Basics Of Financial Mathematics - Class 11 Commerce Applied Mathematics - Extra Questions

Let, A girl deposits  in a bank account per mounth.=$$300rs$$

At one year total deposits money(p)  =$$12\times 300\,rs=3600\,rs$$

Given interest per year=$$7%$$

Interest per month(R) =$$\dfrac{7}{12}$$ month.

T=12 month

Now , money after 12 month ?

Interest in 12 month

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

  $$ \dfrac{3600\times 12\times \dfrac{7}{12}}{100}=\dfrac{3600\times 7}{100} $$

 $$ =252rs $$

Hence money will becomes in 12 month $$3600+252=3852rs$$

For ,
$$ N = 2  years $$
$$ P = Rs  1,200 $$

On interest being compounded for 2 years,

 $$ Amount = P{(1 + \dfrac {R}{100})}^{N} $$
$$ 1,452 =  1,200 \times {(1 + \dfrac {R}{100})}^{2}$$
$$ 1.21  = {(1 + \dfrac {R}{100})}^{2}$$
Taking square root on both sides, 

$$ 1.1 = 1 + \dfrac {R}{100} $$

$$ => R = 10 $$ % 

Compound interest for the second year will be the simple interest gained in the second year, with the amount after first year being the principal for the second year.
For the first year 
$$ P = Rs 8,000 $$ 
$$ N = 1  year $$
$$ R  = 10 $$ % 
We have $$  S.I.  = \dfrac { PNR}{100} = \dfrac {8,000 \times 1 \times 10}{100} =Rs  800 $$
And Amount at the end of first year $$ P + S.I. = Rs 8,000 + Rs 800 = Rs  8,800 $$
Now, for the second year 
$$ P = Rs 8,800 $$ 
$$ N = 1  year $$
$$ R  = 10 $$ % 
We have $$  S.I.  = \dfrac { PNR}{100} = \dfrac {8,800 \times 1 \times 10}{100} =Rs  880 $$
Thus, Compound interest for the second year $$ = Rs  880 $$

$$  {\textbf{Step - 1: Calculating Simple interest}} $$

                 $$  {\text{t = 2 years}} $$

                 $$  {\text{r = 8 %  }} $$

                 $$  {\text{P = Rs 4,000}} $$

                 $$  {\text{Simple interest = }}\dfrac{{ptr}}{{100}} $$

                 $$   = \dfrac{{4000 \times 2 \times 8}}{{100}} $$

                 $$   = 640 $$

$$  {\textbf{Step - 2: Calculating Compound interest}} $$

                 $$  {\text{n = 2 years}} $$

                 $$  {\text{r = 8 %  }} $$        

                 $$  {\text{P = Rs 4,000}} $$ 

                 $$  {\text{And on interest being compounded for 2  years and R = 8 % }} $$

                 $$  {\text{Amount = P}}{\left( {1 + \dfrac{r}{{100}}} \right)^n} $$ 

                 $$   = 4000{\left( {1 + \dfrac{8}{{100}}} \right)^2} $$

                 $$   = 4000{\left( {\dfrac{{108}}{{100}}} \right)^2} $$

                 $$   = 4000 \times {1.08^2} $$

                 $$   = 4665.60 $$

                 $$  {\text{So compound interest is 4665}}{\text{.60 - 4000 = 665}}{\text{.60}} $$

$$  {\textbf{Step - 3: Finding the difference}} $$

                $$  {\text{So the difference is 665}}{\text{.60 - 640}} $$

                $$  {\text{ = 25}}{\text{.60}} $$

$$  {\textbf{ Hence the answer is 25}}{\textbf{.60 Rs}}{\text{.}} $$

We have, $$P_0 $$ $$=$$ Original count of bacteria $$=506000$$,
Rate of increase $$=R =2.5\%$$ per hour,
Time $$=2$$ hours

$$\therefore$$ Bacteria count after $$2$$ hours $$=$$ $$P$$

$$P = P_0(1 +\dfrac{R}{100})^T$$

$$=506000\times {\left(1+\dfrac{2.5}{100}\right)}^{2}$$

$$=506000\times \dfrac{102.5}{100}\times \dfrac{102.5}{100}$$

$$=531616.25=531616$$ (approx)

Given:

Let the original price of the car $$= Rs.\ x$$

Percentage of depreciation per year $$= 20\%$$

Time period $$= 2$$ yrs

Price of a car after two years $$= Rs.\ 208000$$ 

Amount of depreciation $$A=P\bigg( 1-\dfrac{r}{100} \bigg)^n,$$ 

here $$A$$ is the amount after depreciation, $$r$$ is the rate of depreciation and $$n$$ is time.

$$\Rightarrow 208000=x\bigg( 1-\dfrac{20}{100} \bigg)^2$$

$$\Rightarrow 208000=x\bigg( \dfrac{100-20}{100} \bigg)^2$$

$$\Rightarrow 208000=x\bigg( \dfrac{80}{100} \bigg)^2$$

$$\Rightarrow 208000=x\bigg( \dfrac{4}{5} \bigg)^2$$

Current price of the car, $$P_o = Rs 360000$$
Rate of depreciation, 10% for first 2 years, 20% for the third year
Now, Let price after 2 years= $$P$$

Thus, $$P = P_o(1 + \dfrac{R}{100})^T$$

$$P = 360000(1 + \dfrac{-10}{100})^2$$

$$P = 360000\times \dfrac{9\times 9}{10 \times 10}$$

$$P = 291600$$

After the third year, let the price be $$P'$$

$$P' = P (1 + \dfrac{R'}{100})^T$$

$$P' = 291600(1 + \dfrac{-20}{100})^1$$

$$P' = \dfrac{291600\times 80}{100}$$

$$P' = 233280$$

Thus, the car will cost Rs $$233280$$ after three years.

Let the population last year be $$x$$
Now, Current year population = Last year + 10% of last year
$$9900 = x + \dfrac{10}{100} \times x$$
$$9900 = \dfrac{110}{100} x$$
$$x = 9000$$
Last year the population was $$9000$$

Initial Population, $$P_o = 20000$$
Rate of increase $$= 5\%$$
Let say population after $$3$$ years at the end of the year $$2000= P$$

$$P = P_o(1 + \dfrac{R}{100})^T$$

$$P = 200000(1 + \dfrac{5}{100})^3$$

$$P = \dfrac{200000\times 21\times 441}{20\times 400}$$

$$P = 231525$$

Thus, the population after $$3$$ years in $$2000$$ will be $$231525.$$

$${P}_{0}=130,000$$ ,
$$R=4$$%,
$$n=4$$
If the population after 4 years is P,then 
$$P={P}_{0}\times {(1+\dfrac{R}{100})}^{n}$$
$$=130000{(1+\dfrac{4}{100})}^{4}$$

$$=130000\times \displaystyle\dfrac{104}{100}\times \dfrac{104}{100}\times \dfrac{104}{100}\times \dfrac{104}{100}$$
=$$152082$$
Population at the end of 4 years = $$152082$$

Hence, the sum of digits $$ = 1+5+2+0+8+2 = 18$$

Initial worth $$= 10000,$$ Depreciation $$= 5$$%

Value after $$1$$ year = Initial worth – Depreciation $$= 10000 – 5$$% of $$10000$$

Value $$=10000-\left( \cfrac { 5 }{ 100 }  \right) \times 10000=9500$$

the machinery will be worth $$ Rs.9500$$ after a year.

$${\textbf{Step-1:  Using relation between cost and depreciation.}}$$

                  $${\text{Cost of TV = 21000, Depreciation = 5 %.}}$$

                  $${\text{Value after one year = Cost – Depreciation}}$$

                  $$=21000- \dfrac{5}{100} \times 21000$$

                  $$= 19950.$$

$${\textbf{Hence, Value of TV after one year is Rs.19950.}}$$

Let the Initial principle amount $$= P$$

After $$5$$ years the amount received $$ = \left(\cfrac{8}{5}\right)P$$

As the interest applicable is simple interest, the total interest is Final amount received $$ -$$ Initial principle amount.

Interest amount for $$5$$ years $$=\left(\cfrac{8}{5}\right)P - P$$

Interest amount for $$5$$ years $$=\left(\cfrac{3}{5}\right)P$$

With simple interest, interest amount every year is the same.

Thus interest for one year $$=$$ Interest for $$5$$ years $$\div 5$$

Interest for one year $$=\cfrac{\left(\cfrac{3}{5}\right)P}{5}$$

Interest for one year $$=\cfrac{3}{25}P$$

Simple Rate of interest per year $$= \cfrac {Amount\ of\ Interest\ received\ per\ year}{Initial\ Principle\ amount}$$ $$\times 100\%$$

$$\therefore$$ Rate of interest per year $$= \left(\left(\cfrac{3}{25}\right)P \div P\right) \times 100 \%$$

$$\therefore$$ simple rate of interest $$= \cfrac{3}{25}\times 100 \%$$

$$\therefore$$ Simple Rate of interest $$= 12\%$$

The simple interest is given as $${\rm{Rs}}\;829.50$$ with the rate of $$10\% $$ per annum for $$3$$ years.

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

$$829.50 = \dfrac{{P \times 10 \times 3}}{{100}}$$

$$P = \dfrac{{829.50 \times 100}}{{10 \times 3}}$$

$$\Rightarrow P = {\rm{Rs}}\;2765$$

The principle is given as $${\rm{Rs}}\;6300$$ and the simple interest is given as $${\rm{Rs}}\;2100$$ with time as $$4$$ years.

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

$$2100 = \dfrac{{6300 \times R \times 4}}{{100}}$$

$$R = \dfrac{{2100 \times 100}}{{6300 \times 4}}$$

$$R = 8.34\% $$

Let $${\rm{Rs}}\;x$$ be the sum. Then according to the question, the simple interest will be $$\dfrac{3}{5}x$$.

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

$$\dfrac{{3x}}{5} = \dfrac{{x \times R \times 5}}{{100}}$$

$$R = \dfrac{{3x \times 100}}{{5 \times x \times 5}}$$

$$R = 12\% $$

$${\textbf{Step - 1: State the formula}}{\text{.}}$$

                  $${\text{If P is the Principle , R is the rate of interest , T is time period}}{\text{.}}$$

                  $${\text{then Interest , I}} = \dfrac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}}.$$

$${\textbf{Step - 2: Calculate the rate of interest}}{\text{.}}$$

                   $${\text{Given the interest I}} = {\text{Rs 280}}{\text{.}}$$

                   $${\text{P}} = {\text{Rs 56000}}{\text{.}}$$                                   

                   $${\text{Time period}} = {\text{2 years}}$$

                   $${\text{Suppose the rate is R% }}{\text{.}}$$

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

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

                 $$ \Rightarrow {\text{R}} = \dfrac{{100 \times 280}}{{56000 \times 2}}$$

                  $$ \Rightarrow {\text{R}} = \dfrac{{140}}{{560}}$$

                  $$ \Rightarrow {\text{R}} = \dfrac{7}{{28}}$$

                  $$ \Rightarrow {\text{R}} = \dfrac{1}{4}$$

                  $$ \Rightarrow {\text{R}} = 0.25 $$

$${\textbf{Hence, The rate of interest is 0}}{\textbf{.25% per annum}}{\text{.}}$$

Consider the given data.

$$ T=2\dfrac{1}{2}=\dfrac{5}{2}\,years, $$

$$ R=? $$

Let the sum be $$x$$

Since, the interest is $$\dfrac{4}{5}$$ of sum.

So, interest $$=\dfrac{4x}{5}$$

We know that

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

$$ \dfrac{4x}{5}=\dfrac{x\times R\times \dfrac{5}{2}}{100} $$

$$ \dfrac{4}{5}=\dfrac{R\times 5}{2\times 100} $$

$$ R=\dfrac{4\times 2\times 100}{5\times 5} $$

$$ R=\dfrac{4\times 2\times 100}{25} $$

$$ R=32\%$$

= 50000× (1-10/100)^2 
= 50000×9×9/10×10 
=40500

We have,

$$P=18000$$

$$Time=t=\dfrac{3}{2}\times 2=3\left( Half\,yearly \right)$$

Rate $$r=10%\,p.a.$$

Then,

$$ A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}} $$

$$ A=18000{{\left( 1+\dfrac{10}{100} \right)}^{3}} $$

$$ A=18000\times \dfrac{11}{10}\times \dfrac{11}{10}\times \dfrac{11}{10} $$

$$ A=18\times 11\times 11\times 11 $$

$$ A=23958 $$

Hence, this is the answer.