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

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19 Dec 2015, 02:30

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Lets assume r (Speed ) to be multiple of 3, say 30 r=30(Since 4r/3 has 3 in denominator). r=30 miles, 4r/3= 40 miles John's Speed = 30 miles. Tom's Speed= 40 miles Let t be time that Tom has travelled to catch with John and therefore John would have travelled for (t+2) hrs When Tom caught up John, the distance travelled would be the same. Equating distance on both sides

30(t+2) = 40t 30t+60=40t t=6hrs Time travelled by Tom to catch up with John = 6hrs, which is 40*6= 240 miles, which is 8r, since we assumed r as 30miles/hr.

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

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19 Dec 2015, 02:56

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Let's say Tom Cought up with john in x hours.

As John started 2 hours before tom, So in (x + 2) hours john traveled = (x + 2)*r = rx + 2r miles

In x hrs Tom traveled = (4r/3)*x = 4rx/3 miles

we can equate the distances covered (same distance - from starting point till meeting point), Hence: 4rx/3 = rx + 2r, 4rx = 3rx + 6r , rx = 6r , x = 6 hours.

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

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19 Dec 2015, 13:56

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The key thing to understand here is that when Tom catches up with John, they have covered the same distance. What the question really asks for is this distance, which is expressed as \(T * R\) in the answer choices. So our goal is to express the distance covered by John and Tom with rate and time.

Distance covered by John : \(2 * r\)(distance covered during the first 2 hours at r miles) We define a new variable, \(T\) = the time it took John to travel the second portion of the trip. The distance he covers is \(r * T\). Total distance covered by John: \(2r + rT\).

Distance covered by Tom : Tom travels at a speed of \(\frac{4}{3}r\). The total travel time of Tom is equal to the time it takes John to travel the second part of his trip --> \(T\). Total distance covered by Tom: \(\frac{4}{3}rT\)

Since John and Tom have covered the same distance when they meet up, we can solve the following equation for T : \(2r + rT = \frac{4}{3}rT\) \(2r = \frac{1}{3}rT\) \(T = 6\). So it took Tom 6 hours to catch up with John.

And the distance covered is : \(\frac{4}{3}r * 6 = 8r\)

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?

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It is important to note that in speed problems, time passes at the same time.

John traveled 2r miles during the first 2 hours and he traveled rt miles during the next t hours. In case of Tom, he traveled 4rt/3 miles during t hours. The distances they traveled are rt and 4rt/3, respectively,instead of rt1 and 4rt2/3,respectively because both traveled during the same time period. Then, \(2r+rt=\frac{4rt}{3}\) → \(2+t=\frac{4t}{3}\) → t=6. This means that Tom caught up with John after John traveled 6 hours. So we can solve 2r+6r=8r, and the correct answer is C.

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

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23 Aug 2017, 10:05

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