Question 388218: here is another tricky question.
On a certain day two inspectors, A and B, employed by a factory, were inspecting a lot of manufactured articles. In the morning A ha inspected 1500 articles before B started in. during the remainder of the morning they inspected 4000 more, A working 4 hours altogether. In the afternoon A worked 4 hrs. and B 2 hrs., in which time they inspected 5200 articles. Find the average number of articles that each inspected per hour and the number of hrs. that B worked in the morning.
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! Okay this is what I got:
let a = articles/hr done by A
b = articles/hr done by B.
Let x = # hrs done by A in the morning BEFORE B joined him.
Then:
ax = 1500 <------(1)
(a+b)(4-x) = 4000 <------(2)
4a + 2b = 5200, which is the same as 2a + b = 2600<---(3)
We got a system of non-linear equations.
From eq'n (3), b = 2600 - 2a
From eq'n(1), x = 1500/a. Plugging both of these into (2), we get
 ;
 ;
 ;
 ;
 . This is equivalent to
 after multiplying both sides by a and transposition.
(a - 1000)(a - 975) = 0----> a = 1000, or 975.
If a = 1000, b = 600, and x = 1.5 hrs.
If a = 975, b = 650, and x = 20/13 hrs.
So we actually have 2 possible scenarios here, both being viable.
B worked for 4 - x = 4 - 1.5 = 2.5 hours in the morning, OR, equally possible,
4 - 20/13 = 32/13 hours in the morning.
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