Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

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

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

  1. 4%

  2. 3%

  3. 2%

  4. 1%
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $R=4\%$
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

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?

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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  $

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

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

  1. $12.5 \%$
  2. $10.5 \%$
  3. $12.0 \%$
  4. $15.5 \%$
Reveal answer Fill a bubble to check yourself
A Correct answer
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\% $

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 $Rs.\,756$ 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?

  1. $\;11\%\,p.a.$
  2. $\;12\%\,p.a.$
  3. $\;13\%\,p.a.$
  4. $\;14\%\,p.a.$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $1000$?

  1. $4\%$.
  2. $5\%$.
  3. $6\%$.
  4. $7\%$.
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$\displaystyle\frac{1000\times\,r\times\,3}{100}+\displaystyle\frac{1500\times(r+2)\times3}{100}=390$
$\Rightarrow\;30r+45r+90=390$
$\Rightarrow\;75r=300$
$\Rightarrow\;r=4\%$

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

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?

  1. $\;13.4\%$
  2. $\;14.33\%$
  3. $\;14.4\%$
  4. $\;13.33\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $=14.4\%$.

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

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 ?

  1. $6\%$
  2. $8\%$
  3. $10\%$
  4. $12\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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