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.
Sundar said:
9 years ago
Hi, I did it in a way can anyone tell me its right or not.
p = 8, q = 10, r = 12.
Thus, the ratio is 4:5:6.
Work starts at 9.00am and p finish at 11am.
Thus, q can be 12pm and c can be 13pm then the average time will be 13pm, it's 1pm.
Did my assumption is correct? Please give me a suggestion.
p = 8, q = 10, r = 12.
Thus, the ratio is 4:5:6.
Work starts at 9.00am and p finish at 11am.
Thus, q can be 12pm and c can be 13pm then the average time will be 13pm, it's 1pm.
Did my assumption is correct? Please give me a suggestion.
Sini said:
1 decade ago
Written by @Sameer:
P 1hrs work 1/8.
Q 1hrs work 1/10.
R 1hrs work 1/12.
p worked for 2 hrs so 2*1/8.
Suppose Q and r worked for 'x' hrs.
then,
2*1/8+(1/10)*x+(1/12)x = 1.
x=4.09.
Ans 9+4.09 = 1pm.
Sameer, How do we get the step 2*1/8+(1/10)*x+(1/12)x = 1.
i.e how is the total equal to 1?
P 1hrs work 1/8.
Q 1hrs work 1/10.
R 1hrs work 1/12.
p worked for 2 hrs so 2*1/8.
Suppose Q and r worked for 'x' hrs.
then,
2*1/8+(1/10)*x+(1/12)x = 1.
x=4.09.
Ans 9+4.09 = 1pm.
Sameer, How do we get the step 2*1/8+(1/10)*x+(1/12)x = 1.
i.e how is the total equal to 1?
Parveen said:
1 decade ago
Printing Pages in one hour:
P---->12500
Q---->10000
R---->8333
Because p will be closed at 11 A.M so printed pages by P till 11 A.M (12500*2)= 25000
Remaining pages(100000-25000)= 75000
So time taken will be 75000/(10000+8333)= 4 hours(approx.)
9 A.M+ 4 hours= 1.00P.M
P---->12500
Q---->10000
R---->8333
Because p will be closed at 11 A.M so printed pages by P till 11 A.M (12500*2)= 25000
Remaining pages(100000-25000)= 75000
So time taken will be 75000/(10000+8333)= 4 hours(approx.)
9 A.M+ 4 hours= 1.00P.M
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.
Shiva said:
6 years ago
We can also do by this method,
Lcm of 8, 10 and 12 is 120.
Suppose total work is 120.
Then, 120/8 = 15.
120/10 = 12.
120/12 = 10.
Now above are respective capacities and we know that work = capacity x time
Hence we have,
15x2+12x+10x = 120.
22x = 90
x = 4 answer.
Lcm of 8, 10 and 12 is 120.
Suppose total work is 120.
Then, 120/8 = 15.
120/10 = 12.
120/12 = 10.
Now above are respective capacities and we know that work = capacity x time
Hence we have,
15x2+12x+10x = 120.
22x = 90
x = 4 answer.
Sai Vinitha said:
4 years ago
P - 8 d
Q - 10 d
R - 12 d -------now,LCM of p,q,r is 120.
So, P's one day hour work = 15.
Q's =12
R's. =10.
P works for 2 hours.
So, 2*15=30.
120-30=90.
Remaining work is done by Q and R = 90/22 = 4(approx).
9:00 AM + 4 hrs = 1:00 PM.
Q - 10 d
R - 12 d -------now,LCM of p,q,r is 120.
So, P's one day hour work = 15.
Q's =12
R's. =10.
P works for 2 hours.
So, 2*15=30.
120-30=90.
Remaining work is done by Q and R = 90/22 = 4(approx).
9:00 AM + 4 hrs = 1:00 PM.
(39)
Girija said:
1 decade ago
Hi @Ajay,
A, B, C have 2hrs time difference.
In the question they told p machine closed at 11:00 am, that means it worked for only 2 hrs, it doesn't mean that P, Q, are machines have 2hrs time difference.
Could you explain the short cut indeed?
A, B, C have 2hrs time difference.
In the question they told p machine closed at 11:00 am, that means it worked for only 2 hrs, it doesn't mean that P, Q, are machines have 2hrs time difference.
Could you explain the short cut indeed?
Preetithapa said:
8 months ago
p-8hr , Q -10hr R - 12 hr.
Total work (LCM) - 120 unit.
Efficiency of P, Q, R - 15, 12, 10 .
For 2hr - 2(15+12+10) = 74 unit.
The remaining work- 46 units.
The time taken by (Q+R) =2.09 hr ~2hr.
9:00am + 2hr = 11 am.
11am + 2hr = 1 pm.
Total work (LCM) - 120 unit.
Efficiency of P, Q, R - 15, 12, 10 .
For 2hr - 2(15+12+10) = 74 unit.
The remaining work- 46 units.
The time taken by (Q+R) =2.09 hr ~2hr.
9:00am + 2hr = 11 am.
11am + 2hr = 1 pm.
(18)
Rose said:
2 years ago
Total work=120 units(lim of 8;10;12).
Efficiency of p-15.
“ of Q-12.
“Of R-10.
For 2 hours(9-11)all three worked = 2 *27 = 54units.
The remaining 66 units are covered by Q+R=66/22 = 3hrs.
11 am+3 hrs~ = 1 pm.
Efficiency of p-15.
“ of Q-12.
“Of R-10.
For 2 hours(9-11)all three worked = 2 *27 = 54units.
The remaining 66 units are covered by Q+R=66/22 = 3hrs.
11 am+3 hrs~ = 1 pm.
(18)
Kumar said:
1 decade ago
@Sowjanya Hey friends. Can any one suggest why 60/11 is multiplied with 23/60?
Ans:See 23/60 is the Remaining Work and 60/11 is the Time taken to do Remaining work.
By multiplying both we will get total time=23/11.
Hope you Got it.
Ans:See 23/60 is the Remaining Work and 60/11 is the Time taken to do Remaining work.
By multiplying both we will get total time=23/11.
Hope you Got it.
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