Cars J and K are making the trip from City A to City B. Car J departs

Question

Cars J and K are making the trip from City A to City B. Car J departs from City A 15 minutes after Car K does, and both cars travel along the same route. If Car K travels at a constant speed that is 80% the constant speed of Car J, then how many minutes will elapse before Car J catches up to Car K?

A. 20
B. 45
C. 60
D. 75
E. 1230

A. 20
B. 45
C. 60
D. 75
E. 1230

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satya2029 · Answer

Speed of Car J= X
Speed of Car K=0.8X
distance travelled by Car K in 15 minutes= 15*0.8X
=12X
So Car J will cover this 12X distance relative to Car K in time "t"
speed*time=distance
(X-0.8X)*t=12X
t=60 minutes
C:)

nick1816 · Answer

V_J : V_K= 5:4

T_J : T_K= 4:5

T_K - T_J=5x - 4x=15

T_J= 4x= 4*15=60

ShankSouljaBoi · Answer

Then how many minutes will elapse before Car J catches up to Car K

(t+15)K = (t)J and K=(4/5)*J . Hence t is 60 . But we are asked of elapsed time so add 15 minutes to it. Answer = 60+15 = 75.

ScottTargetTestPrep · Answer

Let r = the speed of Car J, so 0.8r = the speed of Car K. Let’s also let t = Car J’s travel time, in hours; this means that (t + ¼) is Car K’s travel time (because 15 minutes = ¼ hour). We can create the equation:

Car K’s distance = Car J’s distance

(0.8r)(t + 1/4) = rt

0.8(t + 0.25) = t

0.8t + 0.2 = t

0.2 = 0.2t

1 = t

Since 1 hour = 60 minutes, choice C is the correct answer.

Answer: C

SPatel1992 · Answer

Why isn't the answer 75? Don't have to count the 15 min delay as well?

metalhead2593 · Answer

S(J) = speed of car J (per hour)
S(K) = speed of car K (per hour)

S(K) = 0.8 S(J)

Car K start 15 mins (1/4 hour) earlier compared to Car J. So by the time Car J starts going, Car K has traveled 1/4*S(K) = 1/4*0.8*S(J) = 0.2 S(J)
Difference in speed = S(J)-S(K) = S(J) - 0.8S(J) = 0.2 S(J)

So Car J will take 0.2S(J)/0.2S(J) = 1 hour = 60 mins to catch up.
IMO C.

Jawad001 · Answer

Cars J and K are making the trip from City A to City B. Car J departs from City A 15 minutes after Car K does, and both cars travel along the same route. If Car K travels at a constant speed that is 80% the constant speed of Car J, then how many minutes will elapse before Car J catches up to Car K?

A. 20
B. 45
C. 60
D. 75
E. 1230

A. 20
B. 45
C. 60
D. 75
E. 1230

Solution:
The speed of Car J = 100 and the speed of Car K = 80
Ratio of speed of Car J and Car K = 100: 80 = 5: 4
Ratio of time of Car J and Car K = 4:5
Let time of Car K and Car J = 5x and 4x

Now, 5x – 4x = 15
Or, x= 15
Time elapsed = 4x = 4*15 = 60
Answer: C

jawediqbal94 · Answer

why can't 75 be the answer? Shouldn't we consider the entire time duration to calculate the overall time?

K-ja · Answer

I AGREE. The question needs to state explicitly

Gambrosio · Answer

Would have been great to know wether the time which elapses counts from the time on where K starts or where J starts...

Cars J and K are making the trip from City A to City B. Car J departs

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Cars J and K are making the trip from City A to City B. Car J departs

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