Tag: introduction to interest payments

Questions Related to introduction to interest payments

On a certain Principal if the Simple interest for two years is $Rs.\ 4800$ and Compound interest for the two years is $Rs.\ 5088$, what is the rate of interest

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. $6\%$

  2. $24\%$

  3. $12\%$

  4. $18\%$

Show answer
Correct Option: C
Explanation:
Given,  $SI=4800$
            $CI=5088$
Say the principal amount was $P$ and rate of interest is $r$% p.a for two years
$\therefore$   $SI=\dfrac { PT }{ 100 } =\dfrac { P\times 2\times r }{ 100 } $
$\Rightarrow 4800=\dfrac { 2Pr }{ 100 } $
$\Rightarrow 240000=Pr\quad \longrightarrow \left( 1 \right) $
Similarly $CI=P{ \left( 1+\dfrac { r }{ 100 }  \right)  }^{ 2 }-P$
$\Rightarrow 5088=P\left[ { \left( 1+\dfrac { r }{ 100 }  \right)  }^{ 2 }-1 \right] \quad \longrightarrow \left( 2 \right) $
Substituting $(1)$ in $(2)$ we get
$5088=\dfrac { 240000 }{ r } \left[ 1+\dfrac { { r }^{ 2 } }{ { 100 }^{ 2 } } +\dfrac { 2r }{ 100 } -1 \right] $
$\Rightarrow \dfrac { 5088 }{ 240000 } =\dfrac { r }{ { 100 }^{ 2 } } +\dfrac { 2 }{ 100 } $
$\Rightarrow 0.0212=\dfrac { r }{ { 100 }^{ 2 } } +0.02$
$\Rightarrow 0.0012=\dfrac { r }{ { 100 }^{ 2 } } $
$\therefore$    $r=12$%

At what rate per cent per annum will a sum of $Rs.\ 7500$ amount to $Rs.\ 8427$ in $2$ years compounded annually?

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. $4$%

  2. $5$%

  3. $6$%

  4. $8$%

Show answer
Correct Option: C
Explanation:
We have,
$P=7500\ Rs$
$A=8427\ Rs$
$T=2$ years
$A=P\left (1+\dfrac {R}{100}\right)^T$

$7500+8427=7500\left (1+\dfrac {R}{100}\right)^T$

$\dfrac {8427}{7500}=\left (1+\dfrac {R}{100}\right)^2$

$\left (1+\dfrac {R}{100}\right)=\dfrac {2809}{2500}=\left (\dfrac {53}{50}\right)$

$\left (1+\dfrac {R}{100}\right)=\dfrac {53}{50}-1$

$\dfrac {R}{100}=\dfrac {53}{50}-1$

$R=\dfrac {3}{50}$

$R \ \% = 6\ \%$ 
Hence, this is the answer.

A sum of money compounded annually amounts to 1375 in 5 years and 1980 in 7 years. Find the annual rate of interest.

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. 12%

  2. 20%

  3. 15%

  4. 10%

Show answer
Correct Option: B
Explanation:

interest is compounded, Amount $ A = P(1+ \frac {R}{100})^n $
So, for the first situation
$ 1375 = P \times (1+ \frac {R}{100})^5 $   - (1)

And for the first situation
$ 1980 = P \times (1+ \frac {R}{100})^7 $   ---- (2)

Dividing eqn 2 by eqn 1, we get
$ \frac {1980}{1375} =   (1+ \frac {R}{100})^2 $ 
$ => \frac {396}{275} =   (1+ \frac {R}{100})^2 $ 
$ => \frac {36}{25} =   (1+ \frac {R}{100})^2 $ 
$ => (1+ \frac {R}{100}) = \frac {6}{5} $
$ => \frac {R}{100} = \frac {1}{5} $
$ => R =20 $ %

The difference between the interest earned under compound interest, interest being compounded annually and simple interest for two years on the same sum and at the same rate of interest is 25.60. Find the sum if the rate of interest is 8% p.a

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. 2000

  2. 2500

  3. 3200

  4. 4000

Show answer
Correct Option: D
Explanation:



Simple Interest $ SI = \frac {PNR}{100} $
So, $ SI = \frac {P \times 2 \times 8}{100} = Rs 0.16P $

When interest is compounded, Amount $ A = P(1+ \frac {R}{100})^n $
So, A $ = P \times (1+ \frac {8}{100})^2 = Rs  1.1664P  $
And $ CI = A - P = 0.1664P $

Si, difference $ CI - SI = Rs 0.1664P - Rs 0.16P = Rs 25.60 $
$ => 0.0064P = 25.60 $
$ => P = Rs  4000 $

If the simple interest on a certain sum of money is $\displaystyle \frac{1}{100}$th of the sum and the rate per cent equals the number of years, then the rate of interest per annum is

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. $2$%

  2. $1$%

  3. $3$%

  4. $4$%

Show answer
Correct Option: B
Explanation:
Given $SI=\cfrac { 1 }{ 100 } \times P,R=T$
$SI=\cfrac { PRT }{ 100 } $
$\cfrac { P }{ 100 } =\cfrac { P{ R }^{ 2 } }{ 100 } \Rightarrow R=T=1$

The interest on a certain sum of money is $0.18$ times of itself in $3$ years. Find the rate of interest.

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. $4$

  2. $5$

  3. $6$

  4. $7$

Show answer
Correct Option: C
Explanation:

$I=Interest$

$P=Principal$
$R=Rate of Interest$
$T=Time(in years)$
$0.18I=\dfrac { I\times R\times 3 }{ 100 } \ \Rightarrow R=\dfrac { 0.06\times 100 }{ 3 } =0.06\times 100\$
 $\Rightarrow R=6$ %
Rate of Interest $= 6$%

Principal $= 2500$, rate $= 6$%, time$= 4$ years. Calculate the interest.

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. Rs.$500$

  2. Rs.$550$

  3. Rs.$600$

  4. Rs.$650$

Show answer
Correct Option: C
Explanation:

$Interest=\dfrac { P\times R\times T }{ 100 } \ =\dfrac { Rs.2500\times 6\times 4 }{ 100 } \ =Rs.600$