Tag: mathematics and statistics

Questions Related to mathematics and statistics

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