Tag: mathematics and statistics

Questions Related to mathematics and statistics

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 } $
$\implies r\times (r/4)=1\ \implies r=2\%$