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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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