Machine M can produce x units in 3/4 of the time it takes machine N to

Okieee..

Now ultimately the speed of every machine is given with respect to Mach O.
So lets assume the speed of O, say 12 hrs to make x units (assuming 6 because we can see we will need to divide by 3 and 4

Mach O makes x units in 12 hrs
So, Mach N = 2/3 of O = 2/3*12 = 8 hrs to make x units
and Mach M = 3/4 of N = 3/4*8 = 6 hrs to make x units

No they are running simultaneously. Lets see how much each mach makes in 1 hr
Mach O = x/12 units
Mach N = x/8 units
Mach M = x/6 units

In 1 hr, together they make - x/12+x/8+x/6 = 3x/8
So what ratio of this has Mach N made?

(x/8) / (3x/8) = 1/3

Ans: B=1/3

We will take all rates in terms of N machine.

Suppose N Machine takes ’n’ Hours
Therefore M machine takes 3/4n hours and O machine takes 3/2 n Hours (n=3/2O)

All machines together in 1 hour will do= 4x/3n+x/n+2x/3n= 3x/n

So all three machines works at the rate of 3x/n and machine N works at the rate of x/n.

Output produced by machine N/Total output= x/n/3x/n= 1/3
Thanks

Another way to look at the problem..

Time ratio for M and N

M:N = 3:4....Efficiency(or work done in the same time) Ratio will be the inverse of this..m:n = 4:3

Similarly..Eff. Ratio for N and O
n:o = 3:2

Overall
m:n:o = 4:3:2

Fraction of work done by N?

\(\frac{3}{4+3+2}\)

\(=\frac{1}{3}\)

(B)

If Machine N takes 120 minutes to produce x units, machine M will take \(\frac{3}{4}*120 = 90\) minutes.
Similarly, as Machine N will take \(\frac{2}{3}\)rd of the time to produce the units machine O produces.
\(120 = \frac{2}{3}*x\) -> \(x = 120*\frac{3}{2} = 180\) minutes

If 360 units are being produced,
machine N will produce 3 units in a minute,
machine M will produce 4 units in a minute,
and machine O will produce 2 units in a minute.

Therefore, machine N will produce \(\frac{3}{3+4+2} = \frac{3}{9} = \frac{1}{3}\) of the total units(Option B)

Let the total work be 24 units so, efficiency of the machines are
Since all the 3 machines are working simultaneously work done by N is \(\frac{3}{(2+3+4)}\) = \(\frac{1}{3}\) , Answer must be (B)

Let t = the time it takes machine O to produce x units; then the time it takes machine N to produce x units = (2/3)t, and the time it takes machine M to produce x units = (3/4)(2/3)t = (1/2)t.

Therefore, the rate of M is x/[(1/2)t] = 2x/t, the rate of N is x/[(2/3)t] = 3x/2t, and the rate of O is x/t. Their combined rate is 2x/t + 3x/2t + x/t = 4x/2t + 3x/2t + 2x/2t = 9x/2t.

If all three machines are working simultaneously, the fraction of the total output that is produced by machine N is:

(3x/2t)/(9x/2t) = 3x/9x = 1/3

Alternate Solution:

Let’s denote the number of units produced by machine O in 12 hours as y.

Since machine N can produce the same number of units in 2/3 of the time, machine N can produce y units in 8 hours.

Since machine M can produce the same number of units in 3/4 of the time necessary for machine N, machine M can produce y units in 6 hours.

Now, let’s suppose all the machines ran for 24 hours. In 24 hours, machine M will produce 4y units; machine N will produce 3y units, and machine O will produce 2y units. The total number of units produced is 4y + 3y + 2y = 9y. The number of units produced by machine N is 3y, which is (3y)/(9y) = 1/3 of all the units produced.

Answer: B

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