Basics of Financial Mathematics Test 19

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Basics of Financial Mathematics Test 19

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Weekly Quiz Competition

Question 1
1 / -0

The present population of a town is $$14000$$. If it increases at the rate of $$10\%$$ per annum, what will be its population after $$4$$ years?

A
$$10497$$

B
$$20497$$

C
$$30497$$

D
$$40497$$

Solution

Population after $$4$$ years $$=$$ $$\text{Present Population}$$ $$\left (1+\dfrac{r}{100}\right)^n$$

Population after $$4$$ years $$=$$ $$14000 \times \left (1+\dfrac{10}{100}\right)^4$$

$$=14000\times1.4641$$

$$=20497$$

Population after $$4$$ years $$= 20497$$
Question 2
1 / -0

The present population of a city is $$15000$$. If it increases at the rate of $$4\%$$ per annum. Find its population after $$2$$ years.

A
$$16224$$

B
$$26224$$

C
$$36224$$

D
$$46224$$

Solution

Given, $$P=15000, r=4\%, n=2$$ years
Population after $$2$$ years $$=$$ $$P[(1+\frac{R}{100})^2]$$
$$=15000\left [(1+\dfrac{4}{100}\right)^2]$$
$$=15000\times 1.0816$$
$$=16224$$
Population after $$2$$ years is $$ 16224$$.
Question 3
1 / -0

The cost of a machine is Rs. $$24500$$. After two year the value of that machine was depreciated by $$6\%$$. Find the value of machine after two year.

A
$$11560$$

B
$$21560$$

C
$$31560$$

D
$$41560$$

Solution

Given: Principal $$= 24500$$, $$r = 6\%, n = 2$$
Reduction $$= 6\%$$ of Rs. $$24500$$ per year
$$= \dfrac{24500\times 6\times 2}{100}=2940$$
Value at the end of $$2$$ years $$= 24500 - 2940 =$$ Rs. $$21560$$
Question 4
1 / -0

A man invests Rs. $$3000$$ for four years at a certain rate of interest, compounded annually. At the end of one year it amounts to Rs. $$4500$$. Calculate the rate of interest per annum.

A
$$10\%$$

B
$$20\%$$

C
$$50\%$$

D
$$30\%$$

Solution

Given, $$P =$$ Rs. $$3000$$
At the end of $$1$$ year, the investment amounts to Rs. $$4500$$.
Thus, $$A = $$ Rs. $$4500$$
Interest $$= A - P$$
Therefore, $$I = 4500 - 3000 = 1500$$, $$T = 1 $$ year.
$$R = \dfrac{1500\times 100}{3000\times 1}$$
$$R = 50\%$$
Therefore, the rate of interest per annum is $$50\%$$.
Question 5
1 / -0

The annual growth of capital of a company is $$12\%$$. At present, the capital is Rs. $$35000$$. What will be the capital after $$4$$ years?

A
$$55500$$

B
$$55073$$

C
$$45500$$

D
$$56430$$

Solution

Given: $$P = 35000, r = 12\%, n = 4$$ years
We know $$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$$
Putting all the given values, we get
$$A = 35000\left [\left (1+\dfrac{12}{100}\right)^4\right]$$
$$A =$$ Rs. $$55073$$
Question 6
1 / -0

A machine depreciates at the rate of $$10\%$$ of its value at the beginning of a year. If the present value of a machine is $$\text{Rs. }4000$$, find its value after $$3$$ years.

A
$$\text{Rs. }1916$$

B
$$\text{Rs. }2916$$

C
$$\text{Rs. }3916$$

D
$$\text{Rs. }4916$$

Solution

Given:-
$$P = \text{Rs. }4000$$
$$r = 10\%$$
$$n = 3$$ years

Now, as the value of the machine depriciates every year by $$10\%,$$
$$\begin{aligned}{}A& = P{\left( {1 - \frac{r}{{100}}} \right)^n}\\ &= 4000{\left( {1 - \frac{{10}}{{100}}} \right)^3}\\& = 4000{\left( {0.9} \right)^3}\\& = \text{Rs. }2916\end{aligned}$$

Hence, the value of the machine after $$3$$ years is $$\text{Rs. }2916.$$
Question 7
1 / -0

The population of a town increases every year at rate of $$10\%$$. Find the population of the town three years hence if it is $$13000$$ now.

A
$$17303$$

B
$$27303$$

C
$$37303$$

D
$$47303$$

Solution

Given, $$P = 13000, r = 10\%, n = 3$$ years
We know, $$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$$
$$\Rightarrow A = 13000\left [\left (1+\dfrac{10}{100}\right)^3\right]$$
$$\Rightarrow A= 17303$$
Hence, the population of the town $$3$$ years is $$17303$$.
Question 8
1 / -0

The population of a town increases evry year at rate of $$10\%$$. The present population of the town is $$12,500$$. Find the population of the town after $$2$$ years.

A
$$15125$$

B
$$25125$$

C
$$35125$$

D
$$45125$$

Solution

Given, $$P = 12500, r = 10\%, n = 2$$ years
We know, $$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$$
$$\Rightarrow A = 12500\left [\left (1+\dfrac{10}{100}\right)^2\right]$$
$$\Rightarrow A=$$ Rs. $$15125$$
Hence, the population after $$2$$ years $$= 15125$$.
Question 9
1 / -0

A company bought a car that had a value of Rs. $$10000$$. each year the value of the car depreciates by $$15\%$$. What is the value of the car at the end of $$2$$ years.

A
$$7225$$

B
$$6225$$

C
$$5225$$

D
$$4225$$

Solution

Given, $$P = 10000, r = 15\%$$ depreciation, $$n = 2$$ years
We know $$A = P\left [\left (1-\dfrac{r}{100}\right)^n\right]$$
$$\Rightarrow A = 10000\left [\left (1-\dfrac{15}{100}\right)^2\right]$$
$$\Rightarrow A =$$ Rs. $$7225$$
Therefore, the value of car after $$2$$ years is Rs. $$7225$$.
Question 10
1 / -0

Grace bought a new van for Rs. $$15000$$. Each year, the value of her car depreciated by $$10\%$$. Calculate the value of car before $$2$$ years.

A
$$12150$$

B
$$22150$$

C
$$32150$$

D
$$42150$$

Solution

Given, $$P = 15000, r = 10\% $$ depreciated, $$n = 2$$ years
We know, $$A = P\left [\left (1-\dfrac{r}{100}\right)^n\right]$$
$$\Rightarrow A = 15000\left [\left (1-\dfrac{10}{100}\right)^2\right]$$
$$\Rightarrow A =$$ Rs. $$12150$$
The value of car before $$2$$ years is Rs. $$12150$$.

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Basics of Financial Mathematics Test 19

Result

Basics of Financial Mathematics Test 19

Score

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SHARING IS CARING

Question 1

$$10497$$

$$20497$$

$$30497$$

$$40497$$

Solution

Question 2

$$16224$$

$$26224$$

$$36224$$

$$46224$$

Solution

Question 3

$$11560$$

$$21560$$

$$31560$$

$$41560$$

Solution

Question 4

$$10\%$$

$$20\%$$

$$50\%$$

$$30\%$$

Solution

Question 5

$$55500$$

$$55073$$

$$45500$$

$$56430$$

Solution

Question 6

$$\text{Rs. }1916$$

$$\text{Rs. }2916$$

$$\text{Rs. }3916$$

$$\text{Rs. }4916$$

Solution

Question 7

$$17303$$

$$27303$$

$$37303$$

$$47303$$

Solution

Question 8

$$15125$$

$$25125$$

$$35125$$

$$45125$$

Solution

Question 9

$$7225$$

$$6225$$

$$5225$$

$$4225$$

Solution

Question 10

$$12150$$

$$22150$$

$$32150$$

$$42150$$

Solution

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