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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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closed as off-topic by Gamow, Len, xnor, McMagister, mdc32 Feb 7 '15 at 16:14

  • This question does not appear to be about creation and solving of puzzles, within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

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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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  • 1
    $\begingroup$ 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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