Question 13 Marks
Roma borrowed Rs. $64000$ from a bank for $1\frac{1}{2}$ years at the rate of $10\%$ per annum. Compute the total compound interest payable by Roma after $1\frac{1}{2}$ years, if the interest is compounded half-yearly.
AnswerGiven: $P = Rs. 64,000 R = 10\%p.a. n = 1.5$ years
Amount after $n$ years: $\text{A}=\text{P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$=64,000\Big(1+\frac{10}{200}\Big)^{3}$
$=64,000(1.05)^{3}$
$=\text{Rs. }74, 088$
Now, $CI = A - P = Rs. 74,088 - Rs. 64,000 = Rs. 10,088$
View full question & answer→Question 23 Marks
Kamal borrowed $Rs. 57600$ from $LIC$ against her policy at $12\frac{1}{2}\%$ per annum to build a house. Find the amount that she pays to the $LIC$ after $1\frac{1}{2}$ years if the interest is calculated half-yearly.
AnswerGiven:
$P = Rs. 57,600$
$R = 12.5\%$ p.a.
$n = 1.5$ years
When the interest is compounded half−yearly, we have:
$\text{A = P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$=\text{Rs. }57,600\Big(1+\frac{12.5}{200}\Big)^{3}$
$=\text{Rs. }57,600(1.0625)^3$
$=\text{Rs. }69,089.06$
Thus, the required amount is Rs. $69,089.06.$
View full question & answer→Question 33 Marks
Mewa Lal borrowed Rs. $20000$ from his friend Rooplal at $18\%$ per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after $2$ years.
AnswerGiven: SI = for Meva Lal $=\frac{\text{PRT}}{100}$
$=\frac{20,000\times18\times2}{100}$ $= Rs. 7,200$
Thus, he has to pay Rs. $7,200$
as interest after borrowing.
$CI$ for Mewa Lal $= A − P =20,000\Big(1+\frac{18}{100}\Big)^{2}-20,000$
$=20,000(1.18)^{2}-20,000$
$=27,848-20,000$
$=\text{Rs. }7,848$
He gained Rs. $7,848$ as interest after lending.
His gain in the whole transaction $= Rs. 7,848 - Rs. 7,200 = Rs. 648$
View full question & answer→Question 43 Marks
Find the rate percent per annum, if Rs. $2000$ amount to Rs. $2315.25$ in an year and a half, interest being compounded six monthly.
AnswerLet the rate percent per annum be $R$.
Because interest is compounded every six months, n will be $3$ for $1.5$ years
Now, $\text{A}=\text{P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{n}}$
$2,315.25=2,000\Big(1+\frac{\text{R}}{200}\Big)^{3}$
$\Big(1+\frac{\text{R}}{200}\Big)^{3}=\frac{2,315.25}{2,000}$
$\Big(1+\frac{\text{R}}{200}\Big)^{3}=1.157625$
$\Big(1+\frac{\text{R}}{200}\Big)^{3}=(1.05)^{3}$
$1+\frac{\text{R}}{200}=1.05$
$\frac{\text{R}}{200}=0.05$
$=10$ Thus, the required rate is $10\%$ per annum.
View full question & answer→Question 53 Marks
Meera borrowed a sum of Rs. $1000$ from Sita for two years. if the rate of interest is $10\%$ compounded annually, find the amount that Meera has to pay back.
AnswerGiven:
$P=\text { Rs. } 1,000$
$R=10 \% \text { p.a. }$
$n=2 \text { years }$
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 $A=P\left(1+\frac{R}{100}\right)^n$.
$\therefore A=1,000\left(1+\frac{10}{100}\right)^2$
$=1,000(1.1)^2$
$=1,210$
Thus, the required amount is Rs. $1,210 .$
View full question & answer→Question 63 Marks
On what sum will the compound interest at $5\%$ per annum for $2$ years compounded annually be Rs. $164?$
AnswerLet the sum be Rs. $x$.
We know that: $CI = A - P = \text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-\text{P}$
$= \text{P}\Big[\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-1\Big]$
$164= \text{x}\Big[\Big(1+\frac{5}{100}\Big)^{2}-1\Big]$
$164= \text{x}\Big[(1.05)^{2}-1\Big]$
$\text{x}=\frac{164}{0.1025}$
$=1,600$
Thus, the required sum is $Rs. 1,600.$
View full question & answer→Question 73 Marks
Sum of money amounts to Rs. $453690$ in $2$ years at $6.5\%$ per annum compounded annually. Find the sum.
Answer$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$453,690=\text{P}\Big(1+\frac{6.5}{100}\Big)^{2}$
$\text{P}(1.065)^{2}=453,690$
${\text{P}}=\frac{453,690}{1.134225}$
$\text{P}=400,000$
Thus, the required sum is Rs. 400,000.
View full question & answer→Question 83 Marks
Find the compound interest on Rs. $8000$ for $9$ months at $20\%$ per annum compounded quarterly.
Answer$P = Rs. 8,000 T = 9$ months $= 3$
quarters $R = 20\%$
per annum $= 5\%$
per quarter $\text{A}=8,000\Big(1+\frac{5}{100}\Big)^{3}$
$=8,000(1.05)^{3}$
$=9,261$
The required amount is Rs. $9,261.$
Now, $CI = A - P = Rs. 9,261 - Rs. 8,000 = Rs. 1,261$
View full question & answer→Question 93 Marks
In what time will Rs. $1000$ amount to Rs. $1331$ at $10\%$ per annum, compound interest?
AnswerLet the time be n years.
Then, $\text{A}=\text{P}\Big(1+\frac{10}{100}\Big)^{\text{n}}$
$1,331=1,000\Big(1+\frac{10}{100}\Big)^{\text{n}}$
$(1.1)^{\text{n}}=\frac{1,331}{1,000}$
$(1.1)^{\text{n}}=1.331$
$(1.1)^{\text{n}}=(1.1)^{3}$
On comparing both the sides,
we get: $n = 3$ Thus, the required time is three years.
View full question & answer→Question 103 Marks
Compute the amount and the compound interest in each of the following by using the formulae when:
Principal $= Rs. 5000$, Rate $= 10$ paise per rupee per annum, Time $= 2$ years
AnswerApplying the rule $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ on the given situation, we get:
$\text{A}=5,000\Big(1+\frac{10}{100}\Big)^2$
$=5,000(1.10)^2$
$= Rs. 6,050$
Now,
$Cl = A - P$
$= Rs. 6,050 - Rs 5,000$
$= Rs. 1,050$
View full question & answer→Question 113 Marks
Rohit deposited Rs. $8000$ with a finance company for $3$ years at an interest of $15\%$ per annum. What is the compound interest that Rohit gets after $3$ years?
AnswerWe know that amount A at the end of n years at the rate of $R\%$ per annum is given
Given: $P = Rs. 4,000 R = 5\%p.a. n = 2$ years
Now, $\text{A}=8,000\Big(1+\frac{15}{100}\Big)^{3}$
$=8,000(1.15)^{3}$
$=\text{Rs. }12,167$
And, $CI = A - P = Rs. 12,167 - Rs. 8,000 = Rs. 4,167$
View full question & answer→Question 123 Marks
In what time will Rs. $4400$ become Rs. $4576$ at $8\%$ per annum interest compounded half-yearly?
AnswerLet the time period be n years. $R = 8\% = 4\%$ (Half−yearly)
Thus, we have: $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$4,576=4,400\Big(1+\frac{4}{100}\Big)^{\text{n}}$
$4,576=4,400(1.04)^{\text{n}}$
$(1.04)^{\text{n}}=\frac{4,576}{4,000}$
$(1.04)^{\text{n}}=1.04$
$(1.04)^{\text{n}}=1.04^{1}$ On comparing both the sides,
we get: $n = 1$ Thus, the required time is half a year.
View full question & answer→Question 133 Marks
Find the compound interest at the rate of $10\%$ per annum for two years on that principal which in two years at the rate of $10\%$ per annum gives Rs. $200$ as simple interest.
Answer$\text{SI}=\frac{\text{PRT}}{100}$
$\therefore\text{ P}=\frac{\text{SI}\times100}{\text{RT}}$
$=\frac{200\times100}{10\times2}$
$=\text{Rs. }1,000$
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=1,000\Big(1+\frac{10}{100}\Big)^{2}$
$=1,000(1.10)^{2}$
$=\text{Rs. }1,210$
Now, $CI = A - P = Rs. 1,210 - Rs. 1,000 = Rs. 210$
View full question & answer→Question 143 Marks
Compute the amount and the compound interest in each of the following by using the formulae when:
Principal $= Rs. 2000$, Rate $= 4$ paise per rupee per annum, Time $= 3$ years
AnswerApplying the rule $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ on the given situation, we get:
$\text{A}=2,000\Big(1+\frac{4}{100}\Big)^3$
$=2,000(1.04)^3$
$= Rs. 2,249.68$
Now,
$Cl = A - P$
$= Rs. 2,249.68 - Rs. 2,000$
$= Rs. 249.68$
View full question & answer→Question 153 Marks
Find the amount of Rs. $2400$ after $3$ years, when the interest is compounded annually at the rate of $20\%$ per annum.
AnswerGiven:
$P = Rs. 2,400$
$R = 20\%$p.a.
$n = 3$ year
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 $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}.$
$\therefore A=2,400(1+20100) 3$
$=2,400(1.2)^3$
$= 4,147.20$
Thus, the required amount is Rs. $4,147.20.$
View full question & answer→Question 163 Marks
Find the compound interest when principal = Rs. $3000$, rate $= 5\%$ per annum and time $= 2$ years.
AnswerPrincipal for the first year $= Rs. 3,000$
Interest for the first year $=\text{Rs.}\Big(\frac{3,000\times5\times1}{100}\Big)$ = Rs. $150$
Amount at the end of the first year $= Rs. 3,000 + Rs. 150 = Rs. 3,150$
Principal for the second year $= Rs. 3,150$
Interest for the second year $=\text{Rs.}\Big(\frac{3,000\times5\times1}{100}\Big)$ $= Rs. 157.50$
Amount at the end of the second year $= Rs. 3,150 + Rs. 157.50 = Rs. 3307.50$
$\therefore$ Compound interest $= Rs. (3,307.50 - 3,000) = Rs. 307.50$
View full question & answer→Question 173 Marks
Rahman lent Rs. $16000$ to Rasheed at the rate of $12\frac{1}{2}\%$ per annum compound interest. Find the amount payable by Rasheed to Rahman after $3$ years.
AnswerGiven:
$P = Rs. 16,000$
$R = 12.5\%$ p.a.
$n = 3$ years
We now that amount A at the end of n years at the rate $R\%$ per annum when the interest is compounded annually is given by $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}.$
$\therefore\text{A}=16,000\Big(1+\frac{12.5}{100}\Big)^3$
$= 16,000(1.125)^3$
$= 22,781.25$
Thus, the required amount is $Rs. 22,781.25.$
View full question & answer→Question 183 Marks
In how much time will a sum of $Rs. 1600$ amount to $Rs. 1852.20$ at $5\%$ per annum compound interest?
Answer$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$1852.20=1600\Big(1+\frac{5}{100}\Big)^{\text{n}}$
$\frac{1852.20}{1600}=(1.05)^{\text{n}}$
$(1.05)^{\text{n}}=1.157625$
$(1.05)^{\text{n}}=(1.05)^{3}$ On comparing both the sides,
we get: $n = 3$
Thus, the required time is three years.
View full question & answer→Question 193 Marks
The present population of a town is $25000$. It grows at $4\%, 5\%$ and $8\%$ during first year, second year and third year respectively. Find its population after $3$ years.
AnswerHere,
$P =$ Initial population $= 25,000$
$R_1 = 4\%$
$R_2 = 5\%$
$R_3 = 8\%$
$n =$ Number of years $= 3$
$\therefore$
Population after three years:
$=\text{P}\Big(1+\frac{\text{R}_{1}}{100}\Big)\Big(1+\frac{\text{R}_{2}}{100}\Big)\Big(1+\frac{\text{R}_{3}}{100}\Big)$
$=25,000\Big(1+\frac{4}{100}\Big)\Big(1+\frac{5}{100}\Big)\Big(1+\frac{8}{100}\Big)$
$=25,000(1.04)(1.05)(1.08)$
$=29,484$
Hence, the population after three years will be $29,484.$
View full question & answer→Question 203 Marks
What sum of money will amount to $Rs. 45582.25$ at $6\frac{3}{4}\%$ per annum in two years, interest being compounded annually?
Answer$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$45,582.25=\text{P}\Big(1+\frac{27}{400}\Big)^{2}$
$\text{P}(1.0675)^{2}=45,582.25$
${\text{P}}=\frac{45,582.25}{1.13955625}$
$\text{P}=40,000$
Thus, the required sum is Rs. $40,000.$
View full question & answer→Question 213 Marks
A certain sum amounts to $Rs. 5832$ in $2$ years at $8\%$ compounded interest. Find the sum.
AnswerLet the sum be $P.$
Thus, we have: $\text{A}=\text{P}(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$5,832=\text{P}\Big(1+\frac{8}{100}\Big)^{2}$
$5,832=1.1664\text{ P}$
$\text{P}=\frac{5,832}{1.1664}$
$=5,000$
Thus, the required sum is $Rs. 5,000.$
View full question & answer→Question 223 Marks
In how much time would Rs. $5000$ amount to Rs. $6655$ at $10\%$ per annum compound interest?
AnswerLet the time period be n years.
Thus, we have: $\text{CI}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-\text{P}$
$6,655=5,000\Big(1+\frac{10}{100}\Big)^{\text{n}}-5,000$
$11,655=5,000(1.10)^{\text{n}}$
$(1.1)^{\text{n}}=\frac{11,655}{5,000}$
$(1.1)^{\text{n}}=2.331$
$(1.1)^{\text{n}}=(1.1)^{3}$
On comparing both the sides,
we get: $n = 3$ Thus, the required time is three years.
View full question & answer→Question 233 Marks
Find the rate percent per annum if Rs. $2000$ amount to Rs. $2662$ in $1\frac{1}{2}$ years, interest being compounded half-yearly?
AnswerLet the rate of interest be $R\%$ Then, $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$2,662=2,000\Big(1+\frac{\text{R}}{100}\Big)^{3}$
$\Big(1+\frac{\text{R}}{100}\Big)^{3}=\frac{2,662}{2,000}$
$\Big(1+\frac{\text{R}}{100}\Big)^{3}=1.331$
$\Big(1+\frac{\text{R}}{100}\Big)^{3}=(1.1)^{3}$
$\Big(1+\frac{\text{R}}{100}\Big)=1.1$
$\frac{\text{R}}{100}=0.1$
$\text{R}=10$ Because the interest rate is being compounded half−yearly,
it is $20\%$ per annum.
View full question & answer→Question 243 Marks
Compute the amount and the compound interest in each of the following by using the formulae when:
Principal $= Rs. 12800$, Rate = Time $= 3$ years
AnswerApplying the rule $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ on the given situation, we get:
$\text{A}=12,800\Big(1+\frac{7.5}{100}\Big)^3$
$= 12,800(1.075)^3$
$= Rs. 15,901.40$
Now,
$Cl = A - P$
$= Rs. 15,901.40 - Rs. 12,800$
$= Rs. 3,101.40$
View full question & answer→Question 253 Marks
Ramu borrowed $Rs. 15625$ from a finance company to buy a scooter. If the rate of interest be $16\%$ per annum compounded annually, what payment will he have to make after $2\frac{1}{4}$ years?
AnswerGiven: $P = Rs. 15,625 R = 16%$
p.a. $\text{n} = 2\frac{1}{4}\text{ years}$
$\therefore$ Amount after $2\frac{1}{4}\text{ years}$
$=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{2}\Big(1+\frac{\frac{1}{4}(\text{R})}{100}\Big)$
$=\text{Rs. }15,625\Big(1+\frac{16}{100}\Big)^{2}\Big(1+\frac{\frac{16}{4}}{100}\Big)$
$=\text{Rs. }15,625(1.16)^{2}(1.04)$
$=\text{Rs. }21,866$
Thus, the required amount is Rs. $21,866.$
View full question & answer→Question 263 Marks
The population of a town increases at the rate of $40$ per thousand annually. If the present population be $175760$, what was the population three years ago.
AnswerPopulation after three years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{3}$
$175,760=\text{P}\Big(1+\frac{40}{1000}\Big)^{3}$
$175,760=\text{P}(1.04)^{3}$
$\text{P}=\frac{175,760}{1.124864}$
$=156,250$ Thus, the population three years ago was $156,250.$
View full question & answer→Question 273 Marks
Ramesh deposited Rs, $7500$ in a bank which pays him $12\%$ interest per annum compounded quarterly. What is the amount which he receives after $9$ months.
AnswerGiven: $P = Rs. 7,500 R = 12\% p. a. = 3\%$
quarterly $T = 9$ months $= 3$ quarters
We know that: $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$\text{A}=7,500\Big(1+\frac{3}{100}\Big)^{3}$
$=7,500(1.03)^{3}$
$=8,195.45$
Thus, the required amount is Rs. $8,195.45.$
View full question & answer→Question 283 Marks
A sum of money deposited at $2\%$ per annum compounded annually becomes Rs. $10404$ at the end of $2$ years. Find the sum deposited.
Answer$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$10,404=\text{P}\Big(1+\frac{2}{100}\Big)^{2}$
$10,404=\text{P}(1.02)^{2}$
$\text{P}=\frac{10,404}{1.0404}$
$\text{P}=10,000$
Thus, the required sum is Rs. $10,000.$
View full question & answer→Question 293 Marks
Find the compound interest on $Rs. 160000$ for one year at the rate of $20\%$ per annum, if the interest is compounded quarterly.
AnswerGiven:
$P = Rs. 1,000$
$R = 8\%$p.a.
$n = 1.5$ years
We know that:
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{400}\Big)^{4\text{n}}$
$=16,000\Big(1+\frac{20}{400}\Big)^{4}$
$=16,000(1.05)^{4}$
$=\text{Rs. }19,448.1$
Now,
$CI = A - P$
$= Rs. 1,124.86 - Rs. 1,000$
$= Rs. 124.86$
View full question & answer→Question 303 Marks
Find the compound interest on Rs. $64000$ for $1$ year at the rate of $10\%$ per annum compounded quarterly.
AnswerTo calculate the interest compounded quarterly,
we have: $\text{A}=\text{P}\Big(1+\frac{\text{R}}{400}\Big)^{\text{4n}}$
$=64,000\Big(1+\frac{10}{400}\Big)^{4\times1}$
$=64,000(1.025)^{4}$
$=70,644.03$
Thus, the required amount is Rs. $70,644.03$
Now, $CI = A - P = Rs. 70,644.25 - Rs. 64,000 = Rs. 6,644.03$
View full question & answer→Question 313 Marks
Swati took a loan of $Rs. 16000$ against her insurance policy at the rate of $12\frac{1}{2}\%$ per annum. Calculate the total compound interest payable by Swati after $3$ years.
AnswerGiven: $P = Rs. 16,000 R = 12.5\%p.a. n = 3$ years
We know that: $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=16,000\Big(1+\frac{12.5}{100}\Big)^{3}$
$=16,000(1.125)^{3}$
$=\text{Rs. }22,781.25$
Now, $CI = A - P = Rs. 22,781.25 - Rs. 16,000 = Rs. 6,781.25$
View full question & answer→Question 323 Marks
Compute the amount and the compound interest in each of the following by using the formulae when:
Principal $= Rs. 3000,$ Rate $= 5\%,$ Time $= 2$ years
AnswerApplying the rule $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ on the given situation, we get:
$\text{A}=3,000\Big(1+\frac{5}{100}\Big)^2$
$= 3,000(1.05)^2$
$= Rs 3,307.50$
Now,
$Cl = A - P$
$= Rs. 3,307.50 - Rs. 3,000$
$= Rs. 307.50$
View full question & answer→Question 333 Marks
The difference between the compound interest and simple interest on a certain sum for $2$ years at $7.5\%$ per annum is $Rs. 360.$ Find the sum.
AnswerLet the sum be $P.$ Thus, we have: $\text{CI}-\text{SI}=360$
$\Big[\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-\text{P}\Big]-\frac{\text{P}\times7.5\times2}{100}=360$
$\text{P}\Big[\Big(1+\frac{7.5}{100}\Big)^{2}-1\Big]-\frac{\text{p}\times7.5\times2}{100}=360$
$\text{P}[1.155625-1]-0.15\text{ P}$
$=3600.155625\text{ P}-0.15\text{ P}$
$=3600.005625\text{ P}=360\text{ P}$
$=\frac{360}{0.005625}\text{ P}=64000$
Thus, the required sum is $Rs. 64.$
View full question & answer→Question 343 Marks
Compute the amount and the compound interest in each of the following by using the formulae when:
Principal $= Rs. 10000,$ Rate $20\%$ per annum compounded half-yearly, Time $= 2$ years
AnswerApplying the rule $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ on the given situation, we get:
$\text{A}=10,000\Big(1+\frac{20}{200}\Big)^4$
$= 10,000(1.1)^4$
$= Rs. 14,641$
Now,
$Cl = A - P$
$= Rs. 14,641 - Rs. 10,000$
$= Rs. 4,641$
View full question & answer→Question 353 Marks
The difference between the $S.I.$ and $C.I.$ on a certain sum of money for $2$ years at $4\%$ per annum is $Rs. 20.$ Find the sum.
AnswerGiven: $\text{CI}-\text{SI}={\text{Rs. }}20$
$\Big[\text{P}\Big(1+\frac{4}{100}\Big)^{2}-\text{P}\Big]-\frac{\text{P}\times4\times2}{100}=20$
$\text{P }[(1.04^{2}-1)]-0.08\text{ P}=20$
$0.0816\text{ P}-0.08\text{ P}=20$
$0.0016\text{ P}=20$
${\text{P}}=\frac{20}{0.0016}$
$=12,500$ Thus, the required sum is $Rs. 12,500.$
View full question & answer→Question 363 Marks
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.
AnswerPopulation after two years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$22,050=\text{P}\Big(1+\frac{50}{1000}\Big)^{2}$
$22,050=\text{P}(1.05)^{2}$
$\text{P}=\frac{22,050}{1.1025}$
$=20,000$
Thus, the population two years ago was $20,000.$
View full question & answer→Question 373 Marks
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$?$
AnswerHere, $P =$ Initial population $= 28,000$
$R =$ Rate of growth of population $= 5\%$ per annum
$n =$ Number of years $= 2$
$\therefore$ Population after two years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=28,000\Big(1+\frac{5}{100}\Big)^{2}$
$=28,000(1.05)^{2}$
$=30,870$
Hence, the population after two years will be $30,870.$
View full question & answer→Question 383 Marks
There is a continuous growth in population of a village at the rate of $5\%$ per annum. If its present population is $9261,$ what it was $3$ years ago$?$
AnswerPopulation after three years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$9,261=\text{P}\Big(1+\frac{5}{100}\Big)^{3}$
$9,261=\text{P}(1.05)^{3}$
$\text{P}=\frac{9,261}{1.157625}$
$=8,000$
Thus, the population three years ago was $8,000.$
View full question & answer→Question 393 Marks
Daljit received a sum of $Rs. 40000$ as a loan from a finance company. If the rate of interest is $7\%$ per annum compounded annually, calculate the compound interest that Daljit pays after $2$ years.
Answer$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=40,000\Big(1+\frac{7}{100}\Big)^{2}$
$=40,000(1.07)^{2}$
$=45,796$ Thus, the required amount is $Rs. 45,796.$
Now, $CI = A - P = Rs. 45,796 - Rs. 40,000 = Rs. 5,796.$
View full question & answer→Question 403 Marks
The annual rate of growth in population of a certain city is $8\%.$ If its present population is $196830,$ what it was $3$ years ago$?$
AnswerPopulation after three years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$196,830=\text{P}\Big(1+\frac{8}{100}\Big)^{3}$
$196,830=\text{P}(1.08)^{3}$
$\text{P}=\frac{196,830}{1.259712}$
$=156,250$
Thus, the population three years ago was $156,250.$
View full question & answer→Question 413 Marks
Compute the amount and the compound interest in each of the following by using the formulae when:
Principal $= Rs. 3000,$ Rate $= 18\%,$ Time $= 2$ years
AnswerApplying the rule $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ on the given situation, we get:
$\text{A}=3,000\Big(1+\frac{18}{100}\Big)^2$
$= 3,000(1.18)^2$
$= Rs. 4,177.20$
Now,
$Cl = A - P$
$= Rs. 4,177.20 - Rs 3,000$
$= Rs. 1,177.20$
View full question & answer→Question 423 Marks
Find the rate at which a sum of money will double itself in $3$ years, if the interest is compounded annually.
AnswerLet the rate percent per annum be $R.$
Then, $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$2\text{P}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{3}$
$\Big(1+\frac{\text{R}}{100}\Big)^{3}=2$
$\Big(1+\frac{\text{R}}{100}\Big)=1.2599$
$\frac{\text{R}}{100}=0.2599$
$\text{R}=25.99$
Thus, the required rate is $25.99\%$ per annum.
View full question & answer→Question 433 Marks
The production of a mixi company in $1996$ was $8000$ mixies. Due to increase in demand it increases its production by $15\%$ in the next two years and after two years its demand decreases by $5\%.$ What will be its production after $3$ years$?$
AnswerProduction after three years $=\text{P}\Big(1+\frac{\text{R}_{1}}{100}\Big)^{2}\Big(1-\frac{\text{R}_{2}}{100}\Big)$$=8,000\Big(1+\frac{15}{1,000}\Big)^{2}\Big(1-\frac{5}{100}\Big)$
$=8,000(1.15)^{2}(0.95)$
$=10,051$
Thus, the production after three years will be $10,051.$
View full question & answer→Question 443 Marks
Ishita invested a sum of $Rs. 12000$ at $5\%$ per annum compound interest. She received an amount of $Rs. 13230$ after $n$ years. Find the value of $n.$
Answer$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$13,230=12,000\Big(1+\frac{5}{100}\Big)^{\text{n}}$
$(1.05)^{\text{n}}=\frac{13,230}{12,000}$
$(1.05)^{\text{n}}=1.1025$
$(1.05)^{\text{n}}=(1.05)^{2}$
On comparing both the sides, we get: $n = 2$ Thus, the value of $n$ is two years.
View full question & answer→Question 453 Marks
Aman 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 his net profit for the entire period of three years.
AnswerAman's profit for three years $=\text{P}\Big(1-\frac{\text{R}_{1}}{100}\Big)\Big(1+\frac{\text{R}_{2}}{100}\Big)\Big(1+\frac{\text{R}_{3}}{100}\Big)$
$=100,000\Big(1-\frac{5}{100}\Big)\Big(1+\frac{10}{100}\Big)\Big(1+\frac{12}{100}\Big)$
$=100,000(0.95)(1.10)(1.12)$
$=117,040$
$\therefore$ Net profit $= RS. 117,040 - RS. 100,000 = RS. 17,040.$
View full question & answer→Question 463 Marks
Rachana borrowed a certain sum at the rate of $15\%$ per annum. If she paid at the end of two years $Rs. 1290$ as interest compounded annually, find the sum she borrowed.
AnswerLet the money borrowed by Rachana be $Rs. x$
Then, we have: $\text{CI}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-\text{P}$
$1,290=\text{x}\Big[\Big(1+\frac{15}{100}\Big)^{2}-1\Big]$
$1,290=\text{x}[0.3225]$ $\text{x}=\frac{1,290}{0.3225}$ $=4,000$
Thus, Rachana borrowed $Rs. 4,000.$
View full question & answer→Question 473 Marks
Find the amount of $Rs. 12500$ for $2$ years compounded annually, the rate of interest being $15\%$ for the first year and $16\%$ for the second year.
AnswerGiven:
$P = Rs. 12,000$
$R_1 = 15\% p.a.$
$R_2= 16\% p.a.$
$\therefore$ Amount after two years $=\text{P}\Big(1+\frac{\text{R}_{1}}{100}\Big)\Big(1+\frac{\text{R}_{2}}{100}\Big)$
$=\text{Rs. }12,500\Big(1+\frac{15}{100}\Big)\Big(1+\frac{16}{100}\Big)$
$=\text{Rs. }12,500(1.15)(1.16)$
$=\text{Rs. }16,675$
Thus, the required amount is $Rs. 16,675.$
View full question & answer→Question 483 Marks
What sum will amount to $Rs. 4913$ in $18$ months, if the rate of interest is $12\frac{1}{2}\%$ per annum, compounded half-yearly$?$
AnswerLet the sum be $Rs. x. $
Given: $A = Rs. 4913$
$R = 12.5\%$
$n = 18$
months $= 1.5$ years
We know that: $\text{A = P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$4,913=\text{P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$4,913=\text{x}\Big(1+\frac{12.5}{200}\Big)^{3}$
$4,913= \text{x}\big[(1.0625)\big]^{3}$
$\text{x }=\frac{4,913}{1,1995}$
$=4,096$ Thus, the required sum is $Rs. 4,096.$
View full question & answer→Question 493 Marks
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.
AnswerValue of the machine after two years $=\text{P}\Big(1-\frac{\text{R}}{100}\Big)^{\text{n}}$
$\Rightarrow100,000\Big(1-\frac{10}{100}\Big)^{2}$
$=100,000(0.90)^{2}$
$=81,000$
Thus, the value of the machine after two years will be $Rs. 81,000$
Depreciation $= Rs. 100,000 − Rs. 81,000 = Rs. 19,000.$
View full question & answer→Question 503 Marks
Surabhi borrowed a sum of $Rs. 12000$ from a finance company to purchase a refrigerator. If the rate of interest is $5\%$ per annum compounded annually, calculate the compound interest that Surabhi has to pay to the company after $3$ years.
Answer$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=12000\Big(1+\frac{5}{100}\Big)^{3}$
$=12,000(1.05)^{3}$
$=13,891.50$
Thus, the required amount is $Rs. 13,891.50$
Now, $CI = A - P = Rs. 13,891.50 - Rs. 12,000 = Rs. 1,891.50$
View full question & answer→Question 513 Marks
Compute the amount and the compound interest in each of the following by using the formulae when:
Principal $= Rs. 160000,$ Rate $= 10$ paise per rupee per annum compounded half-yearly, Time $= 2$ years.
AnswerApplying the rule $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ on the given situation, we get:
$\text{A}=10,000\Big(1+\frac{20}{200}\Big)^4$
$= 10,000(1.1)^4$
$= Rs. 14,641$
Now,
$Cl = A - P$
$= Rs. 14,641 - Rs. 10,000$
$= Rs. 4,641$
View full question & answer→Question 523 Marks
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.
AnswerLet the annual rate of growth be $R.$
$\therefore$ Production of scooters after three years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$46,305=4,000\Big(1+\frac{\text{R}}{100}\Big)^{3}$
$(1+0.01\text{R})^{3}=\frac{46,305}{40,000}$
$(1+0.01\text{R})^{3}=1.157625$
$(1+0.01\text{R})^{3}=(1.05)^{3}$
$1+0.01\text{R} = 1.05$
$0.01\text{R}=0.05$
$\text{R}=5$ Thus, the annual rate of growth is $5\%.$
View full question & answer→Question 533 Marks
Find the rate at which a sum of money will become four times the original amount in $2$ years, if the interest is compounded half-yearly.
AnswerLet the rate percent per annum be $R.$
Then, $\text{A}=\text{P}(1+{\text{R}})^{\text{2n}}$
$4\text{P}=\text{P}\Big(1+\frac{\text{R}}{200}\Big)^{4}$
$\Big(1+\frac{\text{R}}{200}\Big)^{4}=4$
$\Big(1+\frac{\text{R}}{200}\Big)=1.4142$
$\frac{\text{R}}{200}=0.4142$
$\text{R}=82.84$
Thus, the required rate is $82.84\%.$
View full question & answer→Question 543 Marks
What will $Rs. 125000$ amount to at the rate of $6\%$, if the interest is calculated after every $3$ months$?$
AnswerBecause interest is calculated after every $3$ months, it is compounded quarterly Given:
$P = Rs. 125,000 $
$R = 6\%$ p.a.
$=\frac{6}{4}\%$
quarterly $= 1.5\%$ quarterly $n = 4$
So, $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=125,000\Big(1+\frac{1.5}{100}\Big)^{4}$
$=125,000(1.015)^{4}$
$=132,670(\text{approx)}$ Thus, the required amount is $Rs. 132,670.$
View full question & answer→Question 553 Marks
The compound interest on $Rs. 1800$ at $10\%$ per annum for a certain period of time is $Rs. 378.$ Find the time in years.
Answer$\text{CI}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-\text{P}$
$\Rightarrow378=1,800\Big(1+\frac{10}{100}\Big)^{\text{n}}-1,800$
$1,800\Big(1+\frac{10}{100}\Big)^{\text{n}}=2,178$
$\Big(1+\frac{10}{100}\Big)^{\text{n}}=\frac{2,178}{1,800}$
$(1.1)^{\text{n}}=1.21$
$(1.1)^{\text{n}}=(1.1)^{2}$
On comparing both the sides, we get: $n = 2$
Thus, the required time is two years.
View full question & answer→Question 563 Marks
A sum amounts to $Rs. 756.25$ at $10\%$ per annum in $2$ years, compounded annually. Find the sum.
AnswerLet the sum be $Rs. x$
We know that: $CI = A - P$
$\text{A}= \text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$= \text{P}\Big[\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}\Big]$
$756.25= \text{x}\Big[\Big(1+\frac{10}{100}\Big)^{2}\Big]$
$756.25= \text{x}\Big[(1.10)^{2}\Big]$
$\text{x}=\frac{756.25}{1.21}$
$=625$ Thus, the required sum is $Rs. 625.$
View full question & answer→Question 573 Marks
At what rate percent compound interest per annum will $Rs. 640$ amount to $Rs. 774.40$ in $2$ years$?$
AnswerLet the rate of interest be $R\%$
Then, $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$774.40=640\Big(1+\frac{\text{R}}{100}\Big)^{2}$
$\Big(1+\frac{\text{R}}{100}\Big)^{2}=\frac{774.40}{640}$
$\Big(1+\frac{\text{R}}{100}\Big)^{2}=1.21$
$\Big(1+\frac{\text{R}}{100}\Big)^{2}=(1.1)^{2}$
$\Big(1+\frac{\text{R}}{100}\Big)=1.1$
$\frac{\text{R}}{100}=0.1$
$\text{R}=10$
Thus, the required rate of interest is $10\%$ per annum.
View full question & answer→Question 583 Marks
Find the principal if the interest compounded annually at the rate of $10\%$ for two years is $Rs. 210.$
AnswerLet the sum be $Rs. x$ We know that: $CI = A - P$
$= \text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-\text{P}$
$= \text{P}\Big[\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-1\Big]$
$210= \text{x}\Big[\Big(1+\frac{10}{100}\Big)^{2}-1\Big]$
$210= \text{x}\Big[(1.10)^{2}-1\Big]$
$\text{x}=\frac{210}{0.21}$
$=1,000$
Thus, the required sum is $Rs. 1,000.$
View full question & answer→Question 593 Marks
Ms. Cherian purchased a boat for $Rs. 16000.$ If the total cost of the boat is depreciating at the rate of $5\%$ per annum, calculate its value after $2$ years.
AnswerValue of the boat after two years $=\text{P}\Big(1-\frac{\text{R}}{100}\Big)^{\text{n}}$
$\Rightarrow16,000\Big(1-\frac{5}{100}\Big)^{2}$
$=16,000(0.95)^{2}$
$=14,440$ Thus, the value of the boat after two years will be $Rs. 14,440.$
View full question & answer→Question 603 Marks
Anil borrowed a sum of $Rs. 9600$ to install a handpump in his dairy. If the rate of interest is $5\frac{1}{2}\%$ per annum compounded annually, determine the compound interest which Anil will have to pay after $3$ years.
Answer$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=9,600\Big(1+\frac{5.5}{100}\Big)^{3}$
$=9,600(1.055)^{3}$
$=\text{Rs. }11,272.72$
Now, $CI = A - P = Rs. 11,272.72 - Rs. 9,600 = Rs. 1,672.72$
View full question & answer→Question 613 Marks
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$?$
AnswerGiven: $P = Rs. 64,000$
$R = 5\%$ for every six months Value of the plot after two years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$\Rightarrow64,000\Big(1+\frac{5}{200}\Big)^{4}$
$=64,000(1.025)^{4}$
$=706,440.25$
Thus, the value of the plot after two years will be $Rs. 706,440.25.$
View full question & answer→Question 623 Marks
What will be the compound interest on $Rs. 4000$ in two years when rate of interest is $5\%$ per annum$?$
AnswerWe know that amount $A$ at the end of $n$ years at the rate of $R\%$ per annum is given Given:
$P = Rs. 4,000$
$R = 5\% p. a.$
$n = 2$ years Now, $\text{A}=4,000\Big(1+\frac{5}{100}\Big)^{2}$
$=4,000(1.05)^{2}$
$=\text{Rs. }4,410$ And,
$CI = A - P = Rs. 4,410 - Rs. 4,000 = Rs. 410$
View full question & answer→Question 633 Marks
The population of a city is $125000.$ If the annual birth rate and death rate are $5.5\%$ and $3.5\%$ respectively, calculate the population of city after $3$ years.
AnswerHere,
$P =$ Initial population $= 125,000$
Annual birth rate $= R _1=5.5 \%$
Annual death rate $= R _2=3.5 \%$
Net growth rate, $R=\left(R_1-R_2\right)=2 \%$
$n =$ Number of years $= 3$
$\therefore$ Population after three years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=125,000\Big(1+\frac{2}{100}\Big)^{3}$ $=125,000(1.02)^{3}$
$=132,651$
Hence, the population after three years will be $132,651.$
View full question & answer→