Latest Elitmus Aptitude Question SOLUTION: a bus left a point X for point Y.two hrs later a car left point X for Y and arrived at Y at the same time as bus.if the car and the bus left simultaneously from the opposite ends X

let B and C be speed of bus and car respectively
and time taken by car to travel from X to Y be T
then
equating distance

B*(T+2)= C*T= (B+C)*4/3
solving which T= 2hrs
and time taken by bus is T+2 i.e. 4 hrs

bus takes 4 hrs

speed of bus=XY/t ;t=time taken by bus to cover XY
speed of car= XY/(t+2)

distance travelled by bus in 1.33hrs + distance travelled by car in 1.33hrs=XY

XY/t * 1.33 + XY/(t+2) *1.33 = XY

then solve the equation

BUS TAKES 4hrs

how to slove
B*(T+2)= C*T= (B+C)*4/3
and get the value of T

To complete 1 round of XY together they take 4/3 hrs.
To complete 2 round of XY together they must take 8/3 hrs.
therefore t+t-2=8/3
t=7/3=2hrs 20mins taken by bus.

2.66 hr
by relative velocity concept we can cal the velocity of bus
and it is given that a bus take t hour to travel a distance

Choose a convenient value for the dist from x to y: 133 miles
Let b = the bus speed
let c - the car speed
then
133/b = the bus travel time
133/c = the car travel time
:
"Two hours later a car left x for y and arrived at y at the same time as the bus."
therefore
bus time - 2 hrs = car time
133%2Fb - 2 = 133%2Fc
mult equation by bc, resulting in:
133c - 2bc = 133b
c(133-2b) = 133b
c = %28133b%29%2F%28%28133-2b%29%29
:
" If the car and the bus left simultaneously from the opposite ends x and y towards each other, they would meet 1.33 hours after the start. "
car dist + bus dist = 133 mi
1.33b + 1.33c = 133

simplify, divide by 1.33
b + c = 100
replace c with %28133b%29%2F%28%28133-2b%29%29
b + %28133b%29%2F%28%28133-2b%29%29 = 100
multiply by (133-2b) to get rid of the denominator
b(133-2b) + 133b = 100(133-2b)
133b - 2b^2 + 133b = 13300 - 200b
combine like terms to form a quadratic equation
-2b^2 + 133b + 133b + 200b - 13300 = 0
-2b^2 + 466b - 13300 = 0
simplify divide by -2
b^2 - 233b + 6650 = 0
Using the quadratic formula I got two solutions
b = 199.7, not reasonable
and
b = 33.3 mph, as the bus speed
"How much time for the bus to travel from x to y ?"
133/33.3 = 4 hrs
:
:
Check this by finding the car speed, we know: b + c = 100
33.3 + c = 100
c ~ 66.7 mph is the car speed
Find the car time
133/66.7 ~ 2 hrs (started 2 hr later and arrive at the same time)

Let us assume C and B be speed of Car and Bus and X be the time taken by C to cover the distance.

Now

C*x=B(x+2)

Cx/B+C=4/3

B(x+2)x/2Bx+2B = 4/3

Solving it you''ll get x=2

Time taken by Bus=x+2=4