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Expert replies
by phoenix9801 » Fri Jun 01, 2012 8:33 am
Over a period of two days, Widget Co.’s assembly line will fill an order for a certain number of widgets. If the assembly line averages 8 widgets per hour over the course of these two days while producing just these widgets, then the assembly line’s average hourly output on the second day is what percent greater than on the first day if on the first day the assembly line averages 4 widgets per hour?
A) 200
B) 300
C) 400
D) 500
E) It cannot be determined from the information given



Two pipes are transferring water into a water tower. Pipe 1 alone would fill up the tower in 9 hours. Pipe 2 alone would fill up the tower in 6 hours. How many minutes would it take for both pipes to fill up 2/3 of the tower?

A) 2 3/5
B) 10
C) 276
D) 600
E) It cannot be determined from the information given
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Source: — Problem Solving |

by Jim@StratusPrep » Fri Jun 01, 2012 9:24 am
1) For this you can really just assume that they worked 1 hour each day. Thus, (4 + x)/2 = 8 or x =12, which is 300% greater.



2) Always look at these as output/time.

1 tower/9 hours + 1tower/6 hours = ?

2/18 + 3/18 = 5/18

That is 5 towers every 18 hours, or 18/5 of an hour for 1 tower. 2/3 * 18/5 = 12 / 5 or 2 2/5. I assume A is a typo.
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by Birottam Dutta » Fri Jun 01, 2012 9:35 am
Dear Jim,

Shouldn't the answer for the first question be 200% greater?

The assembly line averages 8 widgets per hour for 2 days. If on the first day it averages 4, then on the second it must average 12.

12 is 8 greater than 4 or 200% greater. Correct me if I am mistaken. If the question was "output of second day is how many times that of first?" then answer should be 300.

On the second question, yes it appears as if there is a small typo as answer is 2 2/5.
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by Anurag@Gurome » Fri Jun 01, 2012 10:48 pm
phoenix9801 wrote:Over a period of two days, Widget Co.’s assembly line will fill an order for a certain number of widgets. If the assembly line averages 8 widgets per hour over the course of these two days while producing just these widgets, then the assembly line’s average hourly output on the second day is what percent greater than on the first day if on the first day the assembly line averages 4 widgets per hour?
Average of two days = 8 widgets per hour
Average of 1st day = 4 widgets per hour

Say, average of 2nd is x widgets per hour.
Then, (4 + x)/2 = 8 ---> x = (2*8 - 4) = 12

Required percentage = [(12 - 4)/4]*100 = (8/4)*100 = 200

The correct answer is A.
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by Anurag@Gurome » Fri Jun 01, 2012 10:54 pm
phoenix9801 wrote:Two pipes are transferring water into a water tower. Pipe 1 alone would fill up the tower in 9 hours. Pipe 2 alone would fill up the tower in 6 hours. How many minutes would it take for both pipes to fill up 2/3 of the tower?
In one hour,
  • Pipe 1 alone will fill up 1/9 of the tower
    Pipe 2 alone will fill up 1/6 of the tower

    Together both pipes will fill up [1/9 + 1/6] = [(2 + 3)/18] = 5/18 of the tower
Hence, they will fill up 2/3 of the tower in (18/5)*(2/3) hours = 12/5 hours = 2 and 2/5 hours = (120 + 24) minutes = 144 minutes
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