Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 1.13

A = P (1 + i)n or, 2P = P(1 + 0.05)n = P (1.05)n
or, 2 = (1.05)n or log 2 = n log 1.05

n log 2 0.3010 14.2 years (Approx)

∴ =log 1.05 0.0212= =

∴ (anti-logarithm table is not required for finding time).
[To find sum]
Example 24 : The difference between simple and compound interest on a sum put out for 5 years at 3%
was ` 46.80. Find the sum.
Let P = 100, i = .03, n = 5. From A = P (1 + i)n,
A = 100 (1 + .03)^5 = 100 (1.03)^5
log A = log 100 + 5 log (1.03) = 2 + 5 (.0128) = 2 + .0640 = 2.0640
∴ A = antilog 2.0640 = 115.9 ∴ C.I. = 115.9 – 100 = 15.9
Again S.I. = 3 × 5 = 15. ∴ difference 15.9 – 15 = 0.9

Diff. Capital x 100= ×^4 6.800.9 =5,200

0.9 10 0
46.80 x
∴ original sum = 5,200. [ To find present value] Example 25 : What is the present value of 1,000 due in 2 years at 5% compound interest, according as the
interest is paid (a) yearly, (b) half-yearly?

(a) Here A = ` 1,000, i= 1005 = 0.05, n = 2, P =?

A = P (1 + i)n or 1000 = P (1 + .05)^2 = P(1.05)^2

( )^2

P 1000 1000 907.03

∴ = 1.05 =1.1025=

∴ Present value = ` 907.03

(b) Interest per unit per half-year^12 ×0.05 = 0.025

From A

i 2n

=P 1^ + 2 we find.

1,000

0.052 2

P 1 2

×

= + = P (1 + .025)^4 = P (1.025)^4