A train chasing a cat?

3 Comments

1. Madhukar says:

   March 2, 2010 at 8:44 am

   Answer: b) 4 : 1 worked out as under.

   D / v1 = (3/8) x / v2 … ( 1 )

   ( D + x ) / v1 = (5/8) x / v2
   => D / v1 = (5/8) x / v2 – x / v1 … ( 2 )

   From equns. ( 1 ) and ( 2 ),

   (3/8) x /v2 = (5/8) x / v2 – x / v1
   => 3 / (8v2) = 5 (8v2) – 1/ v1
   => 1 / v1 = ( 5/8 – 3/8) / v2
   => 4v2 = v1
2. Parrocchiano says:

   March 2, 2010 at 8:44 am

   The train is far "x" form the entrance A. The tunnell is AB long.
   I use the equation [time=space/speed], and i name v1 the train speed, and v2 the cat speed.

   In the first case
   x/v1=(3/8)*AB/v2
   v1=x*(8/(3AB))*v2

   In the second case
   (x+AB)/v1=5/8*AB/v2
   v1=(x+AB)*(8/(5AB))*v2

   So
   x*(8/(3AB))*v2=
   =(x+AB)*(8/(5AB))*v2

   x/3AB=(x+AB)/5AB
   x/(3AB)-x/(5AB)=1/5
   2x/(15AB)=3/15
   2x=3AB
   x=3AB/2

   Now i use
   v1=x*(8/(3AB))*v2

   v1=3AB/2*(8/(3AB)*v2
   v1=4*v2

   b) is correct.

   I haven’t used two variables for the two times: they don’t interest in this problem. I just know that in each case train’s and cat’s times are equal, so i can equiparate their rates space/speed in each case.

   Sorry for the english, i don’t know scientific words, i hope to have been useful!
3. falzoon says:

   March 2, 2010 at 8:44 am

   Let AB = 8 units in length.
   Therefore the cat is located at a point
   3 units from A and 5 units from B.

   Using the formula : distance = speed * time, we have :

   FOR THE CAT :

   (1) 3 = CS * T1 (CS = cat’s speed, T1 = time to get to A)
   Therefore, T1 = 3 / CS

   (2) 5 = CS * T2 (T2 = time to get to B)
   Therefore, T2 = 5 / CS

   Adding gives :
   T1 + T2 = 3 / CS + 5 / CS = 8 / CS.

   FOR THE TRAIN :

   Let D = distance of train from entrance when whistle blows

   (3) D = TS * T1 (TS = train speed, T1 is the same, towards A)
   Therefore, T1 = D / TS

   (4) D + 3 + 5 = TS * T2 (T2 is the same, towards B)
   Therefore, T2 = (D + / TS

   Adding gives :
   T1 + T2 = D / TS + (D + / TS = (2D + / TS

   Now we equate both calculations of (T1 + T2) to get :

   8 / CS = (2D + / TS

   Therefore, TS / CS = (2D + / 8 = (D + 4) / 4.

   Now we have to find D :

   Note that T2 = (5/3)*T1, because the distances are
   known and the cat’s speed is purportedly constant.

   In (4), where we have T2 = (D + / TS, we now get :
   (5/3)*T1 = (D + / TS,
   so, T1 = 3(D + / (5*TS)

   Now, equating this T1 with the T1 in (3), we get :
   D / TS = 3(D + / (5*TS)
   TS cancels out and we’re left with D = 12 units.

   Substituting D = 12 into TS / CS = (D + 4) /4,
   as shown above, gives :
   TS / CS = 16 / 4 = 4 / 1
   which is a ratio of 4 : 1, or answer (b).