Question

12 Two stations P and Q are 42 km apart.
A person starts from P, walks at 4 km/h
towards Q and meets another person
coming from Q towards P after 6 hours.
Find the rate at which the second person
is walking, if both the persons start
walking at the same time.​

Answer 1

Answer:

3km/h

Explanation:

in 6h Pearson from p travel 24 km

so, Pearson from q travel 42-24=18

as he travel 18km in 6h,. speed =18/6 =3km/h

Answer 2

Answer:

⋆ DIAGRAM :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\linethickness{0.5mm}\put(7.7,2){\line(1,0){4}}\put(7.65,1.7){\sf P}\put(11.6,1.7){\sf Q}\put(9.4,1.7){\textsf{\textbf{42 km}}}\linethickness{.25mm}\put(7.5,2.2){\vector(1,0){1}}\put(11.89,2.2){\vector(-1,0){1}}\put(7.7,2.34){\textsf{\textbf{4 km/h}}}\put(9.9,2){\circle*{0.15}}\put(9.9,3.2){\vector(0,-1){1}}\put(9,3.5){\sf Meet-up Point of both}\put(9.1,3.25){\sf person after 6 Hours}\end{picture}$

☆ Distance b/w Stations P & Q are 42 km.

☆ Let a Person A starts from Station P with Speed 4 km/h and at the same time Person B starts from Station Q.

☆ Both Person Meets after 6 hours.

$\rule{80}{0.8}$

☢ $\underline{\textsf{Distance travelled by A in 6 hrs :}}$

$:\implies\sf Distance=Speed \times Time\\\\\\:\implies\sf Distance_{(A)} = 4\:km/hr \times 6\:hr\\\\\\:\implies\sf Distance_{(A)}=4\:km \times 6\\\\\\:\implies\sf Distance_{(A)}=24\:km$

⠀

☆ If p has travelled 24 km, then q must have travelled (42 - 24) km = 18 km to meet p after 6 hours.

$\rule{150}{1.5}$

☢ $\underline{\textsf{Rate of speed of second person :}}$

$:\implies\sf Distance=Speed Time\\\\\\:\implies\sf 18\:km=Speed_{(B)} \times 6\:h\\\\\\:\implies\sf \dfrac{18\:km}{6\:h}=Speed_{(B)}\\\\\\:\implies\underline{\boxed{\sf Speed_{(B)}=3\:km/h}}$

⠀

$\therefore\:\underline{\textsf{Second person is walking at rate of \textbf{3 km/h}}}.$