MHB How do I find the fraction of the job done?

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To find the fraction of the job done by a worker washing windows in 6 days, the calculation starts by determining that the worker completes 1/9 of the job per day, given that the total job takes 9 days. Multiplying this daily rate by 6 days results in 6/9, which simplifies to 2/3 of the job completed. Ratios can simplify the understanding of this problem, establishing a proportion of windows washed to days worked. Visual aids like drawings can help some learners, but they are not necessary for solving the problem. Ultimately, in 6 days, the worker can wash 2/3 of the windows.
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If a worker can wash all the glass windows of a building in 9 days, what part of the job can
said workers finish in 6 days?1. ¼
2. 1/3
3. 2/3
4. 3/5
5. ¾
 
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Re: Need EXPLANATION on how to solve this one. Thanks!

Here is a hint: If it takes 9 days to wash ALL the windows, what fraction of the total will the worker wash in ONE day? (What number is 1 part of 9 making a whole)?

Multiply your answer to THAT question by six. You may have to "reduce" your final answer.
 
Re: Need EXPLANATION on how to solve this one. Thanks!

Is it 2/3? I tried to draw a figure to be able to visualize it (and it takes much time). I don't know other ways to solve it. :confused:
 
Re: Need EXPLANATION on how to solve this one. Thanks!

jaytheseer said:
Is it 2/3? I tried to draw a figure to be able to visualize it (and it takes much time). I don't know other ways to solve it. :confused:

No drawings are needed. But your answer is correct, in 6 days one person can wash 2/3 of the windows. The easier way is to use ratios.

Set up a ratio "Proportion of the windows washed : Number of days", in this case it will be 1:9. So if you wanted to know how much in one day, you simply have to divide both sides by 9, giving 1/9 : 1 (in other words, 1/9 of the windows in 1 day), then for six days multiply everything by 6, giving 6/9 : 6, or if you like, 2/3 : 6 (2/3 of all the windows will be done in 6 days).

Can you go from here?
 
Got it! Thanks guys! :)
 
Having said that no drawings were necessary, I'd still be very interested in seeing them, as I love seeing different ways to solve problems and will never discourage a student's creativity...
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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