Printer $A$ requires 4 hours more than Printer $B$... | Free GMAT Practice Question | GMAT Panda

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Question

Printer A requires 4 hours more than Printer B to print p pages at their respective constant rates. Together, in 3 hours, they print \frac{4}{5} p pages. How many hours would Printer A alone require to print 2 p pages?

Answer Choices

A. $15$
B. $18$
C. $20$
D. $24$
E. $30$

Steps

Explanation	Calculations	Help
$We let d be the time for Printer B to print p pages and so Printer A needs d + 4 hours for the same task.$	$d = time for Printer B to print p pages d + 4 = time for Printer A to print p pages$
$We express each printer's rate in pages per hour.$	$r_{B} = \frac{p}{d} r_{A} = \frac{p}{d + 4}$
$We write the combined output in 3 hours giving four fifths of p pages.$	$3 \times (r_{A} + r_{B}) = \frac{4}{5} p$
$We substitute the rates into the joint production equation.$	$3 \times (\frac{p}{d + 4} + \frac{p}{d}) = \frac{4}{5} p$
$We divide both sides by p to simplify.$	$3 \times (\frac{1}{d + 4} + \frac{1}{d}) = \frac{4}{5}$
$We add the fractions by finding a common denominator.$	$\frac{1}{d + 4} + \frac{1}{d} = \frac{d + ( d + 4 )}{d ( d + 4 )} = \frac{2 d + 4}{d ( d + 4 )}$	Theory & Tactics Method Card FDPR0-B Click to view full details
$We clear denominators by multiplying both sides by 5d(d + 4).$	$15 \times (2 d + 4) = 4 d (d + 4)$	Theory & Tactics Method Card EQUA4-A Click to view full details
$We expand and rearrange into a quadratic equation.$	$15 \times (2 d + 4) = 30 d + 60 30 d + 60 = 4 d^{2} + 16 d 4 d^{2} - 14 d - 60 = 0$	Theory & Tactics Method Card EQUA2-A Click to view full details
$We divide both sides by 2 to simplify the coefficients.$	$2 d^{2} - 7 d - 30 = 0$
$We factor the quadratic expression.$	$2 d^{2} - 7 d - 30 = (2 d + 5) (d - 6)$	Theory & Tactics Method Card EQUA2-B Click to view full details
$We solve each factor for d.$	$2 d + 5 = 0 d = - \frac{5}{2} d - 6 = 0 d = 6$	Theory & Tactics Method Card EQUA1-A Click to view full details
$We discard the negative solution since time must be positive.$	$d = 6$
$We find Printer A's time to print p pages.$	$d + 4 = 6 + 4 = 10$
$We find Printer A's time to print 2p pages, cancelling p.$	$r_{A} = \frac{p}{10} Time for 2 p pages = \frac{2 p}{r _{A}} = 2 p \times \frac{10}{p} = \frac{p \times 2 \times 10}{p} = 2 \times 10 = 20$	Theory & Tactics Method Card FDPR2-A Click to view full details

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Final Answer

C

Question

Printer A requires 4 hours more than Printer B to print p pages at their respective constant rates. Together, in 3 hours, they print \frac{4}{5} p pages. How many hours would Printer A alone require to print 2 p pages?

Answer Choices

A. $15$
B. $18$
C. $20$
D. $24$
E. $30$

Steps

Explanation	Calculations	Help
$We let d be the time for Printer B to print p pages and so Printer A needs d + 4 hours for the same task.$	$d = time for Printer B to print p pages d + 4 = time for Printer A to print p pages$
$We express each printer's rate in pages per hour.$	$r_{B} = \frac{p}{d} r_{A} = \frac{p}{d + 4}$
$We write the combined output in 3 hours giving four fifths of p pages.$	$3 \times (r_{A} + r_{B}) = \frac{4}{5} p$
$We substitute the rates into the joint production equation.$	$3 \times (\frac{p}{d + 4} + \frac{p}{d}) = \frac{4}{5} p$
$We divide both sides by p to simplify.$	$3 \times (\frac{1}{d + 4} + \frac{1}{d}) = \frac{4}{5}$
$We add the fractions by finding a common denominator.$	$\frac{1}{d + 4} + \frac{1}{d} = \frac{d + ( d + 4 )}{d ( d + 4 )} = \frac{2 d + 4}{d ( d + 4 )}$	Theory & Tactics Method Card FDPR0-B Click to view full details
$We clear denominators by multiplying both sides by 5d(d + 4).$	$15 \times (2 d + 4) = 4 d (d + 4)$	Theory & Tactics Method Card EQUA4-A Click to view full details
$We expand and rearrange into a quadratic equation.$	$15 \times (2 d + 4) = 30 d + 60 30 d + 60 = 4 d^{2} + 16 d 4 d^{2} - 14 d - 60 = 0$	Theory & Tactics Method Card EQUA2-A Click to view full details
$We divide both sides by 2 to simplify the coefficients.$	$2 d^{2} - 7 d - 30 = 0$
$We factor the quadratic expression.$	$2 d^{2} - 7 d - 30 = (2 d + 5) (d - 6)$	Theory & Tactics Method Card EQUA2-B Click to view full details
$We solve each factor for d.$	$2 d + 5 = 0 d = - \frac{5}{2} d - 6 = 0 d = 6$	Theory & Tactics Method Card EQUA1-A Click to view full details
$We discard the negative solution since time must be positive.$	$d = 6$
$We find Printer A's time to print p pages.$	$d + 4 = 6 + 4 = 10$
$We find Printer A's time to print 2p pages, cancelling p.$	$r_{A} = \frac{p}{10} Time for 2 p pages = \frac{2 p}{r _{A}} = 2 p \times \frac{10}{p} = \frac{p \times 2 \times 10}{p} = 2 \times 10 = 20$	Theory & Tactics Method Card FDPR2-A Click to view full details

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Final Answer

C