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Question Number 101616 by 175 last updated on 03/Jul/20

Commented by Tinku Tara last updated on 03/Jul/20

Please post question where you  yourself doubt.

$$\mathrm{Please}\:\mathrm{post}\:\mathrm{question}\:\mathrm{where}\:\mathrm{you} \\ $$$$\mathrm{yourself}\:\mathrm{doubt}. \\ $$

Answered by mr W last updated on 03/Jul/20

90×((35×(3/4))/(90−35))=42.95 km

$$\mathrm{90}×\frac{\mathrm{35}×\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{90}−\mathrm{35}}=\mathrm{42}.\mathrm{95}\:{km} \\ $$

Commented by 175 last updated on 03/Jul/20

how? Explain

Commented by mr W last updated on 03/Jul/20

as the car started, the motorcycle  had traveled a distance 35×(3/4).  in one hour the car can travel 90−35  =55 km more than the motorcycle.  such that they met, the car needed  ((35×(3/4))/(90−35)) hours. the distance the car  traveled is 90×((35×(3/4))/(90−35)), which is also  the total distance the motorcycle  traveled.

$${as}\:{the}\:{car}\:{started},\:{the}\:{motorcycle} \\ $$$${had}\:{traveled}\:{a}\:{distance}\:\mathrm{35}×\frac{\mathrm{3}}{\mathrm{4}}. \\ $$$${in}\:{one}\:{hour}\:{the}\:{car}\:{can}\:{travel}\:\mathrm{90}−\mathrm{35} \\ $$$$=\mathrm{55}\:{km}\:{more}\:{than}\:{the}\:{motorcycle}. \\ $$$${such}\:{that}\:{they}\:{met},\:{the}\:{car}\:{needed} \\ $$$$\frac{\mathrm{35}×\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{90}−\mathrm{35}}\:{hours}.\:{the}\:{distance}\:{the}\:{car} \\ $$$${traveled}\:{is}\:\mathrm{90}×\frac{\mathrm{35}×\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{90}−\mathrm{35}},\:{which}\:{is}\:{also} \\ $$$${the}\:{total}\:{distance}\:{the}\:{motorcycle} \\ $$$${traveled}. \\ $$

Commented by mr W last updated on 03/Jul/20

excuse me boy or sir!  as i saw this question of you, i thought  you are still in elementary school.  as you said you don′t understand my  answer, i tried to explain in a way that  a child of third class may understand.  but as i saw the other posts of you,  OMG, you must be a least a college  student. many questions from you  are even too high for me, such as  following one. how can somebody  understand integral calculus, but  doesn′t understand basic arithmetic?

$${excuse}\:{me}\:{boy}\:{or}\:{sir}! \\ $$$${as}\:{i}\:{saw}\:{this}\:{question}\:{of}\:{you},\:{i}\:{thought} \\ $$$${you}\:{are}\:{still}\:{in}\:{elementary}\:{school}. \\ $$$${as}\:{you}\:{said}\:{you}\:{don}'{t}\:{understand}\:{my} \\ $$$${answer},\:{i}\:{tried}\:{to}\:{explain}\:{in}\:{a}\:{way}\:{that} \\ $$$${a}\:{child}\:{of}\:{third}\:{class}\:{may}\:{understand}. \\ $$$${but}\:{as}\:{i}\:{saw}\:{the}\:{other}\:{posts}\:{of}\:{you}, \\ $$$${OMG},\:{you}\:{must}\:{be}\:{a}\:{least}\:{a}\:{college} \\ $$$${student}.\:{many}\:{questions}\:{from}\:{you} \\ $$$${are}\:{even}\:{too}\:{high}\:{for}\:{me},\:{such}\:{as} \\ $$$${following}\:{one}.\:{how}\:{can}\:{somebody} \\ $$$${understand}\:{integral}\:{calculus},\:{but} \\ $$$${doesn}'{t}\:{understand}\:{basic}\:{arithmetic}? \\ $$

Commented by mr W last updated on 03/Jul/20

Commented by mr W last updated on 03/Jul/20

to be honest, i was  displeased as i  realised this.

$${to}\:{be}\:{honest},\:{i}\:{was}\:\:{displeased}\:{as}\:{i} \\ $$$${realised}\:{this}. \\ $$

Commented by Tinku Tara last updated on 03/Jul/20

Also same as in q101329.

$$\mathrm{Also}\:\mathrm{same}\:\mathrm{as}\:\mathrm{in}\:\mathrm{q101329}. \\ $$

Answered by Dwaipayan Shikari last updated on 03/Jul/20

Relative velocity =(90−35)((km)/h)  The scooter has crossed (35×(3/4))km during this time .  Time taken by car to reach the motor cyclist=((35×(3/4))/(90−35)) h  Distance covered=90×((35×(3/4))/(90−35))=42.95★km

$${Relative}\:{velocity}\:=\left(\mathrm{90}−\mathrm{35}\right)\frac{{km}}{{h}} \\ $$$${The}\:{scooter}\:{has}\:{crossed}\:\left(\mathrm{35}×\frac{\mathrm{3}}{\mathrm{4}}\right){km}\:{during}\:{this}\:{time}\:. \\ $$$${Time}\:{taken}\:{by}\:{car}\:{to}\:{reach}\:{the}\:{motor}\:{cyclist}=\frac{\mathrm{35}×\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{90}−\mathrm{35}}\:{h} \\ $$$${Distance}\:{covered}=\mathrm{90}×\frac{\mathrm{35}×\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{90}−\mathrm{35}}=\mathrm{42}.\mathrm{95}\bigstar{km} \\ $$

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