Tom, Dick and Harry piled stones into a pyramid to my precise specifications. Fred, an efficiency expert carefully watched them.
When the pyramid was complete, Fred reported that if Tom had worked twice as fast the job would have taken exactly 2 hours less, if Dick had worked three times as fast the job would have taken exactly 3 hours less, and if Harry had worked four times as fast the job would have taken exactly 4 hours less.
How long had the job taken?
This is an attempt at an analytic solution without Wolfram Alpha; so far, I have an expression for t in terms of a,b,c; this formula works correctly with Charlie's numbers.
a,b,c are in units per hour corresponding to Tom, Dick, and Harry.
It is apparent from the text that Harry is the fastest worker, we expect: a < b < c
and that t > 4 (t being the time to do the job when they all worked normally)
W = (a + b + c)t
W = (2a + b + c)(t-2)
W = (a + 3b + c)(t-3)
W = (a + b + 4c)(t-4)
subtract first from each of the others:
(2a + b + c)t - 2(2a + b + c) - (a + b + c)t = 0
(a + 3b + c)t - 3(a + 3b + c) - (a + b + c)t = 0
(a + b + 4c)t - 4(a + b + 4c) - (a + b + c)t = 0
at = 2(2a + b + c)
2bt = 3(a + 3b + c)
3ct = 4(a + b + 4c)
add all 3 new eqns
(a + 2b + 3c)t = 11a + 15b + 21c
t = (11a + 15b + 21c) / (a + 2b + 3c) <--- formula for t
(so I don't have a value for t, just a formula)
When I plug my 4 simultaneous equations into Wolfram, I get the same numbers as Charlie:
t≈7.83815 + 0 i
c≈(0.0443204 + 0 i)
b≈(0.0395547 + 0 i) <-- expect b>a but b<a
a≈(0.043706 + 0 i)
Reassuringly, when I plug these a,b,c values into this formula, I do get the same value for time.
But looking at the values for a,b,c coming from Wolfram,
it is not true that a < b < c
which is puzzling...
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Posted by Larry
on 2023-12-23 14:43:51 |
Beginning of analytic solution; incomplete
