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

| 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 |

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

| 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.

****

More examples to follow ...

Cheers!

Ravi

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