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Maths 3 Mark Question

Compound Interest · Maharashtra Board STD 8 (MSBSHSE - English Medium). 63 questions.

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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.
Answer
Given:
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
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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.
Answer
Given:
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.
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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.
Answer
Given:
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
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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.
Answer
Let 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.
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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.
Answer
Given:
$P=\text { Rs. } 1,000$
$R=10 \% \text { p.a. }$
$n=2 \text { years }$
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 .
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Question 63 Marks
On what sum will the compound interest at 5% per annum for 2 years compounded annually be Rs. 164?
Answer
Let 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.
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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.
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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
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Question 93 Marks
In what time will Rs. 1000 amount to Rs. 1331 at 10% per annum, compound interest?
Answer
Let 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.
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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
Answer
Applying 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,
$Cl=A-P$
$=\text { Rs. } 6,050-\text { Rs } 5,000$
$=\text { Rs. } 1,050$
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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?
Answer
We 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
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Question 123 Marks
In what time will Rs. 4400 become Rs. 4576 at 8% per annum interest compounded half-yearly?
Answer
Let 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.
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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
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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
Answer
Applying 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,
$Cl=A-P$
$=\text { Rs. } 2,249.68-\text { Rs. } 2,000$
$=\text { Rs. } 249.68$
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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.
Answer
Given:
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.
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Question 163 Marks
Find the compound interest when principal = Rs. 3000, rate = 5% per annum and time = 2 years.
Answer
Principal 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
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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.
Answer
Given:
$P=\text { Rs. } 16,000$
$R=12 \cdot 5 \% \text { p.a. }$
$n=3 \text { years }$
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. $22,781.25$.
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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.
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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.
Answer
Here,
$P=\text { Initial population }=25,000$
$R_1=4 \%$
$R_2=5 \%$
$R_3=8 \%$
$n=\text { Number of years }=3$
$\therefore$ 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.
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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.
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Question 213 Marks
A certain sum amounts to Rs. 5832 in 2 years at 8% compounded interest. Find the sum.
Answer
Let 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.
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Question 223 Marks
In how much time would Rs. 5000 amount to Rs. 6655 at 10% per annum compound interest?
Answer
Let 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.
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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?
Answer
Let 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.
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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
Answer
Applying the rule $A = P \left(1+\frac{ R }{100}\right)^{ n }$ on the given situation, we get:
$A=12,800\left(1+\frac{7.5}{100}\right)^3$
$=12,800(1.075)^3$
$=\text { Rs. } 15,901.40$
$\text { Now, }$
$Cl=A-P$
$=\text { Rs. } 15,901.40-\text { Rs. } 12,800$
$\text { = Rs. } 3,101.40$
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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?
Answer
Given:
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.
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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.
Answer
Population 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.
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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.
Answer
Given:
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.
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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.
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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.
Answer
Given:
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
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Question 303 Marks
Find the compound interest on Rs. 64000 for 1 year at the rate of 10% per annum compounded quarterly.
Answer
To 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
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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.
Answer
Given:
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
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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
Answer
Applying 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,
$Cl=A-P$
$=\text { Rs. } 3,307.50-\text { Rs. } 3,000$
$=\text { Rs. } 307.50$
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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.
Answer
Let 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.
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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
Answer
Applying 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,
$Cl=A-P$
$=\text { Rs. } 14,641-\text { Rs. } 10,000$
$=\text { Rs. } 4,641$
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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.
Answer
Given:
$\text{CI}-\text{SI}={\text{Rs. }}20$
$\Big[\text{P}\Big(1+\frac{4}{100}\Big)^{2}-\text{P}\Big]-\frac{\text{P}\times4\times2}{100}=20$
$\text{P }[(1.04^{2}-1)]-0.08\text{ P}=20$
$0.0816\text{ P}-0.08\text{ P}=20$
$0.0016\text{ P}=20$
${\text{P}}=\frac{20}{0.0016}$
$=12,500$
Thus, the required sum is Rs. 12,500.
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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.
Answer
Population 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.
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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?
Answer
Here,
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.
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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?
Answer
Population 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.
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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.
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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?
Answer
Population after three years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$$196,830=\text{P}\Big(1+\frac{8}{100}\Big)^{3}$
$196,830=\text{P}(1.08)^{3}$
$\text{P}=\frac{196,830}{1.259712}$
$=156,250$
Thus, the population three years ago was 156,250.
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Question 413 Marks
Compute the amount and the compound interest in each of the following by using the formulae when:
Principal $=$ Rs. 3000 , Rate $=18 \%$, Time $=2$ years
Answer
Applying the rule $A = P \left(1+\frac{ R }{100}\right)^{ n }$ on the given situation, we get:
$A=3,000\left(1+\frac{18}{100}\right)^2$
$=3,000(1.18)^2$
$=\text { Rs. } 4,177.20$
Now,
$Cl=A-P$
$=\text { Rs. } 4,177.20-\text { Rs } 3,000$
$=\text { Rs. } 1,177.20$
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Question 423 Marks
Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.
Answer
Let the rate percent per annum be R.
Then,
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$2\text{P}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{3}$
$\Big(1+\frac{\text{R}}{100}\Big)^{3}=2$
$\Big(1+\frac{\text{R}}{100}\Big)=1.2599$
$\frac{\text{R}}{100}=0.2599$
$\text{R}=25.99$
Thus, the required rate is 25.99% per annum.
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Question 433 Marks
The production of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will be its production after 3 years?
Answer
Production after three years $=\text{P}\Big(1+\frac{\text{R}_{1}}{100}\Big)^{2}\Big(1-\frac{\text{R}_{2}}{100}\Big)$$=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.
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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.
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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.
Answer
Aman'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.
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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.
Answer
Let 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.
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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.
Answer
Given:
$P=\text { Rs. } 12,000$
$R_1=15 \% \text { p.a. }$
$R_2=16 \% \text { p.a. }$
$\therefore$ 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 .
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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?
Answer
Let 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.
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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.
Answer
Value 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.
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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
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3 Mark Question - Maths STD 8 Questions - Vidyadip