Aptitude Shortcuts Simple Interest and Compound Interest Problems- Download in PDF

Aptitude Shortcut methods and tricks for the problems related to Simple Interest and Compound Interest were given below. Candidates those who are preparing for Bank PO/Clerk and all other competitive Exams can also download this in PDF.
TYPE I:
 If the no. of years given in the question is 2 years: i.e.,N=2yrs only
 
Question:
Find the difference between the compound interest and simple interest on Rs.1200 at 7% per annum for 2 years?
GIVEN:
P=Rs. 1200
R=7%
N=2 yrs
NORMAL METHOD:
S.I. = PNR/100
S.I. = [1200×7×2] / 100 = Rs. 168
C.I. = {P × (1 +[R/100])N} – P
 ={1200×(1+[7/100])2} – 1200
       = Rs. 173.88
Therefore the difference = C.I. – S.I
                                       = 173.88 – 168
                    Difference = Rs. 5.88
SHORTCUT METHOD:
This applies only when the number of years given is 2yrs
Difference between C.I and S.I can be calculated by the following formula:
DIFFERENCE =PR2 / 1002
In this question,
Difference = [1200 × 7 ×7] / [100×100]
                 = Rs. 5.88
TYPE II:
If the number of years given in the question is more than 2 years i.e.,N>2
 
QUESTION:
Find the difference between C.I. and S.I on Rs. 1000 at 10% per annum for 3 years?
GIVEN:
P = Rs. 1000
R= 10%
N= 3years
NORMAL METHOD:
S.I. = PNR/100
S.I. = [1000×10×3] / 100 = Rs. 300
C.I. = {P × (1 +[R/100])N} – P
={1000×(1+[10/100])3} – 1000
       = Rs. 331
Therefore the difference = C.I. – S.I
                                       = 331 – 300
                    Difference = Rs. 31
SHORTCUT METHOD:
Difference between C.I and S.I for N>2yrs can be calculated by the following formulae:
Difference = [PR2/ 1002] × [N + (R/100)]
In the above question
Difference between C.I and S.I
[(1000×10×10) / 1002] × [3 + (10/100)] = 31
Therefore, the difference isRs. 31

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