FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 1.SOLVED EXAMPLES :
Example 10 : Amit deposited ` 1200 to a bank at 9% interest p.a. find the total interest that he will get at the
end of 3 years.
Here P = 1200, i= 1009 =0.09,n = 3, I =?
I = P. i.n = 1200 x 0.09 x 3 = 324.
Amit will get ` 324 as interest.
Example 11 : Sumit borrowed ` 7500 at 14.5% p.a. for^221 years. Find the amount he had to pay after that
period.
P = 7500, i=14.5 100 =0.145, n 2 2.5,A ?= 21 = =
A = P (1+ in) = 7500 (1+ 0.145 x 2.5) = 7500 (1+0.3625)
= 7500 × 1.3625 = 10218.
Reqd. amount = 10218.75. Example 12 : Find the simple interest on 5600 at 12% p.a. from July 15 to September 26, 2013.
Time = number of days from July 15 to Sept. 26
= 16 (July) + 31 (Aug.) + 26 (Sept.)= 73 days.
P = 5600, i= 10012 = 0.12, n= 36573 yr. =^15 yr.
S.I. = P. i.n. = 5600 x 0.12 x^15 =134.
∴ Reqd. S.I. = 134.40. (In counting days one of two extreme days is to be excluded, Usually the first day is excluded). To find Principle : Example 13 : What sum of money will amount to 1380 in 3 years at 5% p.a. simple interest?
Here A = 1380, n = 3, i= 1005 =0.05,P ?=
From A = P (1 + 0.05x3) or, 1380 = P (1+0.15)
Or, 1380 = P (1.15) or, P =^1380 1.15=^1200
∴ Reqd. sum = 1200 Example 14 : What sum of money will yield 1407 as interest in^121 year at 14% p.a. simple interest.
Here S.I. = 1407, n = 1.5, I = 0.14, P =?
S.I. = P. i.n. or, 1407 = P x 0.14 x 1.
Or, P=0.14 1.5 0.21^1 407 1407× = =^6700
∴ Reqd. amount = ` 6700.