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Question

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 \frac{5}{4} w widgets in 3 days, how many days would it take Machine X alone to produce 2 w widgets?

Answer Choices

A. $4$
B. $6$
C. $8$
D. $10$
E. $12$

Steps

Explanation	Calculations	Help
$We first translate ("Machine X takes 2 days longer to produce w widgets than Machine Y") by defining the time variables for each machine.$	$d = time for Machine Y to produce w widgets d + 2 = time for Machine X to produce w widgets$
$We express each machine's production rate in widgets per day.$	$r_{y} = \frac{w}{d} r_{x} = \frac{w}{d + 2}$
$We translate the condition that together in 3 days they produce \frac{5}{4} w widgets.$	$3 (r_{x} + r_{y}) = \frac{5}{4} w$
$We substitute the expressions for r_{x} and r_{y} into the joint production equation.$	$3 (\frac{w}{d + 2} + \frac{w}{d}) = \frac{5}{4} w$
$We divide both sides by w to simplify the equation.$	$3 (\frac{1}{d + 2} + \frac{1}{d}) = \frac{5}{4}$
$We combine the sum of fractions on the left-hand side.$	$\frac{1}{d + 2} + \frac{1}{d} = \frac{2 d + 2}{d ( d + 2 )}$
$We clear denominators by multiplying both sides by 4 d (d + 2) .$	$12 (2 d + 2) = 5 d (d + 2)$	Theory & Tactics Method Card EQUA4-A Click to view full details
$We expand and rearrange to form a quadratic in standard form.$	$12 (2 d + 2) = 24 d + 24 24 d + 24 = 5 d^{2} + 10 d 5 d^{2} - 14 d - 24 = 0$	Theory & Tactics Method Card EQUA2-A Click to view full details
$We factor the quadratic expression.$	$5 d^{2} - 14 d - 24 = (5 d + 6) (d - 4) = 0$	Theory & Tactics Method Card EQUA2-B Click to view full details
$We solve each linear factor to find the possible values of d .$	$5 d + 6 = 0 \Rightarrow d = - \frac{6}{5} d - 4 = 0 \Rightarrow d = 4$	Theory & Tactics Method Card EQUA1-A Click to view full details
$We discard the negative solution since time cannot be negative, accepting d = 4 .$	$d = 4$
$We compute Machine X's time to produce w widgets as d + 2 = 6 days.$	$d + 2 = 6$
$We find Machine X's rate and then calculate the time to produce 2 w widgets.$	$r_{x} = \frac{w}{6} \frac{2 w}{r _{x}} = \frac{2 w}{w /6} = 12$

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Final Answer

12 days

Question

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 \frac{5}{4} w widgets in 3 days, how many days would it take Machine X alone to produce 2 w widgets?

Answer Choices

A. $4$
B. $6$
C. $8$
D. $10$
E. $12$

Steps

Explanation	Calculations	Help
$We first translate ("Machine X takes 2 days longer to produce w widgets than Machine Y") by defining the time variables for each machine.$	$d = time for Machine Y to produce w widgets d + 2 = time for Machine X to produce w widgets$
$We express each machine's production rate in widgets per day.$	$r_{y} = \frac{w}{d} r_{x} = \frac{w}{d + 2}$
$We translate the condition that together in 3 days they produce \frac{5}{4} w widgets.$	$3 (r_{x} + r_{y}) = \frac{5}{4} w$
$We substitute the expressions for r_{x} and r_{y} into the joint production equation.$	$3 (\frac{w}{d + 2} + \frac{w}{d}) = \frac{5}{4} w$
$We divide both sides by w to simplify the equation.$	$3 (\frac{1}{d + 2} + \frac{1}{d}) = \frac{5}{4}$
$We combine the sum of fractions on the left-hand side.$	$\frac{1}{d + 2} + \frac{1}{d} = \frac{2 d + 2}{d ( d + 2 )}$
$We clear denominators by multiplying both sides by 4 d (d + 2) .$	$12 (2 d + 2) = 5 d (d + 2)$	Theory & Tactics Method Card EQUA4-A Click to view full details
$We expand and rearrange to form a quadratic in standard form.$	$12 (2 d + 2) = 24 d + 24 24 d + 24 = 5 d^{2} + 10 d 5 d^{2} - 14 d - 24 = 0$	Theory & Tactics Method Card EQUA2-A Click to view full details
$We factor the quadratic expression.$	$5 d^{2} - 14 d - 24 = (5 d + 6) (d - 4) = 0$	Theory & Tactics Method Card EQUA2-B Click to view full details
$We solve each linear factor to find the possible values of d .$	$5 d + 6 = 0 \Rightarrow d = - \frac{6}{5} d - 4 = 0 \Rightarrow d = 4$	Theory & Tactics Method Card EQUA1-A Click to view full details
$We discard the negative solution since time cannot be negative, accepting d = 4 .$	$d = 4$
$We compute Machine X's time to produce w widgets as d + 2 = 6 days.$	$d + 2 = 6$
$We find Machine X's rate and then calculate the time to produce 2 w widgets.$	$r_{x} = \frac{w}{6} \frac{2 w}{r _{x}} = \frac{2 w}{w /6} = 12$

Scroll horizontally to view all columns

Final Answer

12 days