Question 15 Marks
Find the difference between the compound interest and simpal interest, On a sum of $Rs. 50,000$ at 10% per annum for $2$ years.
AnswerGiven:
$P = Rs. 50,000$
$R = 10% p.a.$
$n = 2$ years
We know that amount $A$ at the end of $n$ years at the rate $R\%$ per annum when the interest is compounded annually is given by $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}.$
$\therefore\text{A = Rs. }50,000\Big(1+\frac{10}{100}\Big)^2$
$ = Rs. 50,000(1.1)^2$
$= Rs. 60,500$
Also,
$CI = A - P$
$= Rs. 60,500 - Rs. 50,000$
$= Rs. 10,500$
We know that:
$\text{SI}=\frac{\text{PRT}}{100}$
$=\frac{50,000\times10\times2}{100}$
$\text{= Rs. 10,000}$ $\therefore$
Difference between $CI$ and $SI = Rs. 10,500 - Rs. 10,000$
$= Rs. 500$
View full question & answer→Question 25 Marks
The difference between the compound interest and simple interest on a certain sum at $15\%$ per annum for $3$ years is $Rs. 283.50$. Find the sum.
AnswerGiven: $CI − SI = Rs. 283.50\ R = 15\%\ n = 3$ years Let the sum be $Rs. x$.
We know that: $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{x}\Big(1+\frac{15}{100}\Big)^{3}$
$=\text{x}(1.15)^{3}\ ...(1)$ Also, $\text{SI}=\frac{\text{PRT}}{100}=\frac{\text{x}(15)(3)}{100}=0.45\text{x}$
$\text{A}=\text{SI}+\text{P}=1.45\text{x}\ ...(2)$
Thus, we have: $\text{x}(1.15)^{3}-1.45\text{x}=283.50$ [From $(1)$ and $(2)$]
$1.523\text{x}-1.45\text{x}=283.50$
$0.070875\text{ x}=283.50$
$\text{x}=\frac{283.50}{0.070875}$
$=4,000$
Thus, the sum is $Rs. 4,000.$
View full question & answer→Question 35 Marks
Romesh borrowed a sum of $Rs. 245760$ at $12.5\%$ per annum, compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest, but compounded semi-annually. Find his gain after $2$ years.
AnswerGiven: $P = Rs. 245,760$ $R = 12.5\%$ $p.a. n = 2$ yearsWhen compounded annually,
we have:
$\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }245,760\Big(1+\frac{12.5}{100}\Big)^{2}$
$=\text{Rs. }311,040$
When the interest is compounded half−yearly, we have
$\text{A = P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$=\text{Rs. }10,000\Big(1+\frac{20}{200}\Big)^{4}$
$=\text{Rs. }10,000(1.1)^4$
$=\text{Rs. }14,641$
Difference $= Rs. 14,641 − Rs. 14,400 = Rs. 241$
View full question & answer→Question 45 Marks
Rakesh lent out $Rs. 10000$ for $2$ years at $20\%$ per annum, compounded annually. How much more he could earn if the interest be compounded half-yearly?
AnswerGiven: $P = Rs. 10,000 R = 20\%\ p.a. n = 2$ years
$\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }10,000\Big(1+\frac{20}{100}\Big)^{2}$
$=\text{Rs. }10,000(1.2)^2$
$=\text{Rs. }14,400$
When the interest is compounded half−yearly,
we have $\text{A = P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$=\text{Rs. }10,000\Big(1+\frac{20}{200}\Big)^{4}$
$=\text{Rs. }10,000(1.1)^4$
$=\text{Rs. }14,641$
Difference $= Rs. 14,641 − Rs. 14,400 = Rs. 241$
View full question & answer→Question 55 Marks
A sum of money was lent for $2$ years at $20\%$ compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is $Rs. 482$ more. Find the sum.
Answer$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
Also, $P = A - CI$ Let the sum of money be $Rs. x.$
If the interest is compounded annually, then: $\text{A}_{1}=\text{x}\Big(1+\frac{20}{100}\Big)^{2}$
$=1.44\text{x}$
$\therefore\ \text{CI}=1.44\text{x}-\text{x}$
$=0.44\text{x}\ ...(1)$
If the interest is compounded half−yearly, then: $\text{A}_{2}=\text{x}\Big(1+\frac{10}{100}\Big)^{4}$
$=1.4641\text{x}$
$\therefore\ \text{CI}=1.4641\text{x}-\text{x}$
$=0.4641\text{x}\ ...(2)$
It is given that if interest is compounded half−yearly, then it will be $Rs. 482$ more
$\therefore\ 0.4641\text{x}=0.44\text{x}+482$ [From $(1)$ and $(2)$]
$0.4641\text{x}-0.44\text{x}=482$
$0.0241\text{x}=482$
$\text{x}=\frac{482}{0.0241}$
$=20,000$
Thus, the required sum is $Rs. 20,000$.
View full question & answer→Question 65 Marks
Find the compound interest at the rate of $5\%$ per annum for $3$ years on that principal which in $3$ years at the rate of $5\%$ per annum gives $Rs. 1200$ as simple interest.
AnswerWe know that: $\text{P}=\frac{\text{SI}\times100}{\text{RT}}$
$\therefore\ \text{P}=\frac{1200\times100}{5\times3}$
$=8,000$
Now, $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=80,000\Big(1+\frac{5}{100}\Big)^{3}$
$=8,000(1.05)^{3}$
$=9,261$
Now, $CI = A - P = 9,261 - 8,000 = 1,261$
Thus, the required compound interest is $Rs. 1,261$.
View full question & answer→Question 75 Marks
Abha purchased a house from Avas Parishad on credit. If the cost of the house is $Rs. 64000$ and the rate of interest is $5\%$ per annum compounded half-yearly, find the interest paid by Abha after one year and a half.
AnswerGiven: $P = Rs. 64,600$ $R = 5% p.a. n = 1.5$ years When the interest is compounded half−yearly,
we have: $\text{A = P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$=\text{Rs. }64,000\Big(1+\frac{5}{200}\Big)^{3}$
$=\text{Rs. }64,000(1.025)^3$
$=\text{Rs. }68,921$
Also, $CI = A - P = Rs. 68,921 - Rs. 64,000 = Rs. 4,921$
Thus, the required interest is $Rs. 4,921$.
View full question & answer→Question 85 Marks
Amit borrowed $Rs. 16000$ at $17\frac{1}{2}\%$ per annum simple interest on the same day, he lent it to Ashu at the same rate but compounded annually what dose he gain at the end of $2$ years
AnswerAmount to be paid by Amit: $\text{SI}=\frac{\text{PRT}}{100}$ $=\frac{16000\times17.5\times2}{100}$ $=\text{Rs. }5,600$
Amount gained by Amit: $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }16,000\Big(1+\frac{17.5}{100}\Big)^2$
$=\text{Rs. }16,000(1.175)^2$ $=\text{Rs. }22,090$
we know that: $Cl = A - P = Rs. 22,090 - Rs. 16,000 = Rs. 6090$
Amit's gain in the whole transaction $= Rs. 6,090 - Rs. 5,600 = Rs. 490$
View full question & answer→Question 95 Marks
Find the amount and the compound interest on $Rs. 8000$ for $1\frac{1}{2}$years at $10\%$ per annum, compounded half-yearly.
AnswerGiven: $P = Rs. 8,000 R = 10\%\ P.a. n = 1.5$ years When compounded half−yearly,
we have: $\text{A = P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$=\text{Rs. }8,000\Big(1+\frac{10}{200}\Big)^{3}$
$=\text{Rs. }8,000(1.05)^3$ $=\text{Rs. }9,261$
Also, $CI = A - P = Rs. 9,261 - Rs. 8,000 = Rs. 1,261$
View full question & answer→Question 105 Marks
Find the compound interest on $Rs. 15625$ for $9$ months, at $16\%$ per annum, compounded quarterly.
AnswerGiven: $P = Rs. 15,625 $
$R = 16%$
$p.a.$ $=\frac{16}{4}=4\%$
quarterly $n = 9$ months $= 3$ quarters We know that: $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }15,625\Big(1+\frac{4}{100}\Big)^{3}$
$=\text{Rs. }15,625(1.04)^{3}$
$=\text{Rs. }17,576$
Also, $CI = A - P = Rs. 17,576 - Rs. 15,625 = Rs. 1,951$
Thus, the required compound interest is $Rs. 1,951$.
View full question & answer→Question 115 Marks
The difference in simple interest and compound interest on a certain sum of money at $6\frac{2}{3}\%$ per annum for $3$ years is Rs. 46 Determine the sum.
AnswerGiven: $\text{CI}-\text{SI}=46$
$\text{P}\Big[\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-1\Big]-\frac{\text{PRT}}{100}=46$
$\text{P}\Big[\Big(1+\frac{20}{300}\Big)^{3}-1\Big]-\frac{\text{P}\times20\times3}{3\times100}=46$
$\frac{4,096}{3,375}\text{ P}-\frac{\text{P}}{5}-\text{P}=46$
$\frac{(4,096-3,375-675)\text{P}}{3,375}=46$
$\text{P}=46\times\frac{3,375}{46}$
$=3,375$
Thus, the required sum is $Rs. 3,375.$
View full question & answer→Question 125 Marks
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 $(i) 2001$ $(ii) 1997.$
Answer$i.$ Population of the city in $2001$ $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{2}$
$=6760000\Big(1+\frac{4}{100}\Big)^{2}$
$=6760000(1.04)^{2}$
$=7311616$
Thus, Population of the city in $2001$ is $7311616$.
$ii.$ Population of the city in $1997$ $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{-2}$
$=6760000\Big(1+\frac{4}{100}\Big)^{-2}$
$=6760000(1.04)^{-2}$
$=6250000$
Thus, Population of the city in $1997$ is $6250000$.
View full question & answer→Question 135 Marks
Kamala borrowed from Ratan a certain sum at a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years she received $Rs. 210$ as compound interest, but paid Rs. 200 only as simple interest. Find the sum and the rate of interest.
AnswerLet the sum be $Rs. P$ and the rate of interest be $R\%$.
We know that Kamla paid $Rs. 200$ as simple interest.
$\therefore\ 200=\frac{\text{PR(2)}}{100}$ $\text{PR}=10,000...(1)$
Also, Kamla received $Rs. 210$ as compound interest.
$\therefore\ 210=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{2}-1$
$210(10,000)=\text{P}(\text{R}^{2}+200\text{R})$
$210\text{ R}=\text{ R}^{2}+200\text{R}$ [from $(1)$]
$R = 10\% p.a$.
Putting the equation in $(1)$,
we get: $P = 1,000$
Thus, the required sum is $Rs. 1,000$ and the rate of interest is $10\%$.
View full question & answer→Question 145 Marks
Find the compound interest on $Rs. 1000$ at the rate of $8\%$ per annum for $1\frac{1}{2}$ years when interest is compounded half-yearly.
AnswerGiven: $P = Rs. 1,000$ $R = 8\%p.a.$ $n = 1.5$ years
We know that: $\text{A}=\text{P}\Big(1+\frac{\text{R}}{200}\Big)^{2\text{n}}$
$=1,000\Big(1+\frac{8}{200}\Big)^{3}$
$=1,000(1.04)^{3}$
$=\text{Rs. }1,124.86$
Now, $CI = A - P = Rs. 1,124.86 - Rs. 1,000 = Rs. 124.86.$
View full question & answer→Question 155 Marks
The interest on a sum of $Rs. 2000$ is being compounded annually at the rate of $4\%$ per annum. Find the period for which the compound interest is $Rs. 163.20$.
AnswerLet the time period be $n$ years.
Then, we have: $\text{CI}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-\text{P}$
$163.20=2,000\Big(1+\frac{4}{100}\Big)^{\text{n}}-2,000$
$2,163.20=2,000(1.04)^{\text{n}}$
$(1.04)^{\text{n}}=\frac{2,163.20}{2,000}$
$(1.04)^{\text{n}}=1.0816$
$(1.04)^{\text{n}}=(1.04)^{2}$
On comparing both the sides, we get: $n = 2$ Thus, the required time is two years.
View full question & answer→Question 165 Marks
The count of bacteria in a culture grows by $10\%$ in the first hour, decreases by $8\%$ in the second hour and again increases by $12\%$ in the third hour. If the count of bacteria in the sample is $13125000$, what will be the count of bacteria after $3$ hours?
AnswerGiven:
$R_1= 10%$
$ R_2= - 8%$
$R_3= 12%$
$P =$ Original count of bacteria $= 13,125,000$
We know that:
$\text{P}\Big(1+\frac{\text{R}_{1}}{100}\Big)\Big(1+\frac{\text{R}_{2}}{100}\Big)\Big(1+\frac{\text{R}_{3}}{100}\Big)$
$\therefore$ Bacteria count after three hours:
$=13,125,000\Big(1+\frac{10}{100}\Big)\Big(1-\frac{8}{100}\Big)\Big(1+\frac{12}{100}\Big)$
$= 13,125,000(1.10)(0.92)(1.12)$
$= 14,876,400 $
Thus, the bacteria count after three hours will be $14,876,400.$
View full question & answer→Question 175 Marks
Jitendra set up a factory by investing $Rs. 2500000$. During the first two successive years his profits were $5\%$ and $10\%$ respectively. If each year the profit was on previous year's capital, compute his total profit.
AnswerProfit at the end of the first year $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)$
$=2,500,000\Big(1+\frac{5}{100}\Big)$
$=2,500,000(1.05)$
$=2,625,000$ Profit at the end of the second year $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)$
$=2,625,000\Big(1+\frac{10}{100}\Big)$
$=2,625,000(1.10)$
$=2,887,500$
Total profit $= Rs. 2,887,500 − Rs. 2,500,000 = Rs. 387,500.$
View full question & answer→Question 185 Marks
At what rate percent per annum will a sum of $Rs. 4000$ yield compound interest of $Rs.410$ in $2$ years?
AnswerLet the rate percent be $R$.
We know that: $\text{CI}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-\text{P}$
$410=4,000\Big(1+\frac{\text{R}}{100}\Big)^{2}-4,000$
$4,410=4,000\Big(1+\frac{\text{R}}{100}\Big)^{2}$
$\Big(1+\frac{\text{R}}{100}\Big)^{2}=\frac{4,410}{4,000}$
$\Big(1+\frac{\text{R}}{100}\Big)^{2}=1.1025$
$\Big(1+\frac{\text{R}}{100}\Big)^{2}=(1.05)^{2}$
$1+\frac{\text{R}}{100}=0.05$
$\frac{\text{R}}{100}=0.05$
$\text{R}=5$
Thus, the required rate percent is $5$.
View full question & answer→Question 195 Marks
Three years ago, the population of a town was $50000$. If the annual increase during three successive years be at the rate of $4\%$, $5\%$ and $3\%$ respectively, find the present population.
AnswerHere, $P =$ Initial population $= 50,000$
$ \mathrm{R}_1=4 \% $
$ \mathrm{R}_2=5 \% $
$ \mathrm{R}_3=3 \%$
$n =$ Number of years $= 3$
$\therefore$ Population after three years:
$=\text{P}\Big(1+\frac{\text{R}_{1}}{100}\Big)\Big(1+\frac{\text{R}_{2}}{100}\Big)\Big(1+\frac{\text{R}_{3}}{100}\Big)$
$=50,000\Big(1+\frac{4}{100}\Big)\Big(1+\frac{5}{100}\Big)\Big(1+\frac{3}{100}\Big)$
$=50,000(1.04)(1.05)(1.03)$
$=56,238$
Hence, the population after three years will be $56,238$.
View full question & answer→Question 205 Marks
Find the amount that David would receive if he invests $Rs. 8192$ for $18$ months at $12\frac{1}{2}\%$ per annum, the interest being compounded half-yearly.
AnswerGiven: $P = Rs. 8,192 R = 12.5\%\ p.a. n = 1.5$ years
When the interest is compounded half−yearly,
we have: $\text{A = P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$=\text{Rs. }8,192\Big(1+\frac{12.5}{200}\Big)^{3}$
$=\text{Rs. }8,192(1.0625)^{3}$
$=\text{Rs. }9,826$
Thus, the required amount is $Rs. 9,826$.
View full question & answer→Question 215 Marks
Find the amount of $Rs. 4096$ for $18$ months at $12\frac{1}{2}\%$ per annum, the interest being compounded semi-annually
AnswerGiven: $P = Rs. 4,096 R = 12.5% p.a. n = 18$ months $= 1.5$ years
We have: $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
When the interest is compounded semi-annually,
we have: $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{2\text{n}}$
$=\text{Rs. }4,096\Big(1+\frac{12.5}{200}\Big)^3$
$=\text{Rs. }4,096(1.0625)^3$
$=\text{Rs. }4,913$
Thus, the required amount is $Rs. 4,913$.
View full question & answer→Question 225 Marks
6400 workers were employed to construct a river bridge in four years. At the end of the first year, $25\%$ workers were retrenched. At the end of 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 $25\%$ at the end of the third year. How many workers were working during the fourth year?
AnswerNumber of workers $= 6,400$ At the end of the first year, $25\%$ of the workers were retrenched.
$\therefore$ $25\%$ of $6,400 = 1,600$
Number of workers at the end of the first year $= 6,400 − 1,600 = 4,800$
At the end of the second year, $25\%$ of those working were retrenched.
$\therefore$ $25\%$ of $4,800 = 1,200$
Number of workers at the end of the second year $= 4,800 − 1,200 = 3,600$
At the end of the third year, $25\%$ of those working increased.
$\therefore$ $25\%$ of $3,600 = 900$
Number of workers at the end of the third year $= 3,600 + 900 = 4,500$
Thus, the number of workers during the fourth year was $4,500$.
View full question & answer→Question 235 Marks
At what rate percent will a sum of $Rs. 1000$ amount to $Rs. 1102.50$ in $2$ years at compound interest?
Answer$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$1102.50=1000\Big(1+\frac{\text{R}}{100}\Big)^{2}$
$\frac{1102.50}{1000}=(1+0.01{\text{ R}})^{2}$
$(1+0.01{\text{ R}})^{2}=1.1025$
$(1+0.01{\text{ R}})^{2}=(1.05)^{2}$
On comparing both the sides, we get: $1 + 0.01R = 1.05 n = 5$
Thus, the required rate percent is $5$.
View full question & answer→Question 245 Marks
Find the compound interest at the rate of $5%$ for three years on that principal which in three years at the rate of $5\%$ per annum gives $Rs. 12000$ as simple interest.
Answer$\text{P}=\frac{\text{SI}\times100}{\text{RT}}$
According to the given values, we have: $=\frac{12,000\times100}{5\times3}$ $=80,000$
The principal is to be compounded annually.
So, $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=80,000\Big(1+\frac{5}{100}\Big)^{3}$
$=80,000(1.05)^{3}$ $=92,610$
Now, $CI = A - P = 92,610 - 80,000 = 12,610$
Thus, the required compound interest is $Rs. 12,610$.
View full question & answer→Question 255 Marks
Simple interest on a sum of money for $2$ years at $6\frac{1}{2}\%$ per annum is $Rs. 5200$. What will be the compound interest on the sum at the same rate for the same period?
Answer$\text{P}=\frac{\text{SI}\times100}{\text{RT}}$
$\therefore\ \text{P}=\frac{5,200\times100}{6.5\times2}$
$=40,000$ Now, $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=40,000\Big(1+\frac{6.5}{100}\Big)^{2}$
$=40,000(1.065)^{2}$
$=45,369$
Also, $CI = A - P = 45,369 - 40,000 = 5,369$
Thus, the required compound interest is $Rs. 5,369$.
View full question & answer→Question 265 Marks
Rekha deposited $Rs. 16000$ in a foreign bank which pays interest at the rate of $20\%$ per annum compounded quarterly, find the interest received by Rekha after one year.
AnswerGiven: $P = Rs. 16,000 R = 20% p.a.$ $n = 1$ year
We know that: $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
When compounded quarterly, we have: $\text{A}=\text{P}\Big(1+\frac{\text{R}}{400}\Big)^{\text{4n}}$
$=\text{Rs. }16,000\Big(1+\frac{20}{400}\Big)^{4}$
$=\text{Rs. }16,000(1.05)^{4}$
$=\text{Rs. }19,448.10$
Also, $CI = A - P = Rs. 19,448.1 - Rs. 16,000 = Rs. 3,448.10$
Thus, the interest received by Rekha after one year is $Rs. 3,448.10$.
View full question & answer→