Excellent questions. Thank you for your help. For the last problem with the biker and the runner. You can just use these two equations and set them equal to each other. The first equation is 10t and the second equation is 5t + 15. You get the same answer. The biker and runner, at some point, will travel the same distance once the biker catches up. So, you can set the equations equal to each other to find the time. Thanks again.
In the second example (lesley/airport) the rate for the whole trip is actually important. Since it can be easily calculated that it is 20m/h for the whole trip, which is also the mean of the 2 section rates, it means that she must have spent the same time for both sections. This means she spent 3h at 25mhp and 3h at 15mph.
Sorry maybe this is really obvious but how can we assume she spent the same time for both sections? Is it because if we find the mean of the two rates, automatically the duration is halved at that rate or?
For the GAP problems you can still use the traditional equations and methods to find the answer. Just use the common distance equations and solve for t. You need to set it up correctly, but you can still find the answer. Set, 10t = 5t + 15.
For the second question about Lesley, you over complicated the problem. Instead what you could is label time as time Lesley traveled at 25 miles per hour as t. Then get the distance she traveled at 25 miles per hour as 25T. Then you know that she traveled at 15mph for the rest of the remaining time, or 6 - T so the second distance was 15(6 - T). Add them together and you get 120 miles. Then the equation becomes so much easier so solve and does not involve fractions, you just calculate the time T, and then plug into 25T to get 75. But that method is equally valid obviously.
Excellent questions. Thank you for your help. For the last problem with the biker and the runner. You can just use these two equations and set them equal to each other. The first equation is 10t and the second equation is 5t + 15. You get the same answer. The biker and runner, at some point, will travel the same distance once the biker catches up. So, you can set the equations equal to each other to find the time. Thanks again.
Great tip!
Super helpful! Thanks a lot for your videos
In the second example (lesley/airport) the rate for the whole trip is actually important. Since it can be easily calculated that it is 20m/h for the whole trip, which is also the mean of the 2 section rates, it means that she must have spent the same time for both sections. This means she spent 3h at 25mhp and 3h at 15mph.
after doing the 25 and 15 average, I felt like I reached a dead end. I assumed 3 hours for both "parts" and reached the exact same answer (75 miles)
Sorry maybe this is really obvious but how can we assume she spent the same time for both sections? Is it because if we find the mean of the two rates, automatically the duration is halved at that rate or?
Excellent examples 👌🏼
Awesome, we're glad they helped!
Thanks Magoosh !
Glad you liked it!
For the GAP problems you can still use the traditional equations and methods to find the answer. Just use the common distance equations and solve for t. You need to set it up correctly, but you can still find the answer. Set, 10t = 5t + 15.
Good examples
Questions in this video - can someone hp me gauge what would be their level on the gmat? I know they're under 700 for sure.
😊
For the second question about Lesley, you over complicated the problem. Instead what you could is label time as time Lesley traveled at 25 miles per hour as t. Then get the distance she traveled at 25 miles per hour as 25T. Then you know that she traveled at 15mph for the rest of the remaining time, or 6 - T so the second distance was 15(6 - T). Add them together and you get 120 miles. Then the equation becomes so much easier so solve and does not involve fractions, you just calculate the time T, and then plug into 25T to get 75. But that method is equally valid obviously.
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Just shut up if you don't need, stop watching this first.
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