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?
A. 4
B. 6
C. 8
D. 10
E. 12
Work problem.
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I think I'm wrong becasue I'm getting a fractional answer but here goes:
Get everything in terms of one work w.
1/(y+2) + 1/y = 1/(5/12)
4/(y+2) = 12/5
y = 1/3
x = (y+2) = 7/3
So it takes 7/3 days for X to finsih 1 w.
For 2 it takes 14/3 days or ~ 4.6 days.
I'll go with 4 days.
Can someone say where I have gone wrong?
Get everything in terms of one work w.
1/(y+2) + 1/y = 1/(5/12)
4/(y+2) = 12/5
y = 1/3
x = (y+2) = 7/3
So it takes 7/3 days for X to finsih 1 w.
For 2 it takes 14/3 days or ~ 4.6 days.
I'll go with 4 days.
Can someone say where I have gone wrong?
Y can produce w widgets in say "a" days=> X will do the same job in (a+2) days.sumittaneja009 wrote: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?
A. 4
B. 6
C. 8
D. 10
E. 12
For (5/4)w widgets, Y will take (5/4)*a days and X will take (5/4)*(a+2) days.
Together they produce (5/4)w widgets in 3 days,
so (1/3) = (4/5)*[(1/a)+(1/(a+2)],
5/12 = (2a+2)/(a^2 + 2a)
Solving for a, we get a =4
This means X finishes w widgets in 4+2 days = 6 days
So X will finish 2 w widgets in 6*2 = 12 dys. Ans (E)
Check out this step: 1/(y+2) + 1/y = 1/(5/12)canuckclint wrote:I think I'm wrong becasue I'm getting a fractional answer but here goes:
Get everything in terms of one work w.
1/(y+2) + 1/y = 1/(5/12)
4/(y+2) = 12/5
y = 1/3
x = (y+2) = 7/3
So it takes 7/3 days for X to finsih 1 w.
For 2 it takes 14/3 days or ~ 4.6 days.
I'll go with 4 days.
Can someone say where I have gone wrong?
The above step is not equal to : 4/(y+2) = 12/5 but equal to
(2y+2)/(y^2+2y) = 12/5

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Thanks for looking into this:muzali wrote:Check out this step: 1/(y+2) + 1/y = 1/(5/12)canuckclint wrote:I think I'm wrong becasue I'm getting a fractional answer but here goes:
Get everything in terms of one work w.
1/(y+2) + 1/y = 1/(5/12)
4/(y+2) = 12/5
y = 1/3
x = (y+2) = 7/3
So it takes 7/3 days for X to finsih 1 w.
For 2 it takes 14/3 days or ~ 4.6 days.
I'll go with 4 days.
Can someone say where I have gone wrong?
The above step is not equal to : 4/(y+2) = 12/5 but equal to
(2y+2)/(y^2+2y) = 12/5
The way I looked at it is add 2 to num and denom
1/(y+2) + 1/y =
1/(y+2) + (1+2) / (y+2) =
1/(y+2) + 3/(y+2) = 4/(y+2)
Can't believe I'm getting stumped on an easy one!
Any suggestions??

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Nevermind I can't add to fractions that way.canuckclint wrote:Thanks for looking into this:muzali wrote:Check out this step: 1/(y+2) + 1/y = 1/(5/12)canuckclint wrote:I think I'm wrong becasue I'm getting a fractional answer but here goes:
Get everything in terms of one work w.
1/(y+2) + 1/y = 1/(5/12)
4/(y+2) = 12/5
y = 1/3
x = (y+2) = 7/3
So it takes 7/3 days for X to finsih 1 w.
For 2 it takes 14/3 days or ~ 4.6 days.
I'll go with 4 days.
Can someone say where I have gone wrong?
The above step is not equal to : 4/(y+2) = 12/5 but equal to
(2y+2)/(y^2+2y) = 12/5
The way I looked at it is add 2 to num and denom
1/(y+2) + 1/y =
1/(y+2) + (1+2) / (y+2) =
1/(y+2) + 3/(y+2) = 4/(y+2)
Can't believe I'm getting stumped on an easy one!
Any suggestions??
Stupid mistake!

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Muzali, would you please explain how you got "Y will take (5/4)*a days and X will take (5/4)*(a+2) days." i assume you are using the RT=D formula? Thank You!muzali wrote:Y can produce w widgets in say "a" days=> X will do the same job in (a+2) days.sumittaneja009 wrote: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?
A. 4
B. 6
C. 8
D. 10
E. 12
For (5/4)w widgets, Y will take (5/4)*a days and X will take (5/4)*(a+2) days.
Together they produce (5/4)w widgets in 3 days,
so (1/3) = (4/5)*[(1/a)+(1/(a+2)],
5/12 = (2a+2)/(a^2 + 2a)
Solving for a, we get a =4
This means X finishes w widgets in 4+2 days = 6 days
So X will finish 2 w widgets in 6*2 = 12 dys. Ans (E)

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I'll answer for Muzali here since this one is simple,penumbra547 wrote:Muzali, would you please explain how you got "Y will take (5/4)*a days and X will take (5/4)*(a+2) days." i assume you are using the RT=D formula? Thank You!muzali wrote:Y can produce w widgets in say "a" days=> X will do the same job in (a+2) days.sumittaneja009 wrote: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?
A. 4
B. 6
C. 8
D. 10
E. 12
For (5/4)w widgets, Y will take (5/4)*a days and X will take (5/4)*(a+2) days.
Together they produce (5/4)w widgets in 3 days,
so (1/3) = (4/5)*[(1/a)+(1/(a+2)],
5/12 = (2a+2)/(a^2 + 2a)
Solving for a, we get a =4
This means X finishes w widgets in 4+2 days = 6 days
So X will finish 2 w widgets in 6*2 = 12 dys. Ans (E)
The question tells you that for one work w, it takes X will take a+2 days
Y will take a days.
For 3 times the work it would be 3*a
So for 5/4 of the work w, just multiply
5/4*a
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If X takes 2 days longer to produce w widgets than machine Y , it takes 4 days longer to produce 2w widgets
X= T days
Y = T4 days
If 5w/4 takes 3 days, 5w=12 days, 2w = 24/5 days
I/t + 1/(t4) = 5/24
t = 12
X= T days
Y = T4 days
If 5w/4 takes 3 days, 5w=12 days, 2w = 24/5 days
I/t + 1/(t4) = 5/24
t = 12
LGTCH

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How do you solve this problem using the box matrix (2 rows, 3 columns)? I can't seem to set up the values correctly.
X: R*T = D
Y: R*T = D
Assume w = 1
X: (1/3) * T = D(x)
y: (1) * T = D(y)
RateX*T + RateY*T = D(x) + D(y) = 5/12
(1/3)*T + T = 5/12
T = 16/5
Did I set this up right so far?
X: R*T = D
Y: R*T = D
Assume w = 1
X: (1/3) * T = D(x)
y: (1) * T = D(y)
RateX*T + RateY*T = D(x) + D(y) = 5/12
(1/3)*T + T = 5/12
T = 16/5
Did I set this up right so far?