Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 10)
10.
A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10 hours while machine R can print them in 12 hours. All the machines are started at 9 A.M. while machine P is closed at 11 A.M. and the remaining two machines complete work. Approximately at what time will the work (to print one lakh books) be finished ?
Answer: Option
Explanation:
| (P + Q + R)'s 1 hour's work = |  |
1 | + | 1 | + | 1 |  |
= | 37 | . |
| 8 | 10 | 12 | 120 |
| Work done by P, Q and R in 2 hours = | |
37 | x 2 | |
= | 37 | . |
| 120 | 60 |
| Remaining work = | |
1 - | 37 | |
= | 23 | . |
| 60 | 60 |
| (Q + R)'s 1 hour's work = | |
1 | + | 1 | |
= | 11 | . |
| 10 | 12 | 60 |
| Now, | 11 | work is done by Q and R in 1 hour. |
| 60 |
| So, | 23 | work will be done by Q and R in | |
60 | x | 23 | |
= | 23 | hours  2 hours. |
| 60 | 11 | 60 | 11 |
So, the work will be finished approximately 2 hours after 11 A.M., i.e., around 1 P.M.
Discussion:
95 comments Page 4 of 10.
Hariteja said:
1 decade ago
Hey guys I used a logic mite be correct or not I don't know!
As we know p works for 8hrs, q works for 10hrs and are works for 12hrs so there is difference of 2hrs. Where p complete the works for only 2hrs in time 9am to 11am. In the same time the remaining q and are also works but we observe by comparing q works more than 2hrs so if the q started at the same time where p started then q finish the work at 11am but where as are who work for 4hrs more than p. So in the time of he started he also completes the work at 11am but remaining 2hrs he need to work so the work will be finished at 1pm i.e. (11am+2hrs).
As we know p works for 8hrs, q works for 10hrs and are works for 12hrs so there is difference of 2hrs. Where p complete the works for only 2hrs in time 9am to 11am. In the same time the remaining q and are also works but we observe by comparing q works more than 2hrs so if the q started at the same time where p started then q finish the work at 11am but where as are who work for 4hrs more than p. So in the time of he started he also completes the work at 11am but remaining 2hrs he need to work so the work will be finished at 1pm i.e. (11am+2hrs).
K.anusha said:
1 decade ago
Please I can't get the answer this last step please friends will explain to me.
How will come to this is 3/4 can you explain to me friends please.
How will come to this is 3/4 can you explain to me friends please.
Zoya said:
1 decade ago
Why we multiplying total time (x) by 1/8, 1/10, 1/12?
Why here we doing x/8+1/10+1/12?
Is there any formula like total time/work done?
Please anyone help me for this method.
Why here we doing x/8+1/10+1/12?
Is there any formula like total time/work done?
Please anyone help me for this method.
Rekha Muragod said:
1 decade ago
Here no need of lengthy calculations.
Just look at time taken by each machine P will takes 8 hrs and Q takes 10 hrs, are takes 12 hrs.
So the common time difference between all 3 machines is = 2 hrs.
So just add this time to P's close time.
11 am+2 = 1 pm.
That's it.
Just look at time taken by each machine P will takes 8 hrs and Q takes 10 hrs, are takes 12 hrs.
So the common time difference between all 3 machines is = 2 hrs.
So just add this time to P's close time.
11 am+2 = 1 pm.
That's it.
Vikash said:
1 decade ago
What about the 0.09 hr left in the answer which doesn't add to the answer?
Basant Kumar said:
1 decade ago
P takes = 8 hrs.
Q takes = 10 hrs.
R takes = 12 hrs.
We assume total work as the LCM of 8, 10, 12 = 120.
P completes 15 unit/hours.
Q = 12 unit/hrs.
R = 10 unit/hrs.
P+Q+R together = 37 unit/hours.
They continue to work for 2 hours i.e. 37x2 = 74 unit is completed.
Remaining work = 120-74 = 46 units.
Remaining work is done by Q and R,
So, P+Q = 22 unit/hours.
They will take 2 hours more to complete remaining work.
22x2 = 46 units.
Work will complete in total 4 hours.
So, at 1:00 PM.
Q takes = 10 hrs.
R takes = 12 hrs.
We assume total work as the LCM of 8, 10, 12 = 120.
P completes 15 unit/hours.
Q = 12 unit/hrs.
R = 10 unit/hrs.
P+Q+R together = 37 unit/hours.
They continue to work for 2 hours i.e. 37x2 = 74 unit is completed.
Remaining work = 120-74 = 46 units.
Remaining work is done by Q and R,
So, P+Q = 22 unit/hours.
They will take 2 hours more to complete remaining work.
22x2 = 46 units.
Work will complete in total 4 hours.
So, at 1:00 PM.
Ankit Jha said:
1 decade ago
I did it other way around.
2*(P+Q+R)'s 1 hour work + (Q+R)'s 1 hour work*X = 1.
37/60+(1/10+1/12)*x = 1.
11/60*x = 23/60.
x = 23/11 which is approximately equal to 2.
Hence Total time taken would be the initial two hours where P also contributed and the rest 23/11 i.e. 2 hours where only Q and are contributed.
Therefore, Total time will be 2+2= 4 Hours.
9:00 AM + 4 hours will be 1:00 PM.
Corrections are most welcome.
2*(P+Q+R)'s 1 hour work + (Q+R)'s 1 hour work*X = 1.
37/60+(1/10+1/12)*x = 1.
11/60*x = 23/60.
x = 23/11 which is approximately equal to 2.
Hence Total time taken would be the initial two hours where P also contributed and the rest 23/11 i.e. 2 hours where only Q and are contributed.
Therefore, Total time will be 2+2= 4 Hours.
9:00 AM + 4 hours will be 1:00 PM.
Corrections are most welcome.
Ashna said:
1 decade ago
P in 1 hr-----1/8.
Q in 1 hr-----1/10.
R in 1 hr-----1/12.
Therefore 2/8+(x+2)/10+(x+2)/12 = 1.
On solving x=23/11=2 hrs. (approx).
Therefore, 11 A.M.+2hr = 1 P.M.
Q in 1 hr-----1/10.
R in 1 hr-----1/12.
Therefore 2/8+(x+2)/10+(x+2)/12 = 1.
On solving x=23/11=2 hrs. (approx).
Therefore, 11 A.M.+2hr = 1 P.M.
Sparton said:
10 years ago
Nice methods explains above thank you friends.
Balu said:
10 years ago
Hello guys, easy solution:
X-2/8 + X/10 + X/12 = 1 (lik A+B+C's 1 Day Work).
By solving we get X = 4.
So, Machines are started at 9.
= 9+4 = 13 (i.e, 1'O clock).
X-2/8 + X/10 + X/12 = 1 (lik A+B+C's 1 Day Work).
By solving we get X = 4.
So, Machines are started at 9.
= 9+4 = 13 (i.e, 1'O clock).
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