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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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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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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.

Final Answer:

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EMPOWERgmatRichC

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 7998 times ]
Am I wrong?
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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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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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Time taken by Canoeist on first day = 12-2=10/2= 5 hr
Time taken by Canoeist on second day= 7 hr
Let the speed on first day be x.
Then speed on the second day will be x+2.

5x+7(x+2)=50
x= 3mph

Ans(B)
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Total time = 12hr
let the first days time be t, so the 2nd days time is t+2.
we know 12 is the total time take:
12=t+(t+2)
12=2+2t
10=2t
t=5

t=5 & t+2=7
let s be the speed for day one and s+2 for day two.
total distance given is 50miles. distance on day 1 be m and in day 2 be m2
m+m2=50
5(s)+7(s+2)=50
here you can start substituting from the ans option the value of s or,
12s+14=50
12s=36
s=36/12
s=3
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Deconstructing the Question

Let the time and speed on day 1 be \(t\) and \(v\).

On day 2, the canoeist rows \(t+2\) hours at speed \(v+2\).

Total time is \(12\) hours and total distance is \(50\) miles.

Step-by-step

Time equation:

\(t + (t+2) = 12\)

\(2t + 2 = 12\)

\(t = 5\)

So day 2 time is \(7\).

Distance equation:

\(5v + 7(v+2) = 50\)

Expand:

\(5v + 7v + 14 = 50\)

\(12v + 14 = 50\)

\(12v = 36\)

\(v = 3\)

Answer B
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