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Question 1194386: Working together, Jenny and Natalie can assemble a desk in 42/13 hours. Working alone Jenny could have done it in 7 hours. How long would it take Natalie working
alone?
hours.
Found 4 solutions by math_tutor2020, ikleyn, greenestamps, MathTherapy: Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
Let's say this desk has 42 pieces. I'm picking 42 because it will cancel out later on with the other 42 from 42/13.
Let x be the combined rate of Jenny and Natalie working together.
The idea is to multiply the number of hours they work together (42/13) with their combined rate (x) and that will tell us how many pieces they're able to assemble (42)
So,
(number of hours)*(pieces per hour) = number of pieces
(42/13)*x = 42
x = 42*(13/42)
x = 13
At this point, it's probably more clear why I picked 42
The cancellation happens at the last step.
If there are 42 pieces total, then their combined rate is 13 pieces per hour.
Another reason why I picked 42 is because it's a multiple of 7.
If Jenny works alone and she takes 7 hours, then her rate is 42/7 = 6 pieces per hour.
This leaves Natalie's rate to be 13 - 6 = 7 pieces per hour.
Therefore, if Natalie works alone, then she will take 42/7 = 6 hours.
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Here's an alternative approach using the standard method to solve rate of work problems.
n = number of hours Natalie needs if she works alone
1/n = Natalie's rate in jobs per hour
In other words, she does 1 job per n hours
Also, Jenny needs 7 hours, so her rate is 1/7 of a job per hour.
1/n + 1/7 = their combined rate in jobs per hour
That simplifies to (7+n)/(7n)
If they work together and get the job done in 42/13 hours, then their rate must be 13/42 jobs per hour.
Set this equal to the (7+n)/(7n) and let's solve for n.
(7+n)/(7n) = 13/42
42(7+n) = 7n*13
294+42n = 91n
91n-42n = 294
49n = 294
n = 294/49
n = 6
Natalie can get the job done in 6 hours if she works alone.
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Answer: 6 hours
Answer by ikleyn(52802) (Show Source): Answer by greenestamps(13200) (Show Source): Answer by MathTherapy(10552) (Show Source):
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