Tag: introduction to interest

Questions Related to introduction to interest

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

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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

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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. $6$%

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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. $30\%$

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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. $77.77\%$

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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. $12$%

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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. $8.25\%$

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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%

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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

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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$%