Question Number 99011 by bemath last updated on 18/Jun/20
Commented by MJS last updated on 18/Jun/20
![strange kind of question the given answers 1, 2, 3 are one possible solution but it′s not possible to find an unique solution (1/a)+(1/b)=(1/(16)) (1/b)+(1/c)=(1/(12)) (2/a)+(7/b)+(x/c)=1 [x=the time c needs to finish the work after a worked 2 days and b worked 7 days] with x as parameter we get a=−((48(x−5))/(x−27)) b=((24(x−5))/(2x−21)) c=((24(x−5))/(11)) restrictions: a>16∧b>16∧c>12 ⇒ ((21)/2)<x<27 if we set x=16 we get a=48∧b=c=24](https://www.tinkutara.com/question/Q99070.png)
$$\mathrm{strange}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{question} \\ $$$$\mathrm{the}\:\mathrm{given}\:\mathrm{answers}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3}\:\mathrm{are}\:\mathrm{one}\:\mathrm{possible} \\ $$$$\mathrm{solution}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{an} \\ $$$$\mathrm{unique}\:\mathrm{solution} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\frac{\mathrm{2}}{{a}}+\frac{\mathrm{7}}{{b}}+\frac{{x}}{{c}}=\mathrm{1} \\ $$$$\left[{x}=\mathrm{the}\:\mathrm{time}\:{c}\:\mathrm{needs}\:\mathrm{to}\:\mathrm{finish}\:\mathrm{the}\:\mathrm{work}\:\mathrm{after}\right. \\ $$$$\left.\:\:{a}\:\mathrm{worked}\:\mathrm{2}\:\mathrm{days}\:\mathrm{and}\:{b}\:\mathrm{worked}\:\mathrm{7}\:\mathrm{days}\right] \\ $$$$\mathrm{with}\:{x}\:\mathrm{as}\:\mathrm{parameter}\:\mathrm{we}\:\mathrm{get} \\ $$$${a}=−\frac{\mathrm{48}\left({x}−\mathrm{5}\right)}{{x}−\mathrm{27}} \\ $$$${b}=\frac{\mathrm{24}\left({x}−\mathrm{5}\right)}{\mathrm{2}{x}−\mathrm{21}} \\ $$$${c}=\frac{\mathrm{24}\left({x}−\mathrm{5}\right)}{\mathrm{11}} \\ $$$$\mathrm{restrictions}: \\ $$$${a}>\mathrm{16}\wedge{b}>\mathrm{16}\wedge{c}>\mathrm{12}\:\Rightarrow\:\frac{\mathrm{21}}{\mathrm{2}}<{x}<\mathrm{27} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{set}\:{x}=\mathrm{16}\:\mathrm{we}\:\mathrm{get}\:{a}=\mathrm{48}\wedge{b}={c}=\mathrm{24} \\ $$
Commented by bemath last updated on 18/Jun/20
$$\mathrm{waw}..\mathrm{thanks}\:\mathrm{mr}\:\mathrm{Mjs} \\ $$
Answered by bobhans last updated on 18/Jun/20
$$\left(\mathrm{v}_{\mathrm{A}} +\mathrm{v}_{\mathrm{B}} \right)×\mathrm{16}\:=\:\mathrm{w} \\ $$$$\left(\mathrm{v}_{\mathrm{B}} +\mathrm{v}_{\mathrm{C}} \right)×\mathrm{12}\:=\:\mathrm{w} \\ $$$$\mathrm{2v}_{\mathrm{A}} +\mathrm{7v}_{\mathrm{B}} +\mathrm{t}.\mathrm{v}_{\mathrm{C}} \:=\:\mathrm{w}\: \\ $$$$\mathrm{information}\:\mathrm{on}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{is}\:\mathrm{incomplete} \\ $$