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Abha30
Would the below understanding be correct?
Therefore Time= 50 / (2y+2x+xy)/2xy = 100xy/(2x+2y+xy). We need to check if this is below 60 mins.
100xy/(2x+2y+xy) < 60
On cross multiplication and further simplification we need to check if xy < 3x+3y.
(1) x<1.5. Let x =1 then is y < 3+3y? which becomes is -3 < 2y? which becomes is -3/2 <y ? Since y is time and can't be negative hence answer is Yes. Suff
(2) Similar logic for statement 2, hence Suff
Answer D
| Rate | Time | Work | |
| K | 1/x | x mins | 1 widget |
| M | 1/y | y mins | 1 widget |
| N | 1/2 | 2 mins | 1 widget |
| K+M+N | 1/x + 1/y + 1/2 = (2y+2x+xy)/ 2xy | 50 widgets |
Therefore Time= 50 / (2y+2x+xy)/2xy = 100xy/(2x+2y+xy). We need to check if this is below 60 mins.
100xy/(2x+2y+xy) < 60
On cross multiplication and further simplification we need to check if xy < 3x+3y.
(1) x<1.5. Let x =1 then is y < 3+3y? which becomes is -3 < 2y? which becomes is -3/2 <y ? Since y is time and can't be negative hence answer is Yes. Suff
(2) Similar logic for statement 2, hence Suff
Answer D
I think we should be careful while reducing word problems. Too much redcution can cause confusion. Here the question has one relative sign and then each option has another. So chances of getting confused are a little higher.
Ideally you should have stopped here.
20 < 60 (1/x + 1/y)
that both x and y togther produces more than 20 widgets in 60min/1hour.
next stop
1/3 < (1/x + 1/y)
both togther produce more than 1 widget every 3 minutes.
x<1.5 means K alone produces 1 widget in less than 1.5 minute. Sufficient.
y<1.2 means M alone produces 1 widget in less than 1.2 minute, Sufficient.
SudsMeister
17 May 2025, 09:09
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Quote:
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
(1) x < 1.5
(2) y < 1.2
Many a times we are unable to remove the algebraic lens and quickly deep dive into using equations and variables to power through and get an answer.
Alas, I also went about the hard way to solve this:
Given: To produce 1 widget alone, K takes x minutes, M takes y minutes, and N takes 2 minutes.
Alone, in 1 minute, K produces \(\frac{1 }{ x}\) widgets, M produces \(\frac{1 }{ y}\) widgets, and N produces \(\frac{1 }{ 2}\) widgets.
Together, in 1 minute they produce = \(\frac{1 }{ x} + \frac{1}{y} + \frac{1}{2}\) widgets = say Z widgets
So, 1 minute --- Z widgets
Then, 1 widget --- \(\frac{1 }{ Z}\) minutes
Together, to produce 1 widget they take \(\frac{1 }{ Z}\) minutes.
Together, to produce 50 widgets they take \(\frac{1 }{ Z}*50\) minutes.
What is asked of us: is \(\frac{1 }{ Z}*50\) minutes \(< 1 \) hour?
Let's simplify this.
\(\frac{1 }{ Z}*50\) minutes \(< 60\) minutes [/m]
\(\frac{1 }{ Z}*50 < 60\)
\(\frac{50}{60} < Z \)
\(\frac{50}{60} < (\frac{1 }{ x} + \frac{1}{y} + \frac{1}{2}) \)
\(\frac{5}{6} - \frac{1}{2} < (\frac{1 }{ x} + \frac{1}{y}) \)
Is, \(\frac{1}{3} < (\frac{1 }{ x} + \frac{1}{y}) \)?
Statement 1: \(x < 1.5\)
\(\frac{1}{1.5} < \frac{1}{x}\)
\(\frac{2}{3} < \frac{1}{x}\)
So, we can confidently say that \(\frac{1}{3} < (\frac{1 }{ x} + \frac{1}{y}) \) because \(\frac{1}{y} \) cannot be negative.
1 is sufficient.
Statement 2: \(y < 1.2\)
\(\frac{1}{1.2} < \frac{1}{y}\)
\(\frac{5}{6} < \frac{1}{y}\)
So, we can confidently say that \(\frac{1}{3} < (\frac{1 }{ x} + \frac{1}{y}) \) because \(\frac{1}{x} \) cannot be negative.
2 is sufficient.
Answer: D.
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