SOLUTION: 3
workers can make 27 identical widgets in
1
5
hours . If all the workers work at the same rate, how long will it take
7
workers to make 1260 such widgets?
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workers can make 27 identical widgets in
1
5
hours . If all the workers work at the same rate, how long will it take
7
workers to make 1260 such widgets?
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Question 1123972: 3
workers can make 27 identical widgets in
1
5
hours . If all the workers work at the same rate, how long will it take
7
workers to make 1260 such widgets? Found 3 solutions by josgarithmetic, MathTherapy, ikleyn:Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website! ---------------------------------------------------------
3
workers can make 27 identical widgets in
1
5
hours . If all the workers work at the same rate, how long will it take
7
workers to make 1260 such widgets?
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Intended description may be this:
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Three workers can make 27 identical widgets in 15 hours . If all the workers work at the same rate, how long will it take 7 workers to make 1260 such widgets?
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r, the rate of work for 1 worker
x, unknown time for 7 workers to make 1260 widgets
You can put this solution on YOUR website!
3
workers can make 27 identical widgets in
1
5
hours . If all the workers work at the same rate, how long will it take
7
workers to make 1260 such widgets?
It's NOT 20 hours. Try working it out yourself.
NEVER ACCEPT an answer from the person who says it's 20 hours.
The SIMPLEST and the MOST STRAIGHTFORWARD way to solve such problems is THIS :
1. From the first part of the condition, the rate of work of one worker is widgets per hour.
2. At the second scenario, the rate of work of one worker is widgets per hour, where x is the unknown time under the question (in hours).
3. Write the proportion (equation) saying that the rate of work is the same in both cases
= .
4. Solve the proportion for x :
x = = 300 hours. (ANSWER)
Solved.
Doing in this way, you will never make a mistake.
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To see other similar problems solved in this way, see the lesson
- Rate of work problems
in this site.