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secondary 4 | E Maths
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Need help,how to get rid of the 100?
Date Posted: 4 years ago
Views: 389
5000(1 + a/100)^8 = 1842.85 + 5000
(1 + a/100)^8 = 1842.85/5000 + 1
1 + a/100 = (1842.85/5000 + 1)^⅛
a/100 = (1842.85/5000 + 1)^⅛ - 1
a = 100(1842.85/5000 + 1)^⅛ - 100
a = 4.000009 ≈ 4
(Actually if we used 4% exactly , the interest accumulated
= $5000(1 + 4/100)^8 - $5000
= $1,842.84525 ≈ $1842.85. and that is probably how this amount came about)
b)
4% per year
So per month it's 4% ÷ 12 = 0.333% or ⅓%(exact)
8 years = 8 x 12 months = 96 months
Total amount to pay after 8 years
= $5000 (1 + ⅓/100)^96
= $ 6,881.9756
= $6881.98 (rounded up to nearest cent)
(1 + a/100)^8 = 1842.85/5000 + 1
1 + a/100 = (1842.85/5000 + 1)^⅛
a/100 = (1842.85/5000 + 1)^⅛ - 1
a = 100(1842.85/5000 + 1)^⅛ - 100
a = 4.000009 ≈ 4
(Actually if we used 4% exactly , the interest accumulated
= $5000(1 + 4/100)^8 - $5000
= $1,842.84525 ≈ $1842.85. and that is probably how this amount came about)
b)
4% per year
So per month it's 4% ÷ 12 = 0.333% or ⅓%(exact)
8 years = 8 x 12 months = 96 months
Total amount to pay after 8 years
= $5000 (1 + ⅓/100)^96
= $ 6,881.9756
= $6881.98 (rounded up to nearest cent)
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