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Question Number 16942 by Tinkutara last updated on 28/Jun/17
A distance of 200 km is to be covered by  car in less than 10 hours. Yash does it  in two parts. He first drives for 150 km  at an average speed of 36 km/hr,  without stopping. After taking rest for  30 minutes, he starts again and covers  the remaining distance non-stop. His  average for the entire journey  (including the period of rest) exceeds  that for the second part by 5 km/hr.  Find the speed at which he covers the  second part.
$$\mathrm{A}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{200}\:\mathrm{km}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{covered}\:\mathrm{by} \\ $$$$\mathrm{car}\:\mathrm{in}\:\mathrm{less}\:\mathrm{than}\:\mathrm{10}\:\mathrm{hours}.\:\mathrm{Yash}\:\mathrm{does}\:\mathrm{it} \\ $$$$\mathrm{in}\:\mathrm{two}\:\mathrm{parts}.\:\mathrm{He}\:\mathrm{first}\:\mathrm{drives}\:\mathrm{for}\:\mathrm{150}\:\mathrm{km} \\ $$$$\mathrm{at}\:\mathrm{an}\:\mathrm{average}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{36}\:\mathrm{km}/\mathrm{hr}, \\ $$$$\mathrm{without}\:\mathrm{stopping}.\:\mathrm{After}\:\mathrm{taking}\:\mathrm{rest}\:\mathrm{for} \\ $$$$\mathrm{30}\:\mathrm{minutes},\:\mathrm{he}\:\mathrm{starts}\:\mathrm{again}\:\mathrm{and}\:\mathrm{covers} \\ $$$$\mathrm{the}\:\mathrm{remaining}\:\mathrm{distance}\:\mathrm{non}-\mathrm{stop}.\:\mathrm{His} \\ $$$$\mathrm{average}\:\mathrm{for}\:\mathrm{the}\:\mathrm{entire}\:\mathrm{journey} \\ $$$$\left(\mathrm{including}\:\mathrm{the}\:\mathrm{period}\:\mathrm{of}\:\mathrm{rest}\right)\:\mathrm{exceeds} \\ $$$$\mathrm{that}\:\mathrm{for}\:\mathrm{the}\:\mathrm{second}\:\mathrm{part}\:\mathrm{by}\:\mathrm{5}\:\mathrm{km}/\mathrm{hr}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{at}\:\mathrm{which}\:\mathrm{he}\:\mathrm{covers}\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{part}. \\ $$
Answered by mrW1 last updated on 29/Jun/17
 ((200)/(((150)/(36))+0.5+((50)/x)))=x+5  ((200×36x)/(168x+1800))=x+5  168x^2 +1800x+840x+9000−7200x=0  168x^2 −4560x+9000=0  7x^2 −190x+375=0  x=((190±(√(190^2 −4×7×375)))/(2×7))=((190±160)/(14))  ⇒x_1 =((190−160)/(14))=2.14 (not suitable)  ⇒x_2 =((190+160)/(14))=25 km/h (ok)
$$\:\frac{\mathrm{200}}{\frac{\mathrm{150}}{\mathrm{36}}+\mathrm{0}.\mathrm{5}+\frac{\mathrm{50}}{\mathrm{x}}}=\mathrm{x}+\mathrm{5} \\ $$$$\frac{\mathrm{200}×\mathrm{36x}}{\mathrm{168x}+\mathrm{1800}}=\mathrm{x}+\mathrm{5} \\ $$$$\mathrm{168x}^{\mathrm{2}} +\mathrm{1800x}+\mathrm{840x}+\mathrm{9000}−\mathrm{7200x}=\mathrm{0} \\ $$$$\mathrm{168x}^{\mathrm{2}} −\mathrm{4560x}+\mathrm{9000}=\mathrm{0} \\ $$$$\mathrm{7x}^{\mathrm{2}} −\mathrm{190x}+\mathrm{375}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{190}\pm\sqrt{\mathrm{190}^{\mathrm{2}} −\mathrm{4}×\mathrm{7}×\mathrm{375}}}{\mathrm{2}×\mathrm{7}}=\frac{\mathrm{190}\pm\mathrm{160}}{\mathrm{14}} \\ $$$$\Rightarrow\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{190}−\mathrm{160}}{\mathrm{14}}=\mathrm{2}.\mathrm{14}\:\left(\mathrm{not}\:\mathrm{suitable}\right) \\ $$$$\Rightarrow\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{190}+\mathrm{160}}{\mathrm{14}}=\mathrm{25}\:\mathrm{km}/\mathrm{h}\:\left(\mathrm{ok}\right) \\ $$
Commented by Tinkutara last updated on 29/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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