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17 Oct 2010, 02:11
Kudos
Bookmarks
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
Step2: Y rate is 1/10th of X. Now the tabular form looks like:
Since, Y rate is 1/120 then the Y time is: 120 as below:
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,
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.
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
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:
Step3:
From the above table, 660 sprockets will be produced by A in 110 hours.
==> 6 sprockets for 1 hour.
****
Cheers!
Ravi
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
| Time | Rate = 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:
| Time | Rate = 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:
| Time | Rate = 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,
| Time | Rate = 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.
| Time | Rate = work/Unit Time 1 day | |
| X | 6 | 1/6 |
| Y | 4 | 1/4 |
| X+Y | 5/12 |
****
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
| Time | Rate = 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:
| Time | Rate = 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.
****
Cheers!
Ravi
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08 Aug 2022, 19:45
Kudos
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Archived Topic
Hi there,
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Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions 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!
