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Introduction to interests - class-VIII

Description: introduction to interests
Number of Questions: 97
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Tags: ratio - applications maths percentage life mathematics managing money money math applications of percentage comparison of quantities interest comparing quantities commercial mathematics simple interest banks and simple interest comparing quantity commercial arithmetic mathematics and statistics
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If the interest on $1700$ rupees is $340$ rupees for $2$ year the rate of interest must be

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

  1. $12\ \%$

  2. $15\ \%$

  3. $4\ \%$

  4. $10\ \%$

Show answer
Correct Option: D
Explanation:

Principle$=Rs1770\quad\quad Time=2years$

$SI=Rs340\quad\quad Rate=?\ \cfrac{P\times R\times T}{100}=340\Rightarrow \cfrac{1770\times R\times 2}{100}=340\ \Rightarrow R=\cfrac{340\times100}{1770\times2}=\cfrac{340\times5}{177}=9.6\%$

The simple interest on a sum money is 4/9 of the principal and the number of years is equal to the rate percent per annum. The rate per annum is :  

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

  1. $5$%

  2. $6\dfrac{2}{3}\%$

  3. $6$%

  4. $7\dfrac{1}{5}\%$

Show answer
Correct Option: B
Explanation:
Let the principal be $P$.
Rate of interest be $R\%$
According to the question, Time$=R$
Simple interest $=\dfrac{4P}{9}$.
$SI =\dfrac{\left(PTR\right)}{100}$
$\Rightarrow \dfrac{4P}{9} =\dfrac{\left(PTR\right)}{100}$
$\Rightarrow \dfrac{4P}{9} =\dfrac{\left(P\times R\times R\right)}{100}$
$\Rightarrow \dfrac{4P}{9} =\dfrac{\left(P\times {R}^{2}\right)}{100}$
$\Rightarrow \dfrac{4}{9} =\dfrac{{R}^{2}}{100}$
$\Rightarrow {R}^{2}=100\times\dfrac{4}{9}$
$\Rightarrow R= 10\times \dfrac{2}{3}=\dfrac{20}{3}$
Therefore, rate of interest is $\dfrac{20}{3}\%$ or  $6\dfrac{2}{3}\%$.

A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is :

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

  1. Rs. 650

  2. Rs. 690

  3. Rs. 698

  4. Rs. 700

  5. Rs. 715

Show answer
Correct Option: C
Explanation:

S.I. for $1$ year $= Rs. (854-815) = Rs. 39$


S.I. for $3$ years = Rs. $39 \times 3=Rs. 117$


Therefore,

Principal $= Rs. 815 - Rs. 117 = Rs. 698$

A four year Indira Vikas certificate with a maturity value of Rs. 700 is purcahsed for Rs. 500. The rate $\%$ p.a. is :

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

  1. $

    9 \dfrac { 1 } { 11 } \%

    $

  2. $10\%$

  3. $11\%$

  4. $

    21 \dfrac { 9 } { 11 } \%

    $

Show answer
Correct Option: B
Explanation:

Interest $= 200.....(700-500=200)$


$SI=\dfrac{P\times N\times R}{100}$

$200=\dfrac{500 \times R \times 4}{100}\implies R=10\%$

At what rate per cent per annum, will Rs.32000 yield a compound interest of Rs.5044 in 9 months interest being compounded quarterly ?

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

  1. 25

  2. 23

  3. 20

  4. 18

Show answer
Correct Option: C
Explanation:
Principal = Rs.32000 
Amount $= Rs.(32000 + 5044) = Rs.37044$
Rate $= r\%$ p.a. or $\displaystyle \cfrac{r}{4}\%$ per quarter 
Time = 9 months = 3 quarters i.e., $n = 3$
$\displaystyle \therefore$ Applying $\displaystyle A=P\left ( 1+\cfrac{r}{100} \right )^{n}$ we have
$\displaystyle 37044=32000\left ( 1+\cfrac{r}{400} \right )^{3}\Rightarrow \cfrac{37044}{32000}=\left ( 1+\cfrac{r}{400} \right )^{3}$
$\displaystyle \Rightarrow \cfrac{9261}{8000}=\left ( 1+\cfrac{r}{400} \right )^{3}\Rightarrow \left ( \cfrac{21}{20} \right )^{3}=\left ( 1+\cfrac{r}{400} \right )^{3}$
$\displaystyle \Rightarrow 1+\cfrac{r}{400}=\cfrac{21}{20}\Rightarrow \cfrac{r}{400}=\cfrac{21}{20}-1=\cfrac{1}{20}\Rightarrow r=\cfrac{400}{20}=20\%p.a.$

At what rate of interest per annum will a sum double itself in 8 years?

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

  1. $25\%$

  2. $6\frac{1}{4} \%$

  3. $12\frac{1}{2} \%$

  4. None

Show answer
Correct Option: C
Explanation:

T = 8 years; N = 2; R = ?
R $\times T$ = 100 $\times (N - 1)$


R $\times 8$= 100 $\times (2 - 1)$

$R\, =\, \displaystyle \frac {100}{8}\, =\, 12\frac{1}{2}\, \%$

Simple interest on Rs.2000 for 4 years is Rs.400. Percent rate of interest is

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

  1. $\displaystyle\frac{2000\times 100}{400\times 4}$

  2. $\displaystyle\frac{400\times 4}{2000\times 100}$

  3. $\displaystyle\frac{400\times 100}{2000\times 4}$

  4. None of these

Show answer
Correct Option: C
Explanation:

Principal = Rs 2000
Time = 4 years
Interest = Rs 400
Now, $Interest = \frac{Principal \times Rate \times Time}{100}$
$400 = \frac{2000\times R\times 4}{100}$
$R = \frac{400 \times 100}{2000 \times 4}$

At what rate percent per annum will the simple interest on a sum of money be 2/5 of the amount in 10 years?

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

  1. $4\frac {1}{2}$%

  2. $5\frac {1}{2}$%

  3. 4%

  4. 5%

Show answer
Correct Option: C
Explanation:

$SI=\frac {2}{5}P, t=10, r=?$
$\frac {Ptr}{100}=\frac {2}{5}P$
or $\frac {10\times r}{100}=\frac {2}{5}$
or $r=\frac {20}{5}=4$%

A person finds that an increase in the rate of interest from $\displaystyle4\frac{7}{8}$% to $\displaystyle5\frac{1}{8}$% per annum increases his yearly income by Rs 30. His capital in rupees is

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

  1. 15,000

  2. 14,000

  3. 13,000

  4. 12,000

Show answer
Correct Option: D
Explanation:

increase in rate of interest

$4\frac { 7 }{ 8 } =\frac { 39 }{ 8 } $
$5\frac { 1 }{ 8 } =\frac { 41 }{ 8 } $
$\frac { 41 }{ 8 } -\frac { 39 }{ 8 } =\frac { 2 }{ 8 } $
$\frac { 2 }{ 8 } $% of income is Rs30 of the capital
$1$% of income is $\frac { 8 }{ 2 } \times 30$ of capital
$100$% of income will be$\frac { 8 }{ 2 } \times 30\times 100=12000$
His capital in rupees is 12000

Anil invests Rs 3,000 for a year and Sunil joins him with Rs 2,000 after 4 months. After the year they receive a return of Rs 2,600. Sunil's share is

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

  1. Rs 800

  2. Rs 1,000

  3. Rs 750

  4. Rs 900

Show answer
Correct Option: A
Explanation:

$Ratio\quad in\quad which\quad they\quad should\quad share\quad their\quad profits=Raio\quad of\quad investments\times \quad Time\quad period$

$=\frac { 3000\times 12 }{ 2000\times 8 } =\frac { 3\times 3 }{ 2\times 2 } =\frac { 9 }{ 4 } $
Sunils share will be$=\frac { 4 }{ 13 } \times 2600=800$
Sunils share will be Rs800,

The time required so that Rs 450 may increase to Rs 576 (the rate of simple interest being 7% per annum) is

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

  1. 2 years

  2. 3 years

  3. 4 years

  4. 6 years

Show answer
Correct Option: C
Explanation:

Simple interest will be(576-450)=Rs126

 Let the Time required  be t
As per formula,
$Simple\quad Interest=\frac { P\times t\times r }{ 100 } $
$126=\frac { 450\times t\times 7 }{ 100 } $
$315t=1260$
$t=\frac { 1260 }{ 315 } =4$
Time required will be 4 years


The _______ is the percentage of a sum of money charged for its use.

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

  1. principal

  2. interest

  3. rate of interest

  4. amount

Show answer
Correct Option: C
Explanation:

The rate of interest is the percentage of a sum of money charged for its use.

If the simple interest on a certain sum of money is $\displaystyle \frac{4}{25}$th of the sum and the rate per cent equals the number of years, then 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. $2$%

  2. $4$%

  3. $5$%

  4. $6$%

Show answer
Correct Option: B
Explanation:

$\Rightarrow$   Let the principal be $Rs.x$.

$\Rightarrow$   Then, Simple Interest = $\dfrac{4}{25}x.$
$\Rightarrow$    Let the rate of interest per annum be $r\%$ then time ( T)= $r$ years.
$\Rightarrow$   $R=\dfrac{100\times S.I.}{P\times T}$

$\Rightarrow$   $r=\dfrac{100\times \dfrac {4x}{25}}{x\times r}$

$\Rightarrow$   $r^2=\dfrac{400}{25}$

$\Rightarrow$   $r=\dfrac{20}{5}=4\%$

John earned Rs. $100$ as simple interest on Rs. $600$ for $6$ months. Find the annual rate of interest.

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

  1. $11.11\%$

  2. $32.11\%$

  3. $33.33\%$

  4. $30\%$

Show answer
Correct Option: C
Explanation:

Here, $S.I.=$ Rs. $100,\,P=$ Rs. $600$ and $T=6$ months $=\dfrac{1}{2}$ year

We know $S.I.=\dfrac{P\times R\times T}{100}$
$\Rightarrow$ $100=\dfrac{600\times R\times 1}{2\times 100}$
$\Rightarrow$ $R=\dfrac{20000}{600}$
$\Rightarrow$ $R=33.33\%$

You invested Rs. $1500$ and received Rs. $5000$ after three years. What had been the interest rate?

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

  1. $111.11\%$

  2. $222.22\%$

  3. $99.99\%$

  4. $77.77\%$

Show answer
Correct Option: D
Explanation:
Here, $P=$ Rs. $1500\,A=$ Rs. $5000$ and $T=3$ years
$\Rightarrow$ $S.I=A-P=$ Rs. $5000-$ Rs. $1500=$ Rs. $3500$
$\Rightarrow$ $S.I.=\dfrac{P\times R\times T}{100}$
$\Rightarrow$ $3500=\dfrac{1500\times R\times 3}{100}$
$\Rightarrow$ $R=\dfrac{3500\times 100}{4500}$
Therefore, $R=77.77\%$

What rate will amount to Rs. $33,080$ in three years, if the principle amount was Rs $10,000$ respectively?

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

  1. $48$%

  2. $49$%

  3. $50$%

  4. $12$%

Show answer
Correct Option: B
Explanation:
$A=P(1+\cfrac{r}{100})^n$
$\implies 33,080=10,000[(1+\cfrac{r}{100})^3-1]\\ \implies 1+\cfrac{r}{100}=(33080/10000)^{1/3}\\ \implies r=49$.
Hence rate of intererst is $49\%$ per annum.

Find rate, when principal = Rs. $30,000$; interest = Rs. $900$; time = $3$ years.

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

  1. $1$%

  2. $2$%

  3. $4$%

  4. $5$%

Show answer
Correct Option: A
Explanation:

Under simple interest,


Interest $= Principal \times rate \times time$

Principal $=$ Rs. $30000$
Rate $= r$
Time $= 3$ years
Interest $=$ Rs. $900$

$\Rightarrow 900 = 30000 \times r \times 3$
$\therefore r = 0.01$ or $1\%$

At which percent per annum simple interest will double a sum of money in 12 years?

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

  1. $8\dfrac { 1 }{ 3 }\%$

  2. $24\%$

  3. $\dfrac { 25 }{ 4 }\%$

  4. $8.25\%$

Show answer
Correct Option: A
Explanation:

Let,

Principle P = 100
Amount A = 200
Time T = 12 years
Interest = Rs. 100
Rate of Interest = $\dfrac{(Interest)}{Time}$ = $\dfrac{100}{12}$ = $\dfrac{25}{3}$ = 8$\dfrac{1}{3}$%
Option A is correct

A man invests half his capital at the rate of l0% per annum, one-third at 9% and the rest at 12% per annum. The average rate of interest per annum which he gets, is

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

  1. 9%

  2. 10%

  3. 10.5%

  4. 12%

Show answer
Correct Option: B
Explanation:

Using the rule of weighted average,
The average rate of interest = $\displaystyle \dfrac{\dfrac{1}{2} \, \times \, 10 \, + \, \dfrac{1}{3} \, \times \, 9 \, + \, \dfrac{1}{6} \, \times \, 12}{\dfrac{1}{2} \, + \, \dfrac{1}{3} \, + \,\dfrac{1}{6}}$
= 5 + 3 + 2 = 10%

If $P=5,000$, $T=1$, $S.I.=$Rs. $300$, R will be.

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

  1. $5\%$

  2. $4\%$

  3. $6\%$

  4. None of the above

Show answer
Correct Option: C
Explanation:

$I = ₹300$ 

$P= ₹5000$
$T= 1$ 
$ I = \cfrac{P\times R \times T }{100}$
$ 300 = \cfrac{5000 \times 1 \times R }{100 }$
$ R= \cfrac{300 }{50} = 6$%

A bank charges Rs. 6 for a loan of Rs. 120. The borrower receives Rs. 114 ' and repays the loan in 12 installments of Rs. 10 a month. The interest rate is approximate.

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

  1. 5%

  2. 6%

  3. 7%

  4. 9%

  5. 15%

Show answer
Correct Option: A
Explanation:

Total sum of money paid = Rs 120

$\therefore 120=P{ \left( 1+\cfrac { r }{ n }  \right)  }^{ nt }$
t = 1 year ($\because$ 1 year = 12 months)
n = 12
P = Rs 114
$\Longrightarrow 120=114{ \left( 1+\cfrac { r }{ 12(100) }  \right)  }^{ 12 }\Longrightarrow { \left( \cfrac { 120 }{ 114 }  \right)  }^{ \cfrac { 1 }{ 12 }  }-1=\cfrac { r }{ 1200 } \Longrightarrow r=5.12\%\ \therefore r\approx 5\%$

At what rate per cent per annum will Rs. $1625$ amount to Rs. $2080$ in $3\dfrac{1}{2}$ years ?

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

  1. $8\%$

  2. $10\%$

  3. $12\%$

  4. $14\%$

Show answer
Correct Option: A
Explanation:

We know that $I=\dfrac{PTR}{100}$


where $I$ is the simple interest

$P$ is the principal amount

$T$ is the time period and

$R$ is the rate of interest

and $A=P+I$

where $A$ is the total amount

Given that $P=1625,A=2080$ and $T=3\dfrac 12years=3.5$

Therefore, $2080=1625+\dfrac{1625(3.5)(R)}{100}$

$\implies 455=\dfrac{5687.5(R)}{100}$

$\implies R=\dfrac{45500}{5687.5}=8\%$

Therefore, the rate of interest is $8\%$

At what rate per cent of simple interest will a sum of money double itself in $12$ years?

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

  1. $7\dfrac{1}{2}\%$

  2. $8\dfrac{1}{3}\%$

  3. $10\%$

  4. $12\%$

Show answer
Correct Option: B
Explanation:

In case of Simple interest, total amount $A$ is,

$A=P(1+\frac{rt}{100})$
where
$P=$Principal
$r=$interest rate
$t=$time (in years)=$12$ (given)

After $12$ years, Sum of money doubles itself,
that is $A=2P$

Now apply the formula,
$A=P(1+\frac{rt}{100})$
$2P=P(1+\frac{12r}{100})$

$2=1+\frac{12r}{100}$
$1=\frac{12r}{100}$

Therefore,
$r=\frac{100}{12}=8\frac{1}{3}$percent


At what rate per cent per annum will the simple interest on Rs. $6720$ be Rs. $1911$ in $3$ years $3$ months?

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

  1. $7\dfrac{3}{4}\%$

  2. $8\dfrac{3}{4}\%$

  3. $10\dfrac{1}{4}\%$

  4. $11\dfrac{2}{3}\%$

Show answer
Correct Option: B
Explanation:

According to question, we have:

$6720\times \cfrac{13}{4}\times \cfrac{r}{100}=1911$
$\Rightarrow r=\cfrac{1911\times 4\times 100}{6720\times 13}$
$\Rightarrow r=\cfrac{34}{4}=8\cfrac{3}{4}\%$

Gopal has a cumulative deposit account and deposits Rs. $900 $per month for a period of $4$ years. If he gets Rs.$ 52,020$ at the time of maturity, find the rate of interest.

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

  1. $5\%$

  2. $2\%$

  3. $10\%$

  4. $12\%$

Show answer
Correct Option: C
Explanation:
Installment per month $\left( P \right) = Rs. 900$
No. of months $\left( n \right) = 4 \text{ years} = 12 \times 4 = 48 \text{ months}$
Let rate of interest be $r \%$ per annum
$t = \cfrac{n \left( n + 1 \right)}{2\times 12} = \cfrac{48 \times 49}{24} = 98$
$\therefore \; S.I. = P \times \cfrac{n \left( n + 1 \right)}{2\times 12} \times \cfrac{r}{100}$
$\Rightarrow \; S.I. = 900 \times \cfrac{48 \left( 48 + 1 \right)}{2\times 12} \times \cfrac{r}{100} = Rs. 882 r$
Maturity value $= Rs. \left(900 \times 48 + 882 r \right) = Rs \left( 43200 + 882 r \right)$
maturity value $= Rs. 52020$
$\therefore \; 43200 + 882 r = 52020$
$\Rightarrow \; 882 r = 52020 - 43200$
$\Rightarrow \; r = \cfrac{8820}{882} = 10 \%$
Hence, rate of interest $10 \%$.

A factory kept increasing its output by the same percentage every year. Find the percentage if it is known that the output is doubled in the last two years.

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

  1. $47.53\%$

  2. $45.26\%$

  3. $43.42\%$

  4. $41.42\%$

Show answer
Correct Option: D
Explanation:

Given that the output is doubled in last two years
Let the output before $2$ years be$= x$
Hence after two years it will be$= 2x$

so $n=2$
now using formula $=A=P(1+\frac{R}{100})^n$
Now put the value on given formula .
=> $2x=x(1+\frac{R}{100})^2$
=>$2=1(1+\frac{R}{100})^2$
=>$\sqrt2=1(1+\frac{R}{100})$
=>$\frac{R}{100}=\sqrt{2}-1=1.4142-1=0.4142$
$=>R=41.42\%$
so option D is correct.

If the compound interest on an amount of $29000$ in two years is $9352.5$, what is the rate of interest?

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

  1. $11\%$

  2. $9\%$

  3. $15\%$

  4. $18\%$

Show answer
Correct Option: C
Explanation:

We know that 


$\Rightarrow Total\space amount=P(1+\dfrac{R}{100})^n$

Here $P=29000; \space n=2;\space interest=9352.5$

$\Rightarrow 29000+9352.5=(29000)(1+\dfrac{R}{100})^2$

$\Rightarrow 38352.5=(29000)(1+\dfrac{R}{100})^2$

$\Rightarrow 1.3225=(1+\dfrac{R}{100})^2$

$\Rightarrow 1+\dfrac{R}{100}=1.15$

$\Rightarrow \dfrac{R}{100}=0.15$

$\Rightarrow R=15$

Therefore, Rate of interest is $15\%$

The difference between simple and compound interest on sum of $10000$ is $64$ for $2$ years. Find the rate of interest.  

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

  1. $8$

  2. $64$

  3. $4$

  4. $2$

Show answer
Correct Option: A
Explanation:
Simple Interest $=\dfrac{PNR}{100}$

Compound Interest $=P\left(1+\dfrac{R}{100}\right)^N-P$
Now,
$P\left(1+\dfrac{R}{100}\right)^N-P$ $-\dfrac{PNR}{100}=64$

$\left[10000\times \left(1+\dfrac{R}{100}\right)^2-10000\right]-\left(\dfrac{10000\times R\times 2}{100}\right)=64$

$\Rightarrow$  $10000\left[\left(1+\dfrac{R}{100}\right)^2-1-\dfrac{2R}{100}\right]=64$

$\Rightarrow$  $10000\left[\dfrac{(100+R)^2}{10000}-1-\dfrac{2R}{100}\right]=64$

$\Rightarrow$  $10000\left[\dfrac{10000+200R+R^2-10000-200R}{10000}\right]=64$

$\Rightarrow$  $R^2=64$

$\Rightarrow$  $R=8$

$\therefore$  $Rate=8\%$

A certain amount of money deposited for compound interest, becomes 3 times in 3 years. In how many years will that amount be 27 times the deposited amount if it is given for the same rate of interest?

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

  1. 9

  2. 6

  3. 12

  4. 8

Show answer
Correct Option: A
Explanation:

$A=3P$

For $t=3$
So, $3P=P(1+\cfrac{R}{100})^3\implies R=(3^{2/3}-1)100$
Now, new amount $=27P$
So, $27P=P(1+\cfrac{R}{100})^t$
So, $\implies 27P=P(1+\cfrac{(3^{2/3}-1)100}{100})^t$
$\implies t=9$ years

A sum of money doubled in $10$ 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. $20\%$

  2. $15\%$

  3. $18\%$

  4. $10\%$

Show answer
Correct Option: A

The simple interest and the compound interest on a certain sum for $2$ years is $Rs.1250$ and $Rs.1475$ respectively. Find the rate of interest.

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

  1. $36%$ per annum

  2. $34%$ per annum

  3. $32%$ per annum

  4. $38%$ per annum

Show answer
Correct Option: A

If simple interest on a sum of money for $3$ years is Rs. $240$ and compound interest on the sum at same rate for $2$ years is Rs. $170$, then the rate $\%$ p.a. is 

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

  1. $16\%$

  2. $8\%$

  3. ${ 12 }\dfrac12\%$

  4. ${ 1 }\dfrac18\%$

Show answer
Correct Option: C
Explanation:
simple interest for one year$=\dfrac{240}{3} =Rs. 80$

simple Interest for two year$=80×2= Rs.160$

Compound interest for two years$=Rs. 170$

Difference for $2$ year$=170−160=Rs.10$

Hence

$Rate( \%)=\dfrac{10}{80}×100$

$=12\dfrac{1}{2}\%$

Manish invested a sum of money at CI. It amounted to Rs 2420 in 2 years and Rs 2662 in 3 years. Find the rate percent per annum.

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

  1. 5%

  2. 10%

  3. 20%

  4. 15%

Show answer
Correct Option: B
Explanation:
Last year interest $= 2662 - 2420 =\ Rs. 242$

Difference between SI and CI for $2$ years, Difference $= P{\left[\dfrac{R}{100}\right]}^{2}$

$\therefore\,$Rate $\%=\dfrac{242\times 100}{2420\times 1}=10\%$

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%

Show answer
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. $15.5 \%$

Show answer
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. $7\%$.

Show answer
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

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

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 ?

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

  1. 20

  2. 15

  3. 12

  4. 10

Show answer
Correct Option: C
Explanation:
Let the rate per cent p.a.be r. Then,
$\displaystyle S.I.=Rs.\left ( 5000\times \cfrac{r}{100}\times 2 \right )=Rs.100r$
$\displaystyle C.I.=Rs.\left [ 5000\left ( 1+\cfrac{r}{100} \right )^{2}-5000 \right ]$
$=Rs.5000\left [ \left ( 1+\cfrac{r}{100} \right )^{2}-1 \right ]$
$=Rs.5000\left [ \left ( 1+\cfrac{r^{2}}{10000}+\cfrac{2r}{100} \right )-1 \right ]$
$\displaystyle =Rs.5000\left ( \cfrac{r^{2}}{10000}+\cfrac{r}{50} \right )=Rs.\cfrac{5000(r^{2}+200r)}{10000}$
$=Rs.\left ( \cfrac{r^{2}}{2}+100r \right )$
$\displaystyle \therefore C.I.-S.I.=72$
$\displaystyle \Rightarrow \cfrac{r^{2}}{2}+100r-100r=72$
$\Rightarrow \cfrac{r^{2}}{2}=72$ 
$\Rightarrow r^{2}=144$
$\Rightarrow r=12\%\: \: p.a.$

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 ?

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

  1. 8

  2. 10

  3. 12

  4. 15

Show answer
Correct Option: B
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 $\displaystyle P\left ( 1+\frac{r}{100} \right )^{3}=5324...........(ii)$
$\displaystyle \Rightarrow \frac{5324}{4840}=\frac{(1+r/100)^{3}}{(1+r/100)^{2}}\Rightarrow 1+\frac{r}{100}=\frac{1331}{1210}$
$\displaystyle \Rightarrow \frac{r}{100}=\frac{1331}{1210}-1=\frac{121}{1210}=\frac{1}{10}\Rightarrow r=\frac{1}{10}\times 100=10\%: p.a.$

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

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

  1. 5%

  2. 7%

  3. 9%

  4. 12%

Show answer
Correct Option: A
Explanation:

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

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.

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

  1. 11%

  2. 12%

  3. 12.5%

  4. 5%

Show answer
Correct Option: A
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 \%$ 

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.

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

  1. $11\%$

  2. $12\%$

  3. $12.5\%$

  4. $5\%$

Show answer
Correct Option: A
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} $

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

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

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

  1. $5\%$

  2. $10\%$

  3. $20\%$

  4. $35\%$

Show answer
Correct Option: D
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 $=8.8\times 4 = 35\%$

Find the rate, when 1,800 earns an interest of Rs 432 in 3 years

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

  1. 7%

  2. 6%

  3. 5%

  4. 8%

Show answer
Correct Option: D
Explanation:

Given $P=1, 800, SI=432, T=3 years$
$R=\dfrac {SI\times 100}{P\times T}=\dfrac {432\times 100}{1800\times 3}=8$%

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?

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

  1. $2\%$

  2. $5\%$

  3. $8\%$

  4. $10\%$

Show answer
Correct Option: A
Explanation:

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

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

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

  1. $9$

  2. $10$

  3. $\displaystyle \frac{21}{2}$

  4. $11$

Show answer
Correct Option: B
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 $\displaystyle 1 + \frac{r}{100} = \frac{11}{10} = 1 + \frac{1}{10} = 1 + \frac{10}{100}$
$\therefore r = 10$%

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

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

  1. $ 8 \displaystyle \frac{1}{3}\%$ 

  2. $ 6 \displaystyle \frac{2}{3}\%$ 

  3. $ 10 \displaystyle \frac{1}{3}\%$ 

  4. $ 9 \displaystyle \frac{2}{3}\%$ 

Show answer
Correct Option: A
Explanation:

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

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$=9440-8000=Rs.1440$

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

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

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

A company pays a dividend of 15% on its ten-rupee shares from which it deducts income tax at the rate of 22%. Find the annual income of a man who owns one thousand shares of this company.

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

  1. 1170

  2. 1270

  3. 1070

  4. 1330

Show answer
Correct Option: A
Explanation:
Annual Income $=Rs.\ \dfrac{(10×1000×15)}{100}$
$=Rs.\ 1500$
Income tax deducted $= \dfrac{(22×1500)}{100}$
$=Rs.\ 330$
So, Total Income$=Rs.\ 1500 - 330$
$= Rs.\ 1170 .$

At what rate percent will a sum of money doubles in $16$ years?

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

  1. $\dfrac{25}{4}\%$

  2. $\dfrac{21}{4}\%$

  3. $5\%$

  4. $15\%$

Show answer
Correct Option: A
Explanation:

Let sum of money be $p$.

Thus, amount will be $2p$
Then S.I $= p$
Rate $=\dfrac{100 \times p}{p \times 16}=\dfrac{25}{4}$ $\%$

You invested Rs. $1500$ and received Rs. $5000$ after three years. What had been the interest rate?

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

  1. $111.11\%$

  2. $222.22\%$

  3. $333.33\%$

  4. $77.77\%$

Show answer
Correct Option: D
Explanation:

Amount invested $= 1500$

Amount received after three years $= 5000$

$\Rightarrow$ Interest earned $= 5000-1500 = 3500$
Let interest rate$\%$ be $r$
$\Rightarrow$ $3500 = \dfrac{1500 \times 3 \times r}{100}$
$\Rightarrow$ $r = 77.77\%$

You borrow Rs. $4,000$ from a loan shark. If you owe Rs. $7,200$ in $4$ years, what would be the simple interest rate?

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

  1. $10\%$

  2. $20\%$

  3. $30\%$

  4. $40\%$

Show answer
Correct Option: B
Explanation:

Given:
Principal $=$ Rs. $ 4,000$
Interest $= 7200 - 4000 = 3200$
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$\Rightarrow R =$ $\dfrac{3200\times 100}{4000\times 4}$
$\Rightarrow R = 20\%$

Thus, the simple interest rate is $20\%$.

John earned Rs. $100$ as simple interest on Rs. $600$ for $6$ months. Find the annual rate of interest.

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

  1. $11.11\%$

  2. $22.22\%$

  3. $33.33\%$

  4. $44.44\%$

Show answer
Correct Option: C
Explanation:
Given, Simple interest $=$ Rs. $100$, Principal $=$ Rs. $600$ and period $6$ months
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$R =$ $\dfrac{100\times 100}{600\times 0.5}$
$R = 33.33\%$
Therefore, the annual rate of interest is $33.33\%$.

Sharmila got a Rs. $1300$ loan for $5$ years. She paid Rs. $100$ in interest. What was the interest rate?

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

  1. $1.1\%$

  2. $1.3\%$

  3. $1.5\%$

  4. $1.7\%$

Show answer
Correct Option: C
Explanation:
Given, principal $=$ Rs. $1300$, Interest $=$ Rs. $100$, period $=5$ years
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$\Rightarrow R =$ $\dfrac{100\times 100}{1300\times 5}$
$\Rightarrow R = 1.5\%$
Therefore, the interest rate is $1.5\%$.

Find rate, when principal $=$ Rs. $3000$; interest $=$ Rs. $400$; time $= 3$ years.

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

  1. $1.11\%$

  2. $2.22\%$

  3. $3.33\%$

  4. $4.44\%$

Show answer
Correct Option: D
Explanation:

Given, Principal $=$ Rs. $3000$, interest $=$ Rs. $400$, time $=3$ years
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$R =$ $\dfrac{400\times 100}{3000\times 3}$
$R = 4.44\%$

Oscar earned Rs. $2400$ as simple interest on Rs. `$4500$ for $3$ months. What is the annual rate of interest?

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

  1. $16.66\%$

  2. $17.77\%$

  3. $18.88\%$

  4. $19.99\%$

Show answer
Correct Option: B
Explanation:
Given, principal $=4500$, Interest $=$ Rs. $2400$, period $=3$ months
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$\Rightarrow R =$ $\dfrac{2400\times 100}{4500\times 3}$
$\Rightarrow R = 17.77\%$
Therefore, the annual rate of interest is $17.77\%$.

To start a grocery shop, a woman borrowed Rs. $1,500$. If the loan was for four years and the amount of interest was Rs. $150$, what simple interest rate was she charged?

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

  1. $1.5\%$

  2. $2.5\%$

  3. $3.5\%$

  4. $4.5\%$

Show answer
Correct Option: B
Explanation:
Given, principal $=$ Rs. $1500$, Interest $=$ Rs. $150$, period $=4$ years
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$\Rightarrow R =$ $\dfrac{150\times 100}{1500\times 4}$
$\Rightarrow R = 2.5\%$
Therefore, the simple interest is $2.5\%$.

Mr. Joshua borrowed Rs. $5000$ for $2$ years to make home improvements. If he repaid a total of Rs. $15,000$, at what interest rate did he borrow the money?

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

  1. $15\%$

  2. $25\%$

  3. $35\%$

  4. $45\%$

Show answer
Correct Option: B
Explanation:

Given:
Principal $= 5,000$
Interest $= 15000 - 5000 = 10000$
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$\Rightarrow R =$ $\dfrac{5000\times 100}{10000\times 2}$
$\Rightarrow R = 25\%$

Peter invests $ $5,000$ at $4$% simple annual interest. How much in his investment worth after $2$ months?

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

  1. 5323.66

  2. 5033.33

  3. 5066.33

  4. 5343.67

Show answer
Correct Option: B
Explanation:

If the principal amount is $p$, annual interest is $r$ $\%$ then for simple interest after $n$ years the amount will be $p\left (1+\dfrac {nr}{100}\right)$.
Here $p=5000$, $n=\dfrac {1}{6}$, $r=4$ $\%$
So, the amount after $2$ months is $5000\left (1+\dfrac {2}{300}\right) = 5000\left (\dfrac {151}{150}\right) = 5033.33$

The difference between compound interest and simple interest on an amount of Rs. $15,000$ for $2$ years is Rs. $96$. 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. $8$

  2. $10$

  3. $12$

  4. Cannot be determined

  5. None of these

Show answer
Correct Option: A
Explanation:

$\left [15000\times \left (1 + \dfrac {R}{100}\right )^{2} - 15000\right ] - \left (\dfrac {15000\times R\times 2}{100}\right ) = 96$
$\Rightarrow 15000 \left [\left (1 + \dfrac {R}{100}\right )^{2} - 1 - \dfrac {2R}{100}\right ] = 96$
$\Rightarrow 15000 \left [\dfrac {(100 + R)^{2} - 10000 - (200\times R)}{10000}\right ] = 96$
$\Rightarrow R^{2} = \left (\dfrac {96\times 2}{3}\right ) = 64$
$\Rightarrow R = 8$.
$\therefore Rate = 8$%.

Ramesh borrowed $Rs \ 14000$ from a bank on simple interest for a period of $5$ years. He returned $Rs \ 6000$ to the bank at the end of three years and $Rs \ 10,900$ at the end of the five years and closed the account. 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. $4 \%$

  2. $4.14 \%$

  3. $6 \%$

  4. $9 \%$

Show answer
Correct Option: B
Explanation:

We have,

$P=Rs.\ 14000$

Ramesh returned $Rs.\ 6000$ to the bank at the end of $3$ years.

And returned $Rs.\ 10, 900$ to the bank at the end of $5$ years.

So, the total amount returned by Ramesh is $6000+10,900=Rs.\ 16, 900$

$T=5\ years$
$R=?$

We know that
$A-P=\dfrac{PRT}{100}$ 

So,

$16900-14000=\dfrac{14000\times R\times 5}{100}$ 

$2900=140\times R\times 5$ 

$R=4.14\%$

Hence, this is the answer.

Find the simple interest rate applied to a principal over $8$ years if the total interest paid equals the borrowed principal.

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

  1. $10\%$

  2. $20\%$

  3. $12.5\%$

  4. $15\%$

Show answer
Correct Option: C
Explanation:

We have given that interest and principal are equal.

Let Rs. $x$ be the interest and principal.
Here $T=8$ years
We know $S.I.=\dfrac{P\times R\times T}{100}$
$\Rightarrow$ $x=\dfrac{x\times R\times 8}{100}$
$\Rightarrow$ $R=\dfrac{100}{8}=12.5\%$
Simple interest rate is $12.5\%$.

Find the simple interest rate applied to a principal over $5$ years if the total interest paid equals the borrowed principal.

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

  1. $10\%$

  2. $20\%$

  3. $30\%$

  4. $40\%$

Show answer
Correct Option: B
Explanation:

We have given that interest and borrowed principal are same.

Let $x$ be the interest and borrowed principal.
We know $S.I.=\dfrac{P\times R\times T}{100}$
$\Rightarrow$ $x=\dfrac{x\times R\times 5}{100}$
$\therefore$ $R=\dfrac{100}{5}=20\%$

What is the simple interest rate applied to a principal over $2.5$ years if the total interest paid equals the borrowed principal?

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

  1. $20\%$ 

  2. $30\%$

  3. $40\%$

  4. $50\%$

Show answer
Correct Option: C
Explanation:

We have given that interest and principal are equal.

Let Rs. $x$ be the interest and principal.
Here $T=2.5$ years
We know $S.I.=\dfrac{P\times R\times T}{100}$
$\Rightarrow$ $x=\dfrac{x\times R\times 2.5}{100}$
$\Rightarrow$ $R=\dfrac{100}{2.5}=40\%$
Therefore, simple interest rate is $40\%$.

Jenn borrowed Rs. $5,000$ for $5$ years and had to pay Rs. $1,500$ simple interest at the end of that time. What rate of interest did she pay?

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

  1. $5$

  2. $6$

  3. $7$

  4. $8$

Show answer
Correct Option: B
Explanation:

$\Rightarrow$  $P=Rs.5000,\,T=5\,years $ and $S.I.=Rs.1500$

$\Rightarrow$  $S.I.=\dfrac{P\times R\times T}{100}$

$\Rightarrow$  $1500=\dfrac{5000\times R\times 5}{100}$

$\Rightarrow$  $R=\dfrac{1500}{250}$

$\therefore$    $R=6\%$

Joshita borrowed Rs. $3,000$ for $3$ years and had to pay Rs.$ 1,000$ simple interest at the end of that time. What rate of interest did she pay?

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

  1. $11.11$

  2. $22.22$

  3. $33.33$

  4. $44.44$

Show answer
Correct Option: A
Explanation:

$\Rightarrow$  $P=Rs.3000,\,T=3\,years$ and $S.I.=Rs.1000$.

$\Rightarrow$  $S.I.=\dfrac{P\times R\times T}{100}$

$\Rightarrow$  $1000=\dfrac{3000\times R\times 3}{100}$

$\Rightarrow$  $R=\dfrac{1000}{90}$

$\therefore$    $R=11.11\%$

What rate will amount to Rs. $1,331$ in three years (compounded annually), if the principal amount was Rs $1,000$ respectively? 

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

  1. $10$%

  2. $11$%

  3. $12$%

  4. $13$%

Show answer
Correct Option: A
Explanation:

We have,

$A=Rs.\ 1331$
$P=Rs.\ 1000$
$T=3\ years$
$R=?$

We know that
$A=P\left(1+\dfrac{R}{100}\right)^T$

Therefore,
$1331=1000\left(1+\dfrac{R}{100}\right)^3$

$\dfrac{1331}{1000}=\left(1+\dfrac{R}{100}\right)^3$

$\dfrac{11}{10}=1+\dfrac{R}{100}$

$\dfrac{R}{100}=\dfrac{1}{10}$

$R=10\%$

Hence, this is the answer.

If the simple interest on a certain sum of money is $\displaystyle \frac{1}{100}$th of the sum and the rate per cent equals 4 times the number of years, then 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. $1$%

  2. $2$%

  3. $3$%

  4. $4$%

Show answer
Correct Option: B
Explanation:

$SI=P/100; r=4t \implies t=r/4$

$SI=\cfrac { P\times r\times t }{ 100 } $
$\implies (P/100)=\cfrac { P\times r\times (r/4) }{ 100 } $
$\implies r\times (r/4)=1\ \implies r=2\%$

At simple interest a sum of money is double in 20 years. What is the rate of interest?

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

  1. 20

  2. 4

  3. 5

  4. 10

Show answer
Correct Option: C
Explanation:

For simple interest, the interest earned is determined by the formula: I = PRT

where I=interest, R=rate, T=time, P=principle
The rate where the interest equals the principle in 20 years. i.e., the amount doubles.
So we know, I=P and T=20 years
so we can solve for R
R=$\dfrac{I}{(P)(T)}$=$\dfrac{I}{P}\times\dfrac{1}{20}$=$1\times\dfrac{1}{20}$=0.05 per year or 5% per year interest.

Choose the correct answer from the alternatives given.
Sumit invested Rs. 24000 in a bank for three years. If the rate of interest for 1st, 2nd and 3rd year are 5%, 10% and 4% respectively and the interest is compounded annually, then how much money will be deposited in his account after three years?

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

  1. Rs. 27,472

  2. Rs. 28,828.80

  3. Rs. 3,45,240

  4. Rs. 5,46,280

Show answer
Correct Option: B
Explanation:

Required sum
= $24000 \left ( 1 + \frac{5}{100} \right )\left ( 1 + \frac{10}{100} \right )\left ( 1 + \frac{4}{100} \right )$
= $Rs.24000 \times 1.05 \times 1.10 \times $ $1.04$
= $Rs. 28828.80$
Hence, Amount after three years is $Rs. 28828.80$

A sum of Rs. $46,875$ was lent out as simple interest and at the end of $1$ year $8$ months the total amount was Rs. $50,000$. Find the rate of interest per cent per annum.

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

  1. $3.5\%$

  2. $4.5\%$

  3. $5\%$

  4. $4\%$

Show answer
Correct Option: D
Explanation:

$A=P(1+it)$
$50,000=46,875\left(1\times i\times 1 \frac{8}{12}\right)$
$i=0.04$; rate $=4\%$

Find the rate of interest if the amount owed after $6$ months is Rs. $1,050$, borrowed amount being Rs. $1,000$.

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

  1. $7\%$

  2. $8\%$

  3. $9\%$

  4. $10\%$

Show answer
Correct Option: D
Explanation:

We know $A=$P+ PIT$
i.e., $1050=1000+1000\times I\times 6/12$
$50=500I$
$I=1/10=10\%$

The simple and compound interest that can be earned in two year at the same rate is $Rs. 1500$ and $Rs. 1575$ respectively. What is the rate (% per annum) of interest?

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

  1. $8$

  2. $10$

  3. $12$

  4. $5$

Show answer
Correct Option: B
Explanation:
SI CI
$1^{st} Year$ $750$ $750$
$2^{nd} Year$ $750$ $825$

Required rate of interest $= \dfrac {75}{750}\times 100 = 10$

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

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

  1. $9\dfrac{1}{11}$ %

  2. $10\%$

  3. $11\%$

  4. $21\dfrac{9}{11}$ %

Show answer
Correct Option: D
Explanation:

Rs. 1000 + S.I. 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.I. on Rs. $100$ for $55$ months
S.I. on Rs. 100 for 55 months
$= Rs. 100$
$\therefore  Rate = (\cfrac{100\times 100\times 12}{100\times 55})$% $=21\cfrac{9}{11}$ %

The interest on a certain sum of money is $0.24$ times of itself in $3$ years .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. $5\%$

  3. $8\%$

  4. $7\%$

Show answer
Correct Option: C
Explanation:
Let p=100
Then I =24
Time=3 years
I=p*r*t/100
R=(24*100)/(100*3)
=8%p.a

On a certain Principal if the Simple interest for two years is $Rs.\ 4800$ and Compound interest for the two years is $Rs.\ 5088$, 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. $24\%$

  3. $12\%$

  4. $18\%$

Show answer
Correct Option: C
Explanation:
Given,  $SI=4800$
            $CI=5088$
Say the principal amount was $P$ and rate of interest is $r$% p.a for two years
$\therefore$   $SI=\dfrac { PT }{ 100 } =\dfrac { P\times 2\times r }{ 100 } $
$\Rightarrow 4800=\dfrac { 2Pr }{ 100 } $
$\Rightarrow 240000=Pr\quad \longrightarrow \left( 1 \right) $
Similarly $CI=P{ \left( 1+\dfrac { r }{ 100 }  \right)  }^{ 2 }-P$
$\Rightarrow 5088=P\left[ { \left( 1+\dfrac { r }{ 100 }  \right)  }^{ 2 }-1 \right] \quad \longrightarrow \left( 2 \right) $
Substituting $(1)$ in $(2)$ we get
$5088=\dfrac { 240000 }{ r } \left[ 1+\dfrac { { r }^{ 2 } }{ { 100 }^{ 2 } } +\dfrac { 2r }{ 100 } -1 \right] $
$\Rightarrow \dfrac { 5088 }{ 240000 } =\dfrac { r }{ { 100 }^{ 2 } } +\dfrac { 2 }{ 100 } $
$\Rightarrow 0.0212=\dfrac { r }{ { 100 }^{ 2 } } +0.02$
$\Rightarrow 0.0012=\dfrac { r }{ { 100 }^{ 2 } } $
$\therefore$    $r=12$%

At what rate per cent per annum will a sum of $Rs.\ 7500$ amount to $Rs.\ 8427$ in $2$ years compounded annually?

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

  1. $4$%

  2. $5$%

  3. $6$%

  4. $8$%

Show answer
Correct Option: C
Explanation:
We have,
$P=7500\ Rs$
$A=8427\ Rs$
$T=2$ years
$A=P\left (1+\dfrac {R}{100}\right)^T$

$7500+8427=7500\left (1+\dfrac {R}{100}\right)^T$

$\dfrac {8427}{7500}=\left (1+\dfrac {R}{100}\right)^2$

$\left (1+\dfrac {R}{100}\right)=\dfrac {2809}{2500}=\left (\dfrac {53}{50}\right)$

$\left (1+\dfrac {R}{100}\right)=\dfrac {53}{50}-1$

$\dfrac {R}{100}=\dfrac {53}{50}-1$

$R=\dfrac {3}{50}$

$R \ \% = 6\ \%$ 
Hence, this is the answer.

A sum of money compounded annually amounts to 1375 in 5 years and 1980 in 7 years. Find the annual rate of interest.

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

  1. 12%

  2. 20%

  3. 15%

  4. 10%

Show answer
Correct Option: B
Explanation:

interest is compounded, Amount $ A = P(1+ \frac {R}{100})^n $
So, for the first situation
$ 1375 = P \times (1+ \frac {R}{100})^5 $   - (1)

And for the first situation
$ 1980 = P \times (1+ \frac {R}{100})^7 $   ---- (2)

Dividing eqn 2 by eqn 1, we get
$ \frac {1980}{1375} =   (1+ \frac {R}{100})^2 $ 
$ => \frac {396}{275} =   (1+ \frac {R}{100})^2 $ 
$ => \frac {36}{25} =   (1+ \frac {R}{100})^2 $ 
$ => (1+ \frac {R}{100}) = \frac {6}{5} $
$ => \frac {R}{100} = \frac {1}{5} $
$ => R =20 $ %

The difference between the interest earned under compound interest, interest being compounded annually and simple interest for two years on the same sum and at the same rate of interest is 25.60. Find the sum if the rate of interest is 8% p.a

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

  1. 2000

  2. 2500

  3. 3200

  4. 4000

Show answer
Correct Option: D
Explanation:



Simple Interest $ SI = \frac {PNR}{100} $
So, $ SI = \frac {P \times 2 \times 8}{100} = Rs 0.16P $

When interest is compounded, Amount $ A = P(1+ \frac {R}{100})^n $
So, A $ = P \times (1+ \frac {8}{100})^2 = Rs  1.1664P  $
And $ CI = A - P = 0.1664P $

Si, difference $ CI - SI = Rs 0.1664P - Rs 0.16P = Rs 25.60 $
$ => 0.0064P = 25.60 $
$ => P = Rs  4000 $

If the simple interest on a certain sum of money is $\displaystyle \frac{1}{100}$th of the sum and the rate per cent equals the number of years, then 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. $2$%

  2. $1$%

  3. $3$%

  4. $4$%

Show answer
Correct Option: B
Explanation:
Given $SI=\cfrac { 1 }{ 100 } \times P,R=T$
$SI=\cfrac { PRT }{ 100 } $
$\cfrac { P }{ 100 } =\cfrac { P{ R }^{ 2 } }{ 100 } \Rightarrow R=T=1$

The interest on a certain sum of money is $0.18$ times of itself in $3$ years. 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. $5$

  3. $6$

  4. $7$

Show answer
Correct Option: C
Explanation:

$I=Interest$

$P=Principal$
$R=Rate of Interest$
$T=Time(in years)$
$0.18I=\dfrac { I\times R\times 3 }{ 100 } \ \Rightarrow R=\dfrac { 0.06\times 100 }{ 3 } =0.06\times 100\$
 $\Rightarrow R=6$ %
Rate of Interest $= 6$%

Principal $= 2500$, rate $= 6$%, time$= 4$ years. Calculate the interest.

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

  1. Rs.$500$

  2. Rs.$550$

  3. Rs.$600$

  4. Rs.$650$

Show answer
Correct Option: C
Explanation:

$Interest=\dfrac { P\times R\times T }{ 100 } \ =\dfrac { Rs.2500\times 6\times 4 }{ 100 } \ =Rs.600$

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