Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 March 10th, 2017, 09:35 PM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,954 Thanks: 132 Math Focus: Trigonometry [ASK] Debit A water tub has 3 water channels. Channel I and II for filling the tub and channel III to empty it. Those 3 channels together fulfill the tub for 1 hour and 20 minutes. Channel I and II together fulfill the tub for 1 hour and 12 minutes from the empty condition. Channel II and III together fulfill the tub for 4 hours from the empty condition. The duration of each channel I and channel II fulfill the tub and channel III empties the tub are... Well, I don't know where to start. But if channel I and II together (without channel III being open) fulfill the tub in 1 hour and 12 minutes, does it mean that channel III can empty the tub just within 8 minutes (because 01.20 - 01.12)?
 March 10th, 2017, 09:45 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs I would begin by defining $C_1,\,C_2,\,C_3$ as the flow rates of the 3 channels, in tubs per minute. From the given information, we know: $\displaystyle C_1+C_2-C_3=\frac{1}{80}\tag{1}$ $\displaystyle C_1+C_2=\frac{1}{72}\tag{2}$ $\displaystyle C_2-C_3=\frac{1}{240}\tag{3}$ Suppose you subtract (3) from (1)...what do you get?
 March 10th, 2017, 09:56 PM #3 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,954 Thanks: 132 Math Focus: Trigonometry $\displaystyle C_1=\frac{1}{120}$ $\displaystyle C_2=\frac{1}{180}$ $\displaystyle C_3=\frac{1}{720}$ Am I right? Thanks from MarkFL
March 10th, 2017, 09:58 PM   #4
Senior Member

Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,155
Thanks: 466

Math Focus: Calculus/ODEs
Quote:
 Originally Posted by Monox D. I-Fly $\displaystyle C_1=\frac{1}{120}$ $\displaystyle C_2=\frac{1}{180}$ $\displaystyle C_3=\frac{1}{720}$ Am I right?
Yes, those are the same values I obtained.

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