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.
AnswerP = 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:
$\text{A}=5,000\Big(1+\frac{10}{100}\Big)^2$
$=5,000(1.10)^2$
$=\text { Rs. } 6,050$
Now,
$Cl=A-P$
$=\text { Rs. } 6,050-\text { Rs } 5,000$
$=\text { Rs. } 1,050$
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:
$\text{A}=2,000\Big(1+\frac{4}{100}\Big)^3$
$=2,000(1.04)^3$
$=\text { Rs. } 2,249.68$
Now,
$Cl=A-P$
$=\text { Rs. } 2,249.68-\text { Rs. } 2,000$
$=\text { Rs. } 249.68$
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 $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}.$
$\therefore\text{A}=2,400\Big(1+\frac{20}{100}\Big)^3$
$=2,400(1.2)^3$
$=4,147.20$
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:
$P=\text { Rs. } 16,000$
$R=12 \cdot 5 \% \text { p.a. }$
$n=3 \text { years }$
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,
$P=\text { Initial population }=25,000$
$R_1=4 \%$
$R_2=5 \%$
$R_3=8 \%$
$n=\text { Number of years }=3$
$\therefore$ Population after three years:
$=P\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)\left(1+\frac{R_3}{100}\right)$
$=25,000\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\left(1+\frac{8}{100}\right)$
$=25,000(1.04)(1.05)(1.08)$
$=29,484$
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 $A = P \left(1+\frac{ R }{100}\right)^{ n }$ on the given situation, we get:
$A=12,800\left(1+\frac{7.5}{100}\right)^3$
$=12,800(1.075)^3$
$=\text { Rs. } 15,901.40$
$\text { Now, }$
$Cl=A-P$
$=\text { Rs. } 15,901.40-\text { Rs. } 12,800$
$\text { = Rs. } 3,101.40$
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 $A = P \left(1+\frac{ R }{100}\right)^{ n }$ on the given situation, we get:
$A=3,000\left(1+\frac{5}{100}\right)^2$
$=3,000(1.05)^2$
$=\operatorname{Rs} 3,307.50$
Now,
$Cl=A-P$
$=\text { Rs. } 3,307.50-\text { Rs. } 3,000$
$=\text { Rs. } 307.50$
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 $A = P \left(1+\frac{ R }{100}\right)^{ n }$ on the given situation, we get:
$A=10,000\left(1+\frac{20}{200}\right)^4$
$=10,000(1.1)^4$
$=\text { Rs. } 14,641$
Now,
$Cl=A-P$
$=\text { Rs. } 14,641-\text { Rs. } 10,000$
$=\text { Rs. } 4,641$
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}}$$196,830=\text{P}\Big(1+\frac{8}{100}\Big)^{3}$
$196,830=\text{P}(1.08)^{3}$
$\text{P}=\frac{196,830}{1.259712}$
$=156,250$
Thus, the population three years ago was 156,250.
View full question & answer→Question 413 Marks
Compute the amount and the compound interest in each of the following by using the formulae when:
Principal $=$ Rs. 3000 , Rate $=18 \%$, Time $=2$ years
AnswerApplying the rule $A = P \left(1+\frac{ R }{100}\right)^{ n }$ on the given situation, we get:
$A=3,000\left(1+\frac{18}{100}\right)^2$
$=3,000(1.18)^2$
$=\text { Rs. } 4,177.20$
Now,
$Cl=A-P$
$=\text { Rs. } 4,177.20-\text { Rs } 3,000$
$=\text { Rs. } 1,177.20$
View full question & answer→Question 423 Marks
Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.
AnswerLet the rate percent per annum be R.
Then,
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$2\text{P}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{3}$
$\Big(1+\frac{\text{R}}{100}\Big)^{3}=2$
$\Big(1+\frac{\text{R}}{100}\Big)=1.2599$
$\frac{\text{R}}{100}=0.2599$
$\text{R}=25.99$
Thus, the required rate is 25.99% per annum.
View full question & answer→Question 433 Marks
The production of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will be its production after 3 years?
AnswerProduction after three years $=\text{P}\Big(1+\frac{\text{R}_{1}}{100}\Big)^{2}\Big(1-\frac{\text{R}_{2}}{100}\Big)$$=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:
$P=\text { Rs. } 12,000$
$R_1=15 \% \text { p.a. }$
$R_2=16 \% \text { p.a. }$
$\therefore$ Amount after two years $= P \left(1+\frac{ R _1}{100}\right)\left(1+\frac{ R _2}{100}\right)$
$=\text { Rs. } 12,500\left(1+\frac{15}{100}\right)\left(1+\frac{16}{100}\right)$
$=\text { Rs. } 12,500(1.15)(1.16)$
$=\text { Rs. } 16,675$
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 $A = P \left(1+\frac{ R }{100}\right)^{ n }$ on the given situation, we get:
$A=10,000\left(1+\frac{20}{200}\right)^4$
$=10,000(1.1)^4$
$=\text { Rs. } 14,641$
Now,
$Cl=A-P$
$=\text { Rs. } 14,641-\text { Rs. } 10,000$
$=\text { Rs. } 4,641$
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 $= P \left(1+\frac{ R }{100}\right)^{ n }$
$=125,000\left(1+\frac{2}{100}\right)^3$
$=125,000(1.02)^3$
$=132,651$
Hence, the population after three years will be 132,651 .
View full question & answer→