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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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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.
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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, 22:15
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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 17 Oct 2016, 23:37
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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 1122 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, 01:05
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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

HTH

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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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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] 05 Dec 2017, 17:32
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