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TIME AND WORK

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by apoorva.srivastva » Mon Aug 10, 2009 5:41 am
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?


i DONT HAVE THE OA FOR THIS
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by ankit1383 » Mon Aug 10, 2009 6:00 am
40 hours

Let Total work: 120 units
Speed of X: 10 units(since his rate is 12 hours to complete the work)
Speed of Y : 1 units per hour(1/10 of X)

Work Finished by X= 120*2/3=80
Work Left= 120-80=40

Tim taken by Y
40/1= 40 hours

This is a dofferent method....

HTH
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by gmatv09 » Tue Aug 11, 2009 12:10 pm
Rate Time Work
X 1/12 12 1 (complete work)

Y 1/120 ? 1
(1/10th of the rate of x)


Since X has completed 2/3 work. Y has to complete 1/3 work

Rate Time Work
X 1/12 12 2/3

Y 1/120 ? 1/3

Total time taken by Y = 40 hrs
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by vittalgmat » Tue Aug 11, 2009 7:59 pm
X takes 12 hrs to finish his work...
At that rate he finished 2/3 of the work. So remainig work is 1/3.

Y's rate is 1/10th of X. ie. Y takes 10 times longer to finish his work.
==> 120 hrs

At the of 120 hrs how long will he take to finish 1/3 of the work ?

= 120/3 = 40 hrs.

HT helps
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by rlovely » Wed Aug 12, 2009 7:49 am
Could someone explain the fault in the following logic (and I'm sure I'm missing something extraordinarily simple-- be gentle, this is my first post):

X takes 12 hours to complete 2/3 of task Z
At that rate of work, X would take 18 hours to complete the entire task [12*3/2]
6 hours of work remain for Y

Y works at 1/10 the speed of X
10*6 = 60

60 hours of work remain

*Edit: Nevermind. Good lesson for me-- carefully read the problem.
Last edited by rlovely on Wed Aug 12, 2009 8:12 am, edited 1 time in total.
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by mohitsharda » Wed Aug 12, 2009 7:56 am
rlovely wrote:Could someone explain the fault in the following logic (and I'm sure I'm missing something extraordinarily simple-- be gentle, this is my first post):

X takes 12 hours to complete 2/3 of task Z
At that rate of work, X would take 18 hours to complete the entire task [12*3/2]
6 hours of work remain for Y

Y works at 1/10 the speed of X
10*6 = 60

60 hours of work remain

You interpreted the question wrongly. He finishes the whole work in 12 hours and not just 2/3rd.

He then walks away after working for 8 hours i.e. completing 2/3rd of the work.
MS
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by rbansal » Mon May 16, 2011 1:00 pm
vittalgmat wrote:X takes 12 hrs to finish his work...
At that rate he finished 2/3 of the work. So remainig work is 1/3.

Y's rate is 1/10th of X. ie. Y takes 10 times longer to finish his work.
==> 120 hrs

At the of 120 hrs how long will he take to finish 1/3 of the work ?

= 120/3 = 40 hrs.

HT helps

If we didn't pick 120, for example I picked 36 for total ours of work than this logic does not work out I only get 4 hours. Can you explain how I could get 40 hours if I didnt pick 120?

Thank you,

RB
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by rbansal » Mon May 16, 2011 1:06 pm
ankit1383 wrote:40 hours

Let Total work: 120 units
Speed of X: 10 units(since his rate is 12 hours to complete the work)
Speed of Y : 1 units per hour(1/10 of X)

Work Finished by X= 120*2/3=80
Work Left= 120-80=40

Tim taken by Y
40/1= 40 hours



Hello when I did the problem, I actually took 2/3 of the total for X and divided it by 12. To get the rate, because the question states X only got 2/3rd work done. ( picked 36 as my total) and as a result I got that it takes Y 60 hours to complete the job. Can you please explain to me why we would receive X's rate from the total and not from 2/3rd.

Thank you in advance I am hoping this explanation will help me alot.

RB

This is a dofferent method....

HTH
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by LalaB » Sun Dec 25, 2011 9:51 am
apoorva.srivastva wrote: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?


i DONT HAVE THE OA FOR THIS
since x takes 2/3 of the work, then Y takes 1/3 (=1-2/3) of the work.

so Y's time -

1/3*x=1/10*1/12

x=40
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