Tag: mathematics and statistics

Questions Related to mathematics and statistics

A sum of money invested at compound interest amount in $3$ years to Rs. $2400$ and in $4$ years to Rs. $2520$. The interest rate per annum is 

mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

  1. $5 \%$

  2. $6 \%$

  3. $10 \%$

  4. $12 \%$

Show answer
Correct Option: A
Explanation:

Let the sum of money invested be Rs. $x$ and interest rate per annum $= r\%$
Then $\displaystyle x\left ( 1+\frac{r}{100} \right )^{3}=$ Rs. $2400$........(i)
and $\displaystyle x\left ( 1+\frac{r}{100} \right )^{4}=$ Rs. $2520$........(ii)
Dividing equation (ii) by (i), we get
$\displaystyle \left ( 1+\frac{r}{100} \right )=\frac{2520}{2400}$

$\Rightarrow \dfrac{r}{100}=\dfrac{2520-2400}{2400}=\dfrac{120}{2400}$
$\Rightarrow r=\dfrac{1}{20}\times 100=5\%$ p.a.

Sanju puts equal amount of money one at $10\%$ per annum compound interest payable half yearly and the second at a certain rate percent annum compound interest payable yearly. If he gets equal amounts after $3$ years what is the value of the second rate percent?

mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

  1. $\displaystyle 10\frac{1}{4}$ $\%$

  2. $10\%$

  3. $\displaystyle 9\frac{1}{2}$ $\%$

  4. $\displaystyle 8\frac{1}{4}$ $\%$

Show answer
Correct Option: A
Explanation:

Let the amount $=x$ in Rs .

In the first case, the payment is made half yearly $@10\%$.
So the number of time in 3yrs=$3\div \dfrac { 1 }{ 2 } =6$ and

the rate $=$ $10 p.c.\div 2=5$ p.c.
So, in the firs case, the amount after $6$ time slots @ $5\%$ is
Principal $={ \left( 1+rate \right)  }^{ time }$
$ =x{ \left( 1+\dfrac { 5 }{ 100 }  \right)  }^{ 6 }$.
In the second case, the time $=3$ yrs,  
The principal $=x$ and let the rate $=y\%$ p.a.

So, the amount after $3$ yrs @ $y\%$
$=$ $x{ \left( 1+\dfrac { y }{ 100 }  \right)  }^{ 3 }$.
$\therefore $ By the given condition, we have
$=x{ \left( 1+\dfrac { y }{ 100 }  \right)  }^{ 3 }=x{ \left( 1+\dfrac { 5 }{ 100 }  \right)  }^{ 6 }\ \Rightarrow { \left( 1+\dfrac { y }{ 100 }  \right)  }^{ 3 }={ \left{ { { \left( 1+\dfrac { 5 }{ 100 }  \right)  } }^{ 2 } \right}  }^{ 3 }\ \Rightarrow { \left( 1+\dfrac { y }{ 100 }  \right)  }={ { \left( 1+\dfrac { 5 }{ 100 }  \right)  } }^{ 2 }=\dfrac { 441 }{ 400 } \ \Rightarrow \dfrac { y }{ 100 } =\dfrac { 441 }{ 400 } -1=\dfrac { 41 }{ 400 } \ \Rightarrow y=\dfrac { 41 }{ 4 } p.c.=10\dfrac { 1 }{ 4 } p.c..$.
So, the second rate $=$ $10\dfrac { 1 }{ 4 }$ p.c..

Calculate the amount and the compound interest on Rs. $12,000$ in $3$ years when the rates of interest for successive years are $8\%$, $10\%$ and $15\%$ respectively. 

mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

  1. Rs.$4294.40$

  2. Rs.$3634.40$

  3. Rs.$5394.40$

  4. None of these

Show answer
Correct Option: D
Explanation:
Sum$=Rs.12000$
Time$=3$ years
Rate of interest$=8\%,10\%,15\%$
Required amount, A = $\displaystyle P\left( 1+\frac { { r } _{ 1 } }{ 100 }  \right) \left( 1+\frac { { r } _{ 2 } }{ 100 }  \right) \left( 1+\frac { { r } _{ 3 } }{ 100 }  \right) $
$\displaystyle \Rightarrow \quad A=Rs.12000\left( 1+\frac { 8 }{ 100 }  \right) \left( 1+\frac { 10 }{ 100 }  \right) \left( 1+\frac { 15 }{ 100 }  \right) $
$\displaystyle \Rightarrow 12000\times \frac{108}{100}\times \frac{110}{100}\times \frac{115}{100}$
$\displaystyle \Rightarrow Rs.\quad 16394.40$
$\displaystyle C.I.=Rs.16394.40-Rs.12000=Rs.4394.40$

What sum will amount to Rs. $6,593.40$ in $2$ years C.I., if the rates are $10$ per cent and $11$ per cent for the successive years?

mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

  1. $5000$

  2. $5600$

  3. $5400$

  4. $6000$

Show answer
Correct Option: C
Explanation:

$\Rightarrow$  Let rate of interest for 2 years will be $R _1=10\%$ and $R _2=11\%$.

$\Rightarrow$  $A=P\times (1+\dfrac{R _1}{100})(1+\dfrac{R _2}{100})$

$\Rightarrow$  $6593.40=P\times (1+\dfrac{10}{100})(1+\dfrac{11}{100})$

$\Rightarrow$  $6593.40=P\times \dfrac{11}{10}\times \dfrac{111}{100}$

$\Rightarrow$  $6593.40=P\times \dfrac{1221}{1000}$

$\Rightarrow$  $P=\dfrac{6593.40\times 1000}{1221}=Rs.5400$

A man lends Rs. $12,500$ at $12$% for the first year, at $15$% for the second year and at $18$% for the third year. If the rates of interest are compounded yearly; find the difference between the C.I. for the first year and the compound interest for the third year.

mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

  1. Rs. $1,498$

  2. Rs. $1,598$

  3. Rs. $1,298$

  4. Rs. $1,398$

Show answer
Correct Option: D
Explanation:
For first year
$P=12500,R=12$%, $T=1$
Interest$\cfrac { PRT }{ 100 } =1500$
Amount$=P+I=14000$
For second year previous amount will be Principle
$P=14000,R=15,T=1$
Interest'$=\cfrac { 14000\times 15\times 1 }{ 100 } =2100$
Similarly for third year
Interest''$=\cfrac { 16100\times 18\times 1 }{ 100 } =2898$
Difference between $C{ I } _{ 3 }$ & $C{ I } _{ 1 }=2898-1500=1398$

Mohit invests Rs. 8,000 for 3 years at a certain rate of interest, compounded annually. At the end of one year it amounts to Rs. 9,440. Calculate: the amount at the end of the second year.

mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

  1. Rs. 15,729.50

  2. Rs. 13.079.80

  3. Rs. 12,367.50

  4. Rs. 11,139.20

Show answer
Correct Option: D
Explanation:

$P=Rs.8000$


Amount after one year $=Rs.9440$

Interest for 1 year$=9440-8000=Rs.1440$

let rate of interest $=R$

C.I for one year=S.I for 1 year$=\dfrac{PRT}{100}$

$\Rightarrow 1440=\dfrac{8000\times R\times 1}{100}$

$\Rightarrow R=\dfrac{1440\times 100}{8000}=18$%

For second year
$P=9440$
$R=18$%
$T=1$ year

$\therefore  Amount=P\left(1+\dfrac{R}{100} \right)^T$

$\Rightarrow 9440 \left(1+\dfrac{18}{100} \right)$

$\Rightarrow 9440\times \dfrac{118}{100}=Rs.  11139.20$

Hence Amount at the end of second year $=Rs.11139.20$

Rohit lends Rs. $50,000$ at C.I. for $3$ years. If the rate of interest for the first two years is $15$% per year and for the third year it is $16$%, calculate the sum Rohit will get at the end of the third year.

mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

  1. Rs.$77705$

  2. Rs.$76705$

  3. Rs.$74705$

  4. Rs.$78705$

Show answer
Correct Option: B
Explanation:

$\Rightarrow$  Here, $P=$Rs.$50,000,\,R _1=15\%$ and $R _2=16\%$


$\Rightarrow$  $A=P\times (1+\dfrac{R _1}{100})^2\times (1+\dfrac{R _2}{100})^1$


$\Rightarrow$  $A=50000\times (1+\dfrac{15}{100})^2\times (1+\dfrac{16}{100})^1$

$\Rightarrow$  $A=50000\times (\dfrac{23}{20})^2\times (\dfrac{29}{25})^1$

$\Rightarrow$  $A=$Rs.$76,705.$


Mohit invests Rs. 8,000 for 3 years at a certain rate of interest, compounded annually. At the end of one year it amounts to Rs. 9,440. Calculate: the interest accured in the third year.

mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

  1. Rs. 2,005.06

  2. Rs. 2,196.06

  3. Rs. 2,207.06

  4. None of these

Show answer
Correct Option: A
Explanation:

$P=Rs.8000$


Amount after one year $=Rs.9440$

Interest for 1 year$=9440-8000=Rs.1440$

let rate of interest=R

C.I for one year=S.I for 1 year$=\dfrac{PRT}{100}$

$\Rightarrow 1440=\dfrac{8000\times R\times 1}{100}$

$\Rightarrow R=\dfrac{1440\times 100}{8000}=18$%

For second year
$P=9440$
$R=18$ %
$T=1$ year

$\therefore  Amount=P \left(1+\dfrac{R}{100} \right)^T$

$\Rightarrow 9440 \left(1+\dfrac{18}{100} \right)$

$\Rightarrow 9440\times \dfrac{118}{100}=Rs.  11139.20$

Hence Amount at the end of second year $=Rs.11139.20$

For the third year
$P=Rs.11139.20$
$R=18$%
$T=1$ year

$Interest=\dfrac{11139.20\times 18\times 1}{100}=Rs. 2005.06$

Hence interest for third year $=Rs.2005.06$

Find the sum that will amount to Rs. $4,928$ in $2$ years at compound interest, if the rates for the successive years are $10$ per cent and $12$ per cent respectively.

mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

  1. Rs. $3000$

  2. Rs. $4000$

  3. Rs. $5000$

  4. Rs. $6000$

Show answer
Correct Option: B
Explanation:

Let the sum be x 

Amount after 2 year=Rs.4928
Rate=10% and 12%
Time=2 years
$Amount=P\left(1+\frac{R}{100}\right)^t$
$\Rightarrow 4928=x(1+\frac{10}{100})(1+\frac{12}{100})$
$\Rightarrow 4928=x\times \frac{110}{100}\times \frac{112}{100}$
$\Rightarrow x=\frac{4928\times 100\times 100}{110\times 112}=Rs.4000$

What sum will amount to Rs. $659340$ in $2$ years C.I., if the rates are $10$ per cent and $11$ per cent for the successive years?

mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

  1. $540000$

  2. $550000$

  3. $560000$

  4. $570000$

Show answer
Correct Option: A
Explanation:

$\Rightarrow$   Here, $A=Rs.659340,\,R _1=10\%,\,R _2=11\%$

$\Rightarrow$   $A=P(1+\dfrac{R _1}{100})^T\times (1+\dfrac{R _2}{100})^T$

$\Rightarrow$  $659340=P\times (1+\dfrac{10}{100})^1\times (1+\dfrac{11}{100})^1$

$\Rightarrow$  $659340=P\times \dfrac{11}{10}\times \dfrac{111}{100}$

$\Rightarrow$  $659340=P\times \dfrac{1221}{1000}$

$\Rightarrow$  $P=540\times 1000$

$\therefore$    $P=Rs.540000.$