Quantitative Aptitude · Commerce Accountancy

Interest and Annuities

638 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 banks and simple interest introduction to interests introduction to interest introduction to interest payments

At what rate per cent per annum will Rs.3000 amount to Rs.3993 in 3 years, if the interest is compounded annually ?

  1. 9 % p.a.

  2. 10 % p.a.

  3. 12 % p.a.

  4. 15 % p.a.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 A = Rs.3993, P = Rs.3000, n = 3, r = ?
$\displaystyle \therefore A=P\left ( 1+\frac{r}{100} \right )\Rightarrow 3993=3000\left ( 1+\frac{r}{100} \right )^{3}\Rightarrow \frac{3993}{3000}=\left ( 1+\frac{r}{100} \right )^{3}\Rightarrow \frac{1331}{1000}=\left ( 1+\frac{r}{100} \right )^{3}$
$\displaystyle \Rightarrow \left ( \frac{11}{10} \right )^{3}=\left ( 1+\frac{r}{100} \right )^{3}\Rightarrow 1+\frac{r}{100}=\frac{11}{10}\Rightarrow \frac{r}{100}=\frac{11}{10}-1=\frac{1}{10}$
$\displaystyle \therefore r=\frac{100}{10}=10\%: : p.a.$

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

Rs. 8000 invested at compound interest gives Rs.1261 as interest after 3 years. The rate of interest per annum is

  1. 25 %

  2. 17.5 %

  3. 10 %

  4. 5 %
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

P = Rs.8000 C.I. = Rs. 1261

$\displaystyle \Rightarrow Amount=Rs.9261, n=3, r=?$
$\displaystyle \therefore 9261=8000\left ( 1+\cfrac{r}{100} \right )^{3}$
$\Rightarrow \left ( 1+\cfrac{r}{100} \right )^{3}=\cfrac{9261}{8000}=\left ( \cfrac{21}{20} \right )^{3}$
$\displaystyle \Rightarrow 1+\cfrac{r}{100}=\cfrac{21}{20}$
$\Rightarrow \cfrac{r}{100}=\cfrac{21}{20}-1=\cfrac{1}{20}$
$\Rightarrow r\cfrac{100}{20}\%=5\%p.a.$

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

The difference between compound interest and simple interest at the same rate on Rs.5000 for 2 years is Rs.72 What is the rate of interest per annum ?

  1. 20

  2. 15

  3. 12

  4. 10
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let the rate per cent p.a.be r. Then,
$\displaystyle S.I.=Rs.\left ( 5000\times \cfrac{r}{100}\times 2 \right )=Rs.100r$
$\displaystyle C.I.=Rs.\left [ 5000\left ( 1+\cfrac{r}{100} \right )^{2}-5000 \right ]$
$=Rs.5000\left [ \left ( 1+\cfrac{r}{100} \right )^{2}-1 \right ]$
$=Rs.5000\left [ \left ( 1+\cfrac{r^{2}}{10000}+\cfrac{2r}{100} \right )-1 \right ]$
$\displaystyle =Rs.5000\left ( \cfrac{r^{2}}{10000}+\cfrac{r}{50} \right )=Rs.\cfrac{5000(r^{2}+200r)}{10000}$
$=Rs.\left ( \cfrac{r^{2}}{2}+100r \right )$
$\displaystyle \therefore C.I.-S.I.=72$
$\displaystyle \Rightarrow \cfrac{r^{2}}{2}+100r-100r=72$
$\Rightarrow \cfrac{r^{2}}{2}=72$ 
$\Rightarrow r^{2}=144$
$\Rightarrow r=12\%\: \: p.a.$
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

A sum of money amounts to Rs.4840 in 2 years and Rs.5324 in 3 years at compound interest compounded annually. What is the rate of interest per annum ?

  1. 8

  2. 10

  3. 12

  4. 15
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 Let the principal be Rs.P and rate of interest p.a. = r% Then
$\displaystyle P\left ( 1+\frac{r}{100} \right )^{2}=4840...........(i)$ and $\displaystyle P\left ( 1+\frac{r}{100} \right )^{3}=5324...........(ii)$
$\displaystyle \Rightarrow \frac{5324}{4840}=\frac{(1+r/100)^{3}}{(1+r/100)^{2}}\Rightarrow 1+\frac{r}{100}=\frac{1331}{1210}$
$\displaystyle \Rightarrow \frac{r}{100}=\frac{1331}{1210}-1=\frac{121}{1210}=\frac{1}{10}\Rightarrow r=\frac{1}{10}\times 100=10\%: p.a.$

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

A certain sum of money amounts to $\displaystyle \frac {5}{4}$ of itself in 5 years. The rate percent per annum is

  1. 5%

  2. 7%

  3. 9%

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

R $\times$ T = 100 $\times$ (N - 1)
$R \times 5 = 100 \times \left ( \displaystyle \frac {5}{4} - 1 \right )$
$R \times 5 = 100 \times \displaystyle \frac {1}{4}$
$R = 5\%$

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

Madhav lent out Rs. 7953 for 2 years and Rs. 1800 for 3 years at the same rate of simple interest. If he got Rs. 2343. 66 as total, then find the percent rate of interest.

  1. 11%

  2. 12%

  3. 12.5%

  4. 5%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that Simple Interest $ = \cfrac {PNR}{100} $
Given,  $ \cfrac {7953 \times 2 \times R}{100}  + \cfrac {1800  \times 3 \times R}{100}  = Rs 2343.66 $
$=> 159.06R + 54R = Rs 2343.66 $
$ => 213.06R = 2343.66 $
$ => R = 11 \%$ 

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

Madhav lent out Rs $7953$ for $2$ years and Rs $1800$ for $3$ years at the same rate of simple interest. If he hot Rs $2343.66$ as total interest then find the percent rate of interest.

  1. $11\%$
  2. $12\%$
  3. $12.5\%$
  4. $5\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$S.I = \dfrac{P\times t\times r}{100}$


Let the rate be $r$ Principal $P$ and time $t$

${S.I} _{1}=\dfrac{7953\times 2\times r}{100}$


${S.I} _{2}=\dfrac{1800\times 3\times r}{100}$
$Total$ $simple$ $interest = {S.I} _{1}+{S.I} _{2} $

$2343.66=\dfrac { 7953\times2\times r }{ 100 } +\dfrac { 1800\times3\times r }{ 100 }$ 
$2343.66 = 213.06\times r$
$r = 11\%$

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

If the interest is payable quarterly, Rs. $1600$ amounts to Rs. $2662$ after $1\dfrac{1}{2}$ years, the annual rate of interest is

  1. $5\%$
  2. $10\%$
  3. $20\%$
  4. $35\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Amount $= Rs. 2662$
Principal $= Rs. 1600$
Time $= 1.5$ years $= 6$ quarters
$A = P\left(1 + \cfrac{R}{100}\right)^T$
$2662 = 1600 \left(1 + \cfrac{R}{100}\right)^6$
$1.088 = 1 + \cfrac{R}{100}$
$R = 8.8\%$
Hence, annual rate of interest $=8.8\times 4 = 35\%$

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

Rajan borrowed Rs. $50,000$ from Rakesh at simple interest. After $3$ years, Rakesh got Rs. $3000$ more than what he had given to Rajan. What was the rate of interest per annum?

  1. $2\%$
  2. $5\%$
  3. $8\%$
  4. $10\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rate $= \displaystyle \left ( \frac{100 \times 300}{5000 \times 3} \right )$% = 2%

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

During a period of two years, a principal of Rs. $100$ amounts to Rs. $121$ at the annual compount rate of $r\%$. The value of $r$ will be

  1. $9$
  2. $10$
  3. $\displaystyle \frac{21}{2}$
  4. $11$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, Sum (P) = Rs. 100,
Amount due (A) = Rs. 121
Time (n) = 2 years, Rate (r) = ?
We know $A = P \displaystyle \left ( 1 + \frac{r}{100} \right )$
$\therefore 121 = 100 \displaystyle \left ( 1 + \frac{r}{100} \right )^2$
or $\displaystyle \left ( 1 + \frac{r}{100} \right )^2 = \frac{121}{100}$
or $\displaystyle 1 + \frac{r}{100} = \frac{11}{10} = 1 + \frac{1}{10} = 1 + \frac{10}{100}$
$\therefore r = 10$%

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

Rajan lent Rs. $1200$ to Rakesh for $3$ years at a certain rate of interest and Rs. $1000$ to Mukesh for the same time at the same rate. If he gets Rs. $50$ more from Rakesh than from Mukesh, then the rate percent

  1. $ 8 \displaystyle \frac{1}{3}\%$
  2. $ 6 \displaystyle \frac{2}{3}\%$
  3. $ 10 \displaystyle \frac{1}{3}\%$
  4. $ 9 \displaystyle \frac{2}{3}\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle \frac{1200 \times R \times 3}{100} - \frac{1000 \times R \times 3}{100} = 50$
or $6 R = 50$
or $R = \displaystyle 8 \frac{1}{3}$%

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

A sum of Rs. $1000$ is lent to be returned in $11$ monthly installments of Rs. $100$ each, interest being simple. The rate of interest

  1. 9$\displaystyle \frac{1}{11}\%$
  2. $10\%$
  3. $11\%$
  4. $21\displaystyle \frac{9}{11}\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Rs. 1000 + S.l. on Rs. 1000 for 11 months = Rs. 1000 + S.I. on Rs. 100 for (1 + 2 + 3 + 4 + ... + 10) months Rs. 1000 S.I. on Rs. 100 for 100 months 
= Rs.1000 + S.l. on Rs. 100 for 55 months
S.l. on Rs. 100 for 55 months = Rs. 100
$\therefore Rate = \displaystyle \left ( \frac{100 \times 100 \times 12}{100 \times 55} \right )$% $21 \displaystyle \frac{9}{11}$%

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

A sum of money at compound interest amounts to Rs. $10580$ in $2$ years and to Rs. $12167$ in $3$ years. The rate of interest per annum is

  1. $12\%$
  2. $14\%$
  3. $15\%$
  4. $\displaystyle 16 \frac{2}{3}$%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Interest on Rs. 10580 for 1 year = Rs. (12167 - 10580) = Rs. 1587
$\therefore Rate = \displaystyle \left ( \frac{100 \times 1587}{10580} \right )$% = 15%

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

The compound interest on a sum of money for two years is Rs 52 and the simple interest for two years at the same rate is Rs 50. Then the rate of interest is

  1. 6%

  2. 8%

  3. 9%

  4. 10%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,
Simple Interest for two years is 50
$\frac { P\times T\times R }{ 100 } =Simple\quad Interest$
$\frac { P\times 2\times R }{ 100 } =50$
 $PR=2500$
 $P=\frac { 2500 }{ R } $................EQ(1)
Compound Interest for two years will be 52
$P{ { { \left( 1+\frac { r }{ 100 }  \right)  }^{ 2 } } }-P=52$
$\Rightarrow P\left( 1+\frac { 2R }{ 100 } +\frac { { R }^{ 2 } }{ 10000 } -1 \right) =52$
 $\Rightarrow P\left( \frac { 2R }{ 100 } +\frac { { R }^{ 2 } }{ 10000 }  \right) =52$
$\Rightarrow P\left( \frac { 200R+{ R }^{ 2 } }{ 10000 }  \right) =52$
$\Rightarrow P\left( \frac { 200R+{ R }^{ 2 } }{ 10000 }  \right) =52$
$\Rightarrow P\left( 200R+{ R }^{ 2 } \right) =520000$
$\Rightarrow 200R+{ R }^{ 2 }=\frac { 520000 }{ P } $
 $\Rightarrow 200R+{ R }^{ 2 }=520000\times \frac { R }{ 2500 } $(TAKING EQUATION FROM SIMPLE INTEREST EQ(1))
 $\Rightarrow 200R+{ R }^{ 2 }=208R$
$\Rightarrow { R }^{ 2 }=(208R-200R)$
 $\Rightarrow { { R }^{ 2 } }=8R$
 $R=8$
Rate will be 8%