FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 1.23

Example 41 : If the difference between T.D. and B.D. on a sum due in 4 months at 3% p.a. is ` 10, find the
amount of the bills.

B.G. = B.D. – T.D. = 10; n 4 /12= = 31 yrs.; i= 1003 =0.03; A =?

B.D. = A ni, T.D. = Pni, B.G. = B.D. – T.D.= A ni – Pni = (A –P) ni

Now, (A – P) ni = 10or (A – P)^13. (0.03) = 10,

or, (A–P) =0.01^10 =^100

or, T.D. = 1000 (as T.D. = A – P.) Again T.D. = P ni,

or, P. 21 (0.03) = 1000 or, P=^1000 0.01

∴ P = 1,00, 000. Now, A = P.V. + T.D. = 1,00,000 + 1000 = 1,01,000
∴ required amount of the bill = ` 1,01,000.

Aliter :
A(ni)^2
B.G.= 1 ni+ ,

1 2

A. .03 3

(^101)
1 3 .03
(^) ×
(^)

• ×

, A = 1,01,000.

Example 42 : The T.D. and B.G. on a certain bill of exchange due after a certain time is respectively ` 50 and
Re. 0.50. Find the face value of the bill.

We know B.G. = Int. on T.D. or, 0,50 = 50 × n × i or, ni=^050 .50=0.01

Now, B.D. = T.D + B.G. or, B.D. = 50 + 0.50 = 50.50
Again B.D. int. on B.V. (i.e., A)

or, 50.50 = A. ni or, 50.50 = A.. (0.01) or, A=^5 0.010.50=^5050

∴ reqd. face value of the bill is 5050. Example 43 : A bill of exchange drawn on 5.1.2013 for 2,000 payable at 3 months was accepted on the
same date and discounted on 14.1.13, at 4% p.a. Find out amount of discount.
Unexpired number of days from 14 Jan to 8 April = 17 (J) + 28 (F) + 31 (M) + 8 (A) = 84 (excluding 14.1.13)

2013 is not a leap year, Feb. is of 28 days. B.D. 2000= ×365 10084 4× =18.41(after reduction)

Hence, reqd. discount = ` 18.41.
Drawn on period 5.1.13
Period 3
Nominally due on 5.4.13
Days of grace 3
Legally due 8.4.13