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

Rajan borrowed Rs. $50,000$ from Rakesh at simple interest. After $3$ years, Rakesh got Rs. $3000$ more than what he had given to Rajan. What was the rate of interest per annum?

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

Rate $= \displaystyle \left ( \frac{100 \times 300}{5000 \times 3} \right )$% = 2%

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

During a period of two years, a principal of Rs. $100$ amounts to Rs. $121$ at the annual compount rate of $r\%$. The value of $r$ will be

  1. $9$
  2. $10$
  3. $\displaystyle \frac{21}{2}$
  4. $11$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, Sum (P) = Rs. 100,
Amount due (A) = Rs. 121
Time (n) = 2 years, Rate (r) = ?
We know $A = P \displaystyle \left ( 1 + \frac{r}{100} \right )$
$\therefore 121 = 100 \displaystyle \left ( 1 + \frac{r}{100} \right )^2$
or $\displaystyle \left ( 1 + \frac{r}{100} \right )^2 = \frac{121}{100}$
or $\displaystyle 1 + \frac{r}{100} = \frac{11}{10} = 1 + \frac{1}{10} = 1 + \frac{10}{100}$
$\therefore r = 10$%

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

Rajan lent Rs. $1200$ to Rakesh for $3$ years at a certain rate of interest and Rs. $1000$ to Mukesh for the same time at the same rate. If he gets Rs. $50$ more from Rakesh than from Mukesh, then the rate percent

  1. $ 8 \displaystyle \frac{1}{3}\%$
  2. $ 6 \displaystyle \frac{2}{3}\%$
  3. $ 10 \displaystyle \frac{1}{3}\%$
  4. $ 9 \displaystyle \frac{2}{3}\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle \frac{1200 \times R \times 3}{100} - \frac{1000 \times R \times 3}{100} = 50$
or $6 R = 50$
or $R = \displaystyle 8 \frac{1}{3}$%

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

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

  1. 9$\displaystyle \frac{1}{11}\%$
  2. $10\%$
  3. $11\%$
  4. $21\displaystyle \frac{9}{11}\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$\therefore Rate = \displaystyle \left ( \frac{100 \times 100 \times 12}{100 \times 55} \right )$% $21 \displaystyle \frac{9}{11}$%

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

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

  1. $12\%$
  2. $14\%$
  3. $15\%$
  4. $\displaystyle 16 \frac{2}{3}$%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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 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

  1. 6%

  2. 8%

  3. 9%

  4. 10%
Reveal answer Fill a bubble to check yourself
B Correct answer
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%

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

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.

  1. $10\%$
  2. $12\%$
  3. $15\%$
  4. $20\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $ %

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

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.

  1. $30\%$
  2. $35\%$
  3. $29\%$
  4. $25\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
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=$\dfrac{1710\times 100}{5700\times 1}=30$%  

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

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.

  1. 12%

  2. 15%

  3. 20%

  4. 25%
Reveal answer Fill a bubble to check yourself
C Correct answer
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$=\dfrac{PRT}{100}$

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

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

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

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 

  1. 11%, Rs. 800

  2. 12%, Rs. 400

  3. 13%, Rs. 700

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
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%
$Amount=P\left(1+\frac{R}{100}\right)^t$
$\Rightarrow 1089=x\left(1+\frac{10}{100}\right)^2$
$\Rightarrow 1089=x\times \frac{110}{100}\times \frac{110}{100}$
$\Rightarrow x=\frac{1089\times 100\times 100}{110\times 110}=Rs.900$