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Two workers A and B manufactured a batch of identical parts. A worked for 2 hours and B worked for 5 hours and they did half the job. Then they worked together for another 3 hours and they had to do (1/20)th of the job. How much time does B take to complete the job, if he worked alone?

  • A) 24 hours
  • B) 12 hours
  • C) 15 hours
  • D) 30 hours

HI can anyone please explain how to solve this. Answer is given as option C

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    $\begingroup$ Is this considered a puzzle? I think this is just standard mathematics. $\endgroup$ – justhalf Jul 15 '14 at 3:26
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    $\begingroup$ looks like homework... $\endgroup$ – klm123 Jul 15 '14 at 5:31
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    $\begingroup$ I'm voting to close this question as off-topic because it is a run-off-the-mill exercise in algebra. $\endgroup$ – Gamow Feb 6 '15 at 17:58
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Let's denote the speed of A (in working) as $x$ and the speed of B (in working) as $y$. Then we have:

$$2x+5y = \frac{1}{2}$$ $$3x+3y = \frac{1}{2} - \frac{1}{20} = \frac{9}{20}$$

Just solve that (details left for exercise) to get the final result:

$$x = \frac{1}{12}, y = \frac{1}{15}$$

Which means B will finish the job alone in 15 hours.

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    $\begingroup$ What do you mean by relationship? The relationship is already shown as the two simultaneous equations of $x$ and $y$. $\endgroup$ – justhalf Jul 16 '14 at 2:23
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    $\begingroup$ When you have two simultaneous linear equations of two variables that are independent, you can figure out the value of each variable. This is basic algebra: tutorial.math.lamar.edu/Classes/Alg/SystemsTwoVrble.aspx $\endgroup$ – justhalf Jul 16 '14 at 3:45
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    $\begingroup$ @BrianTkatch "How does the second equation resolve that?" -- By making all but one of the possible answers to the first equation wrong. $\endgroup$ – Jim Balter Jul 17 '14 at 9:43
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    $\begingroup$ @BrianTkatch From the second equation, x + y = 3/20, y = 3/20 - x, substitute into first equation: 2x + 3/4 - 5x = 1/2, 3x = 3/4 - 1/2, x= 1/12. It's hard for me to comprehend how anyone who ever took algebra doesn't how to do this or, even more fundamentally, could exhibit your confusion as to "how" a second linear equation in two unknowns "resolves" the first one. $\endgroup$ – Jim Balter Jul 17 '14 at 20:34
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    $\begingroup$ @JimBalter I understand the concepts, but i haven't really practiced it in over 25 years. I hate to admit it, but i'm bad at it. :( I was hoping to read a fun example or two explaining it, that's all. Back to the problem. How does x + y = 3/20? I see 3x + 3y = 1/2 - 1/20 = 9/20. 3y = 9/20 - 3x, ah, there it is, y = 1/20 - x. (first time around i tried 19/20, oops!) substitute in first equation, 2x + 5(1/20 - x) = 1/2, 2x + 5/20 - 5x = 1/2, 5/20 - 3x = 1/2, 3x = 1/2 - 5/20, 3x = 5/20, 3x = 1/4. x = 1/12. Which means 5y = 1/2 - 2(1/12), 5y = 1/2 - 2/12, 5y = 4/12 = 1/3, y = 1/15. Bingo! Thanx! $\endgroup$ – please delete me Jul 17 '14 at 23:24

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