Tag: applying compound interest

Questions Related to applying compound interest

Shikkan boy lends Rs.28000 for 4 years at $\displaystyle 6\frac{1}{2}$ per cent per annum and Rs.65500 for 5 years at 8 per cent per annum. Find her total earning from these investments.

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

  1. $33480$

  2. $34440$

  3. $33240$

  4. $12220$

Show answer
Correct Option: A
Explanation:

$\Rightarrow$  Here $P=Rs.28000,\,T=4\,years$ and $R=6\dfrac{1}{2}\%=\dfrac{13}{2}\%$

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

$\Rightarrow$  $S.I.=\dfrac{28000\times 13\times 4}{2\times 100}=Rs.7280$

$\Rightarrow$  Now, $P=Rs.65500,\,T=5\,years$ $R=8\%$

$\Rightarrow$  $S.I.=\dfrac{P\times R\times T}{100}=\dfrac{65500\times 8\times 5}{100}$

$\Rightarrow$  $S.I.=Rs.26,200$.

$\therefore$   Total earning from investment = $Rs.7280+Rs.26,200=Rs.33480$

Calculate the amount and the compound interest on  $5000$ in $2$ years when the rate of interest for successive years is $6\%$ and $8\%$ respectively.

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

  1. $724$

  2. $727$

  3. $317$

  4. $239$

Show answer
Correct Option: A
Explanation:

Interest for the $1^{ st }$ year $=Rs$ $\cfrac { 5000\times 1\times 6 }{ 100 } =Rs$ $300$

Amount after $1^{ st }$ year $=Rs$ $5000+Rs$ $300=Rs$ $5300$
Interest for the $2^{ nd }$ year $=Rs$ $\cfrac { 5300\times 1\times 8 }{ 100 } =Rs$ $424$
Amount after $2^{ nd }$ year $=Rs$ $5300+Rs$ $424=Rs$ $5724$
$\therefore$ Compound Interest $=Rs$ $5724-Rs$ $5000= Rs$ $724$

Prem lends Rs. $100,000$ 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.$155410$

  2. Rs.$153410$

  3. Rs.$125410$

  4. Rs.$135410$

Show answer
Correct Option: B
Explanation:

$\Rightarrow$  $P=Rs.100,000,$ $R _1=15\%$ and $R _2=16\%$

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

$\Rightarrow$  $A=100000\times (1+\dfrac{15}{100})^2\times (1+\dfrac{16}{100})^1$

$\Rightarrow$  $A=100000\times (\dfrac{23}{20})^2\times \dfrac{29}{25}$

$\Rightarrow$  $A=100000\times \dfrac{529}{400}\times \dfrac{29}{25}$

$\Rightarrow$  $A=10\times 529\times 29$

$\therefore$    $A=Rs.153410$

Divide Rs.15,600 into two parts such that the interest on one at 5 percent for 5 years may be equal to that on the other at $\displaystyle 4\frac{1}{2}$ per cent for 6 years.

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

  1. Rs.$7,400$ and Rs.$8,300$

  2. Rs.$7,700$ and Rs.$8,500$

  3. Rs.$8,700$ and Rs.$7,000$

  4. Rs.$8,100$ and Rs.$7,500$

Show answer
Correct Option: D
Explanation:

Let the parts be $ Rs A  ; Rs  15,600  - A $ 

Simple Interest $ I = \dfrac {PNR}{100} $ 

For the first situation, Given, 

$ P = Rs  A $ 

$ N =5 years $ 

$ R = 5  $ % 

So, $ => I =  \dfrac {PNR}{100} $

$ =>I _1 = \dfrac {A \times 5 \times 5 }{100} $

$ =>I _1= \dfrac {25A}{100} $

For the second situation, Given, 

$ P = Rs 15,600  - A $ 

$ N =6 years $ 

$ R = \dfrac {9}{2} $ % 

So, $ => I =  \dfrac {PNR}{100} $

$ =>I _2  = \dfrac {(15,600  - A) \times 6 \times \dfrac {9}{2}}{100} $

$ =>I _2 =  \dfrac {(15,600  - A) \times 27}{100} $

Given, $ I _1 = I _2 $

$ => \dfrac {25A}{100} = \dfrac {(15,600  - A) \times 27}{100} $

$ 25A = (15,600  - A) \times 27 $

$ 25A = 421200 -27A $

$ => 52A = 421200 $ 

$ => A = Rs  8,100 $ 

And other part $ = Rs 15,600 - Rs 8,100 = Rs  7,500 $ 

Calculate the amount and the compound interest on  $17000$ in $3$ years when the rate of interest for successive years is $10\%, 10\%$ and $14\%$ respectively.

mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)
  1. $6578.70$
    6449.80$

  2. $1700$

  3. $5489.8$

  4. $6449.80$

Show answer
Correct Option: D
Explanation:

Interest for the $1^{ st } $year $=Rs$ $\cfrac { 17000\times 10\times 1 }{ 100 } =Rs$ $1700$ 

Amount after $1^{ st }$ year $=Rs$ $17000+Rs$ $1700=Rs$ $18700$
Interest for the $2^{ nd }$ year $=Rs$ $\cfrac { 18700\times 10\times 1 }{ 100 } =Rs$ $1870$ 

Amount after $2^{ nd }$ year $=Rs$ $18700+Rs$ $1870=Rs$ $20570$
Interest for the$3^{ rd }$ year $=Rs$ $\cfrac { 20570\times 14\times 1 }{ 100 } =Rs$ $2879.80$ 

Amount after $3^{ rd }$ year $=Rs$ $20570+Rs$ $2879.80=Rs$ $23449.8$
$\therefore$ Compound Interest $=Rs$ $23449.8-Rs$ $17000= Rs$ $6449.80$

Rakshat borrows Rs.$16,000$ out of which Rs.$9,000$ at $5\%$ and remaining at $6\%$. Find the total interest paid by him in $4$ years.

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

  1. Rs.$3,280$

  2. Rs.$3,380$

  3. Rs.$3,480$

  4. Rs.$3,580$

Show answer
Correct Option: C
Explanation:
He borrowed $9000$ at $5$ % and remaining $7000(16000-9000)$ at $6$%
We know that $SI=\cfrac { PRT }{ 100 } $
For $P=9000,R=5,T=4$
$SI=\cfrac { 9000\times 5 \times 4 }{ 100 } =1800$
For $P=7000,R=6,T=4$
$SI=\cfrac { 7000\times 6 \times 4 }{ 100 } =1680$
Total $SI=1800+1680=3480$
$\therefore $ He has to pay Rs.$3480$ at end of $4$ years.

Calculate the amount and the compound interest on Rs. $12,500$ in $3$ years when the rates of interest for successive years are $8 \%$, $10 \%$ and $10 \%$ 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. $16,335$ and Rs. $3,835$

  2. Rs. $14,853$ and Rs. $2,353$

  3. Rs. $15,664$ and Rs. $3,164$

  4. Rs. $17952$ and Rs.$5,452$


Show answer
Correct Option: A
Explanation:

$P=Rs.12,500$

$T=3$ years
$R=8 \%,10 \%, 10 \%,$ 
$A=?,C.I.=?$
$A=P\left(1+\cfrac{R}{100} \right )^T$

$=12500\left(1+\cfrac{8}{100}\right)\left(1+\cfrac{10}{100}\right)\left(1+\cfrac{10}{100}\right)$
$=12500\times \cfrac{108}{100}\times \cfrac{110}{100}\times \cfrac{110}{100}$
$=\cfrac{125\times 108\times 121}{100}$
$A=Rs.  16,335$
$C.I.=A-P$
$C.I=16335-12500$
$C.I=Rs. 3835$ 

Raj borrows Rs.$16,000$; out of which Rs.$9,000$ at $5\%$ and remaining at $6\%$. Find the total interest paid by him in $4$ years.

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

  1. $1600$

  2. $3480$

  3. $3600$

  4. $3300$

Show answer
Correct Option: B
Explanation:

${ I } _{ 1 }=\dfrac { P\times R\times T1 }{ 100 } =\dfrac { 9000\times 5\times 4 }{ 100 } =Rs.1800$


${ I } _{ 2 }=\dfrac { 7000\times 6\times 4 }{ 100 } =Rs.1680$


$\therefore $ Total Interest $=Rs.1800+Rs.1680=Rs.3480$