Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and,

IMO D

Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?

1/a+1/b-1/c will give the solution.

A. 6 hours 25 minutes
B. 6 hours 30 minutes
C. 18 hours 25 minutes
D. 18 hours 30 minutes
E. 19 hours 18 minutes

Sol:

(1/10)+(1/20)-(1/30)=(7/60)

now this represents a unit of three hours,every three hours this much will be the increase in the water,

from smart picking of numbers lets wee if this unit is repeated 6 times we get (7/60)*6=42/60 then 42/60 +1/4 =57/60

so the let is 1-57/60=3/60 or 1/20

then we know that A will start after this so we know

1/10---- 1 hours
then 1/20----x hours

x=(1/20)/1/10=1/2 hours or 30 min.

so we get to 18 hours 30 min. is the solution.

Say volume of tank =60
Rate of A=60/10=6
Rate of B=60/20=3
Rate of C=-60/30=-2
Now,its 1/4th filled =60/4=15l

In 3 hours, Rate=6+3-2=7l is filled in tank
In every 3 hours 7l is added to tank

In 18 hours= 7*6=42l is added to tank
Tank is filled with 15+42=57 l

Next, when Tap A is opened it'll take 30 min to fill 3L as its speed is 6l/hr

So, complete will be filled in 18 hr 30min
Hence D

Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?

A. 6 hours 25 minutes
B. 6 hours 30 minutes
C. 18 hours 25 minutes
D. 18 hours 30 minutes
E. 19 hours 18 minutes

Given: Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C).
Asked: After how long will the tank begin to overflow?

Tap A can fill a tank in 10 hours => Tap A will fill 1/10 tank in an hour
Tap B can fill it in 20 hours => Tap B will fill 1/20 tank in an hour
An outlet, Tap C, can empty the tank in 30 hours => Tap C will empty 1/30 tank in an hour

Taps start working in alphabetic order (A, B, and then C) => Within 3 hours they together fill \(\frac{1}{10}+\frac{1}{20}-\frac{1}{30} = \frac{7}{60}\) tank

If the tank is \(\frac{1}{4}\)th full
=> After 3x hours it will be \(\frac{1}{4} + \frac{7x}{60} = \frac{15+7x}{60}\) full
\(\frac{15+7x}{60}<1\)
=> 15+7x<60
=>7x<45
=>x=6
After 3x = 18 hours tank will be \(\frac{57}{60} full or \frac{3}{60} = \frac{1}{20}\) empty

Now tank A will be in operation and will fill \(\frac{1}{20}\)tank in \(\frac{1}{2}\) hours (30 minutes)

=> Tank will overflow in 18 hours 30 minutes.

IMO D

Let Ta be the time take for pump A to fill the tank, Tb be time taken for pump B to fill the tank and Tc be time taken for tank C to empty the tank.
Then Ta=10hrs, Tb=20hrs and Tc=30hrs.

From the question, we know that the tank is already 1/4 full. Hence the time taken to fill the remaining 3/4 of the tank would be Ta(3/4), Tb(3/4), and Tc(3/4).
Ta=15/2hrs; Tb=15hrs; and Tc=45/2hrs.

From the question we also know that the pumps are opened one after the other in alphabetical order to work for 1 hour, after which it is closed and the next opened. It means A would start at t=0hr and stop at t=1hr, while B takes over and work from t=1hr to t=2hrs. C then takes over at t=2hrs and stops at t=3hrs.

To simplify the calculation, let's assume that the three pumps work together in order to determine how many times each pump has worked at least before the tank gets full.

Effective Rate, R can be gotten as follows:
1/R=1/Ta + 1/Tb - 1/Tc
Tc is negative because it empties the tank rather than fill it.
=2/15+1/15-2/45 =7/45=1/R
R=45/7=6+3/7
This means each pump would have at least worked 6 times.

Now let's find out what portion of the tank is full after each pump has operated 6 times.
=6*(2/15+1/15-2/45)=12/15+6/15-12/45=14/15

Hence after each pump has operated 6 times, we only need the next pump to operate for 1/15 for the tank to be full. Obviously, the next pump to start the cycle is A.
Since Ra=2/15 per hour, 1/15 is half the hour rate, hence pump A only needs to work for half an hour for the tank to be full and start to overflow.

Total time taken to fill the tank is therefore 6*3+30minutes. Which 18hours 30minutes.

The answer is therefore D.

Posted from my mobile device

Let the volume of a pool be \(x\)

Tap \(A\) fills \(\frac{1}{10}\)th in every hour

Tap \(B\) fills \(\frac{1}{20}\)th in every hour

Tap \(C\) empties \(\frac{1}{30}\)th in every hour

If \(A, B,\) and \(C\) are opened one by one for an hour, then \(\frac{1}{10}+\frac{1}{20}-\frac{1}{30} = \frac{7}{60}\) of the pool will be filled in every 3 hours.

So how many of such 3 hour cycles do we need to fill \(\frac{3}{4}\)th of the pool? If \(\frac{3}{4}\) is divided by \(\frac{7}{60}\) we get \(6,...\) So \(6\) of such cycles will fill \(6*\frac{7}{60}=\frac{7}{10}\) of the pool. The remaining \(\frac{3}{4}-\frac{7}{10}=\frac{1}{20}\) of the pool again match the time when \(A\) begins to fill. If \(A\) fills \(\frac{1}{10}\)th in an hour, then \(A\) must fill \(\frac{1}{20}\) in \(30\) mins.

Thus we need 3 cycles or \(18\) hours and \(30\) minutes to fill the \(\frac{3}{4}\)th of the pool.

Hence D

Assume total work is 60 Units
A does 6 units in 1 hour, B 3 units, C -2 units
Since they are working one by one in 3 hours 7 units amount of work will be done.
After 6 such rounds 42 units amount of work will done in 18 hours
3 units will be done by A in 30 minutes.
Thus, the total duration for the work to get completed will be 18 hours 30 minutes.
Hence, answer is D.

Posted from my mobile device

Tap A can fill a tank in 10 hours, tap B can fill it in 20 hours, and, an outlet, Tap C, can empty the tank in 30 hours. Each tap is opened, one by one, for exactly one hour and then closed. If the tank is 1/4th full, and the taps start working in alphabetic order (A, B, and then C), after how long will the tank begin to overflow?


Let's assume that volume of the tank - 100L
1/4 full = 25L in the tank

Tap A in 10 hours fill 100L ---> speed 1h-10L
Tap B in 20 hours fill 100L ---> speed 1h-5L
Tap C in 30 hours empty 100L ---> speed 1h-3.33L

We that taps are working in alphabetic order A, B, and then C - we take in as 1 cycle:

1-cycle: 25L+10L+5L-3.33L=36.67L (in 3h)
2-cycle: 36.67L+10L+5L-3.33L=48.34L (in 6h)
3-cycle: 48.34L+10L+5L-3.33L=60.01L (in 9h)
4-cycle: 60.01L+10L+5L-3.33L=71.71L (in 12h)
5-cycle: 71.71L+10L+5L-3.33L=83.38L (in 15h)
6-cycle: 83.38L+10L+5L-3.33L=95.08L (in 18h)

So we in 6 cycles - 18h - we filled 95.08L out of 100L of the tank.
Next turn is Tap A, we need to find out in how many minutes Tap A will fill 5L.

10L-10h ---> 5L-30min

Finally in 18h 30min tank will be 100.08L begin to overflow.

A. 6 hours 25 minutes
B. 6 hours 30 minutes
C. 18 hours 25 minutes
D. 18 hours 30 minutes
E. 19 hours 18 minutes


D is the answer.

How can we use this method for reverse type of questions? (Where combined work is given and individual work is asked. Please provide link if discussed somewhere. Thanks for the method)

Every 3 hours' work = 1/10+1/20-1/30

= 7/60

Now, tatal time time to finish work = 24 hours 40 minuts

3/4 work to be completed in (24 hours 40 minutes /4*3)
= 18 hours 30 minutes.

Posted from my mobile device

Let the capacity of the tank be 60 units.
Thus in 1 hour,
Tap A does 6 units work
Tap B does 3 unit work
Tap C does 2 units work-- it is negative work
In 3 hours total work done is 7 units
Since tank is 1/4th filled so work to be done is 45 units
Thus time taken to fill tank is 45/7 = 6 complete cycle and 3 units of work
This remaining 3 units of work is done by A in 3/6 hour = 30 min
Total time after which tank starts overflowing is 6*3 + 30 min = 18hours 30 min

Considering the total work to fill the tank = W.
Tap A can fill the tank in 10 hours, the rate of work of A = \(\frac{W}{10}\)
The rate of work of tap B =\( \frac{W}{20}\)
The rate of the drain of tap C =\( \frac{-W}{30}\)
Considering the total work to be 60 units. ( The volume of the tank is 60 units )
In one hour A fills 6 units, B fills 3 units and C drains 2 units.
One-fourth of the tank is filled and hence of the 60 units, only 45 units need to be filled.
In a cycle of 3 hours, the work done is 6+3-2 = 7 units of the 45 units required.
In the first 5 cycles, a total of 35 units get filled with the required 45 units.
In the 16th hour, 6 more units = 41 units filled, in the 17th hour 44 units filled and in the 18th hour 2 units drain and by the end of 18 hours, 42 units are filled.
Hence after the 18th-hour tap, A needs to fill the remaining 3 units. Since A can fill 6units in one hour it will be filling 3 units in 30 minutes.

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