Problem Solving

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 5w/4 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

I tried to use a different method rather than the one listed on Quan. Review to solve the problem, but somehow I couldn't.
Rather than let x = days need to finish w widgets, here is what I come up with so far:

Let X = X widgets/day,
Y = Y widgets/day

1): w/X = (w/Y) + 2
2): 5w/4 = (X + Y)*3
From 1), we have 3): Y = wX/(w - 2X)
I plugged 3) into 2), but I came out with something really ugly and got stuck.

Is there anyone able to solve these two equations, or is there anything wrong with these two equation? Thank you.

yyz5028 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 5/4w 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

I tried to use a different method rather than the one listed on Quan. Review to solve the problem, but somehow I couldn't. This was what I came up with so far:

Let X = X widgets/day,
Y = Y widgets/day

1)w/X = (w/Y) + 2
2) 5w/4 = (X + Y)*3
From 1), Y = wX/(w - 2X) 3)
I plugged 3) into 2), but I came out with something really ugly and got stuck.

Is there anyone able to solve these two equations, or is there anything wrong with these two equation? Thank you.

This problem is similar to https://www.beatthegmat.com/gmat-prep-qu ... tml#286228. The ability to see the Key principles in all these problems can save a lot of practice time. GMAT will only run variations around these problem and there is only so much variation.

The basic approach is to always remember you need to 1) find their individual rates, and 2) combine those rates to figure how much work they do in some unit of time. UNITS are the KEY to for.

Let Y equal to machine Y's rate. Then you know machine x does it in Y+2

Mac X rate then is

W/(y+2 ) ( This may have been where you made your error)

Mac Y rate is

W/Y.

Now you have their rates COMBINE them.

This will give w(2y+2)/y(y+2) widges/day

Since you are always tracking your unitys u know this fractoin can be multiplied by some number of DAYS. We are given 3 days in the problem and told they produce 5/4 widgets which makes sense since the days cancel.

So w(2y+2)/y(y+2) widgets/day x 3 days = (5w/4) widgets

So you cancel all units ( u don't have to write these units as long as u keep track of them in your mind) and cross multiply, you should end up with

5y^2 -14y-24=0

(5y+6)(y-4)=0

So y=4.

But You know x is slower by 2 days so X rate is 6.

If X can do W in 6 days, then it will take 12 days to do 2w.

I did not understand the step where you combine the two rates. Can you please elaborate?

yyz5028 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 5/4w 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

Let me explain this qn.

Let machine X create x widgets per day
Let machine Y create y widgets per day

Now, its given that they together produce 5w/4 widgets in 3 days.
Which means they produce 5w/12 widgets in 1 day.

Hence x+y = 5w/12

Assume Machine X takes d days to complete w widgets.
xd = w
y(d-2)=w

w/d + w/d-2 = 5w/12
1/d+1/d-2 = 5/12
This is solvable.

Else, plugin half of each option to check.

we get d= 6

Hence for 2w widgets, it takes 2*d = 12 days.

Hope this helps!!

kvcpk:

Can you expand on solving this part of the solution?

1/d+1/d-2 = 5/12

sdotcruz wrote:kvcpk:

Can you expand on solving this part of the solution?

1/d+1/d-2 = 5/12

12(2d-2) = 5d(d-2)
24d-24 = 5d^2-10d
5d^2-34d+24=0
5d^2-30d-4d+24=0
5d(d-6) -4(d-6)=0
(5d-4)(d-6)=0
d=4/5 or 6

Hope this helps!!

let machine Y takes d days to produce w widgets

=>machine X takes d+2 days to produce w widgets( from question machine X takes 2 days more than machine Y)

=> machine X produces w/(d+2) widgets in 1 day
and machine Y produces w/d widgets in 1 day

working together for 1 day produces w/d +w/(d+2) widgets

working together for 3 days = 3[ w/d +w/(d+2)] =5w/4

=>12[1/d +1/(d+2)] =5

=>12(2d+2) =5d(d+2)

=>24d+24 =5d^2+10d

=> 5d^2 -14d -24 =0

=> 5d^2 -20d+6d -24 =0

=> (d-4)(5d+6) =0

=> d =4 or d = -6/5

d =4
machine X takes d+2 days to produce w widgets
=> machine X takes 6 days to produce w widgets
for 2w widgets it takes 12 days

IMO E

Went on solving via options..
2w widgets -> 12 days...
w widgets -> 6 days...
X takes 6 days to produce 6 widgets...
Y takes 4..
Their one day work -> 1/6 + 1/4 = 5/12
So they complete the work in 12/5 days...
For w widgets... they take 12/5 days...
For 5/4w widgets..they would take 3 days..
Answer is E...

Can you tell the question number for this q??
I m not able to find this one...