Three cars leave from A to B in equal time intervals. They reach B

Let the car faster to slower be z, y and x.
The first and the third car meet a point that is 80 miles away from C, which is 240 miles away.
This means x traveled 240-80 and z traveled 240+80..Ratio of distance = 240-80:240+80=160:320=1:2, so their speed will be in the ratio 2:1.
So, if speed of x is X, the speed of z is 2X..
So we are looking for Z-X=2X-X=X

If the distance from A to B is D, then equal intervals means..
\(\frac{D}{Z}-\frac{D}{Y}=\frac{D}{Y}-\frac{D}{X}..........\frac{D}{2X}-\frac{D}{Y}=\frac{D}{Y}-\frac{D}{X}........Y=\frac{4X}{3}\)..
Thus the speed are in ratio \(X:\frac{4X}{3}:2X\)..

Now to find the value of X....The first car arrives at C an hour after the second car.
\(\frac{240}{X}-240/\frac{4X}{3}=1.......\frac{240}{X}-\frac{180}{X}=1.........X=60\)

Answer is 60

A

Let the 1st, 2nd and 3rd cars be denoted by C1, C2 and C3 and their speeds by S1, S2 and S3 respectively. Let the time taken by C1 to move from B to C be 't' hrs.

S1/S2 = (t-1)/t...........(i)
S1/S3 = (240-80)/(240+80) = 1/2 (S3 is twice as fast as S1)

Assume it takes 4 hrs for C1 to move from A to B; so it takes C3 2 hrs to travel the same distance since all three cars reach B at the same time. Therefore, C3 must start from A two hrs after C1. Since the three cars start at equal time intervals, C2 must start 1 hr before C3 i.e. 1 hr after C1. T/4, it takes 3 hrs for C2 to move from A to B.

S1/S2 = 3/4 = (t-1)/t (from (i))...> t=4
S1 = 240/4 = 60 which means S3 = 120.
S3-S1 = 60. ANS:A

Let B's time to reach station B be x hours.

Ratio of A's speed and C's speed is 1:2

Now,

x+1:x-1 = 2:1
=> x = 3

So, the difference between C,s and A's speed = (240/2)-(240/4) km/hr
= 60 km/hr

Posted from my mobile device

Since first car arrives at C one hour after second car, the time interval between cars is 1 hour
If we let the speed of the first, second, third cars as v1,v2,v3 respectively and let the time for first car to reach to B as t
We have v1:v2:v3=(t-2):(t-1):t
Now suppose first car meets third car after s time
Then (240-80)/v1=(240+80)/v3
v1/v3 =1/2 =(t-2)/t
gives t=4
For first car 240=v1*t=v1*4
v1=60
Thus v2 =120
Difference of speed=120-60=60

Hello there i hope you understand question well what is happening now my solution here

Since 1st car arrived after 1 hours from second means 3rd car Would arrived at C before 2 hours from first .
(i)

Now second part about meeting of 1st car and 3rd car :

They meet at 80 miles from C
That means 3rd car would have travelled (240+80) 320 miles at same time 1st car travelled 160 miles so :

160/1st car speed...... Time of meeting

320/ 3rd car speed is time of meeting

Both time should be same for meeting so ratio of speed of 1st and 3rd would be 1:2
.since distance is same for all of it Time of 1st:3nd 2:1

Also from above difference in time if 1st and 3rd is 2 hrs from. (i)

So 1 unit of ratio difference is 2 hrs then Time taken by 1st is 4hrs and 3rd is 2 hrs

Then speed of 3rd 120miles/hr and 1st - 60miles/hr difference is 60 miles as question asked

Posted from my mobile device

Let the Distance between A and B be \(D\). Let speeds of the three cars be \(s_1\), \(s_2\), and \(s_3\). Let the time taken by them to reach B from A be \(t_1\), \(t_2\), and \(t_3\). Let the time intervals at which the cars leave be \(x\).

Now we have,

\(t_1 = \frac{D}{s_1}\) ... (1)

\(t_2 = \frac{D}{s_2} = \frac{D}{s_1} - x\) ... (2)

\(t_3 = \frac{D}{s_3} = \frac{D}{s_1} - 2x\) ... (3)

We also know that in the time Car 1 travels \(240-80=160\) kms, Car 3 travels \(240+80=320\) kms, hence, \(s_3 = 2s_1\)

Plugging this back into (3), we get \(\frac{D}{{2s_1}} = \frac{D}{s_1} - 2x\) or \(2x = \frac{D}{{2s_1}}\) or \(\frac{x}{D} = \frac{1}{4s_1}\) ... (4)

Now we also know that \(\frac{240}{s_1} - 1 = \frac{240}{s_2}\) or \((\frac{1}{s_1} - \frac{1}{s_2}) = (\frac{1}{240})\)

Plugging this back into (2), we get \(\frac{x}{D} = \frac{1}{s_1} - \frac{1}{s_2} = \frac{1}{240}\) ... (5)

Using (4) and (5) we now have \(\frac{1}{4s_1} = \frac{1}{240}\) or \(s_1=60\)

Since we knew that \(s_3 = 2s_1\) therefore, \(s_3 - s_1 = 2s_1 - s_1 = s_1 = 60\)

Hence, option A.

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