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Tabular approach for solving Work problems

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Tabular approach for solving Work problems [#permalink] New post 17 Oct 2010, 01:11
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I found the below tabular approach useful and non-confusing, while solving the Work problems. Let me know what do you think and feel free to correct if i am wrong somewhere.

Problem 1:
Working, independently X takes 12 hours to finish a certain work. he finishes 2/3 of the work . The rest of the work is finished by Y whose rate is 1/10 th of X. In how much time does Y finsh his work?

Given:
X takes 12 hours to finish a certain work
X already finished 2/3 of the work
Y rate is 1/10th of X

Step1:
Since, X takes 12 hours to finish a certain work
TimeRate = work/Unit Time 1 HR
X 12 1/12
Y
X+Y


Step2: Y rate is 1/10th of X. Now the tabular form looks like:
TimeRate = work/Unit Time 1 HR
X 12 1/12
Y (1/10) * (1/12) = 1/120
X+Y


Since, Y rate is 1/120 then the Y time is: 120 as below:
TimeRate = work/Unit Time 1 HR
X 12 1/12
Y 120 1/120
X+Y


Step3: X already finished 2/3 of work. So the left out work is 1/3

From the above table, If Y finishes 1/120 of work in 1 hr then
1/3 of work can be finished by Y in:
1/120 -----> 1 hr
1/3 -----> ??
(1/3)/(1/120) * 1 hr = 40 Hrs

****
Problem2:
Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?

A. 4
B. 6
C. 8
D. 10
E. 12

Given:
X takes 2 days longer to produce W widgets than machine Y
Two machine produce 5/4W widgets in 3 days

Step1:
X takes 2 days longer to produce W widgets than machine Y.
Assume Y takes n days to produce W widgets. Then,
TimeRate = work/Unit Time 1 day
X n+2 1/(n+2)
Y n 1/n
X+Y ((1/(n+2)) + (1/n))


In 1 day, X & Y finish ((1/(n+2)) + (1/n)) widgets
In 3 days, X & Y finish 3 * ((1/(n+2)) + (1/n)) widgets

But given,

3 * ((1/(n+2)) + (1/n)) = 5/3 ---> n = 4, n= -6/5 --> The only valid solution is, n = 4
Now substitute n=4 in the above table.
TimeRate = work/Unit Time 1 day
X 6 1/6
Y 4 1/4
X+Y 5/12

i.e., X takes 6 days to produce W widgets --> X takes 12 days to produce 2W widgets
****
Problem3:

Machine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten hours longer to produce 660 sprockets than machine B. Machine B produces 10% more sprockets per hour than machine A. How many sprockets per hour does machine A produce?

Given:
Machine A and Machine B manufacture 660 sprockets.
A takes 10 hours longer than B
B produces 10% more per hour than A

Step1:
A takes 10 hours longer than B. Assume B takes N hours
TimeRate = work/Unit Time 1 HR
A N+10 1/(N+10)
B N 1/N
A+B


Step2:
B produces 10% more per hour than A
i.e., B rate = 1.1 Arate ==> 1/N = (1.1)*(1/(N+10)) ==> N = 100

Substitute N in the above table:
TimeRate = work/Unit Time 1 HR
A 110 1/110
B 100 1/100
A+B


Step3:

From the above table, 660 sprockets will be produced by A in 110 hours.
==> 6 sprockets for 1 hour.

****

More examples to follow ...

Cheers!
Ravi

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Intern
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Joined: 17 Jul 2010
Posts: 15
Location: United States (AL)
GMAT 1: 720 Q49 V39
Followers: 2

Kudos [?]: 7 [0], given: 4

Re: Tabular approach for solving Work problems [#permalink] New post 24 Oct 2010, 08:31
Thank you - I've always struggled with work problems but the chart definitely helps.
Re: Tabular approach for solving Work problems   [#permalink] 24 Oct 2010, 08:31
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