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Re: Collection of work/rate problems? [#permalink]
30 Mar 2009, 11:11

Thanks for the url there are 30 probs lets solve 3 at a time ==================== 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?

2.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

3.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ? 1:600 2:800 3:1000 4:1200 5:1500 ===============

Re: Collection of work/rate problems? [#permalink]
30 Mar 2009, 11:37

1

This post received KUDOS

I think

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?

Q1->is it 40 hrs?

2.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

Q2 E(12)and

3.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ? A:600 B:800 C:1000 D:1200 E:1500

Re: Collection of work/rate problems? [#permalink]
30 Mar 2009, 11:51

=========Next 3 Qs are========= 4.25 men reap a field in 20 days . when should 15 men leave the work.if the whole field is to be reaped in 37-1/2 days after they leave the work?

A 5 days B 10 days

5.Read the question from the diagram above and (WE CAN AVOID THIS) C 7 days D 7-1/2 days

6.when a certain tree was first planted, it was 4 feet tall, and the height of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increase each year? 3/10 2/5 1/2 2/3 6/5 ============

Re: Collection of work/rate problems? [#permalink]
14 Jun 2009, 11:26

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?

Re: Collection of work/rate problems? [#permalink]
04 Aug 2009, 05:52

I got 60 hours????

MACHINE X

(RATE)(HOURS)= JOB 1/X*12 = 2/3 12/X= 2/3 36=2x x= 1/18 of machine X can finish 1/18 of the job each hour

(which can also easily be seen by the fact that it would take 18 hours for machine x to finish the job)

MACHINE Y

Y's rate is 1/10 of Machine X rate 1/10*1/18--> 1/180

(RATE)(HOURS)= JOB 1/180*X = 1/3 3X= 180 X=60

Machine Y will finish the job in 60 hours???

even if you think about it logically....machine X will take an additional 6 hours to finish the whole job and if Y s rate is 1/10th of that, would it not take 60 hours for it to finish the rest of the job?

Re: Collection of work/rate problems? [#permalink]
05 Aug 2009, 05:03

Q1 - Ans is 40 hrs. I have solved quite a few of them..will post the solutions in a bit

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Re: Collection of work/rate problems? [#permalink]
05 Aug 2009, 08:42

6.when a certain tree was first planted, it was 4 feet tall, and the height of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increase each year? 3/10 2/5 1/2 2/3 6/5

Ans - 2/3

solution

Let x be the constant rate of increment

so 6th yr the tree will be ------> 4 + 6x 4th yr the tree will be ------> 4 + 4x Now the tree was 1/5 taller than it was at the end of the 4th year. so we can denote it as

4 + 6x = 1/5* (4+4x) + (4+4x)

Solving for X ----> 2/3

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Re: Collection of work/rate problems? [#permalink]
05 Aug 2009, 08:43

7.If Jim earns x dollars per hour, it will take him 4 hours to earn exactly enough money to purchase a particular jacket. If Tom earns y dollars per hour, it will take him exactly 5 hours to earn enough money to purchase the same jacket. How much does the jacket cost? (1) Tom makes 20% less per hour than Jim does. (2) x + y = $43.75

Solution From the stem of the question we can get

4x = 5y

Ans B Statement II alone is suff As we have 2 variables and 2 equations to solve

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Re: Collection of work/rate problems? [#permalink]
05 Aug 2009, 08:50

11/18 machines each working at the same constant rate together can complete a 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?

Question 11 and 18 are the same

Solution B) 3

Total work done = 6x12 = 72 units Therefore Per machine/day = 1/72

If one machine/day can perform 1/72 of the work ; In 8 days work can be done by 72/8 = 9 machines Additional machines = 9 -6 (given in the question) = 3

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Re: Collection of work/rate problems? [#permalink]
05 Aug 2009, 08:51

13.Working together at their respective rates, machine A, B, and C can finish a certain work in 8/3 hours. How many hours will it take A to finish the work independently? (1) Working together, A and B can finish the work in 4 hours. (2) Working together, B and C can finish the work in 48/7 hours.

Ans B

Using statement II we can find the hrs needed by A

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Re: Collection of work/rate problems? [#permalink]
05 Aug 2009, 08:53

15. working together at their constant rates , A and B can fill an empty tank to capacity in1/2 hr.what is the constant rate of pump B? 1) A's constant rate is 25LTS / min 2) the tanks capacity is 1200 lts.

Ans A

Statement I - We can find out the rate/time needed by B Statement II does not give us any info

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Re: Collection of work/rate problems? [#permalink]
05 Aug 2009, 09:01

1

This post received KUDOS

14. John can complete a given task in 20 days. Jane will take only 12 days to complete the same task. John and Jane set out to complete the task by beginning to work together. However, Jane was indisposed 4 days before the work got over. In how many days did the work get over from the time John and Jane started to work on it together?

Ans 10

Solution

Let the total no of days worked = n So jane worked n-4 days

Work rate of John = 1/20 Work rate of John = 1/12

setting up the equation R x t = w janes+ johns work + Johns work = total work

(1/20+1/12) x (n-4) + (1/20 *4) = 1 Solve for N

Sorry, i dont know how to use the proper symbols in the post..so its kinda crude!

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Re: Collection of work/rate problems? [#permalink]
05 Aug 2009, 09:07

16.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?

1/3x 3x/x-3 x-3/3x x/x-3 x-3/x

Ans - C

Work done for 20 mins = 1/x ; so for 60 mins it is ---> 3/x or (3 times)

Given Lindsay + Joseph paint the room in -- 1hr

If Lindsay did 3/x of the total work then remaining work = 1-3/x or x-3/3

For 20 mins Joseph did (x-3/3) * 20/60 = x-3/3x

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Re: Collection of work/rate problems? [#permalink]
05 Aug 2009, 09:09

Will post others later...

Could anyone share the solutions for problems -- 2,3,4,8,9

Cheers!

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Re: Collection of work/rate problems? [#permalink]
05 Aug 2009, 23:06

28.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.

Ans C With Statement 1 - we cant find out the total work With Statement II - we can only express the speed of one machine in terms of the other

Both statements are needed for the complete picture

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Re: Collection of work/rate problems? [#permalink]
05 Aug 2009, 23:11

30.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?

A.7/17 B.14/31 C. 7/15 D.17/35 E.1/2

Solution

2/3 were basic so 1/3 will be deluxe Assuming it took 1 hr to complete Basic stereo ; deluxe will take 7/5 X 1 hr

Total time taken (basic radio + deluxe radio) = 2/3 + (1/3 x 7/5) = 17/15

Ratio of hrs needed for deluxe ratio to total hrs = (7/15)/(17/15) = 7/17

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Re: Collection of work/rate problems? [#permalink]
05 Aug 2009, 23:13

1

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25.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?

28 30 33 36 38

Ans 36 (approx)

Formula = Total distance traveled/total Gallons of fuel used for the same

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

gmatclubot

Re: Collection of work/rate problems?
[#permalink]
05 Aug 2009, 23:13