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09 Nov 2021, 03:00
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
70% (02:54) correct
30%
(03:08)
wrong
based on 342
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Two printing presses, press A and press B, working together at their respective constant rates, can print 9p/10 pages in 2 hours. If press A takes 1 hour longer than press B to print p pages, how many hours does it take press B to print p pages?
A. 3
B. 4
C. 5
D. 6
E. 7
A. 3
B. 4
C. 5
D. 6
E. 7
09 Nov 2021, 09:07
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\( a = b + 1\)
Rate of machine A: \(\frac{1}{b+1}\)
Rate of machine B: \(\frac{1}{b}\)
Combined rate: as they do 90% of p in 2 hours the rate becomes-
\(\frac{1}{b+1} + \frac{1}{b} = \frac{\frac{9}{10}}{2}\)
\(\frac{2b+1}{b^2+b} = \frac{9}{20}\)
\(40b + 20 = 9b^2 + 9b \)
\(9b^2 - 31b - 20 = 0\)
\((9b + 5)(b - 4)\)
b = \(\frac{5}{9}\) or b = \(4\)
Answer can only be 4, so B.
Rate of machine A: \(\frac{1}{b+1}\)
Rate of machine B: \(\frac{1}{b}\)
Combined rate: as they do 90% of p in 2 hours the rate becomes-
\(\frac{1}{b+1} + \frac{1}{b} = \frac{\frac{9}{10}}{2}\)
\(\frac{2b+1}{b^2+b} = \frac{9}{20}\)
\(40b + 20 = 9b^2 + 9b \)
\(9b^2 - 31b - 20 = 0\)
\((9b + 5)(b - 4)\)
b = \(\frac{5}{9}\) or b = \(4\)
Answer can only be 4, so B.
General Discussion
10 Nov 2021, 13:40
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9p/10 pages take 2hrs ==> p pages take 20/9 hrs
Plugging in values
(c) 5hrs ==> A takes 6 hrs => p/5 + p/6 = p(11/30) ==> p pages take 30/11 hrs ... this is more than 20/9 which implies that speed of B needs to be faster
(B) 4 hrs ==> A takes 5 hrs ==> p/4 + p/5 = p (9/20) ==> bingo .. thats our answer!
Experts pls comment on method used.. i find plugging in values much easier than forming complex equations in work rate problems
Plugging in values
(c) 5hrs ==> A takes 6 hrs => p/5 + p/6 = p(11/30) ==> p pages take 30/11 hrs ... this is more than 20/9 which implies that speed of B needs to be faster
(B) 4 hrs ==> A takes 5 hrs ==> p/4 + p/5 = p (9/20) ==> bingo .. thats our answer!
Experts pls comment on method used.. i find plugging in values much easier than forming complex equations in work rate problems
