1 hour
2006-06-15 16:36:45
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answer #1
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answered by Blondie 3
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It takes the friend 2 hours to address the envelopes alone.
x + y = 3 (180mins)
y (friend) = 2x
x + 2x = 3
3x = 3
x = 1
y = 2x = 2(1) = 2.
Therefore, 2 hours.
Here's the verdict:
Your friend can do it twice at fast, which means at 1 hour she did as much as you did in 2 hours.
2 hours of your work = 1 hour of your friend's.
so 1 + 1 = 2hours. It is that simple.
2006-06-15 23:38:05
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answer #2
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answered by Anonymous
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It's got to take your friend LONGER to address the envelopes if she is doing it by herself.
I think the answer is 4.5 hours.
I think it's something like:
1/x + 1/2x = 1/3
Simplifying, you would get 6 + 3 = 2x
9 = 2x
4.5 = x
but, it's been awhile. I do know for CERTAIN the answer is not 1 or 2 hours.
2006-06-15 23:39:21
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answer #3
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answered by scotsgirl 2
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4.5 hours
say you have N envelopes.
Rate at which your friend addresses envelopes: y (envelopes/hr)
Rate at which you address envelopes: x envelopes/hr
3 hrs * x + 3 hrs * y = N
Also,
y = 2x
Let H represent how long it would take your friend by herself
y*H = N
Solve for H:
H = N/y
H = (3x + 3y)/y
H = (3 (0.5y) + 3y ) /y substituting x=0.5y
H = (1.5y + 3y )/ y
H = 4.5
Therefore it will take your friend 4.5 hours by herself
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The other way to think about it is that your friend addressed 2 times more envelopes than you did, so if she worked for 3 hours by herself, she would have to do half the work she did so far again, or another 1.5 hours, so it would take her 4.5 hours.
2006-06-15 23:46:18
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answer #4
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answered by karen 2
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2 1/2 hrs. total time
I work at a speed of 1t. My friend works at a speed of 2t.
Together (3t) it took us 3 hrs. My friend got 2 hours worth of work done to my 1 hour. For my friend to do my 1 hour's worth of work, it would take 1/2 hour plus the 2 he already worked. Total 2 1/2 hours.
2006-06-15 23:50:45
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answer #5
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answered by DianeD 4
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f = the time it takes your friend
y = the time it takes you
y = 2f
f + y = 3
f + 2f = 3
3f = 3
f = 1
It would take your friend 1 hour.
2006-06-15 23:37:13
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answer #6
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answered by Anonymous
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1 HOUR
2006-06-15 23:36:09
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answer #7
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answered by John Doe 4
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you would need the number of invitations there are to adress.
2006-06-15 23:36:03
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answer #8
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answered by carebear92wi 2
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how many invitations?
2006-06-15 23:36:25
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answer #9
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answered by scooterbassplyr 2
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