Question

50 points

A takes 3 hrs more than B to walk a distance of 30 km. but if A doubles his speed he is ahead of B by 1.5 hrs . Find the speed of A and B​​

Answer 1

Answer:

$<b>$

Check in attachment

When A doubles his speed then, time taken by A = 30/2x

[Given: time taken by B = Time taken by A + 3/2 h]

30/y = 30/2x + 3/2

30/y - 30/2x = 3/2

30/y - 15/x = 3/2 ……..(2)

Let 1/x= p and 1/y = q

so eqs 1 & 2 become ,then On adding eq 3 & 4

30p - 30q = 3……….(3)

-15p + 30q = 3/2………(4)

---------------------------

15 p = 3 + 3/2

15 p =( 6 + 3)/2 = 9/2

15p = 9/2

p = 9 /(2×15) = 3/10

p = 3/10

On Putting the value of p = 3/10 in eq 3

30p - 30q = 3

30 (3/10) - 30q = 3

9 - 30q = 3

9 - 3 = 30q

6 = 30q

q = 6/30= 1/5

1/y = q [ Let 1/y = q]

1/y = ⅕

y = 5 km/h

1/x = p [Let 1/x= p ]

1/x = 3/10

x = 10/3 km/h

so x = 10/3 km/h. & y = 5 km/h

Hence, the speed of A is 10/3 km/h & B is 5 km/h.

Answer 2

SOLUTION

Speed of A = x

Speed of B= y

Distance= 30km

1st case

$= > \frac{30}{x} = \frac{30}{y} + 3 \\ = > \frac{30}{x} - 3 = \frac{30}{y} ..............(1)$

2nd case

$= > \frac{30}{x} = \frac{30}{y} - \frac{3}{2} \\ = > \frac{30}{x} + \frac{3}{2} = \frac{30}{y} ............(2)$

from (1) &(2),

$= > \frac{30}{x} - 3 = \frac{30}{x} + \frac{3}{2} \: \: \: \: (axiom \: 1) \\ = > \frac{30}{x} - \frac{30}{2x} = \frac{3}{2} + 3 \\ = > \frac{60x - 30x}{ {2x}^{2} } = \frac{3 + 6}{2} \\ = > \frac{30x}{ {x}^{2} } = 9 \\ = > 30x = 9 {x}^{2} \\ = > 9x = 30 \\ = > x = \frac{30}{9} = \frac{10}{3} km \: per \: hr.$

Speed of A= 10/3 km/hrs.

Substitute in eq. (1), we get

$= > \frac{ \frac{30}{10} }{3} - 3 = \frac{30}{y} \\ = > 9 - 3 = \frac{30}{y} \\ = > 6 = \frac{30}{y} \\ = > 6y = 30 \\ \\ = > y = \frac{30}{6} = 5$

Speed of B= 5km/ hrs.

hope it helps ☺️