Emily rows six miles downstream in 1 hour and her friend Ashley, rowing 1 mile per hour faster, completes the return trip in 2 hours. If Emily and Ash ley were rowing separately, who would complete their trip first and by how long? Round to the nearest hundredth, if necessary.

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Let the speed of rowing of Emily be E mph.

So, the speed of rowing of Ashley is (E+1) mph per hour. Also assume that the speed of current in the river is C.

For the downstream trip of Emily, `6/(E+C)=1`

`rArr (E+C)=6` ,

and `E=6-C` ------ (i)

For the return trip of Ashley, `6/(E+C+1)+6/(E-C+1)=2`

Plugging in the values from eqn. (i),

`6/(6+1) + 6/(6-C-C+1)=2`

`rArr 6/7 + 6/(7-2C) = 2`

`rArr C=7/8` mph.

and E = (6-7/8)=41/8 mph.

Total time required by Emily for her return trip

`=6/(E+C)+6/(E-C)=1+6/(34/8) = 2 7/17` hrs.

Total time required by Ashley for her return trip = 2 hrs (given).

Therefore, Ashley would complete the return trip faster by 7/17 hrs. i.e. by **0.41** hrs.

**Sources:**

### 3 Replies | Hide Replies ▲

If the curent is `7/8` mph then Emily's speed will be `6-7/8=5 1/8` mph, not `4 1/8` mph as you have.

That would make Ashley's speed `6 1/8` mph. Then her upstream leg is at the rate of `6 1/8 - 7/8 = 5 1/4` mph and will not take 2 hours to cover 6 miles.

Respected Educator,

I have written 6-7/8 =41/8, and not 4 1/8 as you have shown in your reply. If you express it in the form of mixed fraction 41/8 equals to 5 1/8 and that satisfies all the conditions of the problem.

Ashley's speed upstream = 6 1/8 - 7/8 = 42/8 =5 1/4 mph

that makes her upstream trip 24/21 hrs. long.

Total time of return trip (=round trip?)=24/21+6/(6 1/8+7/8)=2 hrs.(Given, "she completes the return trip in 2 hours").

The main issue, it seems, is that part of the language. If "return trip" refers to the Ashley's total downstream and upstream journey, my logic stands ground. I have started digging into the problem exactly on that promise.

On the contrary, if "return trip" refers to the Ashley's upstream journey (assuming the downstream journey was driven by Emily), your logic holds good.

Thanks and regards,

Sorry, I did indeed read 41/8 as `4 1/8` instead of `41/8` .

I believe since the problem never references Emily returning back upstream that the return trip (by Ashley) should be upstream only. It does not seem to be a very well defined question.

Let Emily's speed be r in mph. Then Asley's speed will be (r+1) in mph.

We use distance equals rate times time or d=rt:

For Emiy's trip downstream she rows with the current; if the speed of the current is c then her rate is r+c.

So 6=(r+c)(1) 88 time is 1 hour ==> c=6-r

For Ashley's trip upstream she rows against the current; again with the speed of the current c we have her rate as r+1-c. (1 mph faster rowing than Emily, but against the current so subtract.)

So 6=(r+1-c)(2) **time here is 2 hours ==> c=r-2

Then 6-r=r-2 ==> r=4.

Thus Emily's rowing speed is 4mph, the current's speed is 2mph, and Ashley's rowing speed is 5mph.

(Check: rowing with the current Emily goes 4+2=6mph; so she takes 1 hour to row 6 miles. Ashley rows against the current so she goes 5-2=3mph and will take 2 hours to row 6 miles.)

If Emily rowed down and back at 4 mph she would travel 6 miles at 6 mph and 6 miles at 4-2=2mph. It would take 1 hour for the downstream leg and 3 hours for the upstream leg for a total of 4 hours.

If Ashley rows down and back at 5mph she will travel 6 miles at 5+2=7mph downstream and 6 miles at 5-2=3mph upstream. The downstream leg will take `6/7` of an hour while the upstream leg will take 2 hours for a total of `2 6/7` hours.

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The current flows at 2mph. Emily's rowing speed is 4mph (without considering current) while Ashley's rowing speed (again without considering current) is 5mph.

Emily's round-trip time is 4 hours.

Ashley's round-trip time is `2 6/7` hours.

Ashley finishes first by `4-2 6/7=1 1/7` hours or 1.14hours.

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