數學科問題............20點(急)

Q1)
20000 x (1 + 8%/4 )^(2x4) - 20000
= 20000 x (1.02)^8 - 20000
= 23433.1876 - 20000
= 3433.19 (2 d.p.)

Q2)
Let $Y be 該筆款項的數額.
Y(1 + 6%x3) + 140 = Y(1+6%)^3
1.18Y + 140 = Y(1.06)^3
140 = 1.191016Y - 1.18Y
140 = 0.011016Y
Y = 140 / 0.011016
Y = 12708.7872
該筆款項的數額 = $12,708.79 (2 d.p.)

Q3) a)
50000 + ( 50000 x 6%/2 ) x4
= 50000 + 50000 x 0.12
= 50000 + 6000
= 56000
在每期的期滿時, 把利息提取... 黃先生所得的本利和 = $56,000
Q3) b)
50000 x (1 + 6%/2)^4
= 50000 x (1.03)^4
= 50000 x 1.12550881
= 56275.4405
在每期的期滿時, 把利息加上本金... 黃先生所得的本利和 = $56,275.44 (2 d.p.)

Q1
2年後所得的複利息 :
20 000(1+8%)^(2*4) - 20000
=$3433.18762

Q2
let the principal be X
compound interest - simple interest = 140
{X(1+6%)^3-X}-X(6%)(3) = 140
0.191016X - 0.18X = 140
X=12708.78722


Q3a)
在每期的期滿時,把利息提取 = (simple interest problem)
黃 先生所得的本利和 :
50 000(6%/2)(4) + 50 000
=6000+50000
=56000

Q3b)
在每期的期滿時,把利息加上本金,用作下期的複利息計算之用 = compound interest problem

黃 先生所得的本利和 :
50000(1+6%/1)^4
=56275.4405