If a car starts from A towards B with some velocity due to
some problem in the engine after travelling 30km.If the car
goes with 4/5 th of its actuval velocity the car reaches B
45min later to the actual time. If the car engine fails
ofter travelling 45km, the car reaches the destination B
36min late to the actual time , what is the initial
velocity of car and what is the distance between A and B in km
let x is distance from A to B
and y is initial speed.
30/y+(x-30)5/4y -x/y = 3/4
=>
4x-12y=120 —-(1).
45/y + (x-45)5/4y -x/y=3/5
=>
5x-12y=225 ——-(2).
From equ (1) and equ (2) we will get.
x=25 and y=105
so initial speed is 25
and Distance From A to B is 105
Ans 25,105
Ans. let x is distance from A to B
and y is initial speed.
30/y+(x-30)5/4y -x/y = 3/4
=> 4x-12y=120 —-(1).
45/y + (x-45)5/4y -x/y=3/5
=> 5x-12y=225 ——-(2).
From equ (1) and equ (2) we will get.
x=25 and y=105
so initial speed is 25 km/hr
and Distance From A to B is 105 km
20 and 130
let x is distance from A to B
and y is initial speed.
30/y+(x-30)5/4y -x/y = 3/4
=>
4x-12y=120 —-(1).
45/y + (x-45)5/4y -x/y=3/5
=>
5x-12y=225 ——-(2).
From equ (1) and equ (2) we will get.
x=25 and y=105
so initial speed is 25
and Distance From A to B is 105
Ans 25,105
Ans. let x is distance from A to B
and y is initial speed.
30/y+(x-30)5/4y -x/y = 3/4
=> 4x-12y=120 —-(1).
45/y + (x-45)5/4y -x/y=3/5
=> 5x-12y=225 ——-(2).
From equ (1) and equ (2) we will get.
x=25 and y=105
so initial speed is 25 km/hr
and Distance From A to B is 105 km