Basics of Financial Mathematics Test 14

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Basics of Financial Mathematics Test 14

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Weekly Quiz Competition

Question 1
1 / -0

An emergency lamp costs Rs. 1100. Every year, its value depreciates by 10%. What is its value at the end of 3 years?

A
Rs.908.90

B
Rs. 809.10

C
Rs. 801.90

D
Rs. 980.90

Solution

Cost of Lamp = Rs 1100
Depreciation rate = 10 %
Now after 3 years, let the cost be C
then, $$C = 1100(1 - \frac{R}{100})^T$$
$$C = 1100(1 - \frac{10}{100})^3$$
$$C = 801.9$$ Rs
Question 2
1 / -0

Mr. Mittal finds that an increase in the rate of interest from $$4 \displaystyle \dfrac{7}{8}\%$$ to $$5 \displaystyle \dfrac{1}{8}\%$$ per annum increases his yearly income by Rs. $$250$$. His investment is

A
Rs. $$1,00,000$$

B
Rs. $$1,20,000$$

C
Rs. $$1,50,000$$

D
Rs. $$2,00,000$$

Solution

Let the investment be $$Rs. x$$. Then
$$x \times \displaystyle \dfrac{41}{8} \times \dfrac{1}{100} - x \dfrac{39}{8} \times \dfrac{1}{100} = 250$$

or $$2x = 200000$$

or $$x = 100000$$
Question 3
1 / -0

Rakesh took a loan for $$7$$ years at the rate of $$6\%$$ p.a. $$S.I$$. If the total interest paid was Rs. $$2100$$, the principal was

A
Rs. $$4400$$

B
Rs. $$4800$$

C
Rs. $$5000$$

D
Rs. $$5200$$

Solution

Simple interest $$S.I=\dfrac{P.R.T}{100}$$

$$\Rightarrow 2100 = \dfrac{P\times 6\times 7}{100}$$

$$\Rightarrow P=\dfrac{2100\times 100}{6\times 7}$$

$$\Rightarrow P=\dfrac{2100\times 100}{42}$$

$$\Rightarrow P=50\times 100$$

$$\Rightarrow P=5000$$
Hence, the principal amount $$P=Rs.5000$$
Question 4
1 / -0

Depreciation refers to

A
Closure of a plant due to lock out

B
Closure of a plant due to labour issues

C
Loss of equipment over time due to wear to tear

D
Destruction of plant in a fire accident

Solution

Depreciation means a decline in the value of fixed assets due to use, the passage of time or obsolescence or diminution in the intrinsic value of the asset due to use and/or lapse of time.
Question 5
1 / -0

X, Y and Z are the three sums of money. X is the simple interest on Y, and Y is the simple interest on Z. The rate percent per year and the time in years being the same in each case. Which one of the following relations in X, Y, Z is correct?

A
$$XYZ=100$$

B
$$Y^2=XZ$$

C
$$Z^2=XY$$

D
$$X^2=YZ$$

Solution

b'$$\displaystyle X=\frac{Y\times r\times t}{100}$$ and $$\displaystyle Y=\frac{Z\times r\times t}{100}$$

$$\displaystyle\frac{X}{Y}=\frac{Y}{Z}$$ i.e., $$Y^2=XZ$$'
Question 6
1 / -0

If the annual decrease in the population of a town by $$5\%$$ and the present population be $$68590$$, what was the population three years ago?

A
$$80000$$

B
$$60000$$

C
$$86000$$

D
$$65000$$

Solution

Present population, $$P = 68590$$
Let $$R = 5$$%
The population three years ago
= $$=\cfrac { P }{ {\left (1-\cfrac { R }{ 100 }\right ) }^{3} } =\cfrac { 68590 }{ {\left (1-\cfrac { 5 }{ 100 }\right) }^{ 3 } } \\ =\cfrac { 68590 }{ {\left (\cfrac { 19 }{ 20 } \right) }^{ 3 } } =\cfrac { 68590\times 20\times 20\times 20 }{ 19\times 19\times 19 } \\ =\cfrac { 548720000 }{ 6859 } \\ =80,000$$
Question 7
1 / -0

A sum of money at compound interest is doubled in $$20$$ years. Then the number of years required to tripple itself are

A
$$32$$ years

B
$$35$$ years

C
$$40 $$years

D
$$45$$ years

Solution

We know that
$$A=P\left(1+\dfrac{r}{100}\right)^T$$

According to the question,
$$2P=P\left(1+\dfrac{r}{100}\right)^{20}$$
$$2=\left(1+\dfrac{r}{100}\right)^{20}$$ $$.............(1)$$

And
$$3P=P\left(1+\dfrac{r}{100}\right)^T$$
$$3=\left(1+\dfrac{r}{100}\right)^T$$ $$...........(2)$$

From equation (1) and (2), we get
$$3^{1/T}=2^{1/20}$$

On taking $$log$$ both sides, we get
$$\dfrac{1}{T}\log\ 3=\dfrac{1}{20}\log\ 2$$
$$T=\dfrac{20log\ 3}{\log\ 2}$$
$$T=31.7$$
$$T\approx 32\ years$$

Hence, this is the answer.
Question 8
1 / -0

Find the compound interest on Rs 1,200 for 2 years at 10% per annum

A
Rs 252

B
Rs 225

C
Rs 300

D
Rs 350

Solution

Compound Interest $$C.I=A-P$$
Formula for amount when compound interest is applied : $$A=P\left (1+\dfrac {r}{100}\right )^n$$
Given $$P=1,200; n=2; r=10$$
$$\therefore A=1,200\left (1+\dfrac {10}{100}\right )^2=1,200\left (\dfrac {11}{10}\right )^2=\dfrac {1,200\times 121}{100}=Rs\ 1,452$$
$$CI=1,452-1,200=Rs\ 252$$
Question 9
1 / -0

Rajan borrowed Rs. $$5000$$ from Rakesh at simple interest. After $$3$$ years, Rakesh got Rs. $$300$$ more than what he had given to Rajan. What was the rate of interest per annum?

A
$$2\%$$

B
$$5\%$$

C
$$8\%$$

D
$$10\%$$

Solution

Given: Rajan borrowed $$Rs.5000$$ from Rakesh

Let the rate of interest be $$R\ \%$$

Then, simple interest after $$3 $$ years $$=\dfrac{5000\times 3 \times R}{100}$$

Given: Rakesh get $$Rs.300$$ more than the interest

$$\Rightarrow \dfrac{5000\times 3\times R}{100}=300$$

$$\Rightarrow R=\dfrac{300\times 100}{5000\times 3}$$

$$\Rightarrow R=2\%$$
Question 10
1 / -0

Find the rate, when $$Rs.\ 1800$$ earns an interest of $$Rs.\ 432$$ in $$3$$ years

A
$$7\%$$

B
$$6\%$$

C
$$5\%$$

D
$$8\%$$

Solution

Given:
Principal, $$P=Rs.\ 1800$$
Interest, $$I=Rs.\ 432$$
Times, $$ T=3$$ years

$$I=\dfrac{P\times R\times T}{100}$$

$$R=\dfrac {I\times 100}{P\times T}$$

$$=\dfrac {432\times 100}{1800\times 3}$$

$$=8\%$$

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Basics of Financial Mathematics Test 14

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Basics of Financial Mathematics Test 14

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SHARING IS CARING

Question 1

Rs.908.90

Rs. 809.10

Rs. 801.90

Rs. 980.90

Solution

Question 2

Rs. $$1,00,000$$

Rs. $$1,20,000$$

Rs. $$1,50,000$$

Rs. $$2,00,000$$

Solution

Question 3

Rs. $$4400$$

Rs. $$4800$$

Rs. $$5000$$

Rs. $$5200$$

Solution

Question 4

Closure of a plant due to lock out

Closure of a plant due to labour issues

Loss of equipment over time due to wear to tear

Destruction of plant in a fire accident

Solution

Question 5

$$XYZ=100$$

$$Y^2=XZ$$

$$Z^2=XY$$

$$X^2=YZ$$

Solution

Question 6

$$80000$$

$$60000$$

$$86000$$

$$65000$$

Solution

Question 7

$$32$$ years

$$35$$ years

$$40 $$years

$$45$$ years

Solution

Question 8

Rs 252

Rs 225

Rs 300

Rs 350

Solution

Question 9

$$2\%$$

$$5\%$$

$$8\%$$

$$10\%$$

Solution

Question 10

$$7\%$$

$$6\%$$

$$5\%$$

$$8\%$$

Solution

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