parkhydel wrote:
Machines K, M, and N, each working alone at its constant rate, produce 1 widget in x, y, and 2 minutes, respectively. If Machines K, M, and N work simultaneously at their respective constant rates, does it take them less than 1 hour to produce a total of 50 widgets?
(1) x < 1.5
(2) y < 1.2
DS98530.02
Rate of m/c K for producing 1 widget= \(\frac{1}{x}\)
Rate of m/c M for producing 1 widget= \(\frac{1}{y}\)
Rate of m/c N for producing 1 widget= \(\frac{1}{2}\)
We can approach this one in two ways:
A) Either check for whether the 3 machines working together can produce 50 widgets in less than 1 hour(as question states)
B) Or check for what number of widgets are produced in 1 hour by the 3 machines working together.
Getting answers of any of the two conditions above will fulfil our requirement.
A) Solving this is tricky as we need values of both x and y - on an initial thought. However, let's try
St. 1: x < 1.5, Checking for the slowest speed of K.
Time taken by K to complete 1 widget(job done alone) = 1.5min
Time taken by K to complete 50 widgets = 50*1.5 = 75min
Time taken by N to complete 1 widget(job done alone) = 2min
Time taken by N to complete 50 widgets = 50*2 = 100min
Thus, in 1 minute, K completes = \(\frac{1}{75}\) of the work
in 1 minute, N completes = \(\frac{1}{100}\) of the work
Therefore, in 1 minute, together they complete = \((\frac{1}{75} + \frac{1}{100})\) of the work(lets say 1/T, where T is the time taken together by K and N)
\(\implies \frac{1}{60} + \frac{1}{100} = \frac{1}{T}\)
T = \(\frac{300}{8}\) ~43 minutes(actually would be less)
With only 2 machines - K and N - 50 widgets are completed in less than 1 hour.
SUFFICIENT.
St. 2: y < 1.2, Checking for the slowest speed of M.(Though this is not required to be done as similar approach as that applied in St. 1 is applicable)
Time taken by M to complete 1 widget(job done alone) = 1.2min
Time taken by M to complete 50 widgets = 50*1.5 = 60min
Time taken by N to complete 1 widget(job done alone) = 2min
Time taken by N to complete 50 widgets = 50*2 = 100min
Thus, in 1 minute, M completes = \(\frac{1}{60}\) of the work
in 1 minute, N completes = \(\frac{1}{100}\) of the work
Therefore, in 1 minute, together they complete = \((\frac{1}{60} + \frac{1}{100})\) of the work(lets say 1/T, where T is the time taken together by M and N)
\(\implies \frac{1}{60} + \frac{1}{100} = \frac{1}{T}\)
T = \(\frac{300}{8}\) ~37.xx minutes(again would be less)
With only 2 machines - M and N - 50 widgets are completed in less than 1 hour.
SUFFICIENT.
Personally, I like second option(B) better.
B) \(60* (\frac{1}{x} + \frac{1}{y} + \frac{1}{2}) < 50\)
St. 1: x < 1.5
Now, any value of x that is less than 1.5 means the machine K works faster - the lesser the value of x the faster the machine K works. Let's check for x = 1.5
\(\frac{60}{1.5} + \frac{60}{y} + \frac{60}{2} < 60 \)
\(40 + \frac{60}{y} + 30\) < 50
No matter what LHS is always > 70.
SUFFICIENT.
St. 2: y < 1.2
Applying same logic and checking for y = 1.2
\(\frac{60}{x} + \frac{60}{1.2} + \frac{60}{2} < 60 \)
\(\frac{60}{x} + 50 + 30\) < 50
Again LHS is always > 80.
SUFFICIENT.
Answer D.
Sachin19 Genoa2000 HTH.
Thank you for explaining this in such a detailed way! Really grateful!