Tag: mathematics and statistics

Questions Related to mathematics and statistics

A sum of Rs. $1000$ is lent to be returned in $11$ monthly installments of Rs. $100$ each, interest being simple. The rate of interest

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

  1. 9$\displaystyle \frac{1}{11}\%$

  2. $10\%$

  3. $11\%$

  4. $21\displaystyle \frac{9}{11}\%$

Show answer
Correct Option: D
Explanation:

Rs. 1000 + S.l. on Rs. 1000 for 11 months = Rs. 1000 + S.I. on Rs. 100 for (1 + 2 + 3 + 4 + ... + 10) months Rs. 1000 S.I. on Rs. 100 for 100 months 
= Rs.1000 + S.l. on Rs. 100 for 55 months
S.l. on Rs. 100 for 55 months = Rs. 100
$\therefore Rate = \displaystyle \left ( \frac{100 \times 100 \times 12}{100 \times 55} \right )$% $21 \displaystyle \frac{9}{11}$%

A sum of money at compound interest amounts to Rs. $10580$ in $2$ years and to Rs. $12167$ in $3$ 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. $12\%$

  2. $14\%$

  3. $15\%$

  4. $\displaystyle 16 \frac{2}{3}$%

Show answer
Correct Option: C
Explanation:

Interest on Rs. 10580 for 1 year = Rs. (12167 - 10580) = Rs. 1587
$\therefore Rate = \displaystyle \left ( \frac{100 \times 1587}{10580} \right )$% = 15%

The compound interest on a sum of money for two years is Rs 52 and the simple interest for two years at the same rate is Rs 50. Then the rate of interest is

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

  1. 6%

  2. 8%

  3. 9%

  4. 10%

Show answer
Correct Option: B
Explanation:

Given,
Simple Interest for two years is 50
$\frac { P\times T\times R }{ 100 } =Simple\quad Interest$
$\frac { P\times 2\times R }{ 100 } =50$
 $PR=2500$
 $P=\frac { 2500 }{ R } $................EQ(1)
Compound Interest for two years will be 52
$P{ { { \left( 1+\frac { r }{ 100 }  \right)  }^{ 2 } } }-P=52$
$\Rightarrow P\left( 1+\frac { 2R }{ 100 } +\frac { { R }^{ 2 } }{ 10000 } -1 \right) =52$
 $\Rightarrow P\left( \frac { 2R }{ 100 } +\frac { { R }^{ 2 } }{ 10000 }  \right) =52$
$\Rightarrow P\left( \frac { 200R+{ R }^{ 2 } }{ 10000 }  \right) =52$
$\Rightarrow P\left( \frac { 200R+{ R }^{ 2 } }{ 10000 }  \right) =52$
$\Rightarrow P\left( 200R+{ R }^{ 2 } \right) =520000$
$\Rightarrow 200R+{ R }^{ 2 }=\frac { 520000 }{ P } $
 $\Rightarrow 200R+{ R }^{ 2 }=520000\times \frac { R }{ 2500 } $(TAKING EQUATION FROM SIMPLE INTEREST EQ(1))
 $\Rightarrow 200R+{ R }^{ 2 }=208R$
$\Rightarrow { R }^{ 2 }=(208R-200R)$
 $\Rightarrow { { R }^{ 2 } }=8R$
 $R=8$
Rate will be 8%

The difference between the C.I. and S.I. on a sum of 7200 for two years is 72. Find the rate of interest per annum.

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

  1. $10\%$

  2. $12\%$

  3. $15\%$

  4. $20\%$

Show answer
Correct Option: A
Explanation:

Simple Interest $ SI = \dfrac {PNR}{100} $


So, $ SI = \dfrac {7200 \times 2 \times R}{100} = Rs 144R $

When interest is compounded, Amount $ A = P(1+ \dfrac {R}{100})^n $

So, A $ = 7200 \times (1+ \dfrac {R}{100})^2 = Rs  7200 \times (1+ \dfrac {R^2}{10000} + \dfrac {2R}{100})  = Rs 7200 + Rs 0.72R^2 + Rs 144R  $

And $ CI = A - P = Rs 0.72R^2 + Rs 144R $

Si, difference $ CI - SI =  Rs 0.72R^2 + Rs 144R - Rs 144R =  Rs  72 $

$0.72R^2 = 72 $

$R^2 = 100 $ 

So, $ R = 10 $ %

A sum is invested at compound interest compounded yearly. If the interest for two successive years be Rs. 5,700 and Rs. 7,410 calculate the rate of interest.

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

  1. $30\%$

  2. $35\%$

  3. $29\%$

  4. $25\%$

Show answer
Correct Option: A
Explanation:

Difference between the C.I of two successive years=$7410-5700=Rs. 1710$


Rate of interest$=\dfrac{Difference  \ between \ the  \ C.I  \ of \ two \ successive \ year \times100 }{C.I \ of \ preceding   \ year\times time}$


$\therefore$Rate of intereast=$\dfrac{1710\times 100}{5700\times 1}=30$%  

A sum of money placed out at compound interest amounts to Rs. 20,160 in 3 years and to Rs. 24,192 in 4 years. Calculate the rate of interest.

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

  1. 12%

  2. 15%

  3. 20%

  4. 25%

Show answer
Correct Option: C
Explanation:

Amount in three year $=Rs. 20160$


Amount in four year $=Rs.24192$

Interest in 1 year $=24192-20160=Rs.4032$

Let the rate of interest $=R$%

C.I fir 1 year=S.I for 1 year$=\dfrac{PRT}{100}$

$\Rightarrow 4032=\dfrac{20160\times R\times 1}{100}$

$\Rightarrow R=\dfrac{4032\times 100}{20160\times 1}=20$%

The compouned interest, calculate yearly, on a certain sum of money for the second year is Rs. 1,089 and for the third year it is Rs. 1,197.90. Calculate the rate of interest and the sum of money 

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

  1. 11%, Rs. 800

  2. 12%, Rs. 400

  3. 13%, Rs. 700

  4. None of these

Show answer
Correct Option: D
Explanation:

Sum after 2 years=Rs.1089

Sum after 3 years=Rs.1197.90
Interest for onr year$=1197.90-1089=108.90$
C.I for 1 year=S.I for 1 year 
$Interest=\frac{PRT}{100}$
$\Rightarrow 108.90=\frac{1089\times R\times 1}{100}$
$\Rightarrow R=\frac{108.90\times 100}{1089}=10$%

Let the sum=Rs.x
Amount=Rs. 1089,Time 2 years,R=10%
$Amount=P\left(1+\frac{R}{100}\right)^t$
$\Rightarrow 1089=x\left(1+\frac{10}{100}\right)^2$
$\Rightarrow 1089=x\times \frac{110}{100}\times \frac{110}{100}$
$\Rightarrow x=\frac{1089\times 100\times 100}{110\times 110}=Rs.900$


A certain sum of money is put at compound interest, compounded half-yearly. If the interest for two successive half-years are Rs. 650 and Rs. 760.50; find the rate of interest. 

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

  1. $34$

  2. $32$

  3. $27$

  4. $38$

Show answer
Correct Option: A
Explanation:

Difference between the C.I of two successive half -years=$760.50-650=110.50$

Time =6 months=$\frac{6}{12}=\frac{1}{2}years$
Rate of interest$=\frac{Difference  between  the  C.I  of  two  successive year\times100 }{C.I  of  preceding   year\times time}$

$\therefore$Rate of intereast=$\frac{110.50\times 100\times 2}{650\times 1}=34$%  

A certain sum amounts to Rs. 5,292 in two years and Rs. 5,556.60 in three years, interest being compounded annually. Find the rate of interest.

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

  1. 9%

  2. 15%

  3. 7%

  4. 5%

Show answer
Correct Option: D
Explanation:

Sum after 2 years $=Rs.5292$


Sum after 3 years $=Rs.5556.60$

Interest for onr year $=5556.60-5292=Rs. 264.60$

C.I for 1 year $=$ S.I for 1 year 

$Interest =\dfrac{PRT}{100}$

$\Rightarrow 264.60=\dfrac{5292\times R\times 1}{100}$

$\Rightarrow R=\dfrac{264.60\times 100}{5292}=5$%

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 rate of interest per annum.

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

  1. 12%

  2. 14%

  3. 18%

  4. 22%

Show answer
Correct Option: C
Explanation:

$P=Rs.8000$


$Time =\ 1 \ year$


$Amount=9440$

Interest for 1 year$=9440-8000=Rs.1440$

$Interest=\dfrac{PRT}{100}$

$1440=\dfrac{8000\times R\times 1}{100}$

$R=\dfrac{1440\times 100}{8000}=18$%