Aptitude - Time and Work - Discussion

Very easy approach.

Work to do (W) -> print 1 lakh books.
Machine P can do that work W in 8 hours.
Machine Q can do that work W in 10 hours.
Machine R can do that work W in 12 hours.


All machines started at 9 AM means all Machines P,Q and R are working together till 11 AM, meaning 2 hours together
So we write W/8 +W/10 +W/12, which together gives us per per-hour work (print) of each machine
->they did that together for two hours, so we multiply it by 2
2(W/8 + W/10 + W/12).

P is switched off at 11 AM means now remaining to machines Q and R will do the remaining work.
(W/10 + W/12)

But for how long these remaining machines are doing this remaining work, let's say x hours.
x(W/10 + W/12).

So total.

2(W/8 +W/10 +W/12) +x (W/10 +W/12) = W = total work.
solving for x , we get;
x ~= 2 hours.
Machines started from 9Am to 9Am + x time. So we get 1 AM as an Answer.

P - 1,00,000 = 8H.
Q - 1,00,000 = 10H.
R - 1,00,000 = 12H.

P, Q, R start at 9.00AM,
P = 11.00AM finished,
R - P = 2H.
Total 2 hrs difference,
11.00AM + 2 = 1:00 PM.
Then Ans : 1.00pm.

Well explained. Thanks all.

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.

P = 1/8 the amount of work in one day.
Q = 1/10 amount of work in one day.
R = 1/12 amount of work in one day.

All machines start at 9.00 AM.
Machine P stopped at 11.00 AM so it worked 2hrs,
11.00AM - 9.00AM = 2hrs.

P's work in 2hrs = 2(1/8) = 1/4.
Remaining work = 1 - (1/4) = 3/4 pending.

Q+R 's work in one day=(1/10)+(1/12)=11/60
to complete 3/4 of the work, Q and R need 'x' Days.
x(11/60) = 3/4
x = (3/4) * (60/11),
x = 45/11,
x = 4.99 (approx. 5hrs)

5 hours from 9AM is 1.00 PM

@All.

According to me, the Time taken for P, Q, R machines together to print everything will be 120/37 ~3.5 hours or about 12:30 pm.
So, if one machine stops working in between then obviously it will take more time than 12:30 and there was only one option satisfying this condition.

Anyone, please clarify this to me clearly.

Thanks for the explanation @Sai Vinitha.

Total unit of work is 240 units.p done per hour 30 units and q done 24 units and r done 20 units per hour.
Now three worked per 2 hr i.e 2(30×24×20)=148 units, remaining=240-148=92 units.
Now p left so only q and r can do in 1 hr (24+20)=44 units .
If they both worked for 2hr.
i.e 44×2=88 units then the total work will finish here 92approx 88,
So, take 2 hr 11 to 2 hr means 1 P.M.

P=80,
Q=10,
R=12.

L.C.M = 120
P=15,Q = 12,R = 10
For 2 hours = 2*15+12+10 = 27units
The remaining 66 units are (Q+R)
Q+R = 12+10 = 22
Q+R = 66/22 = 3hours
11 A.M + 3hours = 1 P.M.

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.