Quantitative Aptitude · Commerce Accountancy

Interest and Annuities

621 Questions

Interest and annuities represent a critical quantitative aptitude section focusing on the mathematical calculation of simple interest, compound interest, and future values of investments. Questions challenge candidates to determine maturity values, compute recurring deposit returns, and calculate prevailing interest rates. Mastery of this topic is essential for scoring high in banking and SSC examinations.

Simple and compound interestFuture value of annuitiesRecurring deposit calculationsInterest rate determinationPresent value formulas

Interest and Annuities Questions

Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

A man deposits $Rs.\ 1200$ in a bank on the $1st$ day of each year. If the bank pays $5\%$ per annum compound interest on deposited sum of money, what will be the amount to his credit on the $10th$ day of the second year?

  1. $Rs.\ 2583$
  2. $Rs.\ 2460$
  3. $Rs.\ 2370$
  4. $Rs.\ 2860$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

The compound interest on Rs. $2000$ in $2$ years, if the rate of interest is $4 \%$ per annum for the first year and $3\%$ per annum for the second year will be 

  1. Rs. $142.40$
  2. Rs. $140.40$
  3. Rs. $141.40$
  4. Rs. $143.40$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $P=$ Rs. $2000$, $\displaystyle r _{1}$ $=4\%$ p.a. for $1^{st}$ year, $\displaystyle r _{2}$ $=3\%$ p.a. for $2^{nd}$ year
$\displaystyle \therefore A=P\left ( 1+\frac{r _{1}}{100} \right )\left ( 1+\frac{r _{2}}{100} \right )$

$=2000\left ( 1+\dfrac{4}{100} \right )\left ( 1+\dfrac{3}{100} \right )$
$\displaystyle =$ Rs. $\left ( 2000\times \dfrac{26}{25}\times \dfrac{103}{100} \right )=$ Rs. $2142.40$
$\displaystyle \therefore$ C.I. $=$ Rs. $2142.40 - $ Rs. $2000=$ Rs. $142.40$

Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

How much will Rs. $9,000$ amount to in $3$ years, at compound interest, if the rates for successive years are $12 \%$, $15 \%$ and $20 \%$ per year respectively?

  1. Rs. $13810.40$
  2. Rs. $13710.40$
  3. Rs. $13910.40$
  4. Rs. $12910.40$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$A=9000\times \left(1+\dfrac{12}{100}\right)\left(1+\dfrac{15}{100}\right)\left(1+\dfrac{20}{100}\right)$

$A=9000\times \dfrac{112}{100}\times \dfrac{115}{100}\times \dfrac{120}{100}$

$A=\dfrac{112\times 115\times 12\times 9}{100}$

$A=13910.40  Rs$

Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

Find the amount on Rs. $ 12,500$ for $2$ years compounded annually, the rate of interest being $15\%$ for the first year and $16\%$ for the second year.

  1. Rs. $16,500$
  2. Rs. $16,750$
  3. Rs. $16,675$
  4. None of these

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $P=$ Rs. $ 12500$,$R _1$ $=15\%$,$R _2 $ $=16\%$
$A = p \left(1 + \displaystyle \frac{R _1}{100} \right) \left(1 + \displaystyle \frac{R _2}{100} \right)$
$A = 12500 \times \left(1 + \displaystyle \frac{15}{100} \right) \times \left(1 \times \displaystyle \frac{16}{100} \right)$
$= 12500 \times \displaystyle \frac{115}{200} \times \frac{116}{100}$
$= $ Rs. $16,675$

Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

A sum of money double itself in 4 years at compound interest.! n how many years it will become eight times at same rate of interest

  1. 12 years

  2. 18 years

  3. 24 years

  4. 16 years

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle A = P \left( 1 + \frac {R}{100} \right)^T$

$\displaystyle 2x = x \left( 1 + \frac {R}{100} \right)^4$

$\displaystyle \left( 1 + \frac {R}{100} \right) = 2^{1/4}$

$\displaystyle 8x=x \left( 1 + \frac {R} {100}\right)^T$

$\displaystyle \because 1 + \frac {R} {100} = 2^{1/4}$

$\displaystyle  \therefore 8=2^{T/4}$

$\displaystyle 2^3 = 2^{T/4}$

$\displaystyle \frac {T}{4}=3$

$T=12\, years$

Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

Sam invested Rs. $15000$ at the rate of $10 \%$ per annum for one year. If the interest is compounded half-yearly, then the amount received by Sam at the end of the year will be:

  1. Rs. $16,500$
  2. Rs. $16,525.50$
  3. Rs. $16,537.50$
  4. Rs. $18,150$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$P =$ Rs. $15000$; $R =10 \%$ p.a $= 5 \%$ per half-year; $T= 1$ year $= 2$ half-years.
$\therefore$ Amount $=\, \left [ 15000\, \times\, \left ( 1\,+\, \displaystyle \frac{5}{100} \right )^2 \right ]$
$=$ Rs. $\left ( 15000\, \times\, \displaystyle \frac{21}{20}\, \times\, \frac{21}{20} \right )$
$= 16,537.50$

Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

There is a question followed by three statements. While answering the question, you may or may not require the data provided in all the statements. You have to read, the question and the three statements and then decide whether the question can be answered with any one or two of the statements or all the three statements are required to answer the question. Mr. Gupta borrowed a sum of money on compound interest. What will be the amount to be repaid if he is repaying the entire amount at the end of $2$ years?

$I$. The rate of interest is $5$ p.c.p.a.
$II$. Simple interest fetched on the same amount in one year is $Rs. 600$.
$III$. The amount borrowed is $10$ times the simple interest in $2$ years.

  1. $I$ only
  2. $III$ only
  3. $I$ or $II$ only
  4. None of these

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given 
I. The rate of interest is 5 p.c.p.a.
II. Simple interest fetched on the same amount in one year is Rs. 600.
III. The amount borrowed is 10 times the simple interest in 2 years.
Now | and || give the sum,
For this sum , C.I and hence amount can be obtained
Thus , ||| is redundant
Again || gives S.I for 2 years =$600\times 2=1200$
Now from ||| ,sum =$10\times 1200=12000$
Thus Rate=$\frac{100\times 1200}{2\times 1200}=5%$p.a
Thus ,C.I for 2 years and therefore ,amount can be obtained.
Thus ,| is redundent
Hence either  |  and II  or II and III are sufficient to answer the question 

Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

A sum of money invested at compound interest amount in $3$ years to Rs. $2400$ and in $4$ years to Rs. $2520$. The interest rate per annum is 

  1. $5 \%$
  2. $6 \%$
  3. $10 \%$
  4. $12 \%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the sum of money invested be Rs. $x$ and interest rate per annum $= r\%$
Then $\displaystyle x\left ( 1+\frac{r}{100} \right )^{3}=$ Rs. $2400$........(i)
and $\displaystyle x\left ( 1+\frac{r}{100} \right )^{4}=$ Rs. $2520$........(ii)
Dividing equation (ii) by (i), we get
$\displaystyle \left ( 1+\frac{r}{100} \right )=\frac{2520}{2400}$

$\Rightarrow \dfrac{r}{100}=\dfrac{2520-2400}{2400}=\dfrac{120}{2400}$
$\Rightarrow r=\dfrac{1}{20}\times 100=5\%$ p.a.

Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

Sanju puts equal amount of money one at $10\%$ per annum compound interest payable half yearly and the second at a certain rate percent annum compound interest payable yearly. If he gets equal amounts after $3$ years what is the value of the second rate percent?

  1. $\displaystyle 10\frac{1}{4}$ $\%$
  2. $10\%$
  3. $\displaystyle 9\frac{1}{2}$ $\%$
  4. $\displaystyle 8\frac{1}{4}$ $\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the amount $=x$ in Rs .

In the first case, the payment is made half yearly $@10\%$.
So the number of time in 3yrs=$3\div \dfrac { 1 }{ 2 } =6$ and

the rate $=$ $10 p.c.\div 2=5$ p.c.
So, in the firs case, the amount after $6$ time slots @ $5\%$ is
Principal $={ \left( 1+rate \right)  }^{ time }$
$ =x{ \left( 1+\dfrac { 5 }{ 100 }  \right)  }^{ 6 }$.
In the second case, the time $=3$ yrs,  
The principal $=x$ and let the rate $=y\%$ p.a.

So, the amount after $3$ yrs @ $y\%$
$=$ $x{ \left( 1+\dfrac { y }{ 100 }  \right)  }^{ 3 }$.
$\therefore $ By the given condition, we have
$=x{ \left( 1+\dfrac { y }{ 100 }  \right)  }^{ 3 }=x{ \left( 1+\dfrac { 5 }{ 100 }  \right)  }^{ 6 }\ \Rightarrow { \left( 1+\dfrac { y }{ 100 }  \right)  }^{ 3 }={ \left{ { { \left( 1+\dfrac { 5 }{ 100 }  \right)  } }^{ 2 } \right}  }^{ 3 }\ \Rightarrow { \left( 1+\dfrac { y }{ 100 }  \right)  }={ { \left( 1+\dfrac { 5 }{ 100 }  \right)  } }^{ 2 }=\dfrac { 441 }{ 400 } \ \Rightarrow \dfrac { y }{ 100 } =\dfrac { 441 }{ 400 } -1=\dfrac { 41 }{ 400 } \ \Rightarrow y=\dfrac { 41 }{ 4 } p.c.=10\dfrac { 1 }{ 4 } p.c..$.
So, the second rate $=$ $10\dfrac { 1 }{ 4 }$ p.c..

Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

Calculate the amount and the compound interest on Rs. $12,000$ in $3$ years when the rates of interest for successive years are $8\%$, $10\%$ and $15\%$ respectively. 

  1. Rs.$4294.40$
  2. Rs.$3634.40$
  3. Rs.$5394.40$
  4. None of these

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Sum$=Rs.12000$
Time$=3$ years
Rate of interest$=8\%,10\%,15\%$
Required amount, A = $\displaystyle P\left( 1+\frac { { r } _{ 1 } }{ 100 }  \right) \left( 1+\frac { { r } _{ 2 } }{ 100 }  \right) \left( 1+\frac { { r } _{ 3 } }{ 100 }  \right) $
$\displaystyle \Rightarrow \quad A=Rs.12000\left( 1+\frac { 8 }{ 100 }  \right) \left( 1+\frac { 10 }{ 100 }  \right) \left( 1+\frac { 15 }{ 100 }  \right) $
$\displaystyle \Rightarrow 12000\times \frac{108}{100}\times \frac{110}{100}\times \frac{115}{100}$
$\displaystyle \Rightarrow Rs.\quad 16394.40$
$\displaystyle C.I.=Rs.16394.40-Rs.12000=Rs.4394.40$
Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

What sum will amount to Rs. $6,593.40$ in $2$ years C.I., if the rates are $10$ per cent and $11$ per cent for the successive years?

  1. $5000$
  2. $5600$
  3. $5400$
  4. $6000$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow$  Let rate of interest for 2 years will be $R _1=10\%$ and $R _2=11\%$.

$\Rightarrow$  $A=P\times (1+\dfrac{R _1}{100})(1+\dfrac{R _2}{100})$

$\Rightarrow$  $6593.40=P\times (1+\dfrac{10}{100})(1+\dfrac{11}{100})$

$\Rightarrow$  $6593.40=P\times \dfrac{11}{10}\times \dfrac{111}{100}$

$\Rightarrow$  $6593.40=P\times \dfrac{1221}{1000}$

$\Rightarrow$  $P=\dfrac{6593.40\times 1000}{1221}=Rs.5400$

Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

A man lends Rs. $12,500$ at $12$% for the first year, at $15$% for the second year and at $18$% for the third year. If the rates of interest are compounded yearly; find the difference between the C.I. for the first year and the compound interest for the third year.

  1. Rs. $1,498$
  2. Rs. $1,598$
  3. Rs. $1,298$
  4. Rs. $1,398$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
For first year
$P=12500,R=12$%, $T=1$
Interest$\cfrac { PRT }{ 100 } =1500$
Amount$=P+I=14000$
For second year previous amount will be Principle
$P=14000,R=15,T=1$
Interest'$=\cfrac { 14000\times 15\times 1 }{ 100 } =2100$
Similarly for third year
Interest''$=\cfrac { 16100\times 18\times 1 }{ 100 } =2898$
Difference between $C{ I } _{ 3 }$ & $C{ I } _{ 1 }=2898-1500=1398$
Multiple choice mathematics and statistics interest applying compound interest compound interest formula with different successive rate of interest compound interest ( for different time period)

Mohit invests Rs. 8,000 for 3 years at a certain rate of interest, compounded annually. At the end of one year it amounts to Rs. 9,440. Calculate: the amount at the end of the second year.

  1. Rs. 15,729.50

  2. Rs. 13.079.80

  3. Rs. 12,367.50

  4. Rs. 11,139.20

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$P=Rs.8000$


Amount after one year $=Rs.9440$

Interest for 1 year$=9440-8000=Rs.1440$

let rate of interest $=R$

C.I for one year=S.I for 1 year$=\dfrac{PRT}{100}$

$\Rightarrow 1440=\dfrac{8000\times R\times 1}{100}$

$\Rightarrow R=\dfrac{1440\times 100}{8000}=18$%

For second year
$P=9440$
$R=18$%
$T=1$ year

$\therefore  Amount=P\left(1+\dfrac{R}{100} \right)^T$

$\Rightarrow 9440 \left(1+\dfrac{18}{100} \right)$

$\Rightarrow 9440\times \dfrac{118}{100}=Rs.  11139.20$

Hence Amount at the end of second year $=Rs.11139.20$