Machine M,N,O working simultaneously

[wawan_andrean]

Machine M can produce x units in 3/4 of the time it takes machine N to produce the same amount of units. Machine N can produce x units in 2/3 the time it takes machine O to produce that amount of 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

[crazy800]

is the answer B?

[Tirpurari]

Here is my take on this:

Let us assume that O takes 1 hour to complete X.
Consequently,
N will take 2/3 hours and
M will take (3/4)*(2/3) hours = 1/2 hours

as work formula:
(1/M)+(1/N)+(1/O) = 1/x where M,N and O represents varibale indicating time taken by individual machines

substitute the value of M,N, and O in above equation, result:
9/2 = 1/x
=>x=2/9

So, if all three machines are working togather then the work is finished in 2/9 hours.

in 2/3 hours N finish x work then in 2/9 hours N will finish:

(3/2)x(2/9) --- here x is multiply symbol
=1/3

The answer is B.


What is OA?

[zoom2suman]

Assume there are 120 units. Also assume that in 1 minute O produces 1 unit

==>In 1 hr, O produces 60 Units
==>In 1 hr, N produces, 40 units(N=2/3O)
==>In 1 hr, M produces, 30 units (M=3/4N)

Total units in 1 hr = 60+40+30 =130
and units produced by N= 40

Answer = 40/130 = 4/13(C) ...Am i makin a mistake???

[Tirpurari]

Originally Posted by zoom2suman

Assume there are 120 units. Also assume that in 1 minute O produces 1 unit

==>In 1 hr, O produces 60 Units
==>In 1 hr, N produces, 40 units(N=2/3O)
==>In 1 hr, M produces, 30 units (M=3/4N)

Total units in 1 hr = 60+40+30 =130
and units produced by N= 40

Answer = 40/130 = 4/13(C) ...Am i makin a mistake???

Suman, in my opinion here we are not asked to find the proportion of work performed in 1 hour by all wrt N.

The question asks: all three machines are working simultaneously, what fraction of the total output is produced by machine N?

It does not talk about 1 hours.

I agree that the the question should have been "all three machines are working simultaneously and completes the work, what fraction of the total output is produced by machine N?"

but still it is a work problem.

Also, you assume that the total unit is 120, so the machines have to stop working after 120 unit of work is completed.

wawan_andrean: can you please post OA, thanks. Expected a Spoiler with the question.

Cheers!
Tirpurari

[solawgmat]

I got C also, using M=12 units in 1 hour

[swethapv]

IMO B....please post the OA....

[genius_in_the_gene]

Imo B

[lizishao]

IMO B

Machine Time consumed Unit Produced
M 3/4T X
N T X
O 3/2T X

If these 3 machines produce unit in the same time, the above table is changed to

M T 4/3X
N T X
O T 2/3X

So, the total unit produced is: 4/3X+X+2/3X=3X

N/total unit produced is x/3x=1/3

Hope my explanation is clear enough.

[selytch]

Originally Posted by lizishao

IMO B

Machine Time consumed Unit Produced
M 3/4T X
N T X
O 3/2T X

If these 3 machines produce unit in the same time, the above table is changed to

M T 4/3X
N T X
O T 2/3X

So, the total unit produced is: 4/3X+X+2/3X=3X

N/total unit produced is x/3x=1/3

Hope my explanation is clear enough.

IMO correct answer, but wrong notation:
4x/3+3x/3+2x/3=3x
9x/3=3x
3=3

Should be:
M=4/3*N (this is units/time, not time/units!)
O=2/3*N

(units/time)
Productivity Time Units
M 3/4*T1 X
N T1 X
O T2 X
N 2/3*T2 X

from N, T2=3/2*T1, so
O 3/2*T1 X
N T1 X
M 3/4*T1 X
or if T1 constant
O T1 2/3*X
N T1 X
M T1 4/3*X

100 units to make
O,N,M will contribute at pO=O/100 pN=N/100 pM=M/100
O+N+M=100, substituting O and M from above
2/3*N+N+4/3*N=100
N*(2/3+3/3+4/3)=100
N*(9/3)=100
3N=100
N=100/3
pN will be (100/3)/100=1/3