Can someone please help me with this problem;
500$ is invested part@ 9% per annum and the rest @ 11%. After one year, interest earned on the 9% investment was 20$ less than the 11% investment. How much was invested @ each rate?
Thanks for any help.
2007-03-25 18:46:13 · 4 answers · asked by Alyssa W 1 in Science & Mathematics ➔ Mathematics
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2007-03-25 18:52:25 · answer #1 · answered by tuf gal 2 · 0⤊ 0⤋
well, assuming the money was invested using simple intrest,
the two parts are $ 175 and $ 325
the formula for simple intrest is P*T*R/1000
where P is the principle
T is the time period
R is the rate of intrest
let one part of the investment be $ x...therefore the othr part is $ 500-x
use the formula given and compute the value of x, and therefore $500-x
ull get the intrest on the 1st part as
x * 9 *1/100
and the second part as
500-x * 11 *1/100
the first part as given is $20 less than the second
therefore
(x * 9 *1/100) = (500-x * 11 *1/100) - 20
i hope u can solve this linear equation
contact me if u hav ne problems
2007-03-26 01:58:59 · answer #2 · answered by kabini 2 · 0⤊ 0⤋
x + y = $500 (1)
0.11x - 0.09y = $20 (2)
You can solve by substitution or elimination. I prefer elination for this one. Multiply (1) by 9 and (2) by 100, giving
9x + 9y = $4500
11x - 9y = $2000
2007-03-26 02:03:35 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
N + E = 500. 0.09 N + 20 = 0.11 E. From here, it's just algebra.
2007-03-26 01:49:39 · answer #4 · answered by Anonymous · 0⤊ 0⤋