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Question Number 4539 by love math last updated on 05/Feb/16

Two tractor, working together, have plowed the field for 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 100 hours. How many hours to plow the field if every tractor working separately?

$${Two}\:{tractor},\:{working}\:{together},\: \\ $$$${have}\:{plowed}\:{the}\:{field}\:{for}\:\mathrm{48}\:{hours}. \\ $$$${If}\:{half}\:{of}\:{the}\:{field}\:{is}\:{plowed}\:{by}\: \\ $$$${one}\:{of}\:{them}\:{and}\:{then}\:{the}\:{remaining} \\ $$$${half}\:{to}\:{the}\:{other},\:{the}\:{work}\:{would} \\ $$$${be}\:{carried}\:{out}\:\mathrm{100}\:{hours}.\:{How}\:{many} \\ $$$${hours}\:{to}\:{plow}\:{the}\:{field}\:{if}\:{every}\:{tractor} \\ $$$${working}\:{separately}? \\ $$

Commented by Yozzii last updated on 07/Feb/16

Let p(1)=area plowed per hour by tractor 1 and p(2) the area plowed per hour by tractor 2. In general, define p=((total area plowed)/(total time taken))=(a/t) (a,t>0) ⇒t=(a/p). For fixed a⇒t∝(1/p). ⇒ for two tractors, t falls since p=p(1)+p(2). ∴ 48(p(1)+p(2))=a........(1) Since t=(a/p), for the second case, 100=((0.5a)/(p(1)))+((0.5a)/(p(2))) ((200)/a)=(1/(p(1)))+(1/(p(2)))=((p(1)+p(2))/(p(1)p(2)))=((a/48)/(p(1)p(2))) p(1)p(2)=(a^2 /(48×200))=(a^2 /(9600)) ⇒p(1)=(a^2 /(9600p(2))).............(2) Substituting p(1) from (2) into (1) ⇒ 48((a^2 /(9600p(2)))+p(2))=a (a^2 /(200p(2)))+48p(2)=a a^2 +9600p(2)^2 =200ap(2) 9600p(2)^2 −200ap(2)+a^2 =0 ⇒p(2)=((200a±(√(40000a^2 −4×9600×a^2 )))/(2×9600)) p(2)=((200a±40a)/(2×10×2×480))=((5a±a)/(480))=(6/(480))a,(4/(480))a p(2)=(a/(80)),(a/(120)) ∴ p(1)=(a^2 /(9600))×((80)/a),(a^2 /(9600))×((120)/a) p(1)=(a/(120)),(a/(80)). So {p(1),p(2)}={(a/(80)),(a/(120))}∨{(a/(120)),(a/(80))} If tractor 1 works alone, t_1 =(a/(a/80))=80hours If tractor 2 works alone, t_2 =(a/(a/120))=120 hours. The symbols for each tractor are assigned arbitrarily so that in all, one of the tractors (1 or 2) takes 80 hours by themselves to plow the field while the other(2 or 1) takes 120 hours.

$${Let}\:{p}\left(\mathrm{1}\right)={area}\:{plowed}\:{per}\:{hour}\:{by} \\ $$$${tractor}\:\mathrm{1}\:{and}\:{p}\left(\mathrm{2}\right)\:{the}\:{area}\:{plowed} \\ $$$${per}\:{hour}\:{by}\:{tractor}\:\mathrm{2}.\:{In}\:{general},\: \\ $$$${define}\:{p}=\frac{{total}\:{area}\:{plowed}}{{total}\:{time}\:{taken}}=\frac{{a}}{{t}}\:\left({a},{t}>\mathrm{0}\right) \\ $$$$\Rightarrow{t}=\frac{{a}}{{p}}.\:\:{For}\:{fixed}\:{a}\Rightarrow{t}\propto\frac{\mathrm{1}}{{p}}.\: \\ $$$$\Rightarrow\:{for}\:{two}\:{tractors},\:{t}\:{falls}\:{since}\: \\ $$$${p}={p}\left(\mathrm{1}\right)+{p}\left(\mathrm{2}\right).\: \\ $$$$\therefore\:\mathrm{48}\left({p}\left(\mathrm{1}\right)+{p}\left(\mathrm{2}\right)\right)={a}........\left(\mathrm{1}\right) \\ $$$${Since}\:{t}=\frac{{a}}{{p}},\:{for}\:{the}\:{second}\:{case}, \\ $$$$\mathrm{100}=\frac{\mathrm{0}.\mathrm{5}{a}}{{p}\left(\mathrm{1}\right)}+\frac{\mathrm{0}.\mathrm{5}{a}}{{p}\left(\mathrm{2}\right)} \\ $$$$\frac{\mathrm{200}}{{a}}=\frac{\mathrm{1}}{{p}\left(\mathrm{1}\right)}+\frac{\mathrm{1}}{{p}\left(\mathrm{2}\right)}=\frac{{p}\left(\mathrm{1}\right)+{p}\left(\mathrm{2}\right)}{{p}\left(\mathrm{1}\right){p}\left(\mathrm{2}\right)}=\frac{{a}/\mathrm{48}}{{p}\left(\mathrm{1}\right){p}\left(\mathrm{2}\right)} \\ $$$${p}\left(\mathrm{1}\right){p}\left(\mathrm{2}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{48}×\mathrm{200}}=\frac{{a}^{\mathrm{2}} }{\mathrm{9600}} \\ $$$$\Rightarrow{p}\left(\mathrm{1}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{9600}{p}\left(\mathrm{2}\right)}.............\left(\mathrm{2}\right) \\ $$$${Substituting}\:{p}\left(\mathrm{1}\right)\:{from}\:\left(\mathrm{2}\right)\:{into}\:\left(\mathrm{1}\right) \\ $$$$\Rightarrow\:\mathrm{48}\left(\frac{{a}^{\mathrm{2}} }{\mathrm{9600}{p}\left(\mathrm{2}\right)}+{p}\left(\mathrm{2}\right)\right)={a} \\ $$$$\frac{{a}^{\mathrm{2}} }{\mathrm{200}{p}\left(\mathrm{2}\right)}+\mathrm{48}{p}\left(\mathrm{2}\right)={a} \\ $$$${a}^{\mathrm{2}} +\mathrm{9600}{p}\left(\mathrm{2}\right)^{\mathrm{2}} =\mathrm{200}{ap}\left(\mathrm{2}\right) \\ $$$$\mathrm{9600}{p}\left(\mathrm{2}\right)^{\mathrm{2}} −\mathrm{200}{ap}\left(\mathrm{2}\right)+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{p}\left(\mathrm{2}\right)=\frac{\mathrm{200}{a}\pm\sqrt{\mathrm{40000}{a}^{\mathrm{2}} −\mathrm{4}×\mathrm{9600}×{a}^{\mathrm{2}} }}{\mathrm{2}×\mathrm{9600}} \\ $$$${p}\left(\mathrm{2}\right)=\frac{\mathrm{200}{a}\pm\mathrm{40}{a}}{\mathrm{2}×\mathrm{10}×\mathrm{2}×\mathrm{480}}=\frac{\mathrm{5}{a}\pm{a}}{\mathrm{480}}=\frac{\mathrm{6}}{\mathrm{480}}{a},\frac{\mathrm{4}}{\mathrm{480}}{a} \\ $$$${p}\left(\mathrm{2}\right)=\frac{{a}}{\mathrm{80}},\frac{{a}}{\mathrm{120}} \\ $$$$\therefore\:{p}\left(\mathrm{1}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{9600}}×\frac{\mathrm{80}}{{a}},\frac{{a}^{\mathrm{2}} }{\mathrm{9600}}×\frac{\mathrm{120}}{{a}} \\ $$$${p}\left(\mathrm{1}\right)=\frac{{a}}{\mathrm{120}},\frac{{a}}{\mathrm{80}}. \\ $$$${So}\:\left\{{p}\left(\mathrm{1}\right),{p}\left(\mathrm{2}\right)\right\}=\left\{\frac{{a}}{\mathrm{80}},\frac{{a}}{\mathrm{120}}\right\}\vee\left\{\frac{{a}}{\mathrm{120}},\frac{{a}}{\mathrm{80}}\right\} \\ $$$${If}\:{tractor}\:\mathrm{1}\:{works}\:{alone},\: \\ $$$${t}_{\mathrm{1}} =\frac{{a}}{{a}/\mathrm{80}}=\mathrm{80}{hours}\: \\ $$$${If}\:{tractor}\:\mathrm{2}\:{works}\:{alone}, \\ $$$${t}_{\mathrm{2}} =\frac{{a}}{{a}/\mathrm{120}}=\mathrm{120}\:{hours}. \\ $$$${The}\:{symbols}\:{for}\:{each}\:{tractor}\:{are}\: \\ $$$${assigned}\:{arbitrarily}\:{so}\:{that}\:{in}\:{all}, \\ $$$${one}\:{of}\:{the}\:{tractors}\:\left(\mathrm{1}\:{or}\:\mathrm{2}\right)\:{takes}\:\mathrm{80} \\ $$$${hours}\:{by}\:{themselves}\:{to}\:{plow}\:{the}\:{field} \\ $$$${while}\:{the}\:{other}\left(\mathrm{2}\:{or}\:\mathrm{1}\right)\:{takes}\:\mathrm{120}\:{hours}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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 (1/x) of field per hour B ploughs (1/y) of field per hour In 48 hours : A ploughs ((48)/x) of field B ploughs ((48)/y) of field ((48)/x)+((48)/y)=1(Whole field).........(i) Half field: A ploughs half field in (x/2) hours B ploughs half field in (y/2) hours (x/2)+(y/2)=100.....................(ii) x+y=200 y=200−x Substituting in (i) ((48)/x)+((48)/(200−x))=1 48(200−x)+48x=x(200−x) 9600−48x+48x=200x−x^2 x^2 −200x+9600=0 x=((−(−200)±(√((−200)^2 −4(1)(9600))))/(2(1))) =((200±40)/2)=120,80 If A ploughs the field in 120 hours ,B ploughs 200−120=80 hours. If A ploughs the field in 80 hours ,B ploughs 200−80=120 hours.

$${Let}\:{tractor}\:{A}\:{can}\:{plough}\:{whole}\:{field} \\ $$$$\:\:\:\:\:\:{in}\:{x}\:{hours}\:\:{AND} \\ $$$$\:\:\:\:\:\:\:{tractor}\:{B}\:{in}\:{y}\:{hours}. \\ $$$${Per}\:{hour}\:: \\ $$$$\:\:{A}\:{ploughs}\:\frac{\mathrm{1}}{{x}}\:{of}\:\:{field}\:{per}\:{hour} \\ $$$$\:\:{B}\:{ploughs}\:\frac{\mathrm{1}}{{y}}\:{of}\:\:{field}\:{per}\:{hour} \\ $$$${In}\:\mathrm{48}\:{hours}\:: \\ $$$$\:\:\:{A}\:{ploughs}\:\frac{\mathrm{48}}{{x}}\:{of}\:{field} \\ $$$$\:\:\:{B}\:{ploughs}\:\frac{\mathrm{48}}{{y}}\:{of}\:{field} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{48}}{{x}}+\frac{\mathrm{48}}{{y}}=\mathrm{1}\left({Whole}\:{field}\right).........\left({i}\right) \\ $$$${Half}\:\:{field}: \\ $$$$\:\:\:{A}\:{ploughs}\:\:{half}\:{field}\:{in}\:\frac{{x}}{\mathrm{2}}\:{hours} \\ $$$$\:\:\:{B}\:{ploughs}\:\:{half}\:{field}\:{in}\:\frac{{y}}{\mathrm{2}}\:{hours} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{x}}{\mathrm{2}}+\frac{{y}}{\mathrm{2}}=\mathrm{100}.....................\left({ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}+{y}=\mathrm{200} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}=\mathrm{200}−{x} \\ $$$${Substituting}\:{in}\:\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{48}}{{x}}+\frac{\mathrm{48}}{\mathrm{200}−{x}}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{48}\left(\mathrm{200}−{x}\right)+\mathrm{48}{x}={x}\left(\mathrm{200}−{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{9600}−\mathrm{48}{x}+\mathrm{48}{x}=\mathrm{200}{x}−{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{200}{x}+\mathrm{9600}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{x}=\frac{−\left(−\mathrm{200}\right)\pm\sqrt{\left(−\mathrm{200}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{9600}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{200}\pm\mathrm{40}}{\mathrm{2}}=\mathrm{120},\mathrm{80} \\ $$$$ \\ $$$${If}\:{A}\:{ploughs}\:{the}\:{field}\:{in}\:\mathrm{120}\:{hours}\:,{B}\:{ploughs} \\ $$$$\mathrm{200}−\mathrm{120}=\mathrm{80}\:{hours}. \\ $$$${If}\:{A}\:{ploughs}\:{the}\:{field}\:{in}\:\mathrm{80}\:{hours}\:,{B}\:{ploughs} \\ $$$$\mathrm{200}−\mathrm{80}=\mathrm{120}\:{hours}. \\ $$

Commented by Yozzii last updated on 07/Feb/16

My answer satisfies your equations. (x,y)=(80,120)

$${My}\:{answer}\:{satisfies}\:{your}\:{equations}. \\ $$$$\left({x},{y}\right)=\left(\mathrm{80},\mathrm{120}\right) \\ $$

Commented by Rasheed Soomro last updated on 07/Feb/16

Yes answers are same.

$${Yes}\:{answers}\:{are}\:{same}. \\ $$

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