Question Number 4539 by love math last updated on 05/Feb/16
$${Two}\:{tractor},\:{working}\:{together},\: \\ $$$${have}\:{plowed}\:{the}\:{field}\:{for}\:\mathrm{48}\:{hours}. \\ $$$${If}\:{half}\:{of}\:{the}\:{field}\:{is}\:{plowed}\:{by}\: \\ $$$${one}\:{of}\:{them}\:{and}\:{then}\:{the}\:{remaining} \\ $$$${half}\:{to}\:{the}\:{other},\:{the}\:{work}\:{would} \\ $$$${be}\:{carried}\:{out}\:\mathrm{100}\:{hours}.\:{How}\:{many} \\ $$$${hours}\:{to}\:{plow}\:{the}\:{field}\:{if}\:{every}\:{tractor} \\ $$$${working}\:{separately}? \\ $$
Commented by Yozzii last updated on 07/Feb/16
$${Let}\:{p}\left(\mathrm{1}\right)={area}\:{plowed}\:{per}\:{hour}\:{by} \\ $$$${tractor}\:\mathrm{1}\:{and}\:{p}\left(\mathrm{2}\right)\:{the}\:{area}\:{plowed} \\ $$$${per}\:{hour}\:{by}\:{tractor}\:\mathrm{2}.\:{In}\:{general},\: \\ $$$${define}\:{p}=\frac{{total}\:{area}\:{plowed}}{{total}\:{time}\:{taken}}=\frac{{a}}{{t}}\:\left({a},{t}>\mathrm{0}\right) \\ $$$$\Rightarrow{t}=\frac{{a}}{{p}}.\:\:{For}\:{fixed}\:{a}\Rightarrow{t}\propto\frac{\mathrm{1}}{{p}}.\: \\ $$$$\Rightarrow\:{for}\:{two}\:{tractors},\:{t}\:{falls}\:{since}\: \\ $$$${p}={p}\left(\mathrm{1}\right)+{p}\left(\mathrm{2}\right).\: \\ $$$$\therefore\:\mathrm{48}\left({p}\left(\mathrm{1}\right)+{p}\left(\mathrm{2}\right)\right)={a}……..\left(\mathrm{1}\right) \\ $$$${Since}\:{t}=\frac{{a}}{{p}},\:{for}\:{the}\:{second}\:{case}, \\ $$$$\mathrm{100}=\frac{\mathrm{0}.\mathrm{5}{a}}{{p}\left(\mathrm{1}\right)}+\frac{\mathrm{0}.\mathrm{5}{a}}{{p}\left(\mathrm{2}\right)} \\ $$$$\frac{\mathrm{200}}{{a}}=\frac{\mathrm{1}}{{p}\left(\mathrm{1}\right)}+\frac{\mathrm{1}}{{p}\left(\mathrm{2}\right)}=\frac{{p}\left(\mathrm{1}\right)+{p}\left(\mathrm{2}\right)}{{p}\left(\mathrm{1}\right){p}\left(\mathrm{2}\right)}=\frac{{a}/\mathrm{48}}{{p}\left(\mathrm{1}\right){p}\left(\mathrm{2}\right)} \\ $$$${p}\left(\mathrm{1}\right){p}\left(\mathrm{2}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{48}×\mathrm{200}}=\frac{{a}^{\mathrm{2}} }{\mathrm{9600}} \\ $$$$\Rightarrow{p}\left(\mathrm{1}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{9600}{p}\left(\mathrm{2}\right)}………….\left(\mathrm{2}\right) \\ $$$${Substituting}\:{p}\left(\mathrm{1}\right)\:{from}\:\left(\mathrm{2}\right)\:{into}\:\left(\mathrm{1}\right) \\ $$$$\Rightarrow\:\mathrm{48}\left(\frac{{a}^{\mathrm{2}} }{\mathrm{9600}{p}\left(\mathrm{2}\right)}+{p}\left(\mathrm{2}\right)\right)={a} \\ $$$$\frac{{a}^{\mathrm{2}} }{\mathrm{200}{p}\left(\mathrm{2}\right)}+\mathrm{48}{p}\left(\mathrm{2}\right)={a} \\ $$$${a}^{\mathrm{2}} +\mathrm{9600}{p}\left(\mathrm{2}\right)^{\mathrm{2}} =\mathrm{200}{ap}\left(\mathrm{2}\right) \\ $$$$\mathrm{9600}{p}\left(\mathrm{2}\right)^{\mathrm{2}} −\mathrm{200}{ap}\left(\mathrm{2}\right)+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{p}\left(\mathrm{2}\right)=\frac{\mathrm{200}{a}\pm\sqrt{\mathrm{40000}{a}^{\mathrm{2}} −\mathrm{4}×\mathrm{9600}×{a}^{\mathrm{2}} }}{\mathrm{2}×\mathrm{9600}} \\ $$$${p}\left(\mathrm{2}\right)=\frac{\mathrm{200}{a}\pm\mathrm{40}{a}}{\mathrm{2}×\mathrm{10}×\mathrm{2}×\mathrm{480}}=\frac{\mathrm{5}{a}\pm{a}}{\mathrm{480}}=\frac{\mathrm{6}}{\mathrm{480}}{a},\frac{\mathrm{4}}{\mathrm{480}}{a} \\ $$$${p}\left(\mathrm{2}\right)=\frac{{a}}{\mathrm{80}},\frac{{a}}{\mathrm{120}} \\ $$$$\therefore\:{p}\left(\mathrm{1}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{9600}}×\frac{\mathrm{80}}{{a}},\frac{{a}^{\mathrm{2}} }{\mathrm{9600}}×\frac{\mathrm{120}}{{a}} \\ $$$${p}\left(\mathrm{1}\right)=\frac{{a}}{\mathrm{120}},\frac{{a}}{\mathrm{80}}. \\ $$$${So}\:\left\{{p}\left(\mathrm{1}\right),{p}\left(\mathrm{2}\right)\right\}=\left\{\frac{{a}}{\mathrm{80}},\frac{{a}}{\mathrm{120}}\right\}\vee\left\{\frac{{a}}{\mathrm{120}},\frac{{a}}{\mathrm{80}}\right\} \\ $$$${If}\:{tractor}\:\mathrm{1}\:{works}\:{alone},\: \\ $$$${t}_{\mathrm{1}} =\frac{{a}}{{a}/\mathrm{80}}=\mathrm{80}{hours}\: \\ $$$${If}\:{tractor}\:\mathrm{2}\:{works}\:{alone}, \\ $$$${t}_{\mathrm{2}} =\frac{{a}}{{a}/\mathrm{120}}=\mathrm{120}\:{hours}. \\ $$$${The}\:{symbols}\:{for}\:{each}\:{tractor}\:{are}\: \\ $$$${assigned}\:{arbitrarily}\:{so}\:{that}\:{in}\:{all}, \\ $$$${one}\:{of}\:{the}\:{tractors}\:\left(\mathrm{1}\:{or}\:\mathrm{2}\right)\:{takes}\:\mathrm{80} \\ $$$${hours}\:{by}\:{themselves}\:{to}\:{plow}\:{the}\:{field} \\ $$$${while}\:{the}\:{other}\left(\mathrm{2}\:{or}\:\mathrm{1}\right)\:{takes}\:\mathrm{120}\:{hours}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 07/Feb/16
$${Let}\:{tractor}\:{A}\:{can}\:{plough}\:{whole}\:{field} \\ $$$$\:\:\:\:\:\:{in}\:{x}\:{hours}\:\:{AND} \\ $$$$\:\:\:\:\:\:\:{tractor}\:{B}\:{in}\:{y}\:{hours}. \\ $$$${Per}\:{hour}\:: \\ $$$$\:\:{A}\:{ploughs}\:\frac{\mathrm{1}}{{x}}\:{of}\:\:{field}\:{per}\:{hour} \\ $$$$\:\:{B}\:{ploughs}\:\frac{\mathrm{1}}{{y}}\:{of}\:\:{field}\:{per}\:{hour} \\ $$$${In}\:\mathrm{48}\:{hours}\:: \\ $$$$\:\:\:{A}\:{ploughs}\:\frac{\mathrm{48}}{{x}}\:{of}\:{field} \\ $$$$\:\:\:{B}\:{ploughs}\:\frac{\mathrm{48}}{{y}}\:{of}\:{field} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{48}}{{x}}+\frac{\mathrm{48}}{{y}}=\mathrm{1}\left({Whole}\:{field}\right)………\left({i}\right) \\ $$$${Half}\:\:{field}: \\ $$$$\:\:\:{A}\:{ploughs}\:\:{half}\:{field}\:{in}\:\frac{{x}}{\mathrm{2}}\:{hours} \\ $$$$\:\:\:{B}\:{ploughs}\:\:{half}\:{field}\:{in}\:\frac{{y}}{\mathrm{2}}\:{hours} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{x}}{\mathrm{2}}+\frac{{y}}{\mathrm{2}}=\mathrm{100}…………………\left({ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}+{y}=\mathrm{200} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}=\mathrm{200}−{x} \\ $$$${Substituting}\:{in}\:\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{48}}{{x}}+\frac{\mathrm{48}}{\mathrm{200}−{x}}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{48}\left(\mathrm{200}−{x}\right)+\mathrm{48}{x}={x}\left(\mathrm{200}−{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{9600}−\mathrm{48}{x}+\mathrm{48}{x}=\mathrm{200}{x}−{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{200}{x}+\mathrm{9600}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{x}=\frac{−\left(−\mathrm{200}\right)\pm\sqrt{\left(−\mathrm{200}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{9600}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{200}\pm\mathrm{40}}{\mathrm{2}}=\mathrm{120},\mathrm{80} \\ $$$$ \\ $$$${If}\:{A}\:{ploughs}\:{the}\:{field}\:{in}\:\mathrm{120}\:{hours}\:,{B}\:{ploughs} \\ $$$$\mathrm{200}−\mathrm{120}=\mathrm{80}\:{hours}. \\ $$$${If}\:{A}\:{ploughs}\:{the}\:{field}\:{in}\:\mathrm{80}\:{hours}\:,{B}\:{ploughs} \\ $$$$\mathrm{200}−\mathrm{80}=\mathrm{120}\:{hours}. \\ $$
Commented by Yozzii last updated on 07/Feb/16
$${My}\:{answer}\:{satisfies}\:{your}\:{equations}. \\ $$$$\left({x},{y}\right)=\left(\mathrm{80},\mathrm{120}\right) \\ $$
Commented by Rasheed Soomro last updated on 07/Feb/16
$${Yes}\:{answers}\:{are}\:{same}. \\ $$