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Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in: A. 7 hours B. 7.5 hours C. 8 hours D. 6 hours The right solution of this is 7 hours but i get 7.05 I don't know what I'm doing wrong can someone help me just in the first steps?

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    Please show the calculation you carried out. – 2017-02-15
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    (1\12 + 1\15)x\2 + (1\12 + 1\20)x\2 =1 I tried this equation but it doesn't work I don't know what I'm doing wrong – 2017-02-15
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    That equation appears to assume that it takes an even number of hours. Hint: Tap $B$ is on more often than tap $C$. – 2017-02-15
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    yeah but that doesn't really help – 2017-02-15
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    It should. If you want to stick to your method, just remark that (at most) $B$ can be on for one more hour than $C$. I'm not crazy about this method, as it seems to presume that the answer is an integer number of hours. But it works here. – 2017-02-15
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    My method: in alternate hours, we complete $\frac 1{12}+\frac 1{15}$ and $\frac 1{12}+\frac 1{20}$. So Just do a running sum until you get a total greater than or equal to $1$. That will tell you which hour the work gets done in, at which point a formula like yours will work. – 2017-02-15
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    (1\12 + 1\15)(x+1) + (1\12 + 1\20)x=1 this gives x=3 and then the total time is x+(x+1) which is equal to 7 – 2017-02-15
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    Right. As I say, this works here because the answer is an integer. It wouldn't work in general. – 2017-02-15
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    Solved or have any doubt? – 2017-02-22
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    yes i solved it – 2017-02-23

1 Answers 1

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Hint: When you let the taps on for 2 hours ($A$ for $2$, $B$ and $C$ for $1$), you fill in $2/12 + 1/15 + 1/20$ of the tank. Multiplying that number by $3$ gives $<1$ and multiplying that number by $4$ gives $>1$, so the answer can be neither $C$ or $D$. The one extra hour to $A$ fills in $1/12+1/15$ of the tank and the one-and-half extra hour to $B$ fills in $1/12\cdot3/2+1/15+1/20\cdot1/2$.