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A canoeist spent two days on a large lake. On the second day, the cano

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A canoeist spent two days on a large lake. On the second day, the cano  [#permalink]

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New post 17 Oct 2016, 22:46
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A canoeist spent two days on a large lake. On the second day, the canoeist rowed 2 hours longer and at an average speed 2 miles per hour faster than he rowed on the first day. If the canoeist traveled a total of 50 miles and spent a total of 12 hours rowing on his trip, what was his average speed on the first day?

a) 2mph
b) 3mph
c) 4mph
d) 5mph
e) 6mph
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Re: A canoeist spent two days on a large lake. On the second day, the cano  [#permalink]

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New post 17 Oct 2016, 23:15
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Let time taken be t1 and t2 for day 1 and 2 respectively.
So t1+t2=12 and t2=t1+2. Substituting we get t1 = 5 and t2=7. Since total distance travelled is 50 we can say 50=5X + 7 (X+2) where X is avg speed of day 1. Solving we get X=3 and hnc the answer.

We can also solve just by looKing at numbers the, avg speed is 50/12 which is 4.1... so the average of 2 days must be around 4 and also satisfy the statement that average speed of day 2 is 2 mph more. 3 and 5 is the right combination.
Plz correct me if am wrong.

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Re: A canoeist spent two days on a large lake. On the second day, the cano  [#permalink]

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New post 18 Oct 2016, 00:37
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alanforde800Maximus wrote:
A canoeist spent two days on a large lake. On the second day, the canoeist rowed 2 hours longer and at an average speed 2 miles per hour faster than he rowed on the first day. If the canoeist traveled a total of 50 miles and spent a total of 12 hours rowing on his trip, what was his average speed on the first day?

a) 2mph
b) 3mph
c) 4mph
d) 5mph
e) 6mph

Please assist with above problem.


Form a table it will be a cake walk -
Attachment:
Screenshot.jpg
Screenshot.jpg [ 10.75 KiB | Viewed 2049 times ]

Here, 2t + 2 = 12

Or, 2t = 10

Or, t = 5

Further -

5*s + ( 5 + 2 ) (s + 2 ) = 50

or, 5s + 7s + 14 = 50
Or, 12s = 36

So, S = 3

Hence answer will be (B) 3
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Re: A canoeist spent two days on a large lake. On the second day, the cano  [#permalink]

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New post 19 Oct 2016, 02:05
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t1 + t2 = 12 and t2 - t1 = 2 solving for t1 and t2, t1 =5 t2 = 7

Total Distance = 50 = S*5 + (S+2)7 Solving for S, S = 3

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Re: A canoeist spent two days on a large lake. On the second day, the cano  [#permalink]

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New post 05 Dec 2017, 18:32
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alanforde800Maximus wrote:
A canoeist spent two days on a large lake. On the second day, the canoeist rowed 2 hours longer and at an average speed 2 miles per hour faster than he rowed on the first day. If the canoeist traveled a total of 50 miles and spent a total of 12 hours rowing on his trip, what was his average speed on the first day?

a) 2mph
b) 3mph
c) 4mph
d) 5mph
e) 6mph


We are given that on a second day the canoeist rowed 2 hours longer and at an average speed 2 miles per hour faster than he rowed on the first day.

If we let the number of hours the canoeist rowed on the first day = t, then the number of hours that he canoeist rowed on the second day = t + 2. Also, if we let the rate of the canoeist on the first day = r, then the rate of the canoeist on the second day is r + 2.

Since we are given that the total time spent rowing was 12 hours, we can determine a value for t.

t + t + 2 = 12

2t = 10

t = 5

Thus, 5 hours was spent rowing on day one, and 7 hours on day 2.

We are also given that the canoeist traveled a total of 50 miles. Since distance = rate x time, we can represent the distance traveled on day one as 5r and the distance traveled on day two as 7(r+2) = 7r + 14. Thus:

5r + 7r + 14 = 50

12r = 36

r = 3

Answer: B
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Re: A canoeist spent two days on a large lake. On the second day, the cano  [#permalink]

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New post 23 Jan 2018, 15:22
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Hi All,

Certain versions of this prompt will require lots of 'math steps' to get to the solution, but this specific prompt can be solved with a bit of logic and just a little math.

To start, we're told that the TOTAL trip was 50 miles and took 12 hours. That's an average of 50/12 = 4 1/6 miles/hour.

We're told that the two speeds differed by 2 miles/hour, and we spent just 2 extra hours at the faster speed. Thus, the slower speed was measurably below 4 1/6 mph and the upper speed was measurably above 4 1/6 mph. Remember - the speeds DIFFER by 2mph. We're asked for the speed on the first day (re: the slower speed). Looking at the answer choices, we can quickly compare what each answer implies to what the math dictates must happen...

Answer A: slow speed = 2mph, fast speed = 4mph. Does not make sense (the faster speed has to be GREATER than 4mph).
Answer B: slow speed = 3mph, fast speed = 5mph. This matches the logic really nicely.
Answer C: slow speed = 4mph, fast speed = 6mph. Here, the slow speed is just barely below the average speed, which doesn't make sense.

The remaining two answers will just get further and further away from the average, so we don't have to think much about them.

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Re: A canoeist spent two days on a large lake. On the second day, the cano  [#permalink]

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New post 06 Feb 2018, 18:52
EMPOWERgmatRichC wrote:
Answer C: slow speed = 4mph, fast speed = 6mph. Here, the slow speed is just barely below the average speed, which doesn't make sense.
Rich


I don't get why answer C doesn't make sense. Would you please see my following reasoning?

If the canoeist covers 44 miles at slower speed 4mph and remaining 6 miles at faster speed 6mph, then the time taken would be 11 hours and 1 hour respectively. Total time would be 12 hours.
Attachment:
Canoeist spent.JPG
Canoeist spent.JPG [ 18.04 KiB | Viewed 701 times ]

Am I wrong?
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Re: A canoeist spent two days on a large lake. On the second day, the cano  [#permalink]

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New post 07 Feb 2018, 12:02
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Hi Mahmud6,

You're overlooking an important piece of information in the prompt: the canoeist spent just 2 hours longer rowing at the FASTER speed (meaning that he spent 5 hours at the slower speed and 7 hours at the faster speed). Your example has the canoeist spending TEN hours longer at the SLOWER speed.

In math-terms, this is essentially a Weighted Average - however, the amount of time spent at each speed isn't that different from the other - so the average won't be weighted too heavily in either direction (meaning that the average speed is NOT 'really close' to either speed). We calculated that the average speed was 4 1/6 mph, so we can think about how weighted averages "work"...

One hour at 3 mph and one hour at 5 mph = (8 miles)/(2 hours) = 4 mph
One hour at 3 mph and two hours at 5 mph = (13 miles)/(3 hours) = 4 1/3 mph

Notice in the second example that we spent DOUBLE the amount of time at the faster speed and ended up with an average that was GREATER than 4 1/6 mph. That ratio is 'too much' relative to what we're told in the prompt - and it led to an average speed that was too high. 3mph and 5mph are perfect speeds for this situation; these examples just don't use the proper ratio.

From the answers, we know that the slower speed is an INTEGER (and from the prompt, we know that the faster speed would also be an INTEGER). Thus, the two speeds would have to be 3mph and 5 mph.

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Re: A canoeist spent two days on a large lake. On the second day, the cano  [#permalink]

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New post 07 Feb 2018, 22:22
EMPOWERgmatRichC wrote:
Hi Mahmud6,

You're overlooking an important piece of information in the prompt: the canoeist spent just 2 hours longer rowing at the FASTER speed (meaning that he spent 5 hours at the slower speed and 7 hours at the faster speed). Your example has the canoeist spending TEN hours longer at the SLOWER speed.

In math-terms, this is essentially a Weighted Average - however, the amount of time spent at each speed isn't that different from the other - so the average won't be weighted too heavily in either direction (meaning that the average speed is NOT 'really close' to either speed). We calculated that the average speed was 4 1/6 mph, so we can think about how weighted averages "work"...

One hour at 3 mph and one hour at 5 mph = (8 miles)/(2 hours) = 4 mph
One hour at 3 mph and two hours at 5 mph = (13 miles)/(3 hours) = 4 1/3 mph

Notice in the second example that we spent DOUBLE the amount of time at the faster speed and ended up with an average that was GREATER than 4 1/6 mph. That ratio is 'too much' relative to what we're told in the prompt - and it led to an average speed that was too high. 3mph and 5mph are perfect speeds for this situation; these examples just don't use the proper ratio.

From the answers, we know that the slower speed is an INTEGER (and from the prompt, we know that the faster speed would also be an INTEGER). Thus, the two speeds would have to be 3mph and 5 mph.

GMAT assassins aren't born, they're made,
Rich


Thank you Rich. Got it.
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Re: A canoeist spent two days on a large lake. On the second day, the cano &nbs [#permalink] 07 Feb 2018, 22:22
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