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?

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

Y can produce w widgets in say "a" days=> X will do the same job in (a+2) days.

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)

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?

Check out this step: 1/(y+2) + 1/y = 1/(5/12)

The above step is not equal to : 4/(y+2) = 12/5 but equal to
(2y+2)/(y^2+2y) = 12/5

muzali wrote:

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?

Check out this step: 1/(y+2) + 1/y = 1/(5/12)

The above step is not equal to : 4/(y+2) = 12/5 but equal to
(2y+2)/(y^2+2y) = 12/5

Thanks for looking into this:

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??

canuckclint wrote:

muzali wrote:

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?

Check out this step: 1/(y+2) + 1/y = 1/(5/12)

The above step is not equal to : 4/(y+2) = 12/5 but equal to
(2y+2)/(y^2+2y) = 12/5

Thanks for looking into this:

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??

Nevermind I can't add to fractions that way.
Stupid mistake!

muzali wrote:

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

Y can produce w widgets in say "a" days=> X will do the same job in (a+2) days.

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)

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!

penumbra547 wrote:

muzali wrote:

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

Y can produce w widgets in say "a" days=> X will do the same job in (a+2) days.

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)

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!

I'll answer for Muzali here since this one is simple,

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

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 = T-4 days

If 5w/4 takes 3 days, 5w=12 days, 2w = 24/5 days

I/t + 1/(t-4) = 5/24

t = 12

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?