annum, what will be its population after $2$ years?

Population $P = 28000$, Rate $R = 5\%$ per annum, $n = 2$ years

Population after $n$ years $= P \Big(1+$ $\frac{R}{100}$ $\Big)^{n}$

Population after $2$ years $= 28000 \Big(1+$ $\frac{5}{100}$ $\Big)^{2} = 30870$

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Question 43: The present population of a town is $25000$. It grows at first $4\%, 5\%$ and $8\%$ during year, second year and third year respectively. Find lts population after $3$ years.

Current Population $P = 25000$, Rate $R_1 = 4\%, R_2=5%, R_3 = 8\%$ per annum, $n = 3$ years

Population after $3^{rd} \ Year = P \Big(1+$ $\frac{R_1}{100}$ $\Big).\Big(1+$ $\frac{R_2}{100}$ $\Big).\Big(1+$ $\frac{R_3}{100}$ $\Big)$

Population after $3^{rd} \ Year = 25000 \Big(1+$ $\frac{4}{100}$ $\Big).\Big(1+$ $\frac{5}{100}$ $\Big).\Big(1+$ $\frac{8}{100}$ $\Big)$

$= 25000 \times 1.04 \times 1.05 \times 1.08 = 29484$

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Question 44: There is a continuous growth in population of a village at the rate of $5\%$ per annum. If its present population is $9261$, what was it $3$ years ago?

Population 3 years ago $= P_{-3}$ Current Population $P_0 = 9261$, Rate $R = 5\%$ per annum, $n = 3$ years

Therefore $P_{0} = P_{-3} \Big(1+$ $\frac{R}{100}$ $\Big)^{n}$

$9261 =P_{-3} \Big(1+$ $\frac{5}{100}$ $\Big)^{3}$

$\Rightarrow P_{-3} = 8000$

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Question 45: In a factory the production of scooters rose to $46305$ from $40000$ in $3$ years. Find the annual rate of growth of the production of scooters.

$P = 40000 \ A = 46305 \ n = 3 \ years, Rate = R\%$

$46305 = 40000 \Big(1+$ $\frac{R}{100}$ $\Big)^{3}$

$1.05 = 1 +$ $\frac{R}{100}$

$\Rightarrow R = 5\%$

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Question 46: The population of a town increases at the rate of $50$ per thousand. Its population after $2$ years will be $22050$. Find its present population.

Rate of Increase $= 5% \ \ A$ (population in 2 years) $= 22050, \ P$ (Initial population) $= P$

$22050 = P \Big(1+$ $\frac{5}{100}$ $\Big)^{2}$

$\Rightarrow P = 20000$

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Question 47: The count of bacteria in a culture grows by $10\%$ in the first hour, decreases by $8\%$ in the second hour and again increase by $12\%$ in third hour. If the count of bacteria in the sample $13125000$, what will be the count of bacteria after $3$ hour.

Current bacteria Population $P = 13125000$, Rate $R_1 = 10\%, R_2=-8%, R_3 = 12\%$ per annum, $n = 3$ hours

Bacteria Population after $3^{rd} \ hours = P \Big(1+$ $\frac{R_1}{100}$ $\Big).\Big(1+$ $\frac{R_2}{100}$ $\Big).\Big(1+$ $\frac{R_3}{100}$ $\Big)$

Bacteria Population after $3^{rd} \ hours = 13125000 \Big(1+$ $\frac{10}{100}$ $\Big).\Big(1+$ $\frac{-8}{100}$ $\Big).\Big(1+$ $\frac{12}{100}$ $\Big)$

$= 13125000 \times 1.10 \times 0.92 \times 1.12 = 14876400$

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Question 48: $6400$ workers were employed to construct a river bridge in four years. At the end First year, $25\%$ workers were retrenched. At the second year $25\%$ of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by $25\%$ the end of the third year. How many workers were working during the fourth year?

Current worker Population $P = 6400$, Rate $R_1 = -25\%, R_2=-25%, R_3 = 25\%$ per annum, $n = 3$ years

Worker Population after $3^{rd} \ years = P \Big(1+$ $\frac{R_1}{100}$ $\Big).\Big(1+$ $\frac{R_2}{100}$ $\Big).\Big(1+$ $\frac{R_3}{100}$ $\Big)$

Worker Population after $3^{rd} \ years = 6400 \Big(1+$ $\frac{-25}{100}$ $\Big).\Big(1+$ $\frac{-25}{100}$ $\Big).\Big(1+$ $\frac{25}{100}$ $\Big)$

$= 64000 \times 0.75 \times 0.75 \times 1.25 = 4500$

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Question 49: A man started a factory with an initial investment of $Rs. \ 100000$. In the first year, he incurred a loss of $5\%$. However, during the second year; he earned a profit of $10\%$ which in the third year rose to $12\%$. Calculate the profit for the entire period of three years.

Initial Investment $P = 100000$, Rate $R_1 = -5\%, R_2=10%, R_3 = 12\%$ per annum, $n = 3$ years

Worker Population after $3^{rd} \ years = P \Big(1+$ $\frac{R_1}{100}$ $\Big).\Big(1+$ $\frac{R_2}{100}$ $\Big).\Big(1+$ $\frac{R_3}{100}$ $\Big)$

Worker Population after $3^{rd} \ years = 100000 \Big(1+$ $\frac{-5}{100}$ $\Big).\Big(1+$ $\frac{10}{100}$ $\Big).\Big(1+$ $\frac{12}{100}$ $\Big)$

$= 100000 \times 0.95 \times 1.10 \times 1.12 = 117040 \ Rs.$

Therefore net profit $= 117040 - 100000 = 17040 \ Rs.$

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Question 50: The population of a city increases each year by $4\%$ of what it had been at the beginning of each year. If the population in 1999 had been $6760000$, find the population of the city in (i) 2001 (ii) 1997.

$P (1999) = 6760000 \ , Rate = 4\%$
$P_{2001} = 6760000 \Big(1+$ $\frac{4}{100}$ $\Big)^{2} = 7311616$
$6760000= P_{1997} \Big(1+$ $\frac{4}{100}$ $\Big)^{2}$
$\Rightarrow P_{1997} = 6250000$