Basics of Financial Mathematics Test 11

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Basics of Financial Mathematics Test 11

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Weekly Quiz Competition

Question 1
1 / -0

What will be the compound interest on a sum of Rs.25000 after 3 years at the rate of 12 per cent p.a.?

A
Rs. 10123.20

B
Rs. 11123.20

C
Rs. 12123.20

D
Rs. 13123.20

Solution

P = Rs. 25000, n = 3 years, r = 12% p.a
$$\displaystyle \therefore$$ Amount $$\displaystyle =P\left ( 1+\frac{r}{100} \right )^{n}=Rs.25000\times \left ( 1+\frac{12}{100} \right )^{3}$$
$$\displaystyle =Rs.25000\times \left ( \frac{112}{100} \right )^{3}=Rs.25000\times \frac{28}{25}\times \frac{28}{25}\times \frac{28}{25}=Rs.35123.20$$
$$\displaystyle \therefore$$ Compound interest $$= Rs. (35123.20-25000)=Rs.10123.20$$
Question 2
1 / -0

A finance company declares that, with compound interest rate, a sum of money deposited by anyone will become $$8$$ times in three years. if the same amount is deposited at the same compound-rate of interest, then in how many years it will become $$16$$ times?

A
$$8$$ yrs

B
$$10$$ yrs

C
$$4$$ yrs

D
$$6$$ yrs

Solution

Let the prnicipal $$=x$$ in Rs.
In the first case Time$$=3$$ yrs, amount $$=8x$$, rate $$=R$$.
Now Amount $$=$$ Principal $$\times { \left( 1+rate \right) }^{ time }$$
$$\therefore x{ \left( 1+\dfrac { R }{ 100 } \right) }^{ 3 }=8x\\ \Longrightarrow { \left( 1+\dfrac { R }{ 100 } \right) }^{ 3 }={ 2 }^{ 3 }\\ \Longrightarrow 1+\dfrac { R }{ 100 } =2\\ \Longrightarrow R=100p.c.$$
In the second case the amount $$=16x$$, Time $$=n$$.
$$\therefore \quad x{ \left( 1+\dfrac { 100 }{ 100 } \right) }^{ n }=16x\\ \Rightarrow { \left( 2 \right) }^{ n }={ 2 }^{ 4 }\\ \Rightarrow n=4$$ yrs
So, after $$4$$ years, the amount will become $$16$$ times of the invested money.
Question 3
1 / -0

The population of a town was 1,60,000 three years ago. If it increased by 3%, 2.5% and 5%, respectively, in the last three years, then what is the present population?

A
166,366

B
177,366

C
166,377

D
177,377

Solution

Present population $$\displaystyle =1,60,000\left ( 1+\frac{3}{100} \right )\left ( 1+\frac{2.5}{100} \right )\left ( 1+\frac{5}{100} \right )$$
$$\displaystyle =160000\times \frac{103}{100}\times \frac{102.2}{100}\times \frac{105}{100}=177366$$
Question 4
1 / -0

The half life of Uranium - $$233$$ is $$160000$$ years i.e., Uranium $$233$$ decays at a constant rate in such a way that it reduces to $$50\%$$ in $$160000$$ years. in how many years will it reduce to $$25 \%$$?

A
$$80000$$ years

B
$$240000$$ years

C
$$320000$$ years

D
$$40000$$ years

Solution

The half life of $${ U }_{ 233 }$$ $$=160000$$ yrs

$$\therefore $$ It becomes $$50\%$$ of the original amount in $$160000$$ yrs

Now $$25\%=$$ $$\dfrac { 1 }{ 2 } \times$$ $$50\%$$

i.e Another half life is needed to reduce the original amount into $$25\%$$

So, another $$160000$$ yrs. will reduce the original amount into $$25\%$$

$$\therefore $$ The required total number of years

$$=160000+160000=320000$$

i.e $$320000$$ yrs
Question 5
1 / -0

In what time will $$Rs.\ 1000$$ amount to $$Rs.\ 1331$$ at $$10\%$$ per annum if interest is compounded anually?

A
$$4$$ years

B
$$3$$ years

C
$$2$$ years

D
$$1$$ years

Solution

Given:
$$A=Rs.\ 1331$$
$$P=Rs.\ 1000$$
$$r=10\%$$

We shall apply the formula for compound interest,
$$A = P{\left( {1 + \dfrac{r}{{100}}} \right)^t}$$
$$\Rightarrow 1331 = 1000{\left( {1 + \dfrac{{10}}{{100}}} \right)^t}$$
$$\Rightarrow \dfrac{{1331}}{{1000}} = \left( {\dfrac{{11}}{{10}}} \right)$$
$$\Rightarrow {\left( {\dfrac{{11}}{{10}}} \right)^t} = {\left( {\dfrac{{11}}{{10}}} \right)^3}$$
$$\Rightarrow t=3$$ years

Hence, option $$B$$ is correct.
Question 6
1 / -0

A sum is invested for $$3$$ years compounded at $$5\% ,10\% $$ and $$20\%$$ respectively. In three years, if the sum amounts to Rs. $$16,632$$, then find the sum.

A
Rs. $$11000$$

B
Rs. $$12000$$

C
Rs. $$ 13000$$

D
Rs. $$ 14000$$

Solution

Let the original amount $$=$$ Rs. $$100$$.
In the first year, it increases by $$5\%$$.
$$\therefore $$ the amount after $$1$$ yr. $$=$$ Rs. $$\left( 100+100\times \dfrac { 5 }{ 100 } \right) =$$ Rs. $$105.$$

In the second year, it increases by $$10\%$$.
$$\therefore $$ the amount after $$2$$ yr. $$=$$ Rs. $$\left( 105+105\times \dfrac { 10 }{ 100 } \right) =$$ Rs. $$115.5.$$

In the third year, it increases by $$20\%$$.
$$\therefore $$ the amount after $$3$$ yr. $$=$$ Rs. $$\left( 115.5+115.5\times \dfrac { 20 }{ 100 } \right) =$$ Rs. $$138.6.$$ $$=138.6\%$$ of Rs. $$100$$

$$\therefore $$ Rs. $$138.6\%$$ of the original money $$=$$ Rs. $$16632$$.
$$\Rightarrow $$ The original money $$=$$ Rs. $$\dfrac { 16632 }{ 138.6 } \times 100=$$ Rs. $$12000.$$
Question 7
1 / -0

The population of a town increases annually by $$25 \% $$ if the preset population is $$1$$ crore, then what was the difference between the population three years ago and two years ago?

A
$$25,00,000$$

B
$$12,80,000$$

C
$$15,60,000$$

D
None of these

Solution

Given that,
Population increases annually by $$25\%$$

Let population $$3$$ years ago be $$x$$
$$\Rightarrow$$ Population two years ago was $$\dfrac{125}{100}x$$

Population last year was $$\dfrac{125\times125}{10000}x$$

Population this year is $$\dfrac{125\times125\times125}{1000000}x = 10000000$$

$$\Rightarrow$$ Population three years ago was $$x = 5120000$$

$$\therefore$$ Difference between population three years ago and two years ago $$= 25\%$$ of $$x$$
$$= 1280000$$
Question 8
1 / -0

The annual increase in the population of a town is 10%. If the present population of the town is 180,000, then what will be its population after two years ?

A
200,000

B
210,000

C
217,800

D
207,800

Solution

Population after 1 year $$= 180000 + 180000\times \dfrac{10}{100} = 198000$$
Population after 2 years $$ = 198000 + 198000\times \dfrac{10}{100}$$
$$=217800$$
Question 9
1 / -0

A machine depreciates in value each year at the rate of $$10\%$$ of its previous value. However every second year there is some maintenance work, so that in that particular year depreciation is only $$5\%$$ of its previous value. If at the end of the fourth year the value of the machine stands at Rs. $$146205$$, then find the value of the machine at the start at the first year?

A
Rs. $$ 190000$$

B
Rs. $$ 200000$$

C
Rs. $$ 195000$$

D
Rs. $$ 210000$$

Solution

Let the value of the machine, at the onset, be $$x$$ when $$x$$ is in Rs.
Here we shall apply the rule
Final value $$=$$ original value $$\times { \left( 1+rate \right) }^{ time }$$.
Here the rate will be negative since the value is depriciating.
For two yrs. the rate of depriciation is $$10\%$$.
$$\therefore $$ the value after $$2$$ yrs.
$$=$$ $$x\times { \left( 1-\dfrac { 10 }{ 100 } \right) }^{ 2 }=x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } $$.
As given, the rate of depriciation for next $$2$$ yrs $$=5\%$$
$$\therefore $$ The value after next $$2$$ yrs
$$=$$ $$x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } \times { \left( 1-\dfrac { 5 }{ 100 } \right) }^{ 2 }$$
$$ =x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } \times \dfrac { 19 }{ 20 } \times \dfrac { 19 }{ 20 } $$
But the final value after $$4$$ yrs $$=$$ Rs. $$146205$$.
$$\therefore \quad x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } \times \dfrac { 19 }{ 20 } \times \dfrac { 19 }{ 20 } =146205$$
$$ \Rightarrow x=\dfrac { 146205\times 40000 }{ 81\times 19\times 19 } =5\times 40000$$
$$\Rightarrow x=$$ Rs. $$200000$$.
So, $$4$$ yrs. back, the value of the machine was Rs. $$200000$$.
Question 10
1 / -0

A sum of money at compound interest (compounded annually) doubles itself in $$4$$ years. In how many years will it amount to eight times of itself ?

A
12 years

B
10 years

C
8 years

D
16 years

Solution

Given,
$$\displaystyle 2P=P\left ( 1+\frac{R}{100} \right )^{4}\\ \Rightarrow \left ( 1+\dfrac{R}{100} \right )^{4}=2 \ \ \ ......(1)$$
Let the time in which it amounts to eight times of itself be $$r$$ years
Then,
$$\displaystyle 8P=P\left ( 1+\frac{R}{100} \right )^{r}\\ \Rightarrow \left ( 1+\dfrac{R}{100} \right )^{r}=8=2^{3}$$
$$ \left ( 1+\dfrac{R}{100} \right )^{r} =2^3=\left ( \left ( 1+\dfrac{R}{100} \right )^{4} \right )^{3}$$ from $$(1)$$
$$\left ( 1+\dfrac{R}{100} \right )^{r}=\left ( 1+\dfrac{R}{100} \right )^{12}$$
Bases are equal, $$\implies r =12 $$ years
$$\displaystyle \therefore r = 12 $$ years

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Basics of Financial Mathematics Test 11

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Question 1

Rs. 10123.20

Rs. 11123.20

Rs. 12123.20

Rs. 13123.20

Solution

Question 2

$$8$$ yrs

$$10$$ yrs

$$4$$ yrs

$$6$$ yrs

Solution

Question 3

166,366

177,366

166,377

177,377

Solution

Question 4

$$80000$$ years

$$240000$$ years

$$320000$$ years

$$40000$$ years

Solution

Question 5

$$4$$ years

$$3$$ years

$$2$$ years

$$1$$ years

Solution

Question 6

Rs. $$11000$$

Rs. $$12000$$

Rs. $$ 13000$$

Rs. $$ 14000$$

Solution

Question 7

$$25,00,000$$

$$12,80,000$$

$$15,60,000$$

None of these

Solution

Question 8

200,000

210,000

217,800

207,800

Solution

Question 9

Rs. $$ 190000$$

Rs. $$ 200000$$

Rs. $$ 195000$$

Rs. $$ 210000$$

Solution

Question 10

12 years

10 years

8 years

16 years

Solution

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