Hello,
I really have troubles understanding how to solve this question, even when I look through the solution, I hope you can help me!
Andy and Frank start out at opposite ends of a 450-mile route, cycling toward each other at their respective constant rates. Frank cycles at 15 miles per hour and Andy cycles at 25 miles per hour. If Andy leaves at 6am and Frank leaves at some point after that, which combination below represents a time at which Frank begins his ride and a time at which the two will meet along the route?
The different options for when Frank begins his ride and when the two meet are the following:
8:00am
9:00am
10:00am
12:00pm
6:30pm
7:00pm
7:30pm
- And so you have to choose two options; one where Frank leaves, and one where they meet each other.
In the suggested solution it is stated that:
For a question like this, it can be helpful to set up your own grid, noting the times that Frank can leave, the distance Andy will have traveled at that point,the distance they'll need to cover together, and the time at which they'll have covered that distance (working at their combined rate of 40 miles per hour):
I really don't understand this approach. So do you have to work on the problem like it is a double distance, (900 miles?)
Many thanks in advance,
Br.
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Integrated reasoning, difficult distance/rate question
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Hi kornerua,
The solution to this question involves TESTing THE ANSWERS and a bit of 'brute force math' (which is what the given explanation suggestions).
The question tells us that the distance BETWEEN the two cyclists is 450 miles, and part of the work that we have to do is determine at what time they will MEET one another. Thus, they don't both have to cycle 450 miles - they have to cycle 450 miles combined. Since their two speeds (15mph and 25mph) are fairly slow relative to a total of 450 miles, it's likely that Frank will start cycling at one of the "early" times and the two will meet sometime in the evening (one of the "later" times). To figure out which times exactly, we just have to TEST them and find the one that matches....
IF... Frank starts at 8am, then Andy will have been cycling for 2 hours by himself (2 hours x 25mph = 50 miles). So at 8am, the distance between them is 400 miles. Combined, they travel a total of 40 mph (Andy travels 25 miles every hour and Frank travels 15 miles every hour), so it would take the two of them 10 hours to travel that 400 miles combined. 10 hours AFTER 8am is 6pm. Unfortunately, THAT answer option is NOT among the choices, so 8am is not correct.
Using this logic, can you finish the work involved? (try seeing what happens if Frank starts at 9am and/or 10am).
GMAT assassins aren't born, they're made,
Rich
The solution to this question involves TESTing THE ANSWERS and a bit of 'brute force math' (which is what the given explanation suggestions).
The question tells us that the distance BETWEEN the two cyclists is 450 miles, and part of the work that we have to do is determine at what time they will MEET one another. Thus, they don't both have to cycle 450 miles - they have to cycle 450 miles combined. Since their two speeds (15mph and 25mph) are fairly slow relative to a total of 450 miles, it's likely that Frank will start cycling at one of the "early" times and the two will meet sometime in the evening (one of the "later" times). To figure out which times exactly, we just have to TEST them and find the one that matches....
IF... Frank starts at 8am, then Andy will have been cycling for 2 hours by himself (2 hours x 25mph = 50 miles). So at 8am, the distance between them is 400 miles. Combined, they travel a total of 40 mph (Andy travels 25 miles every hour and Frank travels 15 miles every hour), so it would take the two of them 10 hours to travel that 400 miles combined. 10 hours AFTER 8am is 6pm. Unfortunately, THAT answer option is NOT among the choices, so 8am is not correct.
Using this logic, can you finish the work involved? (try seeing what happens if Frank starts at 9am and/or 10am).
GMAT assassins aren't born, they're made,
Rich
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I'd think of it algebraically.
Say Andy's rate = 25 and his time = t. Frank's rate = 15 and his time = t - x.
Together the two guys will make up the entire distance, so
Total Distance = Andy's Rate*Andy's Time + Frank's Rate*Frank's Time
450 = 25t + 15(t - x)
450 = 40t - 15x
90 = 8t - 3x
We have a few options at this point, but using the answers is probably easiest. Since we need 8t > 90 (or else 90 = 8t - 3x will be impossible), t > 11 hours. That means they meet at 6:30, 7:00, or 7:30.
Now we could just try some answers. If we choose 6:30 PM, t = 12.5. Plugging that in, we have
90 = 8*12.5 - 3x
10/3 = x
But that would have Frank leaving 3 + (1/3) hours after Andy, at 9:20 AM, which isn't an option.
Now choose 7:00 PM. That means t = 13, and we have
90 = 8*13 - 3x
14/3 = x, again not an option.
We're left with 7:30 PM, and t = 13.5. That gives
90 = 8*13.5 - 3x
x = 6
So Frank leaves 6 hours after Andy, at 12:00 PM, and we're set!
Say Andy's rate = 25 and his time = t. Frank's rate = 15 and his time = t - x.
Together the two guys will make up the entire distance, so
Total Distance = Andy's Rate*Andy's Time + Frank's Rate*Frank's Time
450 = 25t + 15(t - x)
450 = 40t - 15x
90 = 8t - 3x
We have a few options at this point, but using the answers is probably easiest. Since we need 8t > 90 (or else 90 = 8t - 3x will be impossible), t > 11 hours. That means they meet at 6:30, 7:00, or 7:30.
Now we could just try some answers. If we choose 6:30 PM, t = 12.5. Plugging that in, we have
90 = 8*12.5 - 3x
10/3 = x
But that would have Frank leaving 3 + (1/3) hours after Andy, at 9:20 AM, which isn't an option.
Now choose 7:00 PM. That means t = 13, and we have
90 = 8*13 - 3x
14/3 = x, again not an option.
We're left with 7:30 PM, and t = 13.5. That gives
90 = 8*13.5 - 3x
x = 6
So Frank leaves 6 hours after Andy, at 12:00 PM, and we're set!
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Another option is to use number properties.
Once we have
90 = 8t - 3x
we also have
90 + 3x = 8t
or
3 * (30 + x) = 8t
Since 8t is a multiple of 8, the left hand side is a multiple of 8. Looking at the options, we can only have x = 2 (if Frank leaves at 8:00 AM), x = 3 (if he leaves at 9:00 AM), or x = 6 (if he leaves at noon).
3 * (30 + 2) and 3 * (30 + 3) aren't multiples of 8, but 3 * (30 + 6) is, so x = 6, and we're set again!
Once we have
90 = 8t - 3x
we also have
90 + 3x = 8t
or
3 * (30 + x) = 8t
Since 8t is a multiple of 8, the left hand side is a multiple of 8. Looking at the options, we can only have x = 2 (if Frank leaves at 8:00 AM), x = 3 (if he leaves at 9:00 AM), or x = 6 (if he leaves at noon).
3 * (30 + 2) and 3 * (30 + 3) aren't multiples of 8, but 3 * (30 + 6) is, so x = 6, and we're set again!
