Question and Answers Forum

All Questions      Topic List

Logic Questions

Previous in All Question      Next in All Question      

Previous in Logic      Next in Logic      

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!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com