Brain Teasers
Speeding Blues
Car 1 started from Point A towards point B. At the same time Car 2 started from point B towards A. They crossed each other at point C. After which, Car 1 reached point B in another 9 hrs, and Car 2 reached point A in another 4 hrs. If both the cars maintained a constant speed throughout the journey, and the speed of Car 1 was 36 miles/hr(mph), then what was the speed of Car 2?
Answer
54 miles/hr.Here's how to solve:
Speed of Car 1 = 36 mph
Time taken from point C to B = 9 hrs
thus, distance between C to B = 36 * 9 = 324 miles (Distance = Speed * Time)
Now let the distance between point A and C be 'a' miles and let the speed of Car 2 be 'b' mph.
Thus time taken by Car 1 to travel from A to C = a/36 (Time = Distance/Speed).
and the time taken by car 2 to travel from B to C = 324/b (Distance between C and B is 324 miles as explained above)
Since both the cars meet at C, then these two times are equal, i.e.
a/36 = 324/b
or b = 324 * 36/a =11664/a
Now time taken by Car 2 to travel from C to A can be described as a/b or a*a/11664 (writing b in terms of a).
But it is given that Car 2 takes 4 hrs to travel from C to A. Thus, we get
a * a/11664 = 4
or, a * a = 46656
Now taking square root on both sides, we get a = 216. (distance cannot be negative).
Thus we get b = 11664/216 = 54.
Thus speed of Car 2 = 54 miles/hr
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Comments
Fun!
I used to anwer a lot of math problems like this back in high school. Quite happy that I still know how to solve this one!
I used to anwer a lot of math problems like this back in high school. Quite happy that I still know how to solve this one!
Nice One and Happy B-day.
thanks for the advance wishes
This is a nice "encounter" problem... but in the answer you took several steps, calculating other parameters that are not needed. Let me show you a shorter way:
Both A and B will be travelling the same time until they meet, so the ratio of the distances is equal to the ratio of velocities, and the distances can be caltulated with the data after the encounter
Vb / Va = CB/AC = (Va * Tacb)/(Vb * Tbca)
(Tacb: time that A needs to go from C to B)
From the previous equation you can immediately obtain the answer
Vb = Va * (Tacb/Tbca)^(1/2)
Vb = 36 * ( 9 / 4 )^(1/2) = 36 * 3 / 2 = 54
Both A and B will be travelling the same time until they meet, so the ratio of the distances is equal to the ratio of velocities, and the distances can be caltulated with the data after the encounter
Vb / Va = CB/AC = (Va * Tacb)/(Vb * Tbca)
(Tacb: time that A needs to go from C to B)
From the previous equation you can immediately obtain the answer
Vb = Va * (Tacb/Tbca)^(1/2)
Vb = 36 * ( 9 / 4 )^(1/2) = 36 * 3 / 2 = 54
Thanks. Your solution works fine with me. You may post in a correction and see if others also feel the same.
Thanks for your puzzle. I managed to solve it with scip (a linear, mixed integer and nonlinear programming solver), using the following simple model file:
var speed1 >=0; var speed2 >=0; var time_to_point_C >=0;
# total distance A-C-B
subto c1: (time_to_point_C + 9) * speed1 == (time_to_point_C + 4) * speed2;
# the distance from C to B (or from B to C)
subto c2: (9 * speed1) == speed2 * time_to_point_C;
subto c3: speed1 == 36
var speed1 >=0; var speed2 >=0; var time_to_point_C >=0;
# total distance A-C-B
subto c1: (time_to_point_C + 9) * speed1 == (time_to_point_C + 4) * speed2;
# the distance from C to B (or from B to C)
subto c2: (9 * speed1) == speed2 * time_to_point_C;
subto c3: speed1 == 36
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