Tag: mathematics and statistics

Questions Related to mathematics and statistics

The CI on a sum of Rs 625 in 2 years is Rs 51. Find the rate of interest.

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

  1. 4%

  2. 3%

  3. 2%

  4. 1%

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Correct Option: A
Explanation:
We know that $A = C.I + P$
$A=625+51=676$
Using the formula $A=P{\left[1+\dfrac{R}{100}\right]}^{n}$
$676=625{\left[1+\dfrac{R}{100}\right]}^{2}$
$\Rightarrow\,\dfrac{676}{625}={\left[1+\dfrac{R}{100}\right]}^{2}$
$\Rightarrow\,{\left[1+\dfrac{R}{100}\right]}^{2}=\dfrac{676}{625}$
$\Rightarrow\,1+\dfrac{R}{100}=\sqrt{\dfrac{676}{625}}$
$\Rightarrow\,1+\dfrac{R}{100}=\dfrac{26}{25}$
$\Rightarrow\,\dfrac{R}{100}=\dfrac{26}{25}-1=\dfrac{26-25}{25}=\dfrac{1}{25}$
$\Rightarrow\,R=\dfrac{100}{25}=4$
$\therefore\,$Rate of interest $R=4\%$

At what rate per cent of simple interest will the interest on Rs.3,750 be one-fifth of itself in 4 years? To what will it amount in 15 years?

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

  1. 6 % and Rs.6,562.50

  2. 8 % and Rs.6,562.50

  3. 5 % and Rs.6,562.50

  4. 4 % and Rs.6,562.50

Show answer
Correct Option: C
Explanation:

Simple Interest $ I = \cfrac {PNR}{100} $
Given,
$ P = Rs  3,750 $
$ I = \cfrac {1}{5}  \times P =  Rs  750 $
$ N = 4  years $
$ R = ? $
So, $ => I = \cfrac {PNR}{100} $
$ => 750  = \cfrac {3,750 \times 4 \times R }{100} $
$ =>R = 5 $ %
And for $ 15 $ years, interest
$ I =  \cfrac {PNR}{100} $
$ => I  = \cfrac {3,750 \times 15 \times 5 }{100} = Rs. 2,812.5 $
And Amount $ = I + P = Rs.  3,750  + Rs.  2,812.5  = Rs.  6,562.5  $

In a simple interest. at what rate percent per annum will a sum of money double in 8 years?

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

  1. $12.5 \%$

  2. $10.5 \%$

  3. $12.0 \%$

  4. $15.5 \%$

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Correct Option: A
Explanation:

Let the initial sum of money be $P$.

Let time in years be $t$ and rate be $r$.

$Final sum = 2\times P$
$Simple interest = \dfrac {P\times t\times r}{100}$
$Total sum = P+\dfrac {P\times t\times r}{100}$
$2P=P+\dfrac{P\times t\times r}{100}$
$P = \dfrac{P\times t\times r}{100}$
given time = 8 years
$r=100/8=12.5\% $

A certain sum of money amounts to $Rs.\,756$ in $2$ years and to $Rs.\,873$ in $3\displaystyle\frac{1}{2}$ years at a certain rate of simple interest. 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. $\;11\%\,p.a.$

  2. $\;12\%\,p.a.$

  3. $\;13\%\,p.a.$

  4. $\;14\%\,p.a.$

Show answer
Correct Option: C
Explanation:

Amount in $2$ years $=$ Rs. $756$

Amount is $3\dfrac{1}{2}$ years $=$ Rs. $873$
$\therefore$ Interest for $1\dfrac{1}{2}$ years $=$ Rs. $873-$ Rs. $756=$ Rs. $117$
Interest for $2$ years $=\dfrac{117}{\frac{3}{2}}\times 2=\dfrac{117\times 2\times 2}{3}=$ Rs. $156$
Interest for $1$ year $=\displaystyle\frac{117\times2}{3}=$ Rs. $78$
$\therefore$ Principal $=\text{Amount}-\text{Interest}$ ....(for $2$ years) $=$ Rs. $756-$ Rs. $156=$ Rs. $600$
$\therefore$ Rate of interest $=\displaystyle\frac{78\times100}{600\times1}=\,13\%$ p.a.

A man invested Rs. $1000$ on simple interest at a certain rate and Rs. $1500$ at $2\%$ higher rate. The total interest in three years is Rs. $390$. What is the rate of interest for Rs. $1000$?

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

  1. $4\%$.

  2. $5\%$.

  3. $6\%$.

  4. $7\%$.

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Correct Option: A
Explanation:

Let the interest rate at which $Rs.\,1000$ is invested is $r\%$ 
Then Rs. $1500$ is invested at $(r+2)\%$

Then according to the question, we have
$\displaystyle\frac{1000\times\,r\times\,3}{100}+\displaystyle\frac{1500\times(r+2)\times3}{100}=390$
$\Rightarrow\;30r+45r+90=390$
$\Rightarrow\;75r=300$
$\Rightarrow\;r=4\%$

A person lends $40\%$ of his sum of money at $15\%\,p.a.$, $50\%$ of rest at $10\%\,p.a.$ and the rest at $18\%\,p.a.$ rate of interest. What would be the annual rate of interest, if the interest is calculated on the whole sum?

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

  1. $\;13.4\%$

  2. $\;14.33\%$

  3. $\;14.4\%$

  4. $\;13.33\%$

Show answer
Correct Option: C
Explanation:

Let the whole sum be Rs. $100$. 

Then, sum at $15\%$ p.a. $=$ Rs. $40$
 Remaining sum $=$ Rs. $60$
$\therefore$ Sum at $10\%$ p.a. $=50\%$ of Rs. $60=$ Rs. $30$ and sum at  $18\%$ p.a. $=$ Rs. $30$

$\therefore \text {S.I.} $ on Rs. $100$ for $1$  year $=\begin{pmatrix}40\times\displaystyle\frac{15}{100}\times1\end{pmatrix}+\begin{pmatrix}30\times\displaystyle\frac{10}{100}\times1\end{pmatrix}+\begin{pmatrix}30\times\displaystyle\frac{18}{100}\times1\end{pmatrix}$
$ =$ Rs. $(6+3+5.4)=$ Rs. $14.4$
Hence, required rate $=14.4\%$.

The compound interest on a sum for two years is Rs. $832$ and the simple interest on the same sum at the same rate for the same period is Rs. $800$. What is the rate of interest ?

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

  1. $6\%$

  2. $8\%$

  3. $10\%$

  4. $12\%$

Show answer
Correct Option: B
Explanation:

Let the sum be Rs. $P$ and rate of interest per annum be $R\%$
Then $\displaystyle P\left [ \left ( 1+\frac{R}{100} \right )^{2}-1 \right ]-\frac{2PR}{100}=$ Rs. $832-$ Rs. $800=$ Rs. $32$
$\displaystyle \Rightarrow P\left [ 1+\frac{2R}{100}+\frac{R^{2}}{10000}-1 \right ]-\frac{2PR}{100}=32$
$\displaystyle \Rightarrow \frac{PR^{2}}{10000}=32$

$\Rightarrow PR\times R=320000$ ..........(i)
Also $\displaystyle \frac{2PR}{100}=800$ (S.I)
$\Rightarrow PR=40000$ .........(ii)
$\displaystyle \therefore$ From (i) and (ii), we have
$ 40000 \times  R = 320000$  

$\displaystyle \Rightarrow$ $R=8\%$ p.a.

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

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