It is currently 25 Jun 2017, 09:00

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Machine M can produce x units in 3/4 of the time it takes

Author Message
Manager
Joined: 11 Nov 2006
Posts: 144
Machine M can produce x units in 3/4 of the time it takes [#permalink]

### Show Tags

02 May 2007, 09:54
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Machine M can produce x units in 3/4 of the time it takes machine N to produce x units. Machine N can produce x units in 2/3 the time it takes machine O to produce x units. If all three machines are working simultaneously, what fraction of the total output is produced by machine N?
(A) 1/2
(B) 1/3
(C) 4/13
(D) 8/29
(E) 6/33

Please solve it and explain it... thanks!
Senior Manager
Joined: 11 Jun 2006
Posts: 254

### Show Tags

02 May 2007, 13:44
Use machine O as a starting point since it is the slowest machine, all other machines will be a fraction of O.

Time of M = 3/4 * 2/3 = 6/12
Time of N = 2/3 * 4/4 = 8/12
Time of O = 12/12

The recipricol of the times stated above = the rates; since rates are additive, add...

M + N + O = 9/2

Putting N over the total:

3/2 / 9/2 = 1/3

Senior Manager
Joined: 24 Nov 2006
Posts: 349
Re: Machines M, N and O [#permalink]

### Show Tags

02 May 2007, 18:14
M N O
----------------------------
t/2 2t/3 t (How long it takes each to produce x)

In t: 2x 3x/2 x (How much they produce in t (hs, secs, etc))

What fraction of total output is produced by N? (3x/2)/(2x+3x/2+x) = 1/3 => B.

Last edited by Andr359 on 11 May 2007, 14:25, edited 1 time in total.
Manager
Joined: 28 Dec 2006
Posts: 61

### Show Tags

02 May 2007, 19:14

let o=12 hence n=8 and m = 6.

Total hours =26
n=8

Therefore fraction =8/26 =4/13
Manager
Joined: 02 May 2007
Posts: 152
Re: Machines M, N and O [#permalink]

### Show Tags

02 May 2007, 21:27
querio wrote:
Machine M can produce x units in 3/4 of the time it takes machine N to produce x units. Machine N can produce x units in 2/3 the time it takes machine O to produce x units. If all three machines are working simultaneously, what fraction of the total output is produced by machine N?
(A) 1/2
(B) 1/3
(C) 4/13
(D) 8/29
(E) 6/33

Please solve it and explain it... thanks!

(B) it is

M = time for M to produce the total output T:
=> T/M = output M produce in 01 unit of time
M = 3/4N
N = time for N to produce the total output T:
=> T/N = output N produce in 01 unit of time
N = 3/4O
O = time for O to produce the total output T

Output the 03 machines produce in 01 unit of time = T/M + T/N + T/O = T/(3/4N) + T/N + T(3N/2) = 3T/N

Fraction of the total output is produced by machine N = (T/N)/3T/N = 1/3

Last edited by kirakira on 02 May 2007, 22:03, edited 1 time in total.
Manager
Joined: 17 Apr 2007
Posts: 90

### Show Tags

02 May 2007, 21:57
B - 1/3

Let say it takes T time for O to produce 1 unit -> rate of work 1/T
N's rate of work 1/(2/3 * T) = 3/2T
M's rate of work - 1/(2/3 * 3/4 T) = 2/T

so total rate of work - 1/T + 3/2T + 2/T = 9/2T

N's part is (3/2T) / (9/2T) = 1/3
Manager
Joined: 04 Oct 2006
Posts: 76

### Show Tags

03 May 2007, 01:20
- time it takes machine O to do the job = t
- time it takes machine N to do the job = 2t/3
- time it takes machine M to do the job = 3/4 * 2/3 * t = t/2

In one hour O+N+M do 1/t + 3/2t + 2/t = 9/2t of the job.

In one hour N do (3/2t) / (9/2t) = 1/3 of the job.

Manager
Joined: 11 Nov 2006
Posts: 144

### Show Tags

03 May 2007, 04:27
[b]OA[/b] is B

thanks!
VP
Joined: 08 Jun 2005
Posts: 1145

### Show Tags

03 May 2007, 12:59
baer wrote:
Use machine O as a starting point since it is the slowest machine, all other machines will be a fraction of O.

Time of M = 3/4 * 2/3 = 6/12
Time of N = 2/3 * 4/4 = 8/12
Time of O = 12/12

The recipricol of the times stated above = the rates; since rates are additive, add...

M + N + O = 9/2

Putting N over the total:

3/2 / 9/2 = 1/3

can you please explain why 4/4 ?
Senior Manager
Joined: 11 Jun 2006
Posts: 254

### Show Tags

03 May 2007, 13:56
KillerSquirrel wrote:
baer wrote:
Use machine O as a starting point since it is the slowest machine, all other machines will be a fraction of O.

Time of M = 3/4 * 2/3 = 6/12
Time of N = 2/3 * 4/4 = 8/12
Time of O = 12/12

The recipricol of the times stated above = the rates; since rates are additive, add...

M + N + O = 9/2

Putting N over the total:

3/2 / 9/2 = 1/3

can you please explain why 4/4 ?

Just to get a common denominator of 12, so that you can add up the rates in the next step. I was just showing the logic of the problem.
03 May 2007, 13:56
Display posts from previous: Sort by