The interest on a certain sum of money for one year is $31.875. If the rate of interest was decreased by 0.5%, it would be necessary to invest $100 more to produce the same amount of interest. Find the original sum of money and the rate of interest at first.
Please show steps clearly.
maths quadratic eqt
2011-01-02 5:27 am
回答 (2)
2011-01-02 5:41 am
✔ 最佳答案
original sum of money = prate of interest at first = r
pr = 31.875
(p + 100)(r - 0.005) = 31.875
(p + 100)(1000r - 5) = 31875
(p + 100)(31875/p - 5) = 31875
(p + 100)(31875 - 5p) = 31875p
3187500 + 31375p - 5p^2 = 31875p
5p^2 + 500p - 3187500 = 0
p^2 + 100p - 637500 = 0
p = 750
r = 31.875/750 = 4.25%
2011-01-02 5:46 am
let original sum = x, original rate = y
x * (1+y) = 31875 -- 1
interest = 31875 - x
interest produced under 2nd condition = (x - 100) * (y - 0.005) = 31875 - x -- 2
with formula 1 and 2 we can solve the question
x + xy = 31875 -- 1
xy - 100y - 0.005x + 0.5 = 31875 - x --2
xy + x = 31875 +100y + 0.005x - 0.5 --3
formula 1 - 3
0 = 100 y + 0.005x + 0.5
0 = 20000y + x + 100
x = -(20000y + 100) --4
sub 4 into 1
.
.
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obtain a quadratic equation for y and solve for y using quadratic formula, sorry, can't be bother to do the rest.
x * (1+y) = 31875 -- 1
interest = 31875 - x
interest produced under 2nd condition = (x - 100) * (y - 0.005) = 31875 - x -- 2
with formula 1 and 2 we can solve the question
x + xy = 31875 -- 1
xy - 100y - 0.005x + 0.5 = 31875 - x --2
xy + x = 31875 +100y + 0.005x - 0.5 --3
formula 1 - 3
0 = 100 y + 0.005x + 0.5
0 = 20000y + x + 100
x = -(20000y + 100) --4
sub 4 into 1
.
.
.
obtain a quadratic equation for y and solve for y using quadratic formula, sorry, can't be bother to do the rest.
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