Basics of Financial Mathematics Test 13

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Basics of Financial Mathematics Test 13

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Weekly Quiz Competition

Question 1
1 / -0

The present population of a town is $$5000$$. If the population decreases at the rate of $$10\%$$ per year, the population after $$2$$ years will be

A
$$4820$$

B
$$4050$$

C
$$4980$$

D
$$4500$$

Solution

Present population, $$P_o = 5000$$
Rate $$= 10\%$$
Time $$= 2$$ years
Let after $$2$$ years population $$= P$$
Thus, $$P = P_o(1 - \frac{R}{100})^T$$
$$P = 5000 (1 - \frac{10}{100})^2$$
$$P = 4050$$
Thus after two years the population will be $$4050$$
Question 2
1 / -0

A sum of money invested at C.I. amounts to Rs. 800 in 3 years and to Rs. 840 in 4 years. The rate of interest per annum is

A
$$2\, \displaystyle \frac{1}{2}$$%

B
$$4$$%

C
$$5$$%

D
$$6\, \displaystyle \frac{2}{3}$$%

Solution

Simple interest for one year$$=840-800=40 Rs.$$

Rate of interest=$$\dfrac{40\times 100}{800\times 1}=5$$%
Question 3
1 / -0

Madhav lent out Rs. 7953 for 2 years and Rs. 1800 for 3 years at the same rate of simple interest. If he got Rs. 2343. 66 as total, then find the percent rate of interest.

A
11%

B
12%

C
12.5%

D
5%

Solution

We know that Simple Interest $$ = \cfrac {PNR}{100} $$
Given, $$ \cfrac {7953 \times 2 \times R}{100} + \cfrac {1800 \times 3 \times R}{100} = Rs 2343.66 $$
$$=> 159.06R + 54R = Rs 2343.66 $$
$$ => 213.06R = 2343.66 $$
$$ => R = 11 \%$$
Question 4
1 / -0

A scooter was bought at $$\text{Rs. } 42000$$. Its value depreciated at the rate of $$8\%$$ per annum, find its value after one year.

A
$$\text{Rs. } 38640$$

B
$$\text{Rs. } 37740$$

C
$$\text{Rs. } 38145$$

D
$$\text{Rs. } 39640$$
Question 5
1 / -0

The present population of a town is $$35,000$$. If the population becomes $$35,700$$ next year, the rate of growth is

A
$$2\%$$

B
$$20\%$$

C
$$70\%$$

D
$$7\%$$

Solution

Population next year, $$P = 35700$$
Current population, $$P_o = 35000$$
$$P = P_o \left(1 + \cfrac{R}{100}\right)^T$$
$$35700 = 35000 \left(1+ \cfrac{R}{100}\right)^1$$
$$35700 - 35000 = \cfrac{35000 R}{100}$$
$$R = \cfrac{700 \times 100}{35000}$$
$$R = 2\%$$
Question 6
1 / -0

The population of a town increases from $$10000$$ to $$10400$$. The growth rate is

A
$$4\%$$

B
$$40\%$$

C
$$2\%$$

D
$$20\%$$

Solution

Population an year ago, $$P_o = 100000$$
Population after one year $$P = 10400$$
Let the growth rate be R,
then $$ P = P_o \left(1 + \cfrac{R}{100}\right)^T$$
$$10400 = 10000\left(1 + \cfrac{R}{100}\right)$$
$$10400 = 10000 + 100 R$$
$$R = \cfrac{400}{100} = 4\%$$
Thus, growth rate $$= 4\%$$
Question 7
1 / -0

A student purchased a pen for Rs 100. At the end of 8 years it was valued at Rs 20 Assuming the yearly depreciation is a constant amount find the annual depreciation

A
8

B
9

C
10

D
11

Solution

Cost Price of the pen $$=$$$$Rs.100$$
After $$8$$ years, Value of the pen $$=Rs. 20$$
Since, depreciation is yearly and constant
So, Annual depreciation $$=\dfrac{Rs.100-Rs.20}{8}=\dfrac{Rs.80}{8}=Rs.10$$
Hence, C is the correct option.
Question 8
1 / -0

Calculate compound interest on Rs $$2000$$ for $$2$$ years at the rate $$5\% $$

A
$$105$$

B
$$200$$

C
$$205$$

D
$$2205$$

Solution

For the first year, $$P = Rs. 2000, R = 5\%$$
so, I = $$\cfrac { P \times R \times T }{ 100 } =\cfrac { Rs.\quad 2100 \times 5 \times 1 }{ 100 } =Rs.\quad 105$$
The amount after the second year =$$ Rs. 2100 + Rs. 105$$
$$=Rs. 2205$$
Final amount =$$ Rs. 2205$$
Hence, Compound Interest = final amount - original amount
$$= Rs. 2205 - Rs. 2000 = Rs. 205$$
Question 9
1 / -0

In what time will Rs 64000 amount to Rs 68921 at 5% per annum interest being compounded half-yearly?

A
$$\displaystyle 1\frac{1}{2}$$ years

B
$$\displaystyle \frac{2}{3}$$ years

C
$$2$$ years

D
None of these

Solution

$$CA=P\left(1+\frac{R}{100}\right)^{2t}$$
$$69821=64000\left(1+\frac{5}{100}\right)^{2t}$$
$$\frac{69821}{64000}=\left(\frac{41}{40}\right)^{2t}$$
$$\left(\frac{41}{40}\right)^3=\left(\frac{41}{40}\right)^{2t}$$
$$3=2t$$
$$t=\frac{3}{2}$$
Hence The required time is $$1\tfrac{1}{2}$$Years.
Question 10
1 / -0

Find the compound interest on Rs 31250 at 8% per annum for $$\displaystyle 2\frac{3}{4}$$ years

A
Rs 8637

B
Rs 7387

C
Rs 1250

D
None of these

Solution

P=Rs. 31250,R=8%,$$T=2times \frac{3}{4}$$,A=?
$$A=P\left(1+\frac{R}{100}\right)$$
$$\Rightarrow 31250\left(1+\frac{8}{100}\right)^{2\tfrac{3}{4}}$$
$$\Rightarrow 31250\times \frac{108}{100}\times \frac{108}{100}\times \frac{106}{100}$$
$$\Rightarrow Rs. 38637$$
$$C.I=A-P$$
$$=38637-31250=Rs. 7387$$

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Basics of Financial Mathematics Test 13

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Basics of Financial Mathematics Test 13

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SHARING IS CARING

Question 1

$$4820$$

$$4050$$

$$4980$$

$$4500$$

Solution

Question 2

$$2\, \displaystyle \frac{1}{2}$$%

$$4$$%

$$5$$%

$$6\, \displaystyle \frac{2}{3}$$%

Solution

Question 3

11%

12%

12.5%

5%

Solution

Question 4

$$\text{Rs. } 38640$$

$$\text{Rs. } 37740$$

$$\text{Rs. } 38145$$

$$\text{Rs. } 39640$$

Question 5

$$2\%$$

$$20\%$$

$$70\%$$

$$7\%$$

Solution

Question 6

$$4\%$$

$$40\%$$

$$2\%$$

$$20\%$$

Solution

Question 7

8

9

10

11

Solution

Question 8

$$105$$

$$200$$

$$205$$

$$2205$$

Solution

Question 9

$$\displaystyle 1\frac{1}{2}$$ years

$$\displaystyle \frac{2}{3}$$ years

$$2$$ years

None of these

Solution

Question 10

Rs 8637

Rs 7387

Rs 1250

None of these

Solution

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