This is from MGMAT Guide 3. P. 110.
A park ranger travels from his base to a campsite via truck at r miles per hour. Upon arriving, he collects a snowmobile and uses it to return to base. If the campsite is d miles from the park ranger's base and the entire trip took t hours to complete, what was his speed on the snowmobile, in terms of t, d, and r ?
This problem was covered in 2012, but my question wasn't answered. I don't understand how to solve this by picking numbers. The book says, "Say that r=10 and d=20. That way it takes 2 hours for the ranger to reach the camp. To keep things easy, you can say that the whole trip took 4 hours. That would mean that the ranger traveled 10 miles per hour on the snowmobile as well. You've picked numbers for your variables and calculated a target. r=10, d=20, t=4, target is 10"
Why is t=4 - why is t not equal to 2? If t=4, why is d not equal to 40? I don't understand the setup. I do understand the algebraic approach. Thanks in advance!
Rate, distance and time
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For the benefit of others, here's one algebraic approach:aces021 wrote:This is from MGMAT Guide 3. P. 110.
A park ranger travels from his base to a campsite via truck at r miles per hour. Upon arriving, he collects a snowmobile and uses it to return to base. If the campsite is d miles from the park ranger's base and the entire trip took t hours to complete, what was his speed on the snowmobile, in terms of t, d, and r ?
I like to begin these kinds of questions by writing a "word equation"
We're told that the entire trip took t hours to complete, so we can write:
(time in truck) + (time on snowmobile) = t
Let x = the speed of the snowmobile
Note: time = distance/speed
So, time in truck = d/r
And time on snowmobile = d/x
Now take the word equation and replace each part to get:
d/r + d/x = t
Now solve for x. There are several ways to do this. Here's one way.
Multiply both sides by rx (this will eliminate the fractions) to get: dx + dr = trx
Rewrite to get: dr = trx - dx
Factor left side to get: dr = x(tr - d)
Divide both sides by (tr - d) to get: [spoiler]dr/(tr - d)[/spoiler] = x
Cheers,
Brent
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NOTE: If this were an actual GMAT problem, there would be answer choices.aces021 wrote:This is from MGMAT Guide 3. P. 110.
A park ranger travels from his base to a campsite via truck at r miles per hour. Upon arriving, he collects a snowmobile and uses it to return to base. If the campsite is d miles from the park ranger's base and the entire trip took t hours to complete, what was his speed on the snowmobile, in terms of t, d, and r ?
This problem was covered in 2012, but my question wasn't answered. I don't understand how to solve this by picking numbers. The book says, "Say that r=10 and d=20. That way it takes 2 hours for the ranger to reach the camp. To keep things easy, you can say that the whole trip took 4 hours. That would mean that the ranger traveled 10 miles per hour on the snowmobile as well. You've picked numbers for your variables and calculated a target. r=10, d=20, t=4, target is 10"
Why is t=4 - why is t not equal to 2? If t=4, why is d not equal to 40? I don't understand the setup. I do understand the algebraic approach. Thanks in advance!
So, assigning the values r=10, d=20, t=4 yields a SNOWMOBILE SPEED of 10 mph
Let's test whether or not this is true.
time = distance/speed
TOTAL TIME = (time in truck) + (time on snowmobile)
Or TOTAL TIME = (distance/truck speed) + (distance/snowmobile speed)
So, 4 = (20/10) + (20/10)
Evaluate: 4 = 2 + 2
Works perfectly.
This above "solution" is just ONE POSSIBLE way to plug in values that satisfy the given conditions.Why is t=4 - why is t not equal to 2? If t=4, why is d not equal to 40? I don't understand the setup. I do understand the algebraic approach.
We could also say that assigning the values r=10, d=40, t=5 yields a SNOWMOBILE SPEED of 40 mph. These values also satisfy the given conditions.
There are MANY other ways to assign values to the variables that satisfy the given conditions.
Cheers,
Brent
Thanks for your response. In your example of r=10, d=40, t=5 yields a SNOWMOBILE SPEED of 40 mph, why does the rdt work? How were you able to derive them since rt != d (10*5 != 40)? How do you know the snowmobile speed is 40 - you're assuming t=1? (if so, why - why not 2) And the truck rate is 10?Brent@GMATPrepNow wrote:NOTE: If this were an actual GMAT problem, there would be answer choices.aces021 wrote:This is from MGMAT Guide 3. P. 110.
A park ranger travels from his base to a campsite via truck at r miles per hour. Upon arriving, he collects a snowmobile and uses it to return to base. If the campsite is d miles from the park ranger's base and the entire trip took t hours to complete, what was his speed on the snowmobile, in terms of t, d, and r ?
This problem was covered in 2012, but my question wasn't answered. I don't understand how to solve this by picking numbers. The book says, "Say that r=10 and d=20. That way it takes 2 hours for the ranger to reach the camp. To keep things easy, you can say that the whole trip took 4 hours. That would mean that the ranger traveled 10 miles per hour on the snowmobile as well. You've picked numbers for your variables and calculated a target. r=10, d=20, t=4, target is 10"
Why is t=4 - why is t not equal to 2? If t=4, why is d not equal to 40? I don't understand the setup. I do understand the algebraic approach. Thanks in advance!
So, assigning the values r=10, d=20, t=4 yields a SNOWMOBILE SPEED of 10 mph
Let's test whether or not this is true.
time = distance/speed
TOTAL TIME = (time in truck) + (time on snowmobile)
Or TOTAL TIME = (distance/truck speed) + (distance/snowmobile speed)
So, 4 = (20/10) + (20/10)
Evaluate: 4 = 2 + 2
Works perfectly.
This above "solution" is just ONE POSSIBLE way to plug in values that satisfy the given conditions.Why is t=4 - why is t not equal to 2? If t=4, why is d not equal to 40? I don't understand the setup. I do understand the algebraic approach.
We could also say that assigning the values r=10, d=40, t=5 yields a SNOWMOBILE SPEED of 40 mph. These values also satisfy the given conditions.
There are MANY other ways to assign values to the variables that satisfy the given conditions.
Cheers,
Brent
Also, in your response you're differentiating between truck and snowmobile, I don't understand why this is relevant since we're making the numbers up. I guess I'm not understanding the fundamentals of the question. Heh.
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Let's try a different approach.
Let's see what happens if we let r=10, d=20, t=4.
Our goal is to find the SNOWMOBILE SPEED that works with these assigned values for r, d and t.
We'll use the fact that time = distance/speed
So, TOTAL TIME = (time in truck) + (time on snowmobile)
Which means TOTAL TIME = (distance/truck speed) + (distance/snowmobile speed)
Which means t = (d/r) + (d/snowmobile speed)
Plug in the three values to get: 4 = (20/10) + (20/snowmobile speed)
Important: What snowmobile speed will satisfy this equation?
To find out, we'll evaluate to get: 4 = 2 + (20/snowmobile speed)
So, it must be the case that 20/snowmobile speed = 2, which means the snowmobile speed = 10
So, if we let r=10, d=20, t=4, then the snowmobile speed = 10
If we use different values, we'll find that if we let r=10, d=40, t=5, then the snowmobile speed = 40
and so on.
Cheers,
Brent
Let's see what happens if we let r=10, d=20, t=4.
Our goal is to find the SNOWMOBILE SPEED that works with these assigned values for r, d and t.
We'll use the fact that time = distance/speed
So, TOTAL TIME = (time in truck) + (time on snowmobile)
Which means TOTAL TIME = (distance/truck speed) + (distance/snowmobile speed)
Which means t = (d/r) + (d/snowmobile speed)
Plug in the three values to get: 4 = (20/10) + (20/snowmobile speed)
Important: What snowmobile speed will satisfy this equation?
To find out, we'll evaluate to get: 4 = 2 + (20/snowmobile speed)
So, it must be the case that 20/snowmobile speed = 2, which means the snowmobile speed = 10
So, if we let r=10, d=20, t=4, then the snowmobile speed = 10
If we use different values, we'll find that if we let r=10, d=40, t=5, then the snowmobile speed = 40
and so on.
Cheers,
Brent