3 Mark Question

Question 13 Marks

Find the difference between simple interest and compound interest on Rs 20,000 in 2 years at 8 p.c.p.a.

Answer

Here, $P=$ Rs $20,000, R=8$ p.c.p.a., $N=2$ years
i. Simple interest (I)
$
\begin{aligned}
I & =\frac{ PNR }{100} \\
\therefore \quad I & =\frac{20,000 \times 2 \times 8}{100}=₹ 3200
\end{aligned}
$
Simple interest $(I)=$ Rs 3200

ii. Compound Interest (I):
$
\begin{aligned}
& A=P\left[1+\frac{R}{100}\right]^N \\
&=20000\left[1+\frac{8}{100}\right]^2 \\
&=20000\left[\frac{100+8}{100}\right]^2 \\
&=20000\left[\frac{108}{100}\right]^2 \\
&=20000\left[\frac{27 \times 4}{25 \times 4}\right]^2 \\
&=20000\left[\frac{27}{25}\right]^2 \\
&= 20000 \times \frac{27}{25} \times \frac{27}{25} \\
&= 32 \times 27 \times 27 \\
&= \text { Rs } 23,328 \\
& \text { Compound interest (I) } \\
&= \text { Amount (A) }- \text { Principal (P) } \\
&= 23,328-20,000 \\
&= \text { Rs } 3328, . . \text { (ii) }
\end{aligned}
$

iii. Difference
= Compound interest – Simple interest
= 3328 – 3200 … [Form (i) and (ii)]
= Rs 128
∴ The difference between compound interest and simple interest is Rs 128.

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Question 23 Marks

In how many years Rs 700 will amount to Rs 847 at a compound interest rate of 10 p.c.p.a.

Answer

Here, $P=$ Rs $700, R=10$ p.c.p.a., $A=$ Rs 847
$
\begin{array}{ll}
& A=P\left[1+\frac{R}{100}\right]^{ N } \\
\therefore & 847=700\left[1+\frac{10}{100}\right]^{ N } \\
\therefore & 847=700\left[\frac{100+10}{100}\right]^{ N } \\
\therefore \quad & 847=700\left[\frac{110}{100}\right]^{ N } \\
\therefore & 847=700\left[\frac{11}{10}\right]^{ N } \\
\therefore \quad & \frac{847}{700}=\left[\frac{11}{10}\right]^{ N } \\
\therefore \quad & \frac{121 \times 7}{100 \times 7}=\left[\frac{11}{10}\right]^{ N } \\
\therefore \quad & \frac{121}{100}=\left[\frac{11}{10}\right]^{ N } \\
\therefore \quad & {\left[\frac{11}{10}\right]^2=\left[\frac{11}{10}\right]^{ N }} \\
\therefore & N =2
\end{array}
$
$\ldots\left[\right.$ If $a^n=a^m$, then $n=m$ ]
$\therefore$ Rs 700 will amount to Rs 847 in 2 vears.

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Question 33 Marks

The population of a suburb is 16,000. Find the rate of increase in the population if the population after two years is 17,640.

Answer

Here, P = Population of a suburb = 16,000
N = 2 years
A = Increase in the population after 2 years = 17,640
R = Rate of increase in population
\begin{aligned}
& A=P\left[1+\frac{R}{100}\right]^{ N } \\
& \therefore \quad 17640=16000\left[1+\frac{ R }{100}\right]^2 \\
& \therefore \quad \frac{17640}{16000}=\left[1+\frac{ R }{100}\right]^2 \\
& \therefore \quad \frac{1764}{1600}=\left[1+\frac{ R }{100}\right]^2 \\
& \therefore \quad \frac{441 \times 4}{400 \times 4}=\left[1+\frac{ R }{100}\right]^2 \\
& \therefore \quad \frac{441}{400}=\left[1+\frac{R}{100}\right]^2 \\
& \therefore \quad\left[\frac{21}{20}\right]^2=\left[1+\frac{ R }{100}\right]^2 \\
& \therefore \quad \frac{21}{20}=1+\frac{ R }{100} \\
& \text {...[Taking square root on both sides] } \\
& \therefore \quad \frac{21}{20}-1=\frac{ R }{100} \\
& \therefore \quad \frac{21-20}{20}=\frac{ R }{100} \\
& \therefore \quad \frac{1}{20}=\frac{ R }{100} \\
& \therefore \quad \frac{100}{20}=R \\
&
\end{aligned}
∴5 = R
i.e., R = 5%
∴The rate of increase in the population is 5 p.c.p.a.

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Question 43 Marks

A principal amounts to Rs 13,924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.

Answer

Here, $A=$ Rs $13,924, R=18$ p.c.p.a., and $N=2$ years
$
\begin{array}{ll}
& A=P\left[1+\frac{ R }{100}\right]^{ N } \\
\therefore 13924= P \left[1+\frac{18}{100}\right]^2 \\
\therefore 13924= P \left[\frac{100+18}{100}\right]^2 \\
\therefore 13924= P \left[\frac{118}{100}\right]^2 \\
\therefore 13924= P \left[\frac{59 \times 2}{50 \times 2}\right]^2 \\
\therefore 13924= P \left[\frac{59}{50}\right]^2 \\
\therefore 13924= P \times \frac{59}{50} \times \frac{59}{50} \\
\therefore \quad \frac{13924 \times 50 \times 50}{59 \times 59}= P \\
\therefore P =\frac{236 \times 50 \times 50}{59 \times 1} \\
\therefore P = 4 \times 50 \times 50 \\
\therefore P = Rs. 10,000
\end{array}
$
$\therefore$ The principal is Rs 10,000 .

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Question 53 Marks

A loan of Rs 15,000 was taken on compound interest. If the rate of compound interest is 12 p.c.p.a. find the amount to settle the loan after 3 years.

Answer

Here, $P=$ Rs $15,000, R=12$ p.c.p.a, and
$
\begin{aligned}
N =3 & \text { years } \\
A & =P\left[1+\frac{R}{100}\right]^N \\
\therefore \quad A & =15000\left[1+\frac{12}{100}\right]^3 \\
& =15000\left[\frac{100+12}{100}\right]^3 \\
& =15000\left[\frac{112}{100}\right]^3 \\
& =15000\left[\frac{28 \times 4}{25 \times 4}\right]^3 \\
& =15000\left[\frac{28}{25}\right]^3 \\
& =15000 \times \frac{28}{25} \times \frac{28}{25} \times \frac{28}{25} \\
& =\frac{24}{25} \times 28 \times 28 \times 28 \\
& =0.96 \times 21952 \\
& =₹ 21,073.92
\end{aligned}
$
$\therefore$ The amount required to settle the loan after 3 years is Rs $21,073.92$

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Question 63 Marks

Find the compound interest if the amount of a certain principal after two years is Rs 4036.80 at the rate of 16 p.c.p.a.

Answer

Here, $A=$ Rs $4036.80, R=16$ p.c.p.a. and $N=2$ years
$
\begin{array}{ll}
\text { i. } A = P \left[1+\frac{ R }{100}\right]^N \\
\therefore 4036.80= P \left[1+\frac{16}{100}\right]^2 \\
\therefore 4036.80= P \left[\frac{100+16}{100}\right]^2 \\
\therefore 4036.80= P \left[\frac{116}{100}\right]^2 \\
\therefore 4036.80= P \left[\frac{29 \times 4}{25 \times 4}\right]^2 \\
\therefore 4036.80= P \left[\frac{29}{25}\right]^2 \\
\therefore 4036.80= P \times \frac{29}{25} \times \frac{29}{25} \\
\therefore \frac{4036.80 \times 25 \times 25}{29 \times 29}= P
\end{array}
$

ii. Interest $=$ Amount $(A)$ - Principal $(P)$
$=4036.80-3000$
$=$ Rs 1036.80
$\therefore$ The compound interest after 2 years would be Rs 1036.80 .

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Question 73 Marks

The cost price of a machine is Rs 2,50,000. If the rate of depreciation is 10% per year, find the depreciation in price of the machine after two years.

Answer

Here, $P =$ Cost price of machine $=$ Rs $2,50,000$
$R=$ Rate of depreciation per year $=10 \%$
$N =2$ years
$A=$ Depreciated price of the machine after 2 years
$
\begin{aligned}
A & =P\left[1+\frac{R}{100}\right]^N \\
& =2,50,000\left[1+\frac{(-10)}{100}\right]^2 \\
& =2,50,000\left[1-\frac{10}{100}\right]^2 \\
& =2,50,000\left[\frac{100-10}{100}\right]^2 \\
& =2,50,000\left[\frac{90}{100}\right]^2 \\
& =2,50,000\left[\frac{9}{10}\right]^2 \\
& =2,50,000\left[\frac{81}{100}\right] \\
= & 2,500 \times 81 \\
= & \text { Rs } 2,02,500
\end{aligned}
$
Depreciation in price $=$ Cost price $( P )-$ Depreciated price $( A )$ $=2,50,000-2,02,500$
$=$ Rs 47,500
$\therefore$ The depreciation in price of the machine after 2 years would be Rs 47,500 .

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Question 83 Marks

In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.

Answer

Here, $P=$ Present number of trees in the forest $=40,000$
$R=$ Increase in the number of trees per year $=5 \%$
$N =3$ years
$A=$ Number of trees after 3 years
$
\begin{aligned}
A & =P\left[1+\frac{R}{100}\right]^N \\
& =40000\left[1+\frac{5}{100}\right]^3 \\
& =40000\left[\frac{100+5}{100}\right]^3 \\
& =40000\left[\frac{105}{100}\right]^3 \\
& =40000\left[\frac{21 \times 5}{20 \times 5}\right]^3 \\
& =40000\left[\frac{21}{20}\right]^3 \\
& =40000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} \\
= & 5 \times 21 \times 21 \times 21 \\
= & 5 \times 9261 \\
= & 46,305
\end{aligned}
$
$\therefore$ The expected number of trees in the forest after 3 years is 46,305

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Question 93 Marks

A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.

Answer

Here, $P=$ Present number of sheeps $=200$
$R=$ Increase in number of sheeps per year $=8 \%$
$N =3$ years
$A=$ Number of sheeps after 3 years
$A=P\left[1+\frac{R}{100}\right]^N$
$=200\left[1+\frac{8}{100}\right]^3$
$=200\left[\frac{100+8}{100}\right]^3$
$=200\left[\frac{108}{100}\right]^3$
$=200\left[\frac{27 \times 4}{25 \times 4}\right]^3$
$=200\left[\frac{27}{25}\right]^3$
$=200 \times \frac{27}{25} \times \frac{27}{25} \times \frac{27}{25}$

$=8 \times 27 \times \frac{27}{25} \times \frac{27}{25}$

$=\frac{0.32}{25} \times 27 \times 27 \times 27$

$=0.0128 \times 27 \times 27 \times 27$

$=251.9424$

$=252$
$\therefore$ The number of sheeps with the shepherd after 2 years would be 252 (approx).

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Question 103 Marks

On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.

Answer

Here, $P=$ Initial number of workers $=320$
$R=$ Increase in the number of workers per year $=25 \%$
$N=2$ years
$A=$ Number of workers after 2 years
$
\begin{aligned}
A & =P\left[1+\frac{R}{100}\right]^N \\
& =320\left[1+\frac{25}{100}\right]^2 \\
& =320\left[\frac{100+25}{100}\right]^2 \\
& =320\left[\frac{125}{100}\right]^2 \\
& =320\left[\frac{25 \times 5}{25 \times 4}\right]^2 \\
& =320\left[\frac{5}{4}\right]^2 \\
& =320\left[\frac{25}{16}\right] \\
& =20 \times 25 \\
& =500
\end{aligned}
$
$\therefore$ The number of workers after 2 years would be 500 .

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Question 113 Marks

Visit the bank nearer to your house and get the information regarding the different schemes and rates of interests. Make a chart and display in your class.

Answer

SELF

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Question 123 Marks

To start a business Shalaka has taken a loan of Rs 8000 at a rate of $10 \frac{1}{2}$ p.c.p.a. After two years how much compound interest will she have to pay ?

Answer

Here, $P=$ Rs $8000, N=2$ years and
$
\begin{aligned}
R & =10 \frac{1}{2} \%=\frac{21}{2} \%=10.5 \text { p.c.p.a. } \\
A & = P \left[1+\frac{R}{100}\right]^{ N } \\
& =8000\left[1+\frac{10.5}{100}\right]^2 \\
& =8000\left[1+\frac{105}{1000}\right]^2 \\
& =8000\left[\frac{1000+105}{1000}\right]^2 \\
& =8000\left[\frac{1105}{1000}\right]^2 \\
& =8000\left[\frac{221 \times 5}{200 \times 5}\right]^2 \\
& =8000\left[\frac{221}{200}\right]^2 \\
& =8000 \times \frac{221}{200} \times \frac{221}{200} \\
& =\frac{1}{5} \times 221 \times 221 \\
& =0.2 \times 48,841 \\
& =₹ 9768.20
\end{aligned}
$

I = Amount (A) – Principal (P)
= 9768.20 – 8000
= Rs 1768.20
∴ After two years Shalaka will have to pay Rs 1768.20 as compound interest.

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Question 133 Marks

Sameerrao has taken a loan of Rs 12500 at the rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan ?

Answer

Here, $P=$ Rs $12,500, R=12$ p.c.p.a. and $N=3$ years
$
\begin{aligned}
A & =P\left[1+\frac{R}{100}\right]^{ N } \\
& =12500\left[1+\frac{12}{100}\right]^3 \\
& =12500\left[\frac{100+12}{100}\right]^3 \\
& =12500\left[\frac{112}{100}\right]^3 \\
& =12500\left[\frac{28 \times 4}{25 \times 4}\right]^3 \\
& =12500\left[\frac{28}{25}\right]^3 \\
& =12500 \times \frac{28}{25} \times \frac{28}{25} \times \frac{28}{25} \\
& =\frac{20}{25} \times 28 \times 28 \times 28 \\
= & 0.8 \times 28 \times 28 \times 28 \\
= & \text { Rs } 17,561.60
\end{aligned}
$
Sameerrao should pay Rs $17,561.60$ to clear his loan.

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Maths 3 Mark Question

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