Tag: introduction to interest

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Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest 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 ?

  1. $10\%$
  2. $8\%$
  3. $5\%$
  4. $6\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. 9 % p.a.

  2. 10 % p.a.

  3. 12 % p.a.

  4. 15 % p.a.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

Rs. 8000 invested at compound interest gives Rs.1261 as interest after 3 years. The rate of interest per annum is

  1. 25 %

  2. 17.5 %

  3. 10 %

  4. 5 %
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

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 ?

  1. 20

  2. 15

  3. 12

  4. 10
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let the rate per cent p.a.be r. Then,
$\displaystyle S.I.=Rs.\left ( 5000\times \cfrac{r}{100}\times 2 \right )=Rs.100r$
$\displaystyle C.I.=Rs.\left [ 5000\left ( 1+\cfrac{r}{100} \right )^{2}-5000 \right ]$
$=Rs.5000\left [ \left ( 1+\cfrac{r}{100} \right )^{2}-1 \right ]$
$=Rs.5000\left [ \left ( 1+\cfrac{r^{2}}{10000}+\cfrac{2r}{100} \right )-1 \right ]$
$\displaystyle =Rs.5000\left ( \cfrac{r^{2}}{10000}+\cfrac{r}{50} \right )=Rs.\cfrac{5000(r^{2}+200r)}{10000}$
$=Rs.\left ( \cfrac{r^{2}}{2}+100r \right )$
$\displaystyle \therefore C.I.-S.I.=72$
$\displaystyle \Rightarrow \cfrac{r^{2}}{2}+100r-100r=72$
$\Rightarrow \cfrac{r^{2}}{2}=72$ 
$\Rightarrow r^{2}=144$
$\Rightarrow r=12\%\: \: p.a.$
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

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 ?

  1. 8

  2. 10

  3. 12

  4. 15
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\displaystyle P\left ( 1+\frac{r}{100} \right )^{3}=5324...........(ii)$
$\displaystyle \Rightarrow \frac{5324}{4840}=\frac{(1+r/100)^{3}}{(1+r/100)^{2}}\Rightarrow 1+\frac{r}{100}=\frac{1331}{1210}$
$\displaystyle \Rightarrow \frac{r}{100}=\frac{1331}{1210}-1=\frac{121}{1210}=\frac{1}{10}\Rightarrow r=\frac{1}{10}\times 100=10\%: p.a.$

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

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

  1. 5%

  2. 7%

  3. 9%

  4. 12%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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.

  1. 11%

  2. 12%

  3. 12.5%

  4. 5%
Reveal answer Fill a bubble to check yourself
A Correct answer
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 \%$ 

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

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.

  1. $11\%$
  2. $12\%$
  3. $12.5\%$
  4. $5\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
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} $

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

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

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

  1. $5\%$
  2. $10\%$
  3. $20\%$
  4. $35\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $=8.8\times 4 = 35\%$