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n2739178
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19 Sep 2010, 06:51
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Hi all,
Does anyone know why, in Chapter 3's (Ratios) Problem Set from the MGMAT Word Translations book, Question 14, in the alternative answer they provide if you were to calculate the quickest way to dry all 3 loads in the dryers, they came up with the time to finish the full job is 320 / 3 minutes ? I have no idea where they pulled the number 320 from. IF anyone could help that would be great.
Thanks
Does anyone know why, in Chapter 3's (Ratios) Problem Set from the MGMAT Word Translations book, Question 14, in the alternative answer they provide if you were to calculate the quickest way to dry all 3 loads in the dryers, they came up with the time to finish the full job is 320 / 3 minutes ? I have no idea where they pulled the number 320 from. IF anyone could help that would be great.
Thanks
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19 Sep 2010, 07:18
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n2739178
Post your Quant question in proper PS/DS section. You are welcome to post the exact question with the options.
20 Sep 2010, 01:24
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Please post the question so others will know what we are talking about 
In solving the problem, the standard dryer finishes a load, and then remains idle while the "Super" dryer finishes the final load. To minimize the time, you would want both dryers working the entire time. It gives you the ratio of \(5/4\) faster for the Super, so you want it to do \(5/9\) of the work, and the regular to do \(4/9\) of the work.
Standard Dryer: \(1/80\), which represents 1 load in 80 minutes.
Super Jubmo-Tron Dryer-Matic: Faster by a ratio of \(5/4\), so \(1/64\), which represents 1 load in 64 minutes.
Combined they can do \(4/320 + 5/320 = 9/320\). This is 9 loads in 320 minutes. We only have 3 loads to dry, so that would take 320/3 minutes using both simultaneously.
Looking at a chart of when they are used may make this clearer:
Start a load in both the Standard and Super one.
Standard Super Remaining After
Time 0 Load 1 4/9 * 80 = 35.5m -- 5/9
Load 2 -- 5/9 * 64 = 35.5m 4/9
Load 3 -- -- 9/9
Now, put Load 2 in the standard dryer and start load 3 in the Super dryer.
Time 35.5m Load 1 -- -- 5/9
Load 2 4/9 * 80 = 35.5m -- 0
Load 3 -- 5/9 * 64 = 35.5m 4/9
Load 2 is now completed. Load 1 has 5/9 remaining and Load 3 has 4/9 remaining. Put load 1 in the Super dryer, and put Load 3 in the standard dryer.
Time 71m Load 1 -- 5/9 * 64 = 35.5m 0
Load 2 -- -- 0
Load 3 4/9 * 80 = 35.5m -- 0
Approximately another 35.5 minutes have passed, so that is 71 + 35.5 = 106.5 minutes, which is approximately 320/3.
Sorry, I couldn't get this chart to lineup, but hopefully you get the idea!
In solving the problem, the standard dryer finishes a load, and then remains idle while the "Super" dryer finishes the final load. To minimize the time, you would want both dryers working the entire time. It gives you the ratio of \(5/4\) faster for the Super, so you want it to do \(5/9\) of the work, and the regular to do \(4/9\) of the work.
Standard Dryer: \(1/80\), which represents 1 load in 80 minutes.
Super Jubmo-Tron Dryer-Matic: Faster by a ratio of \(5/4\), so \(1/64\), which represents 1 load in 64 minutes.
Combined they can do \(4/320 + 5/320 = 9/320\). This is 9 loads in 320 minutes. We only have 3 loads to dry, so that would take 320/3 minutes using both simultaneously.
Looking at a chart of when they are used may make this clearer:
Start a load in both the Standard and Super one.
Standard Super Remaining After
Time 0 Load 1 4/9 * 80 = 35.5m -- 5/9
Load 2 -- 5/9 * 64 = 35.5m 4/9
Load 3 -- -- 9/9
Now, put Load 2 in the standard dryer and start load 3 in the Super dryer.
Time 35.5m Load 1 -- -- 5/9
Load 2 4/9 * 80 = 35.5m -- 0
Load 3 -- 5/9 * 64 = 35.5m 4/9
Load 2 is now completed. Load 1 has 5/9 remaining and Load 3 has 4/9 remaining. Put load 1 in the Super dryer, and put Load 3 in the standard dryer.
Time 71m Load 1 -- 5/9 * 64 = 35.5m 0
Load 2 -- -- 0
Load 3 4/9 * 80 = 35.5m -- 0
Approximately another 35.5 minutes have passed, so that is 71 + 35.5 = 106.5 minutes, which is approximately 320/3.
Sorry, I couldn't get this chart to lineup, but hopefully you get the idea!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
