Work Rate Questions - I collected from GMAT Club : Quantitative Questions

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Lindsay can paint 1/x of a certain room in 20 minutes. What fraction of the same room can Joseph paint in 20 minutes if the two of them can paint the room in an hour, working together at their respective rates?

Lindsay paint 1/x in 20 min so 3/x in 60 min

Together they paint 3/x + j = 1 in 60 min

J = 1 - 3/x
J = x/x - 3/x
J = x-3/x in 60 min so in 20 min he will only paint a third (20/60 = 1/3)
1/3 * (x-3)/x
answer:
(x-3)/3x

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Machines X and Y run at different constant rates, and machine X can complete a certain job in 9 hours. Machine X worked on the job alone for the first 3 hours and the two machines, working together, then completed the job in 4 more hours. How many hours would it have taken machine Y, working alone, to complete the entire job?

X work rate 1/9

X worked for 3 hours + x+y worked for 4 hours and completed 1 work
3*(1/9) + 4(1/9 + 1/y) = 1
(1/3)*(1/4)+ (1/9 + 1/y) = 1/4
1/9 + 1/y = 3/12 - 1/12 => 1/9 + 1/y = 1/6
1/y = 1/18

So it would take y 18 hours to complete the work

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Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days?

Rate for six machines 1/12 (they do 1/12 of the work in 1 day or 1 work in 12 days)
Rate for one machine 1/12 * 6 = 1/72 (so 6 machines working 1/72 is 6*1/72 or 1/12 which is correct)

To complete the work in 8 days we need a rate of 1/8, so how many machines do we need to get a rate of 1/8 ?
(1/72)*x = 1/8
x = 72/8
x = 9

If 6 can complete the work in 12 days 9 can complete it in 8 days, so three more.

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

A produces x sprockets per hour
B produces x * 1,1 (11/10) sprockets per hour

Let y denote hour
B produces rate * time = work => 1,1x sprockets * y yours = 660 sprockets
It takes A hours more to produces the same amount so
x * (y+10) = 660

Our equations
x * (y+10) = 660
1,1x * y = 660

Solving this will lead to x = 6 and y = 100

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A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?

Rate R 1/36
Rate S 1/18

x = number of machines

So how many machines do we need to complete the work in 2 hours or equivalent, have a rate of 1/2?

x/36 + x/18 = 1/2
3x/36 = 1/2
x/12 = 1/2
x = 6

Six machines of each

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One hour after Yolanda started walking from X to Y, a distance of 45 miiles, Bob started walking along the same road from Y to X. If Yolanda's walking rate was 3 miles /hour and Bob's was 4 miles / hour, how many miles had Bab walked when they met?

Yolanda = 3mph
Bob = 4 mph

Yolanda started one hour before Bob, so you could say that they started at the same time with only 42 miles left

When they meet the will have walked for the same hours x but the distance will differ so

Yolanda will walk (rate * time = distance)
3 mph * x hours = y miles

Bob
4 mph * x hours = (42 - y miles) since he is walking in the opposite direction

1) 3x = y
2) 4x = (42-y)

1) x = y/3
using that in 2)

2) 4y/3 = 42 - y
7y/3 = 42
y = 18 (which is the distance Yolanda have walked so for Bob it is 42-18 which is 24)

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Working alone at its own constant rate, a machine seals k cartons in 8 hours,
and working alone at its own constant rate, a second machine seals k cartons in
4 hours. If the two machines, each working at its own constant rate and for the
same period of time, together sealed a certain number of cartons, what percent
of the cartons were sealed by the machine working at the faster rate?

Kartons per hour would be the measurement
Lets denote the machines A and B (the faster one)

A produce at a rate of k/8 = A
B produce at a rate of k/4 = B

Rearranging:
8A = k
4B = k

8A = 4B
2A = B
Ratio B/A = 2/1 so together they produce (2+1=3) and B solely produces 2/3 of the boxes

This problem could be solved without any calculations at all. If I work twice as fast as someone else and together we produce 1 or 100, the equation will always look like x + 2x = 1 which is 33.33... + 66.66... = 1

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Micheal and Adam can do together a piece of work in 20 days. After they have worked together for 12 days Micheal stops and Adam completes the remaining work in 10 days. In how many days Micheal complete the work separately.

Together they compete the work in 20 days (so 1 work in 20 days or work at a rate of 1/20 together)

1) (1/A + 1/M) = 1/20

The second equation:
Working together for 12 days and then Adam works single for 10 days to complete the work so

2) 12(1/A + 1/M) + 10(1/A) = 1

Substituting the the left side in 1) into 2)

12(1/20) + 10/a = 1
12/20 +10/a = 20/20
10/a = 8/20
200 = 8a
a = 25 which is Adams rate when working alone, put into 1)

1/25 + 1/m = 1/20
4/100 + 1/m = 5/100
1/m = 1/100

So Michaels rate is 1/100 or equivalent, it takes him 100 days to complete the work alone

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Is identical with 23 except that they ask for Adams rate which is 25

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A certain car averages 25 miles per gallon of gasoline when driven in the city
and 40 miles per gallon when driven on the highway. According to these rates,
which of the following is closest to the number of miles per gallon that the car
averages when it is driven 10 miles in the city and then 50 miles on the
highway?

To calculate average miles per gallon we need total miles and total gallons

City 25 mpg or 1/25 gallon per mile
Highway 50 mpg or 1/40 gallon per mile

10 miles in city * 1/25 miles per gallon = 10/25 gallons = 2/5 = 8/20
50 miles on highway * 1/40 miles per gallon = 50/40 gallons = 5/4 = 25/20

Total no of miles 10+50 = 60
Total gallons = 33/20

(60/1)/(33/20) = 60*20/33 = 20*3*20/3*11 = 400/11 which is approximately 36 miles per gallon

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Working together, John and Jack can type 20 pages in one hour. They will be able to type 22 pages in one hour if Jack increases his typing speed by 25%. What is the ratio of Jack's normal typing speed to that of John?

1) Jack + John = 20
2) Jack * 1,25 + John = 22

1-2)

0,25 Jack = 2
Jack = 8
John = 12

Jack/John = 8/12 = 2/3

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Tom, working alone, can paint a room in 6 hours. Peter and John, working independently, can paint the same room in 3 hours and 2 hours, respectively. Tom starts painting the room and works on his own for one hour. He is then joined by Peter and they work together for an hour. Finally, John joins them and the three of them work together to finish the room, each one working at his respective rate. What fraction of the whole job was done by Peter?

Rates:
Tom = 1/6
Peter = 1/3 = 2/6
John = 1/2 = 3/6

1 hours Tom + 1 hour (Tom+Peter) + x hours (Tom+Peter+John) = 1

First we need to solve for X

1 hours (1/6) + 1 hour * (3/6) + x(6/6) = 1 (they will work for <1 hour all three of them)

4/6 + x = 1
x = 2/6

So they will work for 1/3 of an hour or 20 minutes all three of them

Peter will work for 1 hour (with Tom) and 1/3 of an hour with the rest of the guys

In one hour he'll produce 1/3 and in 1/3 of an hour (1/3 * 1/3) = 1/9 of the work

So in total Peter will produce 4/9 of the total work.

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Machines X and Y produced identical bottles at
different constant rates. Machine X, operating alone
for 4 hours, filled part of a production lot; then
machine Y, operating alone for 3 hours, filled the rest
of this lot. How many hours would it have taken
machine X operating alone to fill the entire
production lot?

(1) Machine X produced 30 bottles per minute.
(2) Machine X produced twice as many bottles in 4hours as machine Y produced in 3 hours.

B is sufficient here

rate * time = number of bottles
x working for 4 hours produces z bottles and y working for 3 hours produces only half of that

y * 3 = z bottles
x * 4 = 2z bottles (twice the amount)

3y = 4x/2
6y = 4x
x/y = 6/4 = 3/2

So when X produce 3 Y produce only 2 and together 5, we can calculate the rest if we want to.

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One smurf and one elf can build a treehouse together in two hours, but the smurf would need the help of two fairies in order to complete the same job in the same amount of time. If one elf and one fairy worked together, it would take them four hours to build the treehouse. Assuming that work rates for smurfs, elves, and fairies remain constant, how many hours would it take one smurf, one elf, and one fairy, working together, to build the treehouse?

1) 1/s + 1/e = 1/2
2) 1/s + 2/f = 1/2
3) 1/e + 1/f = 1/4

The question:
4) 1/s + 1/e + 1/f = x ?

1-2)
1/e = 2/f
put in 3)

3f = 1/4 => f=12 so the rate is 1/12

using that + (1/s + 1/e) = 1/12 in 1) in 4) results in

4) 1/2 + 1/12 = x => x = 7/12 so it will take them 12/7 hours to complete the work

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Company S produces two kinds of stereos: basic and deluxe. Of the stereos produced by
Company S last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many
hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of
hours it took to produce the deluxe stereos last month was what fraction of the total
number of hours it took to produce all the stereos?

Basic: 2/3 of the total
Deluxe 1/3 of the total

Basic 5/5 hours to produce
Deluxe 7/5 hours to produce

We only produce basic + deluxe so that is the total and we need deluxe/total to find the fraction

Hours spent on deluxe = 7/5 * 1/3 = 7/15
Hours spent on basic = 5/5 * 2/3 = 2/3 = 10/15
Total = 17/15

deluxe/total
(7/15)/(17/15) = (7*15)/(17*15) = 7/17