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Question Number 44967 by peter frank last updated on 06/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Oct/18

The language of problem is not clear...  pls clarify the function of pipe P and its rate  pipe Q and rate   and pipe R and rate...  by the way..  if two pipe is present one fill the tank  another empty the tank  the solution given below  filling pipe fill tank (1/6)/hour  empty pipe empty tank (1/(12))/hour  in 1 hour tank gets filled=(1/6)−(1/(12))=((2−1)/(12))=(1/(12)) portion  in 4 hours it get filled=(4/(12))=(3/4)portion  now pipe which empty the tank  switched off  remaining portion of tank to fill=1−(3/4)=(1/4)  so time taken to fill (1/(4 ))portion=(1/4)×6=(3/2)hours  so total time to fill the tank=4+(3/2)=((11)/2)hours  no mention of pipe P ,  Q   and R in problem    so pls repost the problem with clear language

$${The}\:{language}\:{of}\:{problem}\:{is}\:{not}\:{clear}... \\ $$$${pls}\:{clarify}\:{the}\:{function}\:{of}\:{pipe}\:{P}\:{and}\:{its}\:{rate} \\ $$$${pipe}\:{Q}\:{and}\:{rate}\:\:\:{and}\:{pipe}\:{R}\:{and}\:{rate}... \\ $$$${by}\:{the}\:{way}..\:\:{if}\:{two}\:{pipe}\:{is}\:{present}\:{one}\:{fill}\:{the}\:{tank} \\ $$$${another}\:{empty}\:{the}\:{tank} \\ $$$${the}\:{solution}\:{given}\:{below} \\ $$$${filling}\:{pipe}\:{fill}\:{tank}\:\frac{\mathrm{1}}{\mathrm{6}}/{hour} \\ $$$${empty}\:{pipe}\:{empty}\:{tank}\:\frac{\mathrm{1}}{\mathrm{12}}/{hour} \\ $$$${in}\:\mathrm{1}\:{hour}\:{tank}\:{gets}\:{filled}=\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{2}−\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{12}}\:{portion} \\ $$$${in}\:\mathrm{4}\:{hours}\:{it}\:{get}\:{filled}=\frac{\mathrm{4}}{\mathrm{12}}=\frac{\mathrm{3}}{\mathrm{4}}{portion} \\ $$$${now}\:{pipe}\:{which}\:{empty}\:{the}\:{tank}\:\:{switched}\:{off} \\ $$$${remaining}\:{portion}\:{of}\:{tank}\:{to}\:{fill}=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${so}\:{time}\:{taken}\:{to}\:{fill}\:\frac{\mathrm{1}}{\mathrm{4}\:}{portion}=\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{6}=\frac{\mathrm{3}}{\mathrm{2}}{hours} \\ $$$${so}\:{total}\:{time}\:{to}\:{fill}\:{the}\:{tank}=\mathrm{4}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{11}}{\mathrm{2}}{hours} \\ $$$${no}\:{mention}\:{of}\:{pipe}\:{P}\:,\:\:{Q}\:\:\:{and}\:{R}\:{in}\:{problem} \\ $$$$\:\:{so}\:{pls}\:{repost}\:{the}\:{problem}\:{with}\:{clear}\:{language} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by peter frank last updated on 07/Oct/18

thanks sir for correcton.i really appriciate your effort

$$\mathrm{t}\boldsymbol{\mathrm{hanks}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{correcton}}.\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{really}}\:\boldsymbol{\mathrm{appriciate}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{effort}} \\ $$

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