Tag: compound interest formula with different successive rate of interest

Questions Related to compound interest formula with different successive rate of interest

The compound interest on Rs. $2000$ in $2$ years, if the rate of interest is $4 \%$ per annum for the first year and $3\%$ per annum for the second year will be 

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

  1. Rs. $142.40$

  2. Rs. $140.40$

  3. Rs. $141.40$

  4. Rs. $143.40$

Show answer
Correct Option: A
Explanation:

Given, $P=$ Rs. $2000$, $\displaystyle r _{1}$ $=4\%$ p.a. for $1^{st}$ year, $\displaystyle r _{2}$ $=3\%$ p.a. for $2^{nd}$ year
$\displaystyle \therefore A=P\left ( 1+\frac{r _{1}}{100} \right )\left ( 1+\frac{r _{2}}{100} \right )$

$=2000\left ( 1+\dfrac{4}{100} \right )\left ( 1+\dfrac{3}{100} \right )$
$\displaystyle =$ Rs. $\left ( 2000\times \dfrac{26}{25}\times \dfrac{103}{100} \right )=$ Rs. $2142.40$
$\displaystyle \therefore$ C.I. $=$ Rs. $2142.40 - $ Rs. $2000=$ Rs. $142.40$

How much will Rs. $9,000$ amount to in $3$ years, at compound interest, if the rates for successive years are $12 \%$, $15 \%$ and $20 \%$ per year 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. $13810.40$

  2. Rs. $13710.40$

  3. Rs. $13910.40$

  4. Rs. $12910.40$

Show answer
Correct Option: C
Explanation:

$A=9000\times \left(1+\dfrac{12}{100}\right)\left(1+\dfrac{15}{100}\right)\left(1+\dfrac{20}{100}\right)$

$A=9000\times \dfrac{112}{100}\times \dfrac{115}{100}\times \dfrac{120}{100}$

$A=\dfrac{112\times 115\times 12\times 9}{100}$

$A=13910.40  Rs$

Find the amount on Rs. $ 12,500$ for $2$ years compounded annually, the rate of interest being $15\%$ for the first year and $16\%$ for 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. $16,500$

  2. Rs. $16,750$

  3. Rs. $16,675$

  4. None of these

Show answer
Correct Option: C
Explanation:

Given, $P=$ Rs. $ 12500$,$R _1$ $=15\%$,$R _2 $ $=16\%$
$A = p \left(1 + \displaystyle \frac{R _1}{100} \right) \left(1 + \displaystyle \frac{R _2}{100} \right)$
$A = 12500 \times \left(1 + \displaystyle \frac{15}{100} \right) \times \left(1 \times \displaystyle \frac{16}{100} \right)$
$= 12500 \times \displaystyle \frac{115}{200} \times \frac{116}{100}$
$= $ Rs. $16,675$

A sum of money double itself in 4 years at compound interest.! n how many years it will become eight times at same rate of interest

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

  1. 12 years

  2. 18 years

  3. 24 years

  4. 16 years

Show answer
Correct Option: A
Explanation:

$\displaystyle A = P \left( 1 + \frac {R}{100} \right)^T$

$\displaystyle 2x = x \left( 1 + \frac {R}{100} \right)^4$

$\displaystyle \left( 1 + \frac {R}{100} \right) = 2^{1/4}$

$\displaystyle 8x=x \left( 1 + \frac {R} {100}\right)^T$

$\displaystyle \because 1 + \frac {R} {100} = 2^{1/4}$

$\displaystyle  \therefore 8=2^{T/4}$

$\displaystyle 2^3 = 2^{T/4}$

$\displaystyle \frac {T}{4}=3$

$T=12\, years$

Sam invested Rs. $15000$ at the rate of $10 \%$ per annum for one year. If the interest is compounded half-yearly, then the amount received by Sam at the end of the year will be:

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

  1. Rs. $16,500$

  2. Rs. $16,525.50$

  3. Rs. $16,537.50$

  4. Rs. $18,150$

Show answer
Correct Option: C
Explanation:

$P =$ Rs. $15000$; $R =10 \%$ p.a $= 5 \%$ per half-year; $T= 1$ year $= 2$ half-years.
$\therefore$ Amount $=\, \left [ 15000\, \times\, \left ( 1\,+\, \displaystyle \frac{5}{100} \right )^2 \right ]$
$=$ Rs. $\left ( 15000\, \times\, \displaystyle \frac{21}{20}\, \times\, \frac{21}{20} \right )$
$= 16,537.50$

There is a question followed by three statements. While answering the question, you may or may not require the data provided in all the statements. You have to read, the question and the three statements and then decide whether the question can be answered with any one or two of the statements or all the three statements are required to answer the question. Mr. Gupta borrowed a sum of money on compound interest. What will be the amount to be repaid if he is repaying the entire amount at the end of $2$ years?

$I$. The rate of interest is $5$ p.c.p.a.
$II$. Simple interest fetched on the same amount in one year is $Rs. 600$.
$III$. The amount borrowed is $10$ times the simple interest in $2$ years.

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

  1. $I$ only

  2. $III$ only

  3. $I$ or $II$ only

  4. None of these

Show answer
Correct Option: D
Explanation:

Given 
I. The rate of interest is 5 p.c.p.a.
II. Simple interest fetched on the same amount in one year is Rs. 600.
III. The amount borrowed is 10 times the simple interest in 2 years.
Now | and || give the sum,
For this sum , C.I and hence amount can be obtained
Thus , ||| is redundant
Again || gives S.I for 2 years =$600\times 2=1200$
Now from ||| ,sum =$10\times 1200=12000$
Thus Rate=$\frac{100\times 1200}{2\times 1200}=5%$p.a
Thus ,C.I for 2 years and therefore ,amount can be obtained.
Thus ,| is redundent
Hence either  |  and II  or II and III are sufficient to answer the question 

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 $R _2=11\%$.

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

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

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

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

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