Tag: compound interest formula with different successive rate of interest

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

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.

  1. Rs. $1,498$
  2. Rs. $1,598$
  3. Rs. $1,298$
  4. Rs. $1,398$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $C{ I } _{ 3 }$ & $C{ I } _{ 1 }=2898-1500=1398$
Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

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.

  1. Rs. 15,729.50

  2. Rs. 13.079.80

  3. Rs. 12,367.50

  4. Rs. 11,139.20
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $=Rs.11139.20$

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

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.

  1. Rs.$77705$
  2. Rs.$76705$
  3. Rs.$74705$
  4. Rs.$78705$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.$76,705.$


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

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.

  1. Rs. 2,005.06

  2. Rs. 2,196.06

  3. Rs. 2,207.06

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=Rs.2005.06$

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

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.

  1. Rs. $3000$
  2. Rs. $4000$
  3. Rs. $5000$
  4. Rs. $6000$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the sum be x 

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

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

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?

  1. $540000$
  2. $550000$
  3. $560000$
  4. $570000$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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

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

$\Rightarrow$  $P=540\times 1000$

$\therefore$    $P=Rs.540000.$

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

What principal will amount to Rs. $9,744$ in two years, if the rates of interest for successive years are $16$% and $20$% respectively?

  1. $5000$
  2. $6000$
  3. $7000$
  4. $8000$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow$  Here $A=Rs.9744,\,T=2\,years,\,R _1=16\%$ and $R _2=20\%$

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

$\Rightarrow$  $9744=P\times (1+\dfrac{16}{100})\times (1+\dfrac{20}{100})$

$\Rightarrow$  $9744=P\times \dfrac{29}{25}\times \dfrac{6}{5}$

$\therefore$    $P=9744\times \dfrac{25}{29}\times \dfrac{5}{6}$

$\therefore$   $P=Rs.7000$

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

Vaibhav lent out Rs.$70,000$ at $6$% and Rs.$95,000$ at $5$%. Find his total income from the interest in 3 years.

  1. $26850$
  2. $25650$
  3. $25950$
  4. $26000$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here $P=Rs.70000,\,T=3\,years$ and $R=6\%$

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

$\Rightarrow$  $S.I.=\dfrac{70000\times 6\times 3}{100}=Rs.12,600$

$\Rightarrow$  Now, $P=Rs.95000,\,T=3\,years$ $R=5\%$

$\Rightarrow$  $S.I.=\dfrac{P\times R\times T}{100}=\dfrac{95000\times 5\times 3}{100}$

$\Rightarrow$  $S.I.=Rs.14,250$.

$\therefore$   Total earning from investment = $Rs.12,600+Rs.14,250=Rs.26,850$

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

What sum will amount to Rs. $65934$ in $2$ years C.I., if the rates are $10$ per cent and $11$ per cent for the successive years?

  1. $54000$
  2. $55000$
  3. $56000$
  4. $57000$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Rightarrow$   Here, $A=Rs.65934,\,R _1=10\%,\,R _2=11\%$

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

$\Rightarrow$  $65934=P\times (1+\dfrac{10}{100})^1\times (1+\dfrac{11}{100})^1$

$\Rightarrow$  $65934=P\times \dfrac{11}{10}\times \dfrac{111}{100}$

$\Rightarrow$  $65934=P\times \dfrac{1221}{1000}$

$\Rightarrow$  $P=54\times 1000$

$\therefore$    $P=Rs.54000.$

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

What principal will amount to Rs. $38,976$ in two years, if the rates of interest for successive years are $16$% and $20$% respectively?

  1. $27000$
  2. $28000$
  3. $29000$
  4. $30000$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$A=P(1+\cfrac{r}{100})^n$
for $r=16$ and $n=1$
$A=P(1+\cfrac{16}{100})^1$
$A=1.16P$
Again for $r=20, n=1,P'=1.16P$
$A=P'(1+\cfrac{r}{100})^n$
$A=1.16P\times 1.2=1.392P\\ A=Rs.38976=1.392P\\ \implies P=28000 $