Running at their respective constant rates, machine X takes 2 days longer to produce w
widgets than machine Y. At these rates, if the two machines together produce 5/4 w
widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?
Can somebody plz post the explanation for this one... Appreciate it!
OA 12
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I typically solve this algebraically, but it may not be the fastest way
let d be the num days machine Y takes to build w widgets
Machine X
days= d+2
widgets= w
rate = w/(d+2)
Machine Y
days= d
widgets= w
rate = w/(d)
Machine X & Y together
days= 3
widgets= 5/4w
rate = (5/4w)/3 = 5/12w
so rate of X & Y = w/(d+2) + w/d = 5/12w
1/(d+2) + 1/d = 5/12
(2d+2)/(d^2+2d) = 5/12
24d+24=5d^2+10d
0=5d^2-14d-24
0=(5d+6)(d-4)
d=-ve or 4
so X take d+2=4+2=6 to make w widgets, so to make 2w widgets it takes 12 days
hope that helps
solving for d
let d be the num days machine Y takes to build w widgets
Machine X
days= d+2
widgets= w
rate = w/(d+2)
Machine Y
days= d
widgets= w
rate = w/(d)
Machine X & Y together
days= 3
widgets= 5/4w
rate = (5/4w)/3 = 5/12w
so rate of X & Y = w/(d+2) + w/d = 5/12w
1/(d+2) + 1/d = 5/12
(2d+2)/(d^2+2d) = 5/12
24d+24=5d^2+10d
0=5d^2-14d-24
0=(5d+6)(d-4)
d=-ve or 4
so X take d+2=4+2=6 to make w widgets, so to make 2w widgets it takes 12 days
hope that helps
solving for d
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x=y+2
w/x+w/y=5/4 * 1/3 days
rate then is:
w/(y+2)+w/y=5/12
solve for y. the w's will cancel out.
0=5y^2-14y-24
0=(5y+6)(y-4)
y has to equal 4.
x=4+2=6
w/6=2w/x
x=12
cheers.
w/x+w/y=5/4 * 1/3 days
rate then is:
w/(y+2)+w/y=5/12
solve for y. the w's will cancel out.
0=5y^2-14y-24
0=(5y+6)(y-4)
y has to equal 4.
x=4+2=6
w/6=2w/x
x=12
cheers.