TNPSC Aptitude and Mental Ability – Compound Interest

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COMPOUND INTEREST | tnpsc maths

FORMULAE :

A = P [1 + (R / 100)]T

Where,
A = Amount
P = Principal
R = Rate of interest
T = Time period

Example :

What will be the compound interest on a sum of 50000 after 3 years at the rate of 12 % p.a?

Solution :

P = 50000, R = 12 %, n = 3 yr
A = P [1 + (R / 100)]T
= 50000 [1 + (12 / 100)]2
= 50000 [(28 / 25) * (28 / 25) * (28 / 25)]
= 3551232 / 5
= 70246.40

Answer => 70246.40

Note :
If the rate of interest for 1st year is r1 %, for 2nd year it is r2 % and for 3rd year it is r3 % then the amount for 3 years?
= P * [1 + (r1 / 100)] [1 + (r2 / 100)] [1 + (r3 / 100)]

Example :

The population of particular Area A of the city is 5000, it increases by 10 % in 1st year, and it is decreased by 20 % in the 2nd year because of some reason. In the 3rd year population increases 30 %, what will be the population of Area A at the end of the 3rd year?

Solution :

P =5000, R1 = 10 %, R2 = -20 %, R3 = 30 %
Population of area = P * [1 + (R1 / 100)] [1 + (R2 / 100)] [1 + (R3 / 100)]
= 5000 * [1 + (10 / 100)] [1 – (20 / 100)] [1 + (30 / 100)]
= 5000 * [11 / 10] [4 / 5] [13 / 10]
= 10 * 11 * 4 * 13
= 5720

Answer => 5720

Note :
Population after “n” yrs = P [1 + (R / 100)]n
Population n year ago = {P / [1 + (R / 100)]n}

Example :

The population of a city increases at the rate of 15 % p.a. If its population was 4000 at the end of year 2013, then what will be its population at the end of the year 2015?

Solution :

R = 15 %, n = 2 year, P = 4000
Population after 2 year = 4000 * [1 + (15 / 100)]2
= 4000 * [(23 / 20) * (23 / 20)]
= 10 * 529
= 5290

Answer => 5290

Example :

The population of a city increases at the rate of 5 % p.a. If the present population of the city is 370440, then what was its population 3 years ago?

Solution :

P = 370440, R = 5 %, n = 3 year
Population 3 years ago = {P / [1 + (R / 100)]n }
= {370440 / [1 + (5 / 100)3]}
= 370440 / [21 / 20]3
= (370440 * 8000) / (9261)
= 40 * 8000
= 320000

Answer => 320000

Note :
Shortcut method for calculating compound interest :

= [x + y + (xy / 100)] % of P
Let R = 10 % for 2 year,
= 10 + 10 + [(10 * 10) / 100]
= 20 + 1
= 21 % of P

Answer => 21 % of P

Note :
If a certain sum at Compound interest becomes A1 in “n” year and A2 in       (n + 1) year, then

Rate of CI = [(A2 – A1) / A1] * 100
Sum = [A1 * (A1 / A2)n]

Example :

A sum of money invested at compound interest amounts to 800 in 2 yr and 840 in 3 year. Find the rate of interest per annum and Sum?

Solution :

SI on 800 for 1 year = 840 – 800 = 40
SI = PRT / 100 = 40
= (800 * R * 1) / 100
R = 5 %
Amount = Sum [1 + (R / 100)]T
800 = x [1 + (5 / 100)]2
800 = (105 / 100) * (105 / 100) * (x)
800 = (21 / 20) * (21 / 20) * (x)
x = (800 * 20 * 20) / (21 * 21)
x = 320000 / 441
x = 725.62

Another method :

A1 = 800 , A2 = 840
Rate of CI = [(A2 – A1) / A1] * 100 %
= [(840 – 800) / 800] * 100 %
= 40 / 8 %
= 5 %
Sum = [A1 (A1 / A2)n]
= [800 (800 / 840)2]
= 800 * (800 / 840) * (800 / 840)
= 320000 / 441
Sum = 725.62

Note :
If the difference between the SI & CI on the same amount at “R %” for “2” year,
= P [r / 100]2
And if the difference between the SI and CI on the same amount at r % for 3 year,
= P [r2 / 1002] [3 + (r / 100)]

Example :

If the difference between the CI and SI on a certain sum for 2 year at 8 % per annum is 32, then the sum?

Solution :

n = 2 , R = 8 % , Difference = 32
Difference = P [r / 100]2
32 = p [82 / 1002]
P = (32 * 100 * 100) / (8 * 8)
P = 5000

Answer => 5000

Example :

If the difference between CI and SI on 1000 for 3 year at 5 % p.a

Solution :

P = 1000 , R = 5 %
= P [r2 / 1002] [3 + (r / 100)]
= 1000 [(5 *5) / (100 * 100)] [3 + (5 / 100)]
= 76.25

Answer = > 76.25

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