FORMULAE ?
R = C + NP + OE
GP = NP + OE
M = NP + OE
BEV = C + OEMAX
MD = R - BEV(SOALAN YG ADA LOSS )
MD =[ NP = L (1 -r) ]
Thursday, September 25, 2008
discount
FORMULAE?
TRADE DISCOUNT NP = L ( 1 - r )
CHAIN DISCOUNT NP = L (1-r1)(1-r2)(1-r3)
SINGLE DISCOUNT EQUIVALENT r= 1- (1-r1)(1-r2)(1-r3)
TRADE DISCOUNT NP = L ( 1 - r )
CHAIN DISCOUNT NP = L (1-r1)(1-r2)(1-r3)
SINGLE DISCOUNT EQUIVALENT r= 1- (1-r1)(1-r2)(1-r3)
Monday, September 22, 2008
INTEREST QUESTION
OCT 2003/MAT 108
Q.5 On 20th June 2003 Laila invested RM2000 in an account that offered
a simple interest rate of 6.5%. What is the accumulated amount in the
account on 26th October 2003 by using Banker’s Rule?
solution↙
20 June → 30 June = 10 days
1 July → 31 July = 31 days
1 August → 31 August = 31 days
1 Sept → 30 Sept = 30 days
1 Oct → 26 Oct =26 days
TOTAL DAYS :128 days
Use the : I = Prt
I = 2000 + (6.5%) + (128⁄360)
I = 2000 + (0.065) + (0.35)
I = RM46.22
The accumulated amount:
S = I + P
=46.22 + 2000
=RM2046.22
Q.5 On 20th June 2003 Laila invested RM2000 in an account that offered
a simple interest rate of 6.5%. What is the accumulated amount in the
account on 26th October 2003 by using Banker’s Rule?
solution↙
20 June → 30 June = 10 days
1 July → 31 July = 31 days
1 August → 31 August = 31 days
1 Sept → 30 Sept = 30 days
1 Oct → 26 Oct =26 days
TOTAL DAYS :128 days
Use the : I = Prt
I = 2000 + (6.5%) + (128⁄360)
I = 2000 + (0.065) + (0.35)
I = RM46.22
The accumulated amount:
S = I + P
=46.22 + 2000
=RM2046.22
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