A sum of $3,500 is invested in two parts. One part brings a return of 5% and the other a return of 8%. The total annual return is $250. Find the amount invested at each rate.
How many pounds of seed worth $0.60 a pound must be mixed with 300 pounds of seed worth $0.35 a pound to produce a mixture worth $0.50 per pound?
ONLY 2 QUESTIONS i WIll give 20 points for the answers and work.?
2007-11-26 3:25 pm
回答 (2)
2007-11-26 3:38 pm
✔ 最佳答案
let 5% part =x, and the other part be yboth parts add to equal 250 i.e. x+y=3500 (1)
the annual returns of each part must add to equal 250
i.e. 0.05x+0.08y=250 (2)
0.08 * (1) ==> 0.08x+0.08y = 3500*0.08 (3)
(3) - (2) ==> 0.03x = 3500*0.08 - 250
x = 1000 put into (1) ==> y=2500
second part of Q:
sum of the cost of the first two lot of seeds must add up to equal the cost of the final mixture of seeds
Cost of seed = price per pound * pounds
let x = pounds of $0.60
0.6x+0.35*300 = 0.5 (x+300)
0.1x=0.5*300-0.35*300
x=450
參考: Myself
2007-11-26 3:46 pm
for prob no.1
the answers are 50 and 200.
let x be the amount invested at 5% then our equation would be
x(.05) + (3500-x).08 = 250
.05x + 280 -.08x = 250
.05x - .08x = 250-280
-.03x = -30
x = 1000
substituting x,
.05x + (3500 - x).08 = 250
.05(1000) + (3500-1000).08 = 250
50 + 200 = 250
thus,
50 was invested at 5% while 200 was invested at 8%
the answers are 50 and 200.
let x be the amount invested at 5% then our equation would be
x(.05) + (3500-x).08 = 250
.05x + 280 -.08x = 250
.05x - .08x = 250-280
-.03x = -30
x = 1000
substituting x,
.05x + (3500 - x).08 = 250
.05(1000) + (3500-1000).08 = 250
50 + 200 = 250
thus,
50 was invested at 5% while 200 was invested at 8%
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