A train chasing a cat?
A train approaches a tunnel AB. Inside the tunnel is a cat located at a point that is 3/8 of the distance AB measured from the entrance A. When the train whistles the cat runs. If the cat moves to the entrance of the tunnel, A, the train catches the cat exactly at the entrance. If the cat moves to the exit, B, the train catches the cat at exactly the exit. The speed of the train is greater than the speed of the cat by what order ?
a) 3 : 1
b) 4 : 1
c) 5 :1
d) none of these
what it means MATHEMATICALLY ??
"……If the cat moves to the entrance of the tunnel, A, the train catches the cat exactly at the entrance. If the cat moves to the exit, B, the train catches the cat at exactly the exit…."
how do i frame eqn mathematically for the above statement ? what it means ?
let D is distance , from tunnel entry A to train location.
let , v1=trains sppped,v2=cat speed ,T1,T2 is the meeting time.
D+(3/8)*x =(v1+v2)*T1
D+(3/8)*x =(v1-v2)*T2
someone, please tell me steps/hints ?
Filed under: Cat Tunnel
Like this post? Subscribe to my RSS feed and get loads more!
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
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!
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)
/ TS
Therefore, T2 = (D +
Adding gives :
/ TS = (2D + 
/ TS
T1 + T2 = D / TS + (D +
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 :
/ TS,
/ (5*TS)
(5/3)*T1 = (D +
so, T1 = 3(D +
Now, equating this T1 with the T1 in (3), we get :
/ (5*TS)
D / TS = 3(D +
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).