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

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New post 18 Jul 2018, 10:08
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  55% (hard)

Question Stats:

53% (02:08) correct 47% (01:49) wrong based on 49 sessions

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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
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Re: Cars J and K are making the trip from City A to City B. Car J departs  [#permalink]

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New post 18 Jul 2018, 10:33
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This looks simple.

let's assume speed of car J as x miles /hour.Therefore speed of K ..is 0.80x miles /he

Now car K has covered 15 mns of distance and that means
Distance=speed *time =0.80x*15/60 =.20x

Now this speed J has the make up travelling in same direction as k..which means Speed J-Speed K will be relative speed and we have the make up for 0.20x

So. Time=.20x/(x-0.80)= 1..

Remember this is ONE hour.. We e assumed x miles per hour ..


SO

x=60 mns

Press Kudos if it helps :)

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

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New post 18 Jul 2018, 13:49
1
franteraoka wrote:
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?

20

45

60

75

1230

Smart numbers
(1) Speeds
Let J's speed = \(\frac{15mi}{15mins}\)

K's speed: \(.8J=\frac{4}{5}J=(\frac{4}{5}*\frac{15mi}{15mins})=\frac{12mi}{15mins}\)

(2) Distance?
K creates the "gap" distance between them
\(D=(r*t)=(\frac{12mi}{15mins}*15mins)=12\) miles

(3) Relative speed

To close the distance gap in a chase problem, subtract slower speed from faster speed to get relative speed

Relative speed: \((J - K)=\)
\((\frac{15mi}{15mins}-\frac{12mi}{15mins})=\frac{3mi}{15mins}=\frac{1mi}{5mins}\)

(4) TIME J needs to catch K?

Time: \(\frac{1mi}{5mins}*t=12\) miles

\(t=(12mi*\frac{5min}{1mi})=60\) minutes

Answer C

Algebra
K's rate = \(k\)
J's rate = \(\frac{5}{4}k\)

K creates distance between them
\((r*t)=D\)
\(D=(k*15)=15k\)

Relative speed = (J - K)
Relative speed: \(r_2=(\frac{5}{4}k-\frac{4}{4}k)=\frac{1}{4}k\)

Time for J to catch K?
\((r_{2}*t)=D\)
\(\frac{1}{4}k*t_{mins}=15k\)
\(t_{mins}=(15k*\frac{4}{1k})=60\) minutes

Answer C
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Cars J and K are making the trip from City A to City B. Car J departs  [#permalink]

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New post 18 Jul 2018, 14:10
franteraoka wrote:
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


let s=K's speed
let t=J's time to catch up with K
5/4*s*t=s*15+s*t
t=60 minutes
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Re: Cars J and K are making the trip from City A to City B. Car J departs  [#permalink]

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New post 22 Jul 2018, 13:11
franteraoka wrote:
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


Let Sj be speed of car j and Sk be speed of car k

ATQ !!

Sk=0.8Sj
also ,
Car J departs from City A 15 minutes after Car K
distance traveled by car K = 15*Sk
15*(0.8Sj)
or, 12Sj

now relative speed (since they are travelling in same direction )
Sj-Sk=Sj-0.8Sj=0.2Sj

time taken =relative distance/relative speed
=12Sj/0.2Sj=60 minutes
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Re: Cars J and K are making the trip from City A to City B. Car J departs  [#permalink]

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New post 22 Jul 2018, 13:58
Let's say Speed of K is 4m/min.
The speed of J would be 5m/min
The difference of Speed 1 m/min.
In 15 minutes, K would travel 15* 4 m
Therefore, the time taken would be 60 minutes
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Re: Cars J and K are making the trip from City A to City B. Car J departs  [#permalink]

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New post 22 Jul 2018, 13:59
Let's say Speed of K is 4m/min.
The speed of J would be 5m/min
The difference of Speed 1 m/min.
In 15 minutes, K would travel 15* 4 m
Therefore, the time taken would be 60 minutes
Re: Cars J and K are making the trip from City A to City B. Car J departs &nbs [#permalink] 22 Jul 2018, 13:59
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