Two pipes are connected to the same tank. When working together, they can fill the tank in 14 hours. The larger pipe working alone, can fill the tank in 21 hrs less time than the samller one. How long would the smaller one take, working alone, to fill the tank?
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Firgure what they could do in one hour; each pipe could do a certain part of the job, and together they could do 1/14 of the job.
1/t +1/(21 +t) =1/14
muliply through by the common denominator:
t(21+t)14 (1/t + 1/(21 +t) =1/14 (t (21+t) 14)
14 (21 +t) + 14t =21 t +t^2
294 +14 t +14 t + 21 t +t^2
294 +28 t+ 21 t + t^2
move everything to the right side put in order:
0=t^2 -7t -294
(t-21) (t+14) =0;
t=21 or t= -14 which won't work.
so the time of the larger pipe is 21 hours, and the smaller pipe is t+21, or 42 hours.
1/t +1/(21 +t) =1/14
muliply through by the common denominator:
t(21+t)14 (1/t + 1/(21 +t) =1/14 (t (21+t) 14)
14 (21 +t) + 14t =21 t +t^2
294 +14 t +14 t + 21 t +t^2
294 +28 t+ 21 t + t^2
move everything to the right side put in order:
0=t^2 -7t -294
(t-21) (t+14) =0;
t=21 or t= -14 which won't work.
so the time of the larger pipe is 21 hours, and the smaller pipe is t+21, or 42 hours.
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larger pipe=x
smaller pipe=y
x+y=14 x-14=y
x=y-21
xy=(x-14).(y-21)
=xy+294
y=294?
not sure just a guess
smaller pipe=y
x+y=14 x-14=y
x=y-21
xy=(x-14).(y-21)
=xy+294
y=294?
not sure just a guess
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Let larger pipe = a
Let pipe two = b
a + b = 14
a - 21 = b
Let pipe two = b
a + b = 14
a - 21 = b
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If b is the larger tank,
a + b = 14
b = a-21
a + b = 14
b = a-21