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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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Intern
Joined: 21 Feb 2018
Posts: 23
Cars J and K are making the trip from City A to City B. Car J departs  [#permalink]

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18 Jul 2018, 09:08
00:00

Difficulty:

65% (hard)

Question Stats:

52% (02:08) correct 48% (01:48) wrong based on 54 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
Manager
Joined: 09 Jun 2014
Posts: 214
Location: India
Concentration: General Management, Operations
Schools: Tuck '19
Re: Cars J and K are making the trip from City A to City B. Car J departs  [#permalink]

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18 Jul 2018, 09:33
1
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

Posted from my mobile device
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Joined: 22 May 2016
Posts: 2201
Cars J and K are making the trip from City A to City B. Car J departs  [#permalink]

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18 Jul 2018, 12:49
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

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

VP
Joined: 07 Dec 2014
Posts: 1128
Cars J and K are making the trip from City A to City B. Car J departs  [#permalink]

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18 Jul 2018, 13:10
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
C
Director
Joined: 20 Feb 2015
Posts: 796
Concentration: Strategy, General Management
Re: Cars J and K are making the trip from City A to City B. Car J departs  [#permalink]

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22 Jul 2018, 12:11
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
Intern
Joined: 07 Dec 2017
Posts: 17
Re: Cars J and K are making the trip from City A to City B. Car J departs  [#permalink]

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22 Jul 2018, 12:58
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
Intern
Joined: 07 Dec 2017
Posts: 17
Re: Cars J and K are making the trip from City A to City B. Car J departs  [#permalink]

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22 Jul 2018, 12:59
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, 12:59
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