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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

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

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

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

Commented by 175 last updated on 03/Jul/20

how? Explain

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

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

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

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

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

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

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

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

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

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

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