Pipe A can fill a certain tank in 6 hours when working alone. Another

Question

Pipe A can fill a certain tank in 6 hours when working alone. Another pipe B can empty the same tank in 4 hours when working alone. If pipe A is opened at 9 am and pipe B is opened 'x' hours after pipe A, the tank empties at 4.30 pm on the same day. What is the value of 'x'?

A. 1
B. 1.5
C. 2
D. 2.5
E. 3

A. 1
B. 1.5
C. 2
D. 2.5
E. 3

Nikhil · Accepted Answer

Pipe A can fill a certain tank in 6 hours when working alone

Rate of Pipe A is  \frac{1}{6}

Pipe B can empty the same tank in 4 hours

Rate of Pipe B is  -\frac{1}{4}  (Since pipe B is emptying the tank)

Combined rate of A and B =  \frac{1}{6} - \frac{1}{4} = -\frac{1}{12}

=> The tank empties at the rate of  \frac{1}{12}  when both Pipe A and pipe B works

Pipe A is opened at 9 am and pipe B is opened 'x' hours after pipe A.

The tank empties at 4.30 pm(16:30 in 24 hr time format = 16 \frac{1}{2} ) on the same day

Total time =  16\frac{1}{2} - 9 = \frac{15}{2}

Pipe A is opened at 9 am and pipe B is opened 'x' hours after pipe A, the tank empties at 4.30 pm

=> whatever the amount of water filled in by Pipe A in 'x' hours is emptied in  \frac{15}{2} - x  hours when pipe A and pipe B work together

=>  x * \frac{1}{6} = (\frac{15}{2} - x) * \frac{1}{12}

=>  4x = 15 - 2x

=>  x = 2.5

Hence  option D

generis · Answer

How much of a tank  A  will fill by 4:30?
How long will it take  B  to empty that amount?

A  works from 9:00 a.m.
Treat  B  as if it worked "backwards" from 4:30 p.m. That's  B 's start time.

Pipe A's rate of fill =  \frac{1pool}{6hrs} 
 (r*t)=W  finished

At a rate of  (\frac{1pool}{6hrs}*6hrs)=1  tank is filled by Pipe A at 3 p.m.

Pipe A works for 1.5 more hours until 4:30. 
Pipe A fills another  (\frac{1}{6}*\frac{3}{2})=\frac{3}{12}=\frac{1}{4}  tank

Total filled by Pipe A:  1+\frac{1}{4}=\frac{5}{4}  of a tank

How long will it take Pipe B to empty  \frac{5}{4}  of a tank? Pipe B's empty rate: \frac{1pool}{4hrs} 
 \frac{W}{r}=t

\frac{(\frac{5}{4})}{(\frac{1}{4})}=(\frac{5}{4}*4)=5  hours for Pipe B to empty

4:30 p.m. = 16:30 (hours)
B's start time: (16:30 - 5 hours) = 11:30 a.m.

Pipe A started at 9:00. Pipe B starts at 11:30. 
Pipe B therefore starts 
 x=2.5  hours later

Answer D

PKN · Answer

Rate*time=Work

Work done by pipe A in filling the tank=1/6 * x

Work done by Pipe A & B in  emptying  the tank in (7.5-x) hrs=Work done by pipe A in filling the tank
Or, -(resultant rate of filling & emptying)*(7.5-x)= \frac{1}{6}*x 
Or,  -(\frac{1}{6}-\frac{1}{4})*(7.5-x) = \frac{1}{6}*x 
Or,  \frac{1}{12}*(7.5-x) = \frac{1}{6}*x 
Or, 3x=7.5
Or, x=2.5 hrs

Ans. (D)

BullRunner · Answer

Assume the tank is of 36l.A can fill 6l per hr and B can empty 9l per hour.
By 4:30,A would have filled 45l if working alone.
This means B emptied 45l in total by 4:30pm.
It would take B 5hrs to empty 45l,therefore it must have started 5hrs before 4.30,i.e 11:30am,that is 2.5hrs from 9am.
Ans =D

yoannnesme · Answer

When work is not started at the same time, it can be confusing to work with rate formulas and figure out the exact equations for someone that is not used to this on a daily basis (99% of GMAT applicants?). So it is useful to pick numbers and/or try the answers.

If A can fill the tank in 6 hours and B can empty it in 4 hours, let's choose a capacity of 12 liters for the tank (least common multiple) to make the math as easy as possible.

This means that A pumps 2 liters / hour to fill it in 6 hours, and B takes out 3 liters / hour to empty it in 4 hours.

It is now clear that, working together, 1 liter / hour is removed from the tank.

Now let's take a look at the answers. Strategy says that we should start with answer B or D, because we have a higher chance of getting it right in the first shot (see note below). But we can also pick the easiest number to work with and see what happens, so let's take 2 (we could take 1 but I'll go to the answer in the middle. Starting by trying A or E is the worst option).

After 2 hours, the tank has 4 liters. A and B together would need 4 more hours to empty it. That is a total of 6 hours, but we want a total of 7h30 (9 am to 4:30 pm).

Let's try answer D, 2.5 hours. After this time, the tank has 5 liters. Together A and B will need 5 more hours to empty it. This makes it a total of 7.5 hours.

Answer D.

Note:
1) Let's say you choose to try B. If it's not the correct solution but you know it must be a lower value, the answer will be A. 
2) Let's say you choose to try D. If it's not the correct solution but you know it must be a higher value, the answer will be E. 
This is better than choosing C: if you know it's a lower value you still have to try A or B. If you know it's a higher value, you still have to try D or E.

Neha2404 · Answer

IMO -D

Pipe Efficiency. Time. Total work
A. 2. 6. 12
B. -3. -4. 12
Here total work is LCM of 6 &4
Efficiency =work/time
Here tank is empty So total work is equal to =0

Total work = work by (pipeA+pipeB)
0=2*7.5+(-3*B)
B=5 hour
Pipe B worked for 5 hour it means that it is started after 
2.5 hour (7.5-5)

Posted from my mobile device

rounakkedia172 · Answer

Let us assume the capacity of the tank = 12 units ( LCM of 6 and 4) pipe A's rate = 2 units/hour Pipe B's rate = -3 units/hour Tank emptied after 7.5 hours since Pipe A's opening. Pipe A's work = 7.5 hours*2units/hour = 15 units Pipe B's work will also be the same as the final output = 0 (empty tank) So, Time taken by Pipe B = 15units/ 3 units per hour = 5 hour So Pipe B is opened after 2.5 hours, i.e, the value of x. I hope it helped.

omavsp · Answer

A: 1/6 
B: -(1/4)

A worked for 7.5 hours, then B worked for X hours until the tank emptied:

7.5*(1/6) - X*(1/4) = 0

7.5/6 = X/4

X = 5, B worked for 5 hours.

7.5 - 5 =  2.5 .

ayushidhiman · Answer

Pipe A opens at  9:00 am .
Pipe B opens  x hours later .
Tank is empty at  4:30 pm .
So total elapsed time = 7.5 hours.

From  9:00 am to (9:00+x) : only Pipe A works.
 
 Contribution = x/6  
From  (9:00+x)  to 4:30 pm = (7.5−x) hours: both A and B work.
 
 Net rate = 1/6 - 1/4 = - 1/12 tank per hour 
 Contribution = (7.5−x)(−1/12)  
At 4:30 pm, tank is  empty , so net effect = 0:

x/6 - (7.5−x)(1/12)=0
2x−(7.5−x)=0 
2x−7.5+x=0 
3x=7.5
x=2.5

Pipe A can fill a certain tank in 6 hours when working alone. Another

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