Shortcut -Compound Interest
Compound Interest:
Amount after nth term =$\dfrac{100+R}{100}$ Amount after (n-1)th term
Question:
What is the compound interest obtained for Rs. 16000 at 2% per annum for three years?
Answer: Amount at the end of 1st year = $\dfrac{102}{100}$ x 16000 = 16320
Amount at the end of 2nd year = $\dfrac{102}{100}$ x 16320 = 16646.4
Amount at the end of 3rd year = $\dfrac{102}{100}$ x 16646.4 = 16979.328
Interest obtained at the end of three years:
Interest = Amount – Principle
Interest = 16979.328 – 16000 = 979.328
Note:
In simple interest, the interest is calculated only for the principle. In compound interest, the interest is calculated for the principle as well as the interest obtained till the last term.
Question
What will be the ratio of simple interest earned by certain amount at the same rate of interest for 6years and that for 9years?
Answer:
For T=6,SI1=(P*R*6)/100
For T=9,SI2=(P*R*9)/100
SI1:SI2=6:9=2:3
Question
A person invested in all Rs.2600 at 4%,6% and 8% per annum simple interest.At the end of the year,he got the same interest in all the three cases.The money invested at 4% is:
Answer:At rate of interest 4% ,Let Principle=p1,SI1=x,Time =1year
x=(P1 * 4*1)/100=4P1/100--------> (1)
At rate of interest 6%,Let Principle =p2,SI2=x,Time=1year
x=(p2*6*1)/100=6P2/100----------->(2)
At rate of interest 8%.Let Principle =p3,SI3=x,Time=1year
x=(p3*8*1)/100=8p3/100------------>(3)
From equ(1) and (2)
p1/p2=3/2
p2=(2/3) *p1
From equ(1) and equ (3)
p1/p3=8/4=2
=>p3=(1/2)*p1
p3=2600-p1-p2
Substituting values of p2 and p3 we get p1=1200