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another ratio problem

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by rahul.s » Mon Feb 08, 2010 2:20 am
it takes the average dryer 80 minutes to dry one full load of bob's laundry. bob has one average dryer, and also decided to try one super fast dryer, which is faster than the average dryer by a 5:4 ratio. bob has 3 loads of laundry, and both machines are so precise that he can set them to the minute. how many minutes does it take for him to dry his 3 loads? (assume that no time passes between loads and that he can use both machines concurrently. assume also that he never runs a machine with less than a full load. finally, assume that bob does not partially dry a load in one machine and then switch it to the other)

OA: 128 minutes

source: mgmat

i got the part up to where the super fast dryer takes 64 minutes to dry one load. can't seem to solve any further.
Last edited by rahul.s on Fri Feb 12, 2010 8:56 am, edited 1 time in total.
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by shashank.ism » Mon Feb 08, 2010 2:39 am
rahul.s wrote:it takes the average dryer 80 minutes to dry one full load of bob's laundry. bob has one average dryer, and also decided to try one super fast dryer, which is faster than the average dryer by a 5:4 ratio. bob has 3 loads of laundry, and both machines are so precise that he can set them to the minute. how many minutes does it take for him to dry his 3 loads? (assume that no time passes between loads and that he can use both machines concurrently. assume also that he never runs a machine with less than a full load. finally, assume that bob does not partially dry a load in one machine and then switch it to the other)

OA: 128 minutes

i got the part up to where the super fast dryer takes 64 minutes to dry one load. can't seem to solve any further.
The average dryer takes 80 minutes to dry full load.
Since super fast dryer is 5:4 times faster than average dryer. so time taken by fast dryer = 4/5x80 = 64 minutes.

So part of load dried in 1min by av. dryer = 1/80
So part of load dried in 1min by fast dryer = 1/64

Let to complete 3 full load total time taken = x min
so total work done by av. dryer =x/80
total work done by fast dryer =x/64

so 3 = x/80+x/64 --> x=3.80.64/(80+64) = 106.667 min.Ans.
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by deepakdewani » Fri Feb 12, 2010 8:51 am
Let to complete 3 full load total time taken = x min
so total work done by av. dryer =x/80
total work done by fast dryer =x/64

so 3 = x/80+x/64 --> x=3.80.64/(80+64) = 106.667 min.Ans.
Doesn't the OA and solution offered by MGMAT differ from this? It is something to do with the assumptions given towards the end of this question. Even I couldn't comprehend the official solution, but the above is definitely incorrect because of the accompanying assumptions.
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by rahul.s » Fri Feb 12, 2010 8:55 am
deepakdewani wrote:Doesn't the OA and solution offered by MGMAT differ from this?
yeah, they do differ. the OE is quite obscure though.
Last edited by rahul.s on Fri Feb 12, 2010 9:04 am, edited 1 time in total.
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by lsommer28 » Fri Feb 12, 2010 9:02 am
What I am thinking with this question is:
Bob puts 2 loads in the dryers (1 in the 80min and 1 in the 64min). Since the 64 min dryer is going to finish first he will place the third load into it. Mean while the load in the 80min dryer is finished 16 min after the 3rd load is placed in the 64min dryer, SO he will have to wait for the last load in the 64 min dryer to finish. 64 + 64 + = 128. I think......
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by sars72 » Fri Feb 12, 2010 9:06 am
rahul.s wrote:it takes the average dryer 80 minutes to dry one full load of bob's laundry. bob has one average dryer, and also decided to try one super fast dryer, which is faster than the average dryer by a 5:4 ratio. bob has 3 loads of laundry, and both machines are so precise that he can set them to the minute. how many minutes does it take for him to dry his 3 loads? (assume that no time passes between loads and that he can use both machines concurrently. assume also that he never runs a machine with less than a full load. finally, assume that bob does not partially dry a load in one machine and then switch it to the other)
let a = average dryer
and b = super fast dryer

time taken by A = 80 -->

time taken by B --> 4/5 * 80 = 320/5 = 64 minutes

since we can use the machines concurrently, and have to use the full loads,

we start off with one load in A and one load in B. So, A will take 80 mins and B will take 64 mins to complete. 2 loads done. Since B finishes faster, we load the 3rd load into B as soon as it is done. Thus, the two loads for B will complete in 64+64 = 128 mins whereas A would have completed in 80 minutes. Since the third load will finish @ 128 minutes, the total time is 128 minutes.

hope this helps.
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