|
Author |
Message |
|
TAGS:
|
|
|
Intern
Joined: 12 Aug 2004
Posts: 10
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Running at their respective constant rates, machine X takes [#permalink]
 06 Nov 2004, 00:42
Question Stats:
100% (01:04) correct
0% (00:00) wrong based on 0 sessions
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
Please explain your answer. Thank you.
|
|
|
|
|
|
|
Director
Joined: 16 Jun 2004
Posts: 929
Followers: 1
Kudos [?]:
4
[0], given: 0
|
I am getting 12. E. Hope havent done any calculation errors..
approach..
let y=no. of days taken by Y to do w widgets. Then X will take y+2 days.
1/(y+2) +1/y = 5/12 (5/12 is because (5/4)w widgets are done in 3 days. So, x widgets will be done in 12/5 days or 5/12 th of a widget in a day)
Solving, we have y = 4
=>X takes 6 days to doing x widgets. So, he will take 12 days to doing 2w widgets.
|
|
|
|
|
|
Intern
Joined: 25 Oct 2004
Posts: 7
Followers: 0
Kudos [?]:
0
[0], given: 0
|
I also get 12 (E).
My approach was a little more complicated.
Work formula : COMBINED TIME =(A*B) : (A+B) , where A- time for 1st machine , B - time for 2nd machine.
(2+z) days will take machine A to produce 1 widg. and
z days will take machine to produce 1 widg.
Thus, (z*(z+2)) : (z+z+2)*5/4 =3 (days)
5z^2 +10z - 24 =0
z=4 ; z=-1,2 ( negative is not appropriate so we rid of it )
Consequently,machine A takes (2+z)days = 2+4= 6 (days to produce one widg.); and (6*2) = 12 (days to produce two widg.).
B/R
Oh Gal Len
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
Running at their respective constant rates, machine X takes
|
falcor |
2 |
17 Sep 2004, 12:43 |
|
|
|
|
Running at their respective constant rates, machine X takes
|
duttsit |
6 |
07 Dec 2005, 15:43 |
|
|
|
|
Running at their respective constant rates, machine X takes
|
ricokevin |
2 |
11 Mar 2007, 18:07 |
|
|
|
|
Running at their respective constant rates, machine X takes
|
gluon |
7 |
22 Sep 2007, 07:52 |
|
|
|
|
Running at their respective constant rates, machine X takes
|
vivektripathi |
3 |
24 Sep 2008, 07:40 |
|
|
|
|
|
|
close
