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Question Number 34850 by mondodotto@gmail.com last updated on 11/May/18

Answered by MJS last updated on 11/May/18

$$A.......\frac{1}{3}\mathrm{tank}/\min .$$$$B.......\frac{1}{4}\mathrm{tank}/\min .$$$$(\frac{1}{3}+\frac{1}{4})\times t_1\min .=\frac{1}{2}\mathrm{tank}$$$$\frac{7}{12}{t}_1=\frac{1}{2}$$$$t_1=\frac{6}{7}\min .$$$$\frac{1}{4}\times t_2\min .=\frac{1}{2}\mathrm{tank}$$$$\frac{1}{4}{t}_2=\frac{1}{2}$$$$t_2=2\min .$$$$t=t_1+t_2=\frac{20}{7}\min .\approx 2\min 51\mathrm{s}$$

Answered by candre last updated on 11/May/18

$$\mathrm{let}\mathrm{call}\mathrm{the}\mathrm{tank}\mathrm{capacity}\mathrm{T}$$$$v_A=T/3$$$$v_B=T/4$$$$v\left(t\right)=(v_A+v_B)t\left[tmin\right]$$$$(\frac{T}{3}+\frac{T}{4})t=\frac{T}{2}$$$$\frac{7T}{12}t=\frac{T}{2}$$$$t=\frac{T}{2}\cdot \frac{12}{7T}=\frac{6}{7}$$$$\mathrm{the}\mathrm{time}\mathrm{B}\mathrm{fills}\mathrm{the}\mathrm{half}\mathrm{is}$$$$v_{B}t=\frac{T}{2}$$$$t=\frac{T}{2}\cdot \frac{4}{T}=2$$$$\mathrm{the}\mathrm{total}\mathrm{time}\mathrm{is}$$$$2+\frac{6}{7}=\frac{20}{7}\approx 2,85\min$$$$\approx 2\min 51,43s$$

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