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Math Revolution and GMAT Club Contest! John drove on a highway at a

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Math Revolution and GMAT Club Contest Starts!



QUESTION #12:

John drove on a highway at a constant speed of r miles per hour in 13:00. Then, 2 hours later, Tom drove on the same highway at a constant speed of 4r/3 miles per hour in 15:00. If both drivers maintained their speed, how many did John drive on a highway, in miles, when Tom caught up with John?

A. 3r
B. 5r
C. 8r
D. 9r
E. 10r



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Re: Math Revolution and GMAT Club Contest! John drove on a highway at a [#permalink]

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John drove on a highway at a constant speed of r miles per hour in 13:00. Then, 2 hours later, Tom drove on the same highway at a constant speed of 4r/3 miles per hour in 15:00. If both drivers maintained their speed, how many did John drive on a highway, in miles, when Tom caught up with John?

A. 3r
B. 5r
C. 8r
D. 9r
E. 10r
Explanation:-
Suppose Tom drove for time t speed 4r/3(given) And John for time t+2 speed r ( given)
Since both distance are equal hence t.4r/3=(t+2).r-->t=6
There fore John travelled 8r distance ANS-C

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Re: Math Revolution and GMAT Club Contest! John drove on a highway at a [#permalink]

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Relative speed of tom with reference to john=(4r/3)-r=r/3
The distance john would have travelled in 2 hours=2r
Time taken by tom to catch up with john=2r/(r/3)=6 hours
When tom catches up with john,
total Distance travelled by john=total distance travelled by tom
2r+6r= 6*(4r/3)
=8r
Answer C

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Re: Math Revolution and GMAT Club Contest! John drove on a highway at a [#permalink]

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When Tom started driving, the distance the John had already traveled is: 2r miles
Each hour the distance between Tom and John shortened by (4r/3-r) = r/3 miles
Then it will take Tom \(\frac{2r}{(r/3)}=6\) hours to catch up John, which means till Tom caught up with John, John had been driving for 8 hours => The distance John drove is 8r
Answer C.
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Last edited by HieuNguyenVN on 16 Dec 2015, 08:34, edited 2 times in total.

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Re: Math Revolution and GMAT Club Contest! John drove on a highway at a [#permalink]

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rel =4r/3 - r =r/3
time they met = 2r/(r/3)=6
d=s*t= 6r+2r = 8r
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Re: Math Revolution and GMAT Club Contest! John drove on a highway at a [#permalink]

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Speed of John =r miles per hour and he started at 13:00
speed of Tom =4r/3 miles per hour and he started at 15:00
by the time Tom Started John covered r*2 miles
relative speed =4r/3-r=r+r/3-r=r/3
d=2r
time taken =2r/(r/3)=6 hours
total time John was on road =6+2 =8 hours
therefore John drove (8r)

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Re: Math Revolution and GMAT Club Contest! John drove on a highway at a [#permalink]

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QUESTION #12:

John drove on a highway at a constant speed of r miles per hour in 13:00. Then, 2 hours later, Tom drove on the same highway at a constant speed of 4r/3 miles per hour in 15:00. If both drivers maintained their speed, how many did John drive on a highway, in miles, when Tom caught up with John?

A. 3r
B. 5r
C. 8r
D. 9r
E. 10r


Solution:
The distance traveled by both John and Tom are same when they meet.
Distance traveled by John = Distance traveled by Tom
(Speed of John)*(2hr + time) = (Speed of Tome)*(time)
r*(t+2) = (4r/3)*t
t = 6
So, distance traveled = 8r

Answer (C)

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Firstly we notice that John and Tom will have travelled the same distance when they meet, John will just have been travelling for longer.

If we define \(x\) as the amount of time John spent travelling until he met Tom, then the amount of time Tom spent travelling would be \(x-2\). The distance John travelled would be \(x*r\) and the distance Tom travelled would be \((x-2)*\frac{4r}{3}\).

Since we know that they travelled the same distance, we can write \(xr=(x-2)*\frac{4r}{3}\) and solve for \(x\) to give \(x=8\).

If \(x=8\) and we know that the distance travelled by John is \(x*r\) then we know the distance is \(8r\), the answer is C.
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John - r m/hr, starts at 1 PM. Two hours later, John has travelled 2r miles.
Tom - 4r/3 m/hr, starts at 3 PM.

First find what time does Tom meet John.

2r + r*x = 4r/3*x. x=6 hours in this case. So basically 6 hours after Tom starts, he catches up with John.

John drove 2r + r*6 = 8r. Answer is C
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At 15:00, The lead is 2r for John
This lead of 2r has to be covered at a relative speed of ((4r/3) - r) to catch up with John
This means 6 hours from 15:00, the 2 will meet
But John started at 13:00 i.e. he has driven for 8 hours @ rmph
Hence John's distance is 8r when they meet

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for 2 hrs, J travelled 2r miles.. now tom starts with 4r/3 speed.. suppose they meet after time T.. distance travelled by tom = 4rT/3

also John will travell rT miles in this time..
now.. 2r+rT = 4rT/3
2r=rT/3
T=6

Therefore John travels = 2r +6r = 8r miles
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for 2 hrs, J travelled 2r miles.. now tom starts with 4r/3 speed.. suppose they meet after time T.. distance travelled by tom = 4rT/3

also John will travell rT miles in this time..
now.. 2r+rT = 4rT/3
2r=rT/3
T=6

Therefore John travels = 2r +6r = 8r miles
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Re: Math Revolution and GMAT Club Contest! John drove on a highway at a [#permalink]

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QUESTION #12:
John drove on a highway at a constant speed of r miles per hour in 13:00. Then, 2 hours later, Tom drove on the same highway at a constant speed of 4r/3 miles per hour in 15:00. If both drivers maintained their speed, how many did John drive on a highway, in miles, when Tom caught up with John?
A. 3r
B. 5r
C. 8r
D. 9r
E. 10r

Solution:

V=s/t
r=d/t

For Tom: 4r/3=d/t-2
or, r=3d/4(t-2)

d/t=3d/4(t-2)
or, 3t=4t-8
or,t=8

Since, r=d/t
r=d/8
or d=8r

Answer: "C"
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Re: Math Revolution and GMAT Club Contest! John drove on a highway at a [#permalink]

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Speed of John = r mph
Speed of Tom = 4r/3 mph

Relative speed = (4r/3) - r = r/3 mph

Tom leaves 2 hrs after John --> Distance covered by John in 2 hrs = 2r miles

So Tom has to cover this additional distance to meet John.

Time taken by Tom to meet John = Relative Distance/Relative speed = (2r)/(r/3) = 6 hours.

Since John drives for 2 more hours than Tom, time taken by John to meet Tom = 6 + 2 = 8 hours

Distance driven by John = Speed * Time = 8 * r = 8r miles

Answer: C

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Re: Math Revolution and GMAT Club Contest! John drove on a highway at a [#permalink]

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QUESTION #12:

John drove on a highway at a constant speed of r miles per hour in 13:00. Then, 2 hours later, Tom drove on the same highway at a constant speed of 4r/3 miles per hour in 15:00. If both drivers maintained their speed, how many did John drive on a highway, in miles, when Tom caught up with John?

A. 3r
B. 5r
C. 8r
D. 9r
E. 10r

In 2 hours (from 13:00 to 15:00), the distance John traveled \(= d_r=2*r\) (miles)
Relative speed of Tom and John \(= S_r=\frac{4r}{3}-r=\frac{r}{3}\) (miles per hour) - meaning: in 1 hour, Tom can cover distance of r/3 in relative with distance that John traveled.
Tom caught up with John only when Tom covered the 2r distance that John traveled --> the time Tom need to cover \(2r = \frac{2r}{r/3} = 6\)(hrs)
--> In total, John traveled: \(6+2= 8\) hrs or \(8*r=8r\) miles

Answer C

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Using the Eqn \(D = RT\)

\(D(John) = R * T\)
\(D(Tom) = 4/3 * R * (T - 2)\) (Since Tom started 2 hrs late)

When Tom Caught up with John

\(D(John) = D(Tom) => R*T = 4/3 R (T-2) => 3T = 4 (T-2) => T = 8\)

Therefore John Drove R*8 Miles before Tom Caught up with him

Answer is C

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Since they catch up after driving on the same highway, the two distances are the same:

\(Distance John = Distance Tom\)

\(rt = \frac{4r}{3}*(t-2)\)

\(3rt =4rt -8r\)

\(t= 8\) so the distance for John is rt or 8r

Same distance for Tom \(\frac{4r}{3}*6 =8r\)
Answer C
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option C.
This can be solved using the concept of relative speed.

We are given that Tom started driving 2 hours after John. So, by this time John must have covered distance = r*2 = 2r

Relative speed of Tom with respect to John = 4r/3 - r => r/3
Now to catch up with john this distance of 2r needs to be travelled by Tom relative to John.
So time required to travel this distance = 2*r/relative speed => 2r/(r/3) => 6hrs.

Thus it will take Tom 6 hours to catch up with John.

Now, by this time the distance travelled by John will be = r*6 = 6r.
Also, John had already travelled a distance of 2r before Tom started.

Thus total distance travelled by John will be 2r+6r = 8r. => option C.
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speed equlas to distance travelled time so
John => r=m1/t1----(eq1);
Tom => 4*r/3 = m2/t2---(eq2);
as John should meet tom m1 should be equal to m2
so m1=r*t1 from eq1 & m2= 4*r/3*t2 which on solving both results t1=4/3 t2;
As tom has started 2 hrs late so t2-t1=2; solving for t1 we get t1=8 and m1=8*r

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Re: Math Revolution and GMAT Club Contest! John drove on a highway at a [#permalink]

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QUESTION #12:

John drove on a highway at a constant speed of r miles per hour in 13:00. Then, 2 hours later, Tom drove on the same highway at a constant speed of 4r/3 miles per hour in 15:00. If both drivers maintained their speed, how many did John drive on a highway, in miles, when Tom caught up with John?

A. 3r
B. 5r
C. 8r
D. 9r
E. 10r

---

say time taken by John when John & Tom catch up with each other is "t" hours.
Since Tom has started 2 hours "later", Tom would have taken "t-2" hours (2 hours lesser than John) when they meet up.

Total Distance that John would've traveled in t hours : D1 = r*t
Total Distance that Tom would've traveled in (t-2) hours : D2 = (4r/3)*(t-2)

They would've both covered the same distance when they meet , so D1 = D2

r*t = (4r/3)*(t-2)
=> 3rt = 4rt -8r
=> rt = 8r
=> r(t-8) = 0
we know r>0 , otherwise both John and Tom would've been stationary , so are not moving at all and can never meet.
so t = 8;

Going by the previous equation we've setup ,
Total Distance traveled by John when they meet = r*y = 8r

Correct Answer : C
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