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3 people, Amy, Beth, and Cassie, have speeds of 3 mph

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3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post Updated on: 21 Jan 2014, 04:22
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3 people, Amy, Beth, and Cassie, have speeds of 3 mph, 4 mph and 6 mph respectively. They run a race in which Beth gives Amy a head start of 2 hrs. If both Beth and Cassie overtake Amy at the same time, what head start did Cassie give Amy?

A. 3 hours
B. 4 hours
C. 5 hours
D. 9 miles
E. 10 miles

Originally posted by guerrero25 on 20 Jan 2014, 10:31.
Last edited by guerrero25 on 21 Jan 2014, 04:22, edited 1 time in total.
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Re: 3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 20 Jan 2014, 11:14
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guerrero25 wrote:
3 people, Amy, Beth, and Cassie, have speeds of 3 mph, 4 mph and 6 mph respectively. They run a race in which Beth gives Amy a head start of 2 hrs. If both Beth and Cassie overtake Amy at the same time, what head start did Cassie give Amy?

A. 3 hours
B. 4 hours
C. 5 hours
D. 9 miles
E. 10 miles


I do not have the OA right now . I will post the answer this evening .


Beth gives Amy a head start of 2 hours, thus Beth gives Amy a head start of 2*3 = 6 miles.

To overtake Amy, Beth will need (time) = (distance)/(relative rate) = 6/(4-3) = 6 hours.

From above we have that Amy runs for 2 + 6 = 8 hours, thus runs total of 8*3 = 24 miles.

To run 24 miles Cassie needs 24/6 = 4 hours. Therefore Cassie gave Amy a head start of (time of Amy) - (time of Cassie) = 8 - 4 = 4 hours.

Answer: B.
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Re: 3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 21 Jan 2014, 03:23
head start Beth gave to Amy = 2 hours => 3mph *2 =>6 mile
Head start Cassie gave to Amy = x mile

in time t distance travelled
Cassie- x+6 miles = 3t [ relative speed of Cassie w.r.t to Amy * t]
Beth - 6 miles = 1t [ relative speed of Cassie w.r.t to Amy * t]
equations: x+6=3t and 6=1t
x = 12 miles
Head start = x/ speed of Amy => 12/3 = 4 hours
Answer B
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Re: 3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 21 Jan 2014, 22:23
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guerrero25 wrote:
3 people, Amy, Beth, and Cassie, have speeds of 3 mph, 4 mph and 6 mph respectively. They run a race in which Beth gives Amy a head start of 2 hrs. If both Beth and Cassie overtake Amy at the same time, what head start did Cassie give Amy?

A. 3 hours
B. 4 hours
C. 5 hours
D. 9 miles
E. 10 miles



The most important clue here is this: If both Beth and Cassie overtake Amy at the same time

This means that all A, B and C meet together i.e. at one instant, they all had covered the same distance.
For easy calculations, let's assume that that distance was 12 miles. A would have been running for 4 hrs, B would have been running for 3 hrs and C would have been running for 2 hrs.
But actually, B started 2 hrs after A so the distance must have been 24 miles. A would have been running for 8 hrs, B would have been running for 6 hrs and C would have been running for 4 hrs.
So C gave A a head start of 4 hrs.

Answer (B)
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Re: 3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 07 Feb 2014, 20:17
guerrero25 wrote:
3 people, Amy, Beth, and Cassie, have speeds of 3 mph, 4 mph and 6 mph respectively. They run a race in which Beth gives Amy a head start of 2 hrs. If both Beth and Cassie overtake Amy at the same time, what head start did Cassie give Amy?

A. 3 hours
B. 4 hours
C. 5 hours
D. 9 miles
E. 10 miles


I used following approach:

( A,B and C denote Amy,Beth and Cassie respectively)
Let the total distance be x

then time of A= x/3 , B= X/4

Given that, x/3 - x/4 = 2
So, x = 24

Now time taken by A = 8 h

B= 6h

C= 4h

Thus, C gave head start to A for 8-4= 4h
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Re: 3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 18 Aug 2014, 09:02
VeritasPrepKarishma wrote:
guerrero25 wrote:
3 people, Amy, Beth, and Cassie, have speeds of 3 mph, 4 mph and 6 mph respectively. They run a race in which Beth gives Amy a head start of 2 hrs. If both Beth and Cassie overtake Amy at the same time, what head start did Cassie give Amy?

A. 3 hours
B. 4 hours
C. 5 hours
D. 9 miles
E. 10 miles



The most important clue here is this: If both Beth and Cassie overtake Amy at the same time

This means that all A, B and C meet together i.e. at one instant, they all had covered the same distance.
For easy calculations, let's assume that that distance was 12 miles. A would have been running for 4 hrs, B would have been running for 3 hrs and C would have been running for 2 hrs.
But actually, B started 2 hrs after A so the distance must have been 24 miles. A would have been running for 8 hrs, B would have been running for 6 hrs and C would have been running for 4 hrs.
So C gave A a head start of 4 hrs.

Answer (B)


Hi Karishma,
Do you have any blog on this concept of race and circular tracks? I have trouble understanding them.
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Re: 3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 18 Aug 2014, 20:42
maggie27 wrote:
VeritasPrepKarishma wrote:
guerrero25 wrote:
3 people, Amy, Beth, and Cassie, have speeds of 3 mph, 4 mph and 6 mph respectively. They run a race in which Beth gives Amy a head start of 2 hrs. If both Beth and Cassie overtake Amy at the same time, what head start did Cassie give Amy?

A. 3 hours
B. 4 hours
C. 5 hours
D. 9 miles
E. 10 miles



The most important clue here is this: If both Beth and Cassie overtake Amy at the same time

This means that all A, B and C meet together i.e. at one instant, they all had covered the same distance.
For easy calculations, let's assume that that distance was 12 miles. A would have been running for 4 hrs, B would have been running for 3 hrs and C would have been running for 2 hrs.
But actually, B started 2 hrs after A so the distance must have been 24 miles. A would have been running for 8 hrs, B would have been running for 6 hrs and C would have been running for 4 hrs.
So C gave A a head start of 4 hrs.

Answer (B)


Hi Karishma,
Do you have any blog on this concept of race and circular tracks? I have trouble understanding them.


I have a post on circular motion: http://www.veritasprep.com/blog/2012/08 ... n-circles/
But not one on races. I will put something up in the next few weeks.
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3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 14 Sep 2015, 15:46
Let A=Amy's time to meeting
Let A-2=Beth's time to meeting
Let A-C=Cassie's time to meeting
3A=4(A-2)
A=8
8(3)= 4(6)=24 miles to meeting
6(8-C)=24 miles
C=4 hours head start given by Cassie to Amy
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Re: 3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 06 Aug 2016, 11:33
I'm unable to understand this question. Are they three running together or what? This is a really confusing question.
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Re: 3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 06 Aug 2016, 12:05
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sudhirgupta93 wrote:
I'm unable to understand this question. Are they three running together or what? This is a really confusing question.


What is the confusion?

Question clearly states that A started first. B started after 2 hours and we need to determine When did C start.

As per the question, we know that All three meet at the same point.

So, we can determine when did A and B meet. The point they will meet would be the same at which C will be present at that time.

So,

A and B will meet after t = 6/(4-3) = 6 hours.

In these 6 hours, B would have covered 4*6=24 miles.

=> C would also have covered 24 miles and that would be in 24/6=4 hours.

Thus, C started 2 hours after B or 4 hours after A. hence, answer is B.
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Re: 3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 06 Aug 2016, 12:12
abhimahna wrote:
sudhirgupta93 wrote:
I'm unable to understand this question. Are they three running together or what? This is a really confusing question.


What is the confusion?

Question clearly states that A started first. B started after 2 hours and we need to determine When did C start.

As per the question, we know that All three meet at the same point.

So, we can determine when did A and B meet. The point they will meet would be the same at which C will be present at that time.

So,

A and B will meet after t = 6/(4-3) = 6 hours.

In these 6 hours, B would have covered 4*6=24 miles.

=> C would also have covered 24 miles and that would be in 24/6=4 hours.

Thus, C started 2 hours after B or 4 hours after A. hence, answer is B.


I've been doing distance-rate questions since morning. I guess my glass is full. Got it from your explanation. Thanks..
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3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 10 Aug 2018, 00:45
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My Approach:

x 1h 2h 3h 4h 5h 6h 7h 8h
A 3 6 9 12 15 18 21 24
B 0 0 4 8 12 16 20 24
C 0 0 0 0 6 12 18 24

-we start filling out the distance travelled per hour by A,
-Since B gives A, 2 hours headstart, we have B starting 3 hours
-at the end of 8 hour A and B catch up.
-In order that C also catches up at the same time we start filling at the last column and start reducing at 6 per hour(speed of C)
-Doing so C shoud start at 5th hour. Hence C must give 4 hours head start.

This is alternative approach. Might work for some.
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3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 22 Jan 2019, 07:04
guerrero25 wrote:
3 people, Amy, Beth, and Cassie, have speeds of 3 mph, 4 mph and 6 mph respectively. They run a race in which Beth gives Amy a head start of 2 hrs. If both Beth and Cassie overtake Amy at the same time, what head start did Cassie give Amy?

A. 3 hours
B. 4 hours
C. 5 hours
D. 9 miles
E. 10 miles

Excellent opportunity for RELATIVE VELOCITY (speed) and UNITS CONTROL , two powerful tools covered in our course!

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\(\left( * \right)\,\,\,2{\rm{h}}\,\, \cdot \,\,{{3\,\,{\rm{miles}}} \over {1\,\,{\rm{h}}}}\,\,\, = \,\,\,6\,\,{\rm{miles}}\,\,\,\,\,\,\,\,\,\,\left[ {\,{\rm{distance}}\,\,{\rm{A}}\,\,{\rm{starts}}\,\,{\rm{ahead}}\,\,{\rm{of}}\,\,{\rm{B}}\,} \right]\)

\({{\rm{V}}_{{\rm{B}} \to {\rm{A}}}} = {{4 - 3\,\,{\rm{miles}}} \over {1\,\,{\rm{h}}}}\,\,\, = \,\,\,{{6\,\,{\rm{miles}}} \over {{T_B}}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{T_B} = 6\,{\rm{h}}\,\,\,\,\,\,\,\,\,\left[ {\,{\rm{B}}\,\,{\rm{to}}\,\,{\rm{overtake}}\,\,A\,} \right]\)


\(? = x\,\,{\rm{h}}\)


\(\left( {**} \right)\,\,\,x\,\,{\rm{h}}\,\, \cdot \,\,{{3\,\,{\rm{miles}}} \over {1\,\,{\rm{h}}}}\,\,\, = \,\,\,3x\,\,{\rm{miles}}\,\,\,\,\,\,\,\,\,\,\left[ {\,{\rm{distance}}\,\,{\rm{A}}\,\,{\rm{starts}}\,\,{\rm{ahead}}\,\,{\rm{of}}\,\,{\rm{C}}\,} \right]\)

\({{\rm{V}}_{{\rm{C}} \to {\rm{A}}}} = {{6 - 3\,\,{\rm{miles}}} \over {1\,\,{\rm{h}}}}\,\,\, = \,\,\,{{3x\,\,{\rm{miles}}} \over {{T_C}}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{T_C} = x\,{\rm{h}}\,\,\,\,\,\,\,\,\,\left[ {\,{\rm{C}}\,\,{\rm{to}}\,\,{\rm{overtake}}\,\,A\,} \right]\)


\({\rm{Stem}}\,\,\,\, \Rightarrow \,\,\,\,\, {T_C} + x = {T_B} + 2\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = x = 4\)


We follow the notations and rationale taught in the GMATH method.

Regards,
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Re: 3 people, Amy, Beth, and Cassie, have speeds of 3 mph  [#permalink]

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New post 27 Jan 2019, 19:29
guerrero25 wrote:
3 people, Amy, Beth, and Cassie, have speeds of 3 mph, 4 mph and 6 mph respectively. They run a race in which Beth gives Amy a head start of 2 hrs. If both Beth and Cassie overtake Amy at the same time, what head start did Cassie give Amy?

A. 3 hours
B. 4 hours
C. 5 hours
D. 9 miles
E. 10 miles



We can let t = the time Amy has run before Beth and Cassie overtake her. Thus, we can create the following equation:

4t = 3(2 + t)

4t = 6 + 3t

t = 6

Thus, Amy has run 3(2 + 6) = 24 miles in 8 hours before Beth and Cassie overtake her. Since Cassie only needs 4 hours to run 24 miles then she has to give Amy 4 hours for the head start.

Answer: B
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Re: 3 people, Amy, Beth, and Cassie, have speeds of 3 mph   [#permalink] 27 Jan 2019, 19:29
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