Quantitative Aptitude

Simple and Compound Interest

3,394 Questions
All Simple interest and its inverse forms 768 Compound interest with annual compounding 742 Simple-interest growth and split sums 626 Difference between compound and simple interest 479 Half-yearly and quarterly compounding 252 Changing rates and fractional years 201 Installments (CI based) 190 Ratio of principal to interest 118 Population growth and depreciation 16

Simple and Compound Interest Questions

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

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

  1. 2 years

  2. 3 years

  3. 4 years

  4. 6 years
Reveal answer Fill a bubble to check yourself
C Correct answer
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


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

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

  1. $2$%
  2. $4$%
  3. $5$%
  4. $6$%
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
$\Rightarrow$   $R=\dfrac{100\times S.I.}{P\times T}$

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

$\Rightarrow$   $r^2=\dfrac{400}{25}$

$\Rightarrow$   $r=\dfrac{20}{5}=4\%$

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

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

  1. $11.11\%$
  2. $32.11\%$
  3. $33.33\%$
  4. $30\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

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

  1. $111.11\%$
  2. $222.22\%$
  3. $99.99\%$
  4. $77.77\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $R=77.77\%$
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

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

  1. $1$%
  2. $2$%
  3. $4$%
  4. $5$%
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $1\%$

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

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

  1. $8\dfrac { 1 }{ 3 }\%$
  2. $24\%$
  3. $\dfrac { 25 }{ 4 }\%$
  4. $8.25\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

  1. 9%

  2. 10%

  3. 10.5%

  4. 12%
Reveal answer Fill a bubble to check yourself
B Correct answer
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%

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

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.

  1. 5%

  2. 6%

  3. 7%

  4. 9%

  5. 15%
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$\Longrightarrow 120=114{ \left( 1+\cfrac { r }{ 12(100) }  \right)  }^{ 12 }\Longrightarrow { \left( \cfrac { 120 }{ 114 }  \right)  }^{ \cfrac { 1 }{ 12 }  }-1=\cfrac { r }{ 1200 } \Longrightarrow r=5.12\%\ \therefore r\approx 5\%$

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

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

  1. $7\dfrac{1}{2}\%$
  2. $8\dfrac{1}{3}\%$
  3. $10\%$
  4. $12\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
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


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

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

  1. $7\dfrac{3}{4}\%$
  2. $8\dfrac{3}{4}\%$
  3. $10\dfrac{1}{4}\%$
  4. $11\dfrac{2}{3}\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to question, we have:

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

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

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.

  1. $5\%$
  2. $2\%$
  3. $10\%$
  4. $12\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $10 \%$.
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

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

  1. $11\%$
  2. $9\%$
  3. $15\%$
  4. $18\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $15\%$

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

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

  1. $8$
  2. $64$
  3. $4$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Simple Interest $=\dfrac{PNR}{100}$

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

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

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

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

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

$\Rightarrow$  $R^2=64$

$\Rightarrow$  $R=8$

$\therefore$  $Rate=8\%$
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

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?

  1. 9

  2. 6

  3. 12

  4. 8
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

A sum of money doubled in $10$ years. The rate of interest per annum is?

  1. $20\%$
  2. $15\%$
  3. $18\%$
  4. $10\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the principal be P. Since the sum doubles in 10 years, the simple interest equals the principal (SI = P). Using the simple interest formula SI = (P * R * T) / 100, we substitute P = (P * R * 10) / 100, which simplifies to 100 = 10 * R, yielding R = 10%.