Tag: banks and simple interest

Questions Related to banks and simple interest

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