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

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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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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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

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

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

Check out the blog posts on the link given in my signature below.
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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

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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I've been doing distance-rate questions since morning. I guess my glass is full. Got it from your explanation. Thanks..

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.

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




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

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


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


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

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


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


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

Regards,
Fabio.
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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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Since all three runners start from the same point and meet at a certain point, they run the same distance but take different times to do so. Therefore, we can use the ratio approach based on the fact that the ratios of the speeds of two moving objects are inversely proportional to the ratios of the times taken by them to cover the same distance. Let the time taken by Amy to reach the meeting point be 't' hours and 'x' hours the head start that Cassie gives Amy. Then:

Amy's speed:Beth's speed=Time taken by Beth:Time taken by Amy...> 3/4=(t-2)/t...> t=8 hrs.
Amy's speed:Cassie's speed=Time taken by Cassie:Time taken by Amy...> 3/6=(8-x)/8...> x=4
ANS: B

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