Tag: introduction to interest payments

Questions Related to introduction to interest payments

The population of a village was $20,000$ and after $2$ years it become $22050$. What is the rate of increase per annum ?

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

  1. $10\%$

  2. $8\%$

  3. $5\%$

  4. $6\%$

Show answer
Correct Option: C
Explanation:

We know that

Final population $=$ (original population)$\times { \left( 1+\frac { Rate }{ 100 }  \right)  }^{ time }.$.........(i)
Here original population $=20000$,
final population $=22050$,
Time $=2$ yrs,
rate=?
Let the rate $=R$.
Substituting the values of the given parameters in (i),
$20000{ \left( 1+\dfrac { R }{ 100 }  \right)  }^{ 2 }=22050$
$ \Rightarrow { \left( 1+\dfrac { R }{ 100 }  \right)  }^{ 2 }=\dfrac { 22050 }{ 20000 } =1.1025$
$ \Rightarrow \left( 1+\dfrac { R }{ 100 }  \right) =\sqrt { 1.1025 } =1.05\  $
i.e $R=5\%$
Ans- Option C.

At what rate per cent per annum will Rs.3000 amount to Rs.3993 in 3 years, if the interest is compounded annually ?

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

  1. 9 % p.a.

  2. 10 % p.a.

  3. 12 % p.a.

  4. 15 % p.a.

Show answer
Correct Option: B
Explanation:

 A = Rs.3993, P = Rs.3000, n = 3, r = ?
$\displaystyle \therefore A=P\left ( 1+\frac{r}{100} \right )\Rightarrow 3993=3000\left ( 1+\frac{r}{100} \right )^{3}\Rightarrow \frac{3993}{3000}=\left ( 1+\frac{r}{100} \right )^{3}\Rightarrow \frac{1331}{1000}=\left ( 1+\frac{r}{100} \right )^{3}$
$\displaystyle \Rightarrow \left ( \frac{11}{10} \right )^{3}=\left ( 1+\frac{r}{100} \right )^{3}\Rightarrow 1+\frac{r}{100}=\frac{11}{10}\Rightarrow \frac{r}{100}=\frac{11}{10}-1=\frac{1}{10}$
$\displaystyle \therefore r=\frac{100}{10}=10\%: : p.a.$

Rs. 8000 invested at compound interest gives Rs.1261 as interest after 3 years. 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. 25 %

  2. 17.5 %

  3. 10 %

  4. 5 %

Show answer
Correct Option: D
Explanation:

P = Rs.8000 C.I. = Rs. 1261

$\displaystyle \Rightarrow Amount=Rs.9261, n=3, r=?$
$\displaystyle \therefore 9261=8000\left ( 1+\cfrac{r}{100} \right )^{3}$
$\Rightarrow \left ( 1+\cfrac{r}{100} \right )^{3}=\cfrac{9261}{8000}=\left ( \cfrac{21}{20} \right )^{3}$
$\displaystyle \Rightarrow 1+\cfrac{r}{100}=\cfrac{21}{20}$
$\Rightarrow \cfrac{r}{100}=\cfrac{21}{20}-1=\cfrac{1}{20}$
$\Rightarrow r\cfrac{100}{20}\%=5\%p.a.$

The difference between compound interest and simple interest at the same rate on Rs.5000 for 2 years is Rs.72 What is the rate of interest per annum ?

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

  1. 20

  2. 15

  3. 12

  4. 10

Show answer
Correct Option: C
Explanation:
Let the rate per cent p.a.be r. Then,
$\displaystyle S.I.=Rs.\left ( 5000\times \cfrac{r}{100}\times 2 \right )=Rs.100r$
$\displaystyle C.I.=Rs.\left [ 5000\left ( 1+\cfrac{r}{100} \right )^{2}-5000 \right ]$
$=Rs.5000\left [ \left ( 1+\cfrac{r}{100} \right )^{2}-1 \right ]$
$=Rs.5000\left [ \left ( 1+\cfrac{r^{2}}{10000}+\cfrac{2r}{100} \right )-1 \right ]$
$\displaystyle =Rs.5000\left ( \cfrac{r^{2}}{10000}+\cfrac{r}{50} \right )=Rs.\cfrac{5000(r^{2}+200r)}{10000}$
$=Rs.\left ( \cfrac{r^{2}}{2}+100r \right )$
$\displaystyle \therefore C.I.-S.I.=72$
$\displaystyle \Rightarrow \cfrac{r^{2}}{2}+100r-100r=72$
$\Rightarrow \cfrac{r^{2}}{2}=72$ 
$\Rightarrow r^{2}=144$
$\Rightarrow r=12\%\: \: p.a.$

A sum of money amounts to Rs.4840 in 2 years and Rs.5324 in 3 years at compound interest compounded annually. What is the rate of interest per annum ?

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

  1. 8

  2. 10

  3. 12

  4. 15

Show answer
Correct Option: B
Explanation:

 Let the principal be Rs.P and rate of interest p.a. = r% Then
$\displaystyle P\left ( 1+\frac{r}{100} \right )^{2}=4840...........(i)$ and $\displaystyle P\left ( 1+\frac{r}{100} \right )^{3}=5324...........(ii)$
$\displaystyle \Rightarrow \frac{5324}{4840}=\frac{(1+r/100)^{3}}{(1+r/100)^{2}}\Rightarrow 1+\frac{r}{100}=\frac{1331}{1210}$
$\displaystyle \Rightarrow \frac{r}{100}=\frac{1331}{1210}-1=\frac{121}{1210}=\frac{1}{10}\Rightarrow r=\frac{1}{10}\times 100=10\%: p.a.$

A certain sum of money amounts to $\displaystyle \frac {5}{4}$ of itself in 5 years. The rate percent per annum is

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

  1. 5%

  2. 7%

  3. 9%

  4. 12%

Show answer
Correct Option: A
Explanation:

R $\times$ T = 100 $\times$ (N - 1)
$R \times 5 = 100 \times \left ( \displaystyle \frac {5}{4} - 1 \right )$
$R \times 5 = 100 \times \displaystyle \frac {1}{4}$
$R = 5\%$

Madhav lent out Rs. 7953 for 2 years and Rs. 1800 for 3 years at the same rate of simple interest. If he got Rs. 2343. 66 as total, then find the percent rate of interest.

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

  1. 11%

  2. 12%

  3. 12.5%

  4. 5%

Show answer
Correct Option: A
Explanation:

We know that Simple Interest $ = \cfrac {PNR}{100} $
Given,  $ \cfrac {7953 \times 2 \times R}{100}  + \cfrac {1800  \times 3 \times R}{100}  = Rs 2343.66 $
$=> 159.06R + 54R = Rs 2343.66 $
$ => 213.06R = 2343.66 $
$ => R = 11 \%$ 

Madhav lent out Rs $7953$ for $2$ years and Rs $1800$ for $3$ years at the same rate of simple interest. If he hot Rs $2343.66$ as total interest then find the percent rate of interest.

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

  1. $11\%$

  2. $12\%$

  3. $12.5\%$

  4. $5\%$

Show answer
Correct Option: A
Explanation:

$S.I = \dfrac{P\times t\times r}{100}$


Let the rate be $r$ Principal $P$ and time $t$

${S.I} _{1}=\dfrac{7953\times 2\times r}{100}$


${S.I} _{2}=\dfrac{1800\times 3\times r}{100}$
$Total$ $simple$ $interest = {S.I} _{1}+{S.I} _{2} $

$2343.66=\dfrac { 7953\times2\times r }{ 100 } +\dfrac { 1800\times3\times r }{ 100 }$ 
$2343.66 = 213.06\times r$
$r = 11\%$

If the interest is payable quarterly, Rs. $1600$ amounts to Rs. $2662$ after $1\dfrac{1}{2}$ years, the annual rate of interest is

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

  1. $5\%$

  2. $10\%$

  3. $20\%$

  4. $35\%$

Show answer
Correct Option: D
Explanation:

Amount $= Rs. 2662$
Principal $= Rs. 1600$
Time $= 1.5$ years $= 6$ quarters
$A = P\left(1 + \cfrac{R}{100}\right)^T$
$2662 = 1600 \left(1 + \cfrac{R}{100}\right)^6$
$1.088 = 1 + \cfrac{R}{100}$
$R = 8.8\%$
Hence, annual rate of interest $=8.8\times 4 = 35\%$

Find the rate, when 1,800 earns an interest of Rs 432 in 3 years

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

  1. 7%

  2. 6%

  3. 5%

  4. 8%

Show answer
Correct Option: D
Explanation:

Given $P=1, 800, SI=432, T=3 years$
$R=\dfrac {SI\times 100}{P\times T}=\dfrac {432\times 100}{1800\times 3}=8$%