Question-204158

Question Number 204158 by siyathokoza last updated on 07/Feb/24

Commented by siyathokoza last updated on 07/Feb/24

HI PROF... Please explain this word problem

$$\boldsymbol{\mathrm{HI}}\:\boldsymbol{\mathrm{PROF}}… \\ $$$$\boldsymbol{\mathrm{Please}}\:\boldsymbol{\mathrm{explain}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{word}}\:\boldsymbol{\mathrm{problem}} \\ $$

Commented by siyathokoza last updated on 07/Feb/24

$$ \\ $$

Answered by AST last updated on 07/Feb/24

s_m =(d_m /t_m )=((65)/t_m );s_s =s_m −2=(d_s /t_s )=((65)/(t_m +(2/3)))=((65×3)/(3t_m +2)) ⇒((65)/t_m )=((65×3+6t_m +4)/(3t_m +2))⇒3t_m ^2 +2t_m −65=0⇒t_m =((13)/3) t_m =((13)/3)hrs⇒s_m =15km/hr⇒s_s =13km/hr

$${s}_{{m}} =\frac{{d}_{{m}} }{{t}_{{m}} }=\frac{\mathrm{65}}{{t}_{{m}} };{s}_{{s}} ={s}_{{m}} −\mathrm{2}=\frac{{d}_{{s}} }{{t}_{{s}} }=\frac{\mathrm{65}}{{t}_{{m}} +\frac{\mathrm{2}}{\mathrm{3}}}=\frac{\mathrm{65}×\mathrm{3}}{\mathrm{3}{t}_{{m}} +\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{65}}{{t}_{{m}} }=\frac{\mathrm{65}×\mathrm{3}+\mathrm{6}{t}_{{m}} +\mathrm{4}}{\mathrm{3}{t}_{{m}} +\mathrm{2}}\Rightarrow\mathrm{3}{t}_{{m}} ^{\mathrm{2}} +\mathrm{2}{t}_{{m}} −\mathrm{65}=\mathrm{0}\Rightarrow{t}_{{m}} =\frac{\mathrm{13}}{\mathrm{3}} \\ $$$${t}_{{m}} =\frac{\mathrm{13}}{\mathrm{3}}{hrs}\Rightarrow{s}_{{m}} =\mathrm{15}{km}/{hr}\Rightarrow{s}_{{s}} =\mathrm{13}{km}/{hr} \\ $$

Answered by Rasheed.Sindhi last updated on 07/Feb/24

Speed of Soul: x km/hour (say) ⇒Speed of Mark: (x+2) km/hour Time taken in 65 kms: Saul: t=(s/v)=((65)/x) hours Mark: t=((65)/(x+2)) hours ((65)/x)−((65)/(x+2))=(2/3) [40 minutes=(2/3) hours] 65(3)(x+2)−65(3)(x)=2(x)(x+2) 195x+390−195x=2x^2 +4x 2x^2 +4x−390=0 x^2 +2x−195=0 (x−13)(x+15)=0 x=13 Soul′s speed: 13km/h Mark′s speed : 13 +2=15 km/h

$${Speed}\:{of}\:{Soul}:\:{x}\:{km}/{hour}\:\left({say}\right) \\ $$$$\Rightarrow{Speed}\:{of}\:{Mark}:\:\left({x}+\mathrm{2}\right)\:{km}/{hour} \\ $$$$\: \\ $$$${Time}\:{taken}\:{in}\:\mathrm{65}\:{kms}: \\ $$$${Saul}:\:{t}=\frac{{s}}{{v}}=\frac{\mathrm{65}}{{x}}\:{hours} \\ $$$${Mark}:\:{t}=\frac{\mathrm{65}}{{x}+\mathrm{2}}\:{hours} \\ $$$$\frac{\mathrm{65}}{{x}}−\frac{\mathrm{65}}{{x}+\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{3}}\:\:\:\:\left[\mathrm{40}\:{minutes}=\frac{\mathrm{2}}{\mathrm{3}}\:{hours}\right] \\ $$$$\mathrm{65}\left(\mathrm{3}\right)\left({x}+\mathrm{2}\right)−\mathrm{65}\left(\mathrm{3}\right)\left({x}\right)=\mathrm{2}\left({x}\right)\left({x}+\mathrm{2}\right) \\ $$$$\mathrm{195}{x}+\mathrm{390}−\mathrm{195}{x}=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x} \\ $$$$ \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{390}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{195}=\mathrm{0} \\ $$$$\left({x}−\mathrm{13}\right)\left({x}+\mathrm{15}\right)=\mathrm{0} \\ $$$${x}=\mathrm{13} \\ $$$${Soul}'{s}\:{speed}:\:\mathrm{13}{km}/{h} \\ $$$${Mark}'{s}\:{speed}\::\:\mathrm{13}\:+\mathrm{2}=\mathrm{15}\:{km}/{h} \\ $$

Commented by Rasheed.Sindhi last updated on 07/Feb/24

$$\mathcal{T}{ypo}\:{sir},\:{corrected}\:{now},\:\mathcal{T}{hanks}! \\ $$

Commented by AST last updated on 07/Feb/24

Substitute the speed back, you won′t get ((65)/(x(Saul′s speed)))−((65)/((x+2)(Mark′s speed)))=(2/3)

$${Substitute}\:{the}\:{speed}\:{back},\:{you}\:{won}'{t}\:{get} \\ $$$$\frac{\mathrm{65}}{{x}\left({Saul}'{s}\:{speed}\right)}−\frac{\mathrm{65}}{\left({x}+\mathrm{2}\right)\left({Mark}'{s}\:{speed}\right)}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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