SI & CI is basically a continuation of the application of percentage calculations. A well understanding and a rigorous practice will make you able to tackle the SI & CI questions easily and interestingly. Questions from SI and CI is common in most placement tests, bank PO, MAT,CAT and XAT competitive exams. This article provides important formulas, concepts, tips and tricks to solve aptitude questions on SI & CI.

This is one among the most calculation required area. So you should practice well in your basic computation skills and speed maths tricks. No need to by heart many formulae, even though the topic consisting a bit of formulae related approaches.

- Basic terminologies
- Simple Interest
- Net rate of Simple Interest
- Compound Interest

## Basic Terminologies

First of all, you have to familiar with some important and frequently usable terminologies. Most of them are familiar to you, because we are frequently using those terms in our day to day situations.

** Principal**: An amount of money that you lend to somebody or invest to earn interest.

** Interest**: The extra money that you pay back when you borrow money or that you receive when you invest money.

** Time / Period**: Time or period for which is borrowed or invested denoted by 'n'

** Rate of interest**: Rate at which interest is calculated on principal. Normally it is a percentage value, denoted by 'r'

** Amount**: Principal + Interest (for a given period at a given rate of interest)

Denoted by 'A'

## Simple interest (SI)

This is one of the interest forms, when interest is calculated only on the principal and calculating uniformly through the intervals then the interest is called simple interest.

Simple Interest= pnr/100 |

Where p = principal

n = number of periods

r = rate of interest

Amount under SI = P + I

= P + pnr/100

=P [1 + nr/100 ]

Amount at the end of n^{th} year under Simple Interest = P [1 + nr/100 ] |

**Example: **

Find the simple interest on Rs. 6000 at 5% per annum for 4 yrs?

Solution:

P= Rs 6000

N=4 yrs

R =5% p.a(per annum)

SI = 6000 * 4 * 5/100

= 1200 Rs

**Quick calculation method (**Application of *net rate of simple interest***)**

SI on a particular principal at r% p .a for 'n' yrs is 'nr'% of principal

i.e.From the above example

P = Rs 6000

R = 5% p.a

N = 4 yrs

SI = (5 * 4)% of 6000

= 20 % of 6000

= Rs. 1200

In the given example, Rs. 1200 is 20% of Rs. 6000.

Hence, the net rate of SI is 20%.

Net rate of Simple Interest = (n * r) % |

## Compound interest

The idea of Compound Interest is basically the concept of Successive Variation. In a compound interest calculation, interest being calculated on the then amount, means in each period of time the base of the interest calculation is varying. This is the basic concept of successive variation.

**Illustrative Example**

Let P = Rs.100

r=10%

Let A_{n} is representing the amount at the end of n^{th} year.

A1 = P[1+10/100]

=100[1+10/100]

= 100 * 110/100

A2 = A1[110/100]

= [100 * 110/100] * 110/100

A3 = A2[110/100]

= [100 * 110/100 * 110/100] * 110/100

And so on.

Therefore the amount after 'n' years = p[1+r/100]n

Please note the following tabular illustration

Years | (P) Principal in each period | (CI) Interest (10%) | (P+CI) Amount at the end of the year |
---|---|---|---|

1st year
2nd year 3rd year 4th year |
100
110 121 133.1 |
10
11 12.1 13.31 |
110
121 133.1 146.41 |

When we are observing the increase in the Amount under compound interest in each year, there is also an increment of r% (i.e. 10%) in the interest amount per year, compare with the previous year's interest and the principal in each year also shows the same rate of increase.

i.e.Under the compound interest, the principal at the beginning of each year, compound interest reckoned in each year and the amount at the end of each year are increasing successively at the same rate.

**Useful results**

Let p=principal

R=rate of interest compound annually/ half yearly/ quarterly etc

N=number of periods

Amount after 'n' years = P[1+r/100]^{n} |

Compound Interest after 'n' years = Amount - principal

= P[1+r/100]^{n} - P

This section is simply a continuation of the first module, contains most of the compound interest related concepts.

The main objectives;

- Compounding more than once a year
- Compounded infinitely (continuously)
- Difference between SI and CI for two years
- Difference between SI and CI for three years
- Installments
- Application of Pascal's Triangle

## Compounding more than once a year

If the amount under compound interest is compounding more than once a year is slightly differ from the per annum compounding method.

P = principal

r = rate of interest per annum

If the interest being compounded half yearly, then the compounding is done twice a year. Then the rate of interest is r/2% in each half year.

If it is quarterly, then the compounding is done 4 times a year. Then the rate of interest is r/4 % in each quarter.

If it is per month, then the compounding is 12 times a year .Then rate of interest is r/12% in each month.

In general, if compounding is done by 't' times a year (ie, compounding in each 12/t months) at the rate of r% per annum(interest rate is given in annual basis) then in 'n' years the principal 'P' will be amounted to; A= p[1 + r/t/100] ^{nt} |

## Compounded continuously / infinitely / every moment

Normally an amount compounded in a finite period of time only. If the quantity is successively increased or decreased at a uniform rate, for an infinite period of times, then the calculation will have a convergence manner. In this calculation form the final quantity that we can find approximately.

If the compounding is done in a period is increased infinitely, then we can say that the compounding is done 'Every Moment'.

Suppose a principal (P) is compounded 'k' times in 'n' years, the amount becomes

Amount = p[1+(r/k)/100]^{nk}

= p[1 + 1/k * (r/100)/(r/100)]^{(k (nr))/100}

= p[1+(1/k)]^{knr/100}

Binomial Expansion:
When k → ∞,then; limk → ∞(1+1/k) |

=p.e^{nr/100}

Result:
A principal p compounded infinitelyfor 'n' years at r% p.a, Amount after 'n' years = |

**Example:**

**A sum amounted to Rs.27180 in 10 years at 10% p.a, interest is compounded continuously (infinitely) .Find the principal amount? **

- Rs.15000
- Rs.7500
- Rs.10000
- Rs.12000

Solution:

Amount = p.e^{nr/100}

p.e^{10 * 10/100} = 27180

p.e = p(2.718) = 27180

p=27180/2.718 = 10000

=Rs.10000

Ans: C

## Difference between SI and CI for 2 years

If a principal amount is given, then it is possible to find the difference between the SI fetch on the given principal for 2 yrs and the CI fetch on the same principal for 2 yrs at the same rate of interest .

The detailed method is given below.

Let principal = P

Rate of interest = r% p.a

For convenience in calculation assume R= r/100

SI for first year = PR

SI for 2^{nd} year = PR

Therefore the SI for two years together = 2PR

[Note: Under SI, interest amount is same in all the period]

CI for two years = P[1+R]^{2} - P

= P[1 + 2R + R^{2}] - P

= P + 2PR + PR^{2} - P

= 2PR + PR^{2}

Difference between the SI and CI for 2 years on the same principal at same rate of interest

= 2PR + PR^{2} - 2PR

= PR^{2}

=P[r/100]^{2}

Result: |

Difference between SI and CI for 2 years = **P[r/100] ^{2}**

## Difference between SI and CI for 3 years

Similar to the previous illustration:

SI for 3 years = 3PR

CI for 3 years = P[1+R]^{3} - P

= P[1+3R+3R^{2}+R^{3}] - P

(a + b)^{3}= a^{3} + 3a^{2}b + 3ab^{2} + b^{3} |

= P+3PR+3PR^{2} +PR^{3}- P

= 3PR+3PR^{2} + PR^{3}

Difference between SI and CI for 3 years

= 3PR+3PR^{2} + PR^{3} - 3PR

= 3PR^{2} + PR^{3}

= PR^{2}[3+R]

= P(r/100)^{2} [3+(r/100)]

Result: |

Difference between SI and CI on a certain sum for 3 years at the same rate of interest is ,

P(r/100)^{2} [3+(r/100)]

## Installments

**Basic concepts and Terminologies**

To increase the business shops are providing products under installment scheme. One of the most familiar installment facilities is called EMI (Equated monthly installments). In this facility customer can pay the purchasing price of the product through equal monthly payments. Normally this total payment is more than the actual cost price of the product because the addition of interest amount through the installment periods.

**Cash Price:**

Purchasing price of the article or the price at which the product is available in a lump sum manner.

**Down payment:** This is the amount which a customer has to pay initially.

**Installment:** This is the prefixed portion of the payment as per an installment scheme. This calculated monthly, quarterly, half yearly etc.

**Case I:****Installment under Simple Interest**

**Situation I:**

*Find the rate of interest, if the installment amount is given*

**Concept Review Questions:**

**An article is sold for Rs.1000 lump sum or Rs.500 cash down payment and 6 monthly installments of Rs.100 each. Find the rate of interest charged per month?**

- 25%
- 75%
- 80%
- 90%

Solution:

Cash price = Rs.1000

Down payment = 500

Installment = Rs.100 / month for 6 months

Total amount paid through installments = 6 * 100 = 600

Total amount paid = 500 + 600 = Rs.1100

Interest =1100 - 1000 = Rs.100

After the initial down payment the principal for the 1st month =1000 - 500 = Rs.500

After the first installment of Rs.100, the principal for 2nd month = 500 - 100 = Rs.400

Similarly

Principal for 3rd month = 400 - 100 = Rs.300

Principal for 4th month =300 - 100 = Rs.200

Principal for 5th month = 200 - 100 = Rs.100

Principal for 6th month =100 - 100 = Rs. 0

Total principal =Rs [500+400+300+200+100] = Rs.1500

The interest per month on Rs.1500 is Rs.100

SI =p * n * r/100

r= SI * 100/p * n

= 100 * 100/1500 * (1/12)

= 1200/15

= 80%

Hence the rate of interest per month = 80%

Ans: C

**A table is available for Rs.15000 cash lump sum or a down payment of Rs.5000 and 7 monthly installments of Rs.1500 each. Find the rate of interest charged per month?**

- 14.27%
- 13.25%
- 11%
- 15.58%

Solution:

Cash price (Amount at lump sum) = Rs.15000

Cash down payment = Rs.5000

Amount paid through 7 installments = 7 * 1500 = Rs.10500

Total amount paid as per installment plan = (5000+10500)

= Rs.15500

Interest charged = (15500-10000) = Rs.500

P1=Rs(15000-5000)=Rs.10000

P2=Rs(10000-1500)=Rs.8500

P3=Rs(8500-1500)=Rs.7000

P4=Rs(7000-1500)=Rs.5500

P5=Rs(5500-1500)=Rs.4000

P6=Rs(4000-1500)=Rs.2500

P7 =Rs(2500-1500)=Rs.1000

**Note**: The last installment of Rs.1500 consisting the last principal (Rs.1000) and interest (Rs.500)

Total principal = Rs.38500

Interest=Rs.500

Period of investment = 1 month or 1/12 years

SI=p * n * r/100

r=SI * 100/p * n

=500 * 100/38500 * (1/12)

=15.58%

Hence the rate of interest = 15.58%

Ans: D

**Situation II: **

**Principal/amount, rate of interest and period of installments are given. Find each installment.**

If the principal amount or cash price (purchasing lump sum amount) amounted to 'A' after 'n' years (periods). If the amount 'A' is repaid in 'n' equal annual (periodical) installments at 'r%' simple interest p.a (per period).

Then;

Let 'x' be the amount of each installment.

First installment 'x' paid after one year of purchase, throughout the remaining (n-1) year 'x' will be amounted to

x + x(n-1)r/100 = x[100 + (n-1)r/100]

Second installment of 'x' paid after 2 years, will amounted in (n-2) years to

x[100+(n-2)r/100]

Similarly third installment amounted to

x[100+(n-2)r/100] and so on

Then the nth(last) installment amounted to

x[100+(n-n)r/100] =x itself

These all amounts derived from the installments are favor to the customer. Sum of these individual amounts is equal to the actual amount that he needs to pay through installments.

A= x[100+(n-1)r 100+100+(n-2)r 100+ ... + 1] |

Or

**A=** x[100n+(n-1)n2 r/100]

x = 200A/200n+n(n-1)r

r = 200(A-nx)/x * n * (n-1)

**Case II: Installment under Compound Interest (CI)**

For understanding the method of installment calculation under CI, its require to understand the concept of **'***present value of an amount'.*

**Present**** value of a amount under CI**

If a principal 'P' amounted to 'A' in 'n' years at r% p.a compound interest, then

A = P[1+r/100]^{n}

the present value of 'A' is

**P =**A1+(r/100)n

**Installment (principal based method)**

If a sum 'P' is borrowed and repaid in 'n' equal installment at a compound interest of r% per period of installments of 'x' each

Then the principal amount P = sum of all present values of x

The present value of first installment of 'x' after 1 year = x(1+(r/100))^{1}

Similarly, present value of second installment=x(1+(r/100))^{2}

and so on

Then the present value of nth (last ) installment = x/(1+(r/100))^{n}

P=x(1+(r/100))^{1} +x(1+(r/100))^{2} + ------------------- + x(1+(r/100))^{n}

P = x[(100/100+r)^{1} + (100/100+r)^{2} + ---- + (100/100+r)^{n}] |

x =p/(100/100+r)^{1} + (100/100+r)^{2} + ---- + (100/100+r)^{n}

Or

We can simplify the above result further

Let us take the series separately

i.e. (100/100+r)^{1} + (100/100+r)^{2}+ ---- + (100/100+r)^{n}

this is the sum of 'n' terms of a GP with common ratio 100/100+r and first term 100/100+r

sum of 'n' terms = a[1-r/n1-r]

=100/100 + r/[(1 - (100/100+r)^{n})/(1-(100/100+r))]

= 100/100 + r * [(1 - (100/100 + r)^{n})/r/100+r]

=100/r * [1-(100/100+r)^{n}]

x = 100/r * [1-(100/100+r)^{n}]

Each installment = P * r / 100/ r * [1-(100/100+r)^{n}] Or P = x * 100/r * [1-(100/100+r) ^{n}] |

**Alternate method (Amount based method)**

An amount 'A' discharged through 'n' equal installments of 'x' each .if the interest is compounded 'r%' per period of installments, then;

First installment 'x' amounts in (n-1) years to = x * (1+(r/100))^{n-1}

Second installment 'x' amounts in n-2 years to = x * (1+(r/100))^{n-2}

And so on

Finally the last or nth installment 'x' amounts in n years to = x * (1+(r/100))^{n-n} = x

A=x+x(1 + (r/100))^{1}+ x(1+(r/100))2+ -----+x(1+(r/100))^{n-1}

A= x[1 + (1+(r/100))^{1} + (1+(r/100))^{2}+ -----+ (1 + (r/100))^{n-1}]

A = x[1+(100+r/100)^{1} + (100+r/100)^{2}+ -----+(100+r/100)^{n-1}] |

Or

A =100 x/r * [(100+r/100)^{n} - 1] |

X = Ar/[(100+r/100)n- 1]100

R = x/A * [(100+r/100)^{n} -1]

## Application of Pascal's triangle

*Pascal's triangle is a magical triangle* in the area of compound interest. There are lots applications of Pascal's triangle related to the computation of Compound Interest. It is very interesting and helpful computational tool.

*What is Pascal's triangle?*

It is a triangular array of binomial coefficients. It is named after French mathematician Blaise Pascal.

The triangle is developed in the following manner.

Each row starts and ends with '1' and rest numbers in each row is the sum of its two preceding row values (marked in the picture).

This triangle is basically developed for finding the relationship between the coefficients of the Binomial Expansion of multiple variables.

Example:

(a + b)^{0} = **1**

(a + b)^{1} = **1**a + **1** b

(a + b)^{2} = **1**a^{2} + **2** ab + **1** b^{2}

(a + b)^{3} = **1** a^{3} + **3** a^{2}b + **3** ab^{2} + **1** b^{3}

(a + b)^{4} = **1** a^{4} + **4** a^{3}b + **6** a^{2}b^{2} + **4** ab^{3} + **1** b^{4}

And so on.

*How we can use Pascal's triangle for solving Compound Interest related problems?*

Refer the following illustrated example:

A sum of Rs. 10,000 is invested at 8% per annum compound interested, interest is being compounded annually.

**Requirement 1: Find the amount at the end of two years.**

Solution:

P = Rs. 10,000

r = 8% p.a

Here we have to find out the Amount at the end of 2 years.

So, write the 2nd row of the Pascal's triangle, with sufficient space in between.

1 | 2 | 1 |
---|---|---|

P = 10,000 | 8% of 10,000 = 800 | 8% of 800 = 64 |

1 * 10,000 = 10,000 | 2 * 800 = 1,600 | 1 * 64 = 64 |

Hence the total amount at the end of 2 years = 10,000 + 1,600 + 64

= Rs. 11,664.

**Requirement 2: Find the interest reckoned in 2 years.**

For finding the total interest amount, add the second and third column total of the above table.

i.e. Total interest = 1,600 + 64

= Rs. 1,664

**Requirement 3: Find the difference between Si and CI for two years at Rs. 10,000 at the same rate of 8% per annum.**

For finding this difference in the interest amounts, just simply refer the third column total of the above table.

i.e. The required difference = Rs. 64

**Example 2: **

A sum of Rs. 20,000 invested at 10% p.a compound interest for three years, interested being calculated annually.

Arrange the 3rd row of the Pascal's triangle.

1 | 3 | 3 | 1 |
---|---|---|---|

P = 20,000 | 10% of 20,000 = 2000 | 10% of 2000 = 200 | 10% of 200 = 20 |

1 * 20,000 = 20,000 | 3 * 2000 = 6,000 | 3 * 200 = 600 | 1 * 20 = 20 |

**Requirement 1:** Total amount at the end of three years = 20,000 + 6,000 + 600 + 20

= Rs. 26,620

{Sum of all the column totals}

**Requirement 2:** Total Compound interest earned in three years = 6000 + 600 + 20

= Rs. 6620

{Sum of all the column totals except the first column}

**Requirement 3:** Difference between SI and CI for three years = 600 + 20

= Rs. 620

{Sum of all the column totals except the first two columns}