Question

A takes 1 hour more than B to walk 40Km. But if A doubles his speed, he is
ahead of B by 1 1/2ℎ. Find their speeds of walking.??

Answer 1

If speed dauble he takes = less 11/2+1 h from his normal speed
Then in his normal speed he takes time = 11/2+1x 2 = 13/2x2=13 h
A takes time =13h
B takes time =13-1=12h
Then A speed=40/13=3¹/13 km/h
B speed = 40/12=10/3=3⅓ km/h ans

Answer 2

HELLO !!!.....


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Let the speed of A and B be x km/hr and ykm/hr respectively.

We know that,

$time \: = \frac{distance}{speed}$
So, time taken by A to cover 40 km = 40/x

time taken by B to cover 40 km = 40/y

Given, time taken by A =  time taken by B + 1 hr

∴ Time taken by A – 1 hr = Time taken by B

$\frac{40}{x} - 1 = \frac{40}{y} \\ \\ = > \frac{40 - x}{x} = \frac{40}{y} \\ \\ = > 40y - xy = \frac{40}{y} \\ \\ = > 40x - 40y = - xy \: \: \: \: \: - - - - - - (1)$


When A doubled his speed , then

Time taken by A + 11/2 hrs = time taken by B. ( Given)

$\frac{40}{2x} + \frac{11}{2} = \frac{40}{y} \\ \\ = > \frac{40 + 11x}{2x} = \frac{40}{y} \\ \\ = > 40y + 11xy = 80x \\ \\ = > 80x - 40y = 11xy \: \: \: \: - - - - - - (2)$


● Solving (1) and (2), we get


40x - 40y = -xy

80x - 40y = 11xy
- + -
______________

-40x = -12 xy

$y = \frac{ - 40x}{ - 12x} = \frac{10}{3}$


Putting the value of y in equation (1), we get

$40x - 40( \frac{10}{3} ) = - ( \frac{10}{3} )x \\ \\ = > 40x - \frac{400}{3} = - \frac{10x}{3} \\ \\ = > \frac{40x}{1} + \frac{10x}{3} = \frac{400}{3} \\ \\ = > \frac{120x + 10x}{3} = \frac{400}{3} \\ \\ = > \frac{130x}{3} = \frac{400}{3} \\ \\ = > x = \frac{400}{130} = \frac{40}{13}$



Thus, the speed of A is 40/13 km/hr  and speed of B is 10/3 km/hr.


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HOPE IT HELPS !!!

# NIKKY