3 Marks 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 $\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ 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,
$Cl = A - P$
$= Rs. 14,641 - Rs. 10,000$
$= Rs. 4,641$

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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 $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=125,000\Big(1+\frac{2}{100}\Big)^{3}$ $=125,000(1.02)^{3}$
$=132,651$
Hence, the population after three years will be $132,651.$

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MATHS 3 Marks Question