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Joined: 20 Feb 2017
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Running at their respective constant rates, Machine X takes 2 days
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20 Nov 2021, 01:00
20 Nov 2021, 01:00
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Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?
(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
Re: Running at their respective constant rates, Machine X takes 2 days
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20 Nov 2021, 02:12
20 Nov 2021, 02:12
3
Let \(r_x\) and \(r_y\) be rates of working of Machine \(X\) and \(Y\) respectively
Machine \(X\) take 2 days more than Machine \(Y\),
\(t_x = t_y + 2 \Rightarrow t_y = t_x - 2\), where \(t_x\) and \(t_y\) are time taken by \(X\) and \(Y\) to produce \(w\) widgets alone respectively
Together, they produce \(\frac{5}{4}w\) widgets in \(3\) days,
\((r_x + r_y)3 = \frac{5}{4}w\)
\((\frac{w}{t_x} + \frac{w}{t_y})3 = \frac{5}{4}w\)
\((\frac{w}{t_x} + \frac{w}{t_x - 2})3 = \frac{5}{4}w\)
\((\frac{1}{t_x} + \frac{1}{t_x - 2})3 = \frac{5}{4}\)
\((\frac{t_x + t_x - 2}{t_x(t_x-2)})3 = \frac{5}{4}\)
\(\frac{6t_x - 6}{{t_x}^2 - 2t_x)} = \frac{5}{4}\)
\(24t_x - 24 = 5{t_x}^2 - 10t_x\)
\(5{t_x}^2 - 34t_x + 24 = 0\)
Using quadratic formula, \(x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}\)
\(t_x = \frac{-(-34) \pm\sqrt{(-34)^2 - 4*5*24}}{2*5} = \frac{34 \pm\sqrt{1156 - 480}}{10} = \frac{34 \pm26}{10} = \frac{34+26}{10} AND \frac{34-26}{10} = \textbf{6} AND \textbf{0.8}\)
0.8 days is not valid since \(X\) took 2 days more than \(Y\), so \(t_x > 2 days\)
Therefore, \(t_x = 6\) days
Time taken for Machine \(X\) to produce \(2w\) widgets alone = \(\frac{2w}{r_x} = \frac{2w}{\frac{w}{t_x}} = \frac{2w}{\frac{w}{6}} = \frac{12w}{w} = \textbf{12}\)
Hence, Answer is E
Machine \(X\) take 2 days more than Machine \(Y\),
\(t_x = t_y + 2 \Rightarrow t_y = t_x - 2\), where \(t_x\) and \(t_y\) are time taken by \(X\) and \(Y\) to produce \(w\) widgets alone respectively
Together, they produce \(\frac{5}{4}w\) widgets in \(3\) days,
\((r_x + r_y)3 = \frac{5}{4}w\)
\((\frac{w}{t_x} + \frac{w}{t_y})3 = \frac{5}{4}w\)
\((\frac{w}{t_x} + \frac{w}{t_x - 2})3 = \frac{5}{4}w\)
\((\frac{1}{t_x} + \frac{1}{t_x - 2})3 = \frac{5}{4}\)
\((\frac{t_x + t_x - 2}{t_x(t_x-2)})3 = \frac{5}{4}\)
\(\frac{6t_x - 6}{{t_x}^2 - 2t_x)} = \frac{5}{4}\)
\(24t_x - 24 = 5{t_x}^2 - 10t_x\)
\(5{t_x}^2 - 34t_x + 24 = 0\)
Using quadratic formula, \(x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}\)
\(t_x = \frac{-(-34) \pm\sqrt{(-34)^2 - 4*5*24}}{2*5} = \frac{34 \pm\sqrt{1156 - 480}}{10} = \frac{34 \pm26}{10} = \frac{34+26}{10} AND \frac{34-26}{10} = \textbf{6} AND \textbf{0.8}\)
0.8 days is not valid since \(X\) took 2 days more than \(Y\), so \(t_x > 2 days\)
Therefore, \(t_x = 6\) days
Time taken for Machine \(X\) to produce \(2w\) widgets alone = \(\frac{2w}{r_x} = \frac{2w}{\frac{w}{t_x}} = \frac{2w}{\frac{w}{6}} = \frac{12w}{w} = \textbf{12}\)
Hence, Answer is E
General Discussion
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Joined: 10 Apr 2015
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Re: Running at their respective constant rates, Machine X takes 2 days
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20 Nov 2021, 06:12
20 Nov 2021, 06:12
1
GeminiHeat wrote:
Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?
(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
One approach is to assign a nice value to the job (w)
Let's say that w = 12.
GIVEN: Running at their respective constant rates, machine X takes 2 days longer to produce 12 widgets than machine Y
Let t = time for machine Y to produce 12 widgets
So, t+2 = time for machine X to produce 12 widgets
RATE = output/time
So, machine X's RATE = 12 widgets/(t + 2 days) = 12/(t+2) widgets per day
And machine Y's RATE = 12 widgets/(t days) = 12/t widgets per day
The two machines together produce 5w/4 widgets in 3 days
In other words, The two machines together produce 5(12)/4 widgets in 3 days
Or the two machines together produce 15 widgets in 3 days
This means the COMBINED RATE = 5 widgets per day
So, we can write: 12/(t+2) + 12/t = 5
Multiply both sides by (t+2)(t) to get: 12t + 12t + 24 = 5(t+2)(t)
Simplify: 24t + 24 = 5t² + 10t
Rearrange: 5t² - 14t - 24 = 0
Factor to get: (5t + 6)(t - 4) = 0
So, EITHER t = -6/5 OR t = 4
Since the time cannot be negative, it must be the case that t = 4
If t = 4, then it takes Machine Y 4 days to produce 12 widgets
And it takes Machine X 6 days to produce 12 widgets
How many days would it take machine X alone to produce 2w widgets?
In other words, how many days would it take machine X alone to produce 24 widgets? (since w = 12)
If it takes Machine X 6 days to produce 12 widgets, then it will take Machine X 12 days to produce 24 widgets
Answer: E
Cheers,
Brent
