3 Mark Question

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.

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 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.

Answer

Let 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%.

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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.

Answer

Let 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%.

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Question 543 Marks

What will Rs. 125000 amount to at the rate of 6%, if the interest is calculated after every 3 months?

Answer

Because 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.

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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.

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Question 563 Marks

A sum amounts to Rs. 756.25 at 10% per annum in 2 years, compounded annually. Find the sum.

Answer

Let 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.

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Question 573 Marks

At what rate percent compound interest per annum will Rs. 640 amount to Rs. 774.40 in 2 years?

Answer

Let 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.

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Question 583 Marks

Find the principal if the interest compounded annually at the rate of 10% for two years is Rs. 210.

Answer

Let the sum be Rs. x
We know that:
CI = A - P
$= \text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-\text{P}$
$= \text{P}\Big[\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-1\Big]$
$210= \text{x}\Big[\Big(1+\frac{10}{100}\Big)^{2}-1\Big]$
$210= \text{x}\Big[(1.10)^{2}-1\Big]$
$\text{x}=\frac{210}{0.21}$
$=1,000$
Thus, the required sum is Rs. 1,000.

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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.

Answer

Value 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.

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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

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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?

Answer

Given:
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.

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Question 623 Marks

What will be the compound interest on Rs. 4000 in two years when rate of interest is 5% per annum?

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,
$\text{A}=4,000\Big(1+\frac{5}{100}\Big)^{2}$
$=4,000(1.05)^{2}$
$=\text{Rs. }4,410$
And,
CI = A - P
= Rs. 4,410 - Rs. 4,000
= Rs. 410

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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.

Answer

Here,
$P =$ 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 .

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Maths 3 Mark Question