GRE Time and Work

Problem on GRE Time and Work may appear in any of the quantitative reasoning format and these problems on GRE time and work may cause anxiety for test takers, but by going through our chapter on GRE Time and Work with a slightly clever ability to understand you can attempt each and every problem accurately.

The concept of time & work is another important topic for the quantitative aptitude exam.

We have to understand the following basic concepts of this chapter:

If A does a work in a days, his rate = \(\frac { 1 }{ a }\) work/day

If B does a work in b days, his rate = \(\frac { 1 }{ b }\) work/day

If A & B work together, then their combined rate = \(\frac { 1 }{ a } +\frac { 1 }{ b }\)  work/day

= \((\frac { a+b }{ ab } )\) Work/day

The equation that applies to Time & Work problems is

Work Rate × Time = Work Done

 

Problem 1) A can build a wall in 10 days & B van build it in 5 days, if they both start working at the same time, in how many days will the wall be built?

Sol.  A can build a wall in 10 days, his rate = \({ (\frac { 1 }{ 10 } ) }^{ th }\) wall/days

B can build a wall in 5 days, his rate = \({ (\frac { 1 }{ 5 } ) }^{ th }\) wall/days

If A & B work together, than their combined rate = \( { (\frac { 1 }{ 10 } +\frac { 1 }{ 5 } ) }^{ th }\) wall/day

= \({ (\frac { 3 }{ 10 } ) }^{ th }\) wall/day

Work Rate × Time = Work Done

\({ (\frac { 3 }{ 10 } ) }^{ th }\) Wall/day × Time = 1 wall

Time = \((\frac { 10 }{ 3 } )\) days

When numbers of workers are working, equation becomes

(Individual work rate × Number of workers × Time = Work done)

 

Problem 2) To service a single device in 12 seconds, 700 Nano robots are required with all Nano robots working at the same constant rate. How many hours would it take for a single Nano robot to service 12 devices?

Sol.   700 Nano robots can service 1 device in 12 seconds.

Here ‘work’ is 1 devic.

Rata at which Nano robot is working =

\((\frac { 1 }{ 700\times 12 } )\) Nano robots/sec

Using this equation:

Individual work rate × Number of workers × Time = Work

\((\frac { 1 }{ 700\times 12 } )\times 1\times Time\quad =\quad 12\\ Time\quad =\quad (700\times 12\times 12)\quad seconds\\ \quad \quad \quad \quad =\quad (\frac { 700\times 12\times 12 }{ 60\times 60 } )\quad hours\\ \quad \quad \quad \quad =\quad (\frac { 700 }{ 25 } )\quad =\quad 28\quad hours\)

 

Problem 3) Twelve workers pack boxes at a constant rate of 60 boxes in 9 minutes. How many minutes would it take 27 workers to pack 180 boxes, if all workers pack boxes at the same constant rate?

Sol.   Twelve workers boxes at a constant rate of 60 boxes in 9 minutes.

1 worker can pack box at rate of \((\frac { 60 }{ 12 } )\) boxes in 9 minute.

Rate at which workers work = \((\frac { 60 }{ 12\times 9 } )\)  boxes/minute

Using this equation:

Individual work rate × Number of workers × Time = Work

\((\frac { 60 }{ 12\times 9 } )\times 27\times time\quad =\quad 180\\ time\quad =\quad (\frac { 180\times 12\times 9 }{ 60\times 27 } )\\ =\quad 12\quad minutes\)

 

.Problem 4) Machine A, which produces 15 golf clubs per hour, fills a production lot in 6 hours. Machine B fills the same production lot in 1.5 hours. How many golf clubs does machine B produce per hour?

Sol.   First, calculate the size of a production lot machine A works at a rate of 15 golf clubs per hour & fills a production lot in 6 hours.

Using, Rate × Time = Work

15 golf clubs per hour × 6 hours = 90 golf clubs

Machine A produces 90 golf clubs in 6 hours & fills a production lot.

Therefore, 90 golf clubs are required to fill a production lot.

Machine B produces 90 golf clubs in 1.5 hours.

Using equation:

Rate × Time = Work

Rate × 1.5 hours = 90 golf clubs

Rate = \(\frac { 90 }{ 1.5 } =(\frac { 90 }{ 15 } \times 10)=60\quad\) golf clubs/hour

 

Problem 5) A standard machine fills paint cans at a constant rate of 1 gallon every 4 minutes. A deluxe machine fills gallons of point at twice the rate of a standard machine. How many hours will it take a standard machine & a deluxe machine, working together, to fill 135 gallons of paint?

a) 1

b) 1.5

c) 2

d) 2.5

e) 3

Sol.   Rate at which standard machine fill paint cans = 1 gallon/4 minutes

=  \(\frac { 1 }{ 4 } \quad\) gallon/minute

Rate at which deluxe machine fills = \(\frac { 1 }{ 2 } \quad\) gallon/minute

Combined rate = \((\frac { 1 }{ 2 } +\frac { 1 }{ 4 } )\quad\)  gallon/minute = \(\frac { 1 }{ 2 } \quad\)  gallon/minute

Using equation,

Rate × Time = Work

\((\frac { 3 }{ 4 } )\quad \times \quad Time\quad =\quad 135\\ Time\quad =\quad (\frac { 135\times 4 }{ 3 } )minute\\ =\quad (\frac { 135\times 4 }{ 3\times 60 } )\quad hour\quad =\quad (3)\quad hour\)

 

Problem 6) With 4 identical servers working at a constant rate, a new internet search provider processes 9,600 search requests per hour. If the search provider adds 2 more identical servers & server work rate never varies, the search provider can process 216,000 search requests in how many hours?

Sol.   With 4 identical servers working at a constant rate, a new internet search provider processes 9,600 search requests per hour.

1 sever work at a rate of \(\frac { 9600 }{ 4 }\) search request per hour.

Using equation:

Individual rate × Number of servers × Time = Work

2400 × 6 × Time = 216,000

Time =

\((\frac { 216,000 }{ 2400\times 6 } )\quad =\quad 15\quad hours\)

 

Problem 7) A can do a piece of work in 10 days & B can do in 12 days. Both simultaneously start working.

Quantity A:         Time taken to complete the work, if B leaves 2 days before the actual completion of work.

Quantity B:         Time taken to completes the work, if B leaves 2 days before the scheduled completion of the work.

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.   A can do a piece of work in 10 days & B can do in 12 days.

If B leaves 2 days before the actual completion of the work: In this case, the actual completion of the work is after 2 days of B’s leaving. This means that A has worked alone for the last 2 days to complete the work.

Rate at which A works =  \((\frac { 1 }{ 10 } )\)/day

Rate at which B works =  \((\frac { 1 }{ 12 } )\)/day

Work done by A in last two days = \({ (\frac { 2 }{ 10 } ) }^{ th }\) of work

A & B work together at a rate of

\((\frac { 1 }{ 10 } +\frac { 1 }{ 12 } )/day\\ =\quad (\frac { 11 }{ 60 } )/day\)

Work done by combined by A & B

\((1-\frac { 2 }{ 10 } )\\ =\quad { (\frac { 8 }{ 10 } ) }^{ th }\) of work

Rate × Time = Work

\((\frac { 11 }{ 60 } )\times time\quad =\quad (\frac { 8 }{ 10 } )\\ time\quad =\quad (\frac { 8 }{ 10 } \times \frac { 60 }{ 11 } )\\ =\quad (\frac { 48 }{ 11 } )days\quad =\quad 4.3636\quad days\)

Total time taken to complete the work = Quantity A = 6.3636 days

Now let’ s calculate time taken to complete the work if B leaves 2 days before the scheduled completion of work.

A & B work together at a rate of \(\quad (\frac { 11 }{ 60 } )\)/days

Rate × time = work

\((\frac { 11 }{ 60 } )\times time\quad =\quad 1\\ time\quad =\quad (\frac { 60 }{ 11 } )\quad days\quad =\quad 5.45\quad days\)

A & B work together for \((\frac { 60 }{ 11 } -2)\quad days\quad =\quad (\frac { }{ } )\quad days\)

Let’s use equation:

Rate × time = work

\(\frac { 11 }{ 60 } \times \frac { 38 }{ 11 } ={ (\frac { 38 }{ 60 } ) }^{ th }of\quad work\)

Work done by A alone =

\((1-\frac { 38 }{ 60 } )\quad =\quad { (\frac { 22 }{ 60 } ) }^{ th }of\quad work\)

A work at a rate of \({ (\frac { 1 }{ 10 } ) }^{ th }of\quad work/day\\ (\frac { 1 }{ 10 } )\times time\quad =\quad \frac { 22 }{ 60 } \\ time\quad =\quad (\frac { 22 }{ 6 } )\quad days\\\)

Total time taken to complete the project =

\( =\quad (\frac { 22 }{ 6 } +\frac { 38 }{ 11 } )\\ =\quad 7.12\quad days\\\)

Therefore, Quantity B is greater.

 

Problem 8) Phil collects virtual gold in an online computer game & then sells the virtual gold for real dollars. After playing 10 hours a day for 6 days, he collected 540,000 gold pieces. If he immediately sold this virtual gold at a rate of $1 per 1000 gold pieces, what were his average earnings per hour in real dollars?

Sol.   After playing 60 hour, he collected 540,000 gold pieces.

Rate × time = work

Rate × 60 = 540,000

Rate = \((\frac { 540,000 }{ 60 } )\\\) gold pieces /hour = 9000 gold piece /hour

He sold his virtual gold at a rate of $1 per 1000 gold pieces.

1 gold pieces = \((\frac { 1 }{ 1000 } )\\\) dollars

His average earning per hour = \((\frac { 9000 }{ 1000 } )\\\) dollars/hour = 9 dollars/ hour.

 

 

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