Empty that Tank

Here's an algebra-free solution. This works with any tank size, as long as pipe 1 drains twice as fast as pipe 2, so let's assume some convenient numbers.

Set the tank at 12 gallons.

Then Pipe 1 clears 12 gallons in an hour, or 2 gallons every 10 minutes (one-sixth of an hour).

Then Pipe 2 clears 6 gallons in an hour, or 1 gallon every 10 minutes (one-sixth of an hour).

So Pipe 1 + Pipe 2 clears $2 + 1 = 3$ gallons every 10 minutes.

Since $12/3 = 4 ,$ this means the 12 gallon pipe will be clear in $4 \times 10 = 40$ minutes.

Let $t$ be the time required to empty the tank if both pipes are open.

In one hour, pipe 1 can empty $\dfrac{1}{1}$ of the tank or the full tank. In one hour pipe 2 can empty $\dfrac{1}{2}$ of the tank. Hence, the fractional part of the tank which can be emptied in $t$ hours by pipe 1 is $\dfrac{t}{1}$ and by pipe 2 is $\dfrac{t}{2}$ . Since the whole tank is emptied in $t$ hours, the sum of the fractional parts that can be emptied by the two pipes is equal to $1$ .

$\dfrac{t}{1}+\dfrac{t}{2}=1$

$\dfrac{2t+t}{2}=1$

$t=\dfrac{2}{3}~\text{hours}$

or

$t=\dfrac{2}{3} \times 60 = \boxed{40~\text{minutes}}$ .

Unitary-method

Let $V$ be the volume of tank.

Pipe 1 : In 60 mins, it can empty $V$ volume. In 1 mins, it can empty $\frac{V}{60}$ .

Pipe 2: In 120mins, it can empty $V$ volume. In 1min, it can empty $\frac{V}{120}$ .

Now, $P_1 + P_1$ can empty $\frac{V}{60}+ \frac{V}{120} = \frac{3V}{120}$ in 1 min.

So, Together can empty $V$ volume in $\frac{120}{3V}\times V = 40$ mins.

Many people used the volume of the tank as the starting point. Good work. I took a different approach.

Since pipe 1 clears the tank twice as fast as pipe 2, I reframed the problem to be three pipe 2’s each that clear the tank in 2hrs. Therefore if you open pipe 1.1 and 1.2 together it will clear the tank in 1 hr just like pipe 1 in the original question.

Now they are all the same, if you open all three pipes it will simply be 2hrs (flow rate)/3 (number of pipes flowing) = 40mins

Since we have a multiple choice here, this can be reasoned without all the math.

If pipe 1 and pipe 2 were the same size then the tank would be emptied in half the time of pipe 1 alone (30 minutes). Since pipe 2 is smaller than pipe 1 it will take longer to empty the tank than 30 minutes, but less time than it would take for pipe 1 alone to empty the tank. The only answer that fits that criteria is 40 minutes.

Pie 1 :

Volume emptied by it in 60 minutes = 1

Volume emptied by it in 1 minute = $\frac{1}{60}$

Pipe 2 :

Volume emptied by it in 120 minutes = 1

Volume emptied by it in 1 minute = $\frac{1}{120}$

When combined the volume emptied by them in 1 minute = $\frac{1}{60} + \frac{1}{120} = \frac{3}{120} = \frac{1}{40}$

So, they both combined empty $\frac{1}{40}$ volumes of water in 1 minute . So, they both combined will take 40 minutes to empty the whole tank .

🤔🤔🤔🤔🤔🤔🤔🤔🤔

Pipe 1 drains at twice the rate of pipe 2, so will drain 2/3 of the tank when both are open.

It takes 60 minutes for pipe1 to drain the whole tank so will take 40 minutes to drain 2/3.

I said make the tank 120 litres so i could count on my fingers..;)

Pipe A in 60 mins = 2 litres per min

Pipe B in 120mins= 1 litres per min

A+B = 3 litres per min(lpm)

120 litres/3lpm = 40 mins

As pipe 1 drains twice as fast as pipe 2, we can substitute pipe 1 with 2 pipes of type 2.

Therefore, if we open all 3 pipes of type 2 we will empty the tank in 1/3 of the time of a singe pipe of type 2, which is:

$\frac{1 }{3}\ \times 120$ minutes = $40$ minutes

I know that the answer for this is 40, but I was wondering this. Do we know for sure that when they are both open, their speed of drainage will remain the same for both pipes and that they won't affect one another?

Pipe 2 drains water half as fast as pipe 1. Together they drain water 1.5 times as fast as pipe 1. 1 over 1.5 is 2 over 3 or two thirds. Two thirds of one hour is 40 minutes.

Let's give the tank of water a measurement so we can figure out the rate at which the pipes pump out water. Let's say the tank is 100ml, it doesn't matter what the measurement is, it just needs to be the same for both holes.

First hole: 150ml/60 mins Second hole 150ml/120 mins or 75ml/60mins

The second hole does 75ml in 1 hour, so if both holes were open then after one hour 225ml would be out. However, the tank contains 150ml.

First hole: 25ml/10 mins Second hole: 12.5/10mins

Now it's 37.5 ml per 10 mins. Now let's divide that to see how much 10mins we need.

150/37.5 = 4

This means it takes 40 minutes to empty the tank.

Let the volume of the tank be $V$ , and the flow rates of pipe 1 and pipe 2 be $r_1$ and $r_2$ respectively. Then $r_1 = \dfrac V1$ and $r_2 = \dfrac V2$ . The time it takes to drain the tank when both pipes are used $t = \dfrac V{r_1+r_2} = \dfrac V{V+\frac V2} = \dfrac 23 \text{ hour}$ or $\dfrac 23 \times 60 = \boxed{40 \text{ mins}}$ .

Disclaimer: This solution is confusing. But this is what worked for me, so here it is.

Assume that Pipe 2 leaks 1 minute per minute. Then Pipe 2 will empty the tank in 2 hours if the time tank had 120 minutes of time in it. (60 min per hour * 2 = 120)

• 120 min of time in the tank
• Pipe 2 leaks 1 min

Then for Pipe 1 to leak 120 min in 1 hour, Pipe 1 must be leaking 2 minutes. (120 min / 60 min = 2 min)

• Pipe 1 leaks 2 min

If Pipe 1 and Pipe 2 leaks at the same time, then:

• Leak Rate = 1m (Pipe 1) + 2m (Pipe 2)
• Leak Rate = 3m

And then...

• 120m / 3m = 40m
• Thus 40 minutes.

I took a different logical approach to solving this problem given the answer choices available. Since pipe 1 can empty the tank in 1 hour, right off the bat you can eliminate the bottom 3 answer choices, leaving you with just 30 min and 40 min. Then the logic comes into play. Start with the first answer choice, 30 min. At 30 min, pipe 2 can clear 1/4 of the tank (1/4 at 30 min each = 2 hours). At the same time pipe 1 can clear 1/2 the tank (1/2 at 30 min each = 1 hour). If pipe 1 and pipe 2 are opened up at the same time and pipe 1 can clear 1/2 of the tank and pipe 2 can clear 1/4 of the tank in 30 min, that means 3/4 of the tank will be cleared in 30 min, leaving 1/4 of the tank left to be cleared. That means you need a little bit more time than 30 mins to clear the tank. The only other answer choice left is the 40 min answer.

A simple solution would use the fact that with both open, it would drain 1.5x the speed of pipe 1. Since pipe 1 does it in 1 hour, we multiply 60 minutes by $\frac{1}{1.5}$ to get $40$ minutes.

$\frac{1}{x}$ = $\frac{1}{60}$ + $\frac{1}{120}$ x = $\frac{(60 * 120)}{120+60}$ = 40 minutes

If in one hour Pipe 1 empties 1 tank, in 30 minutes it'll empty 1/2 a tank.

Analogously, Pipe 2 will empty 1/4 a tank in half an hour.

So, if both are open, they'll empty 3/4 of the tank in thirty minutes.

As this is a math and not a fluid dynamics problem (if vice versa it wouldn't be basic), I'll assume the rate of drain is constant.

This causes that time in this problem is directly proportional to the amount of empty tank there is at that time. Hence:

$\frac{\frac{3}{4}}{30} = \frac{1}{t} => t = \frac{30}{\frac{3}{4}} = 40$

Where $t$ is the amount of time in minutes required to empty the tank. Hence, the answer is $\boxed{40}$

Pipe 1 empties 1 tank per hour. Pipe 2 empties 1/2 tanks per hour. Pipes 1 and 2 together empty 1.5 tanks per hour, which equates to 1 tank per 40 minutes.

Not gonna post a complete solution, but here is a quick method for those who are going for SAT where time is an issue: It’s basically product over sum, so (2x1)/(2+1)=2/3 hours or 40 minutes.

My solution is definitely less technical than everyone else's. Given the choices for the problem, one can eliminate 1 hour and 30 minutes and 3 hours because two pipes cannot take longer than one pipe.

One can infer that Pipe 2 takes twice as long as Pipe 1 so having both pipes open cannot bring the time down to 30 minutes. It would take two Pipe 1's to accomplish that. Therefore, the only possible solution that is sensible given the choices is 40 minutes.

When the problem is stated in TIME PER JOB, as this one is, then the quick formula is xy/(x+y), where x and y are the two times. It's a simplified form of the formula 1/timeA + 1/timeB = 1/time.together.

For three times, you can use xyz/(xy+xz+yz), which is a simplified version of the formula 1/timeA + 1/timeB + 1/timeC = 1/time.together.

In this case, the math is: 1x2/(1+2) = 2/3 hrs or 40 min.

Figure this out in minutes. Pipe 1 takes 60 minutes. Pipe 2 takes 120 minutes. Pipe 1 drains at 1/60th tank per minute. Tank 2 does 1/120th tank per minute. Together: 1/60 + 1/120=1/n 2 + 1/ 120=1/n..... 3n=120..... n=40 minutes for the tank to drain, using both pipes ☺☺☺☺

Here is yet another solution with a bit more physical formalism. Let $\dot{V}_1$ and $\dot{V}_2$ be the volumetric flow rates through the each pipe while the other is closed and $\dot{V}$ the flow rate when both are open. Similarly, let $\tau_1, \tau_2$ and $\tau$ be the periods of time required for the entire tank to empty and let $V$ be the total volume of the tank. Then:

$\dot{V}=\dot{V}_1+\dot{V}_2=V(\frac 1 \tau_1 + \frac 1 \tau_2)$

However, $\dot{V}$ can also be expressed as follows:

$\dot{V}=\frac V \tau\implies \frac V \tau =V(\frac 1 \tau_1 + \frac 1 \tau_2)$

Whence, $\tau=\frac{\tau_1\tau_2}{\tau_1+\tau_2}=\boxed{40\min}$

let. the amount of water in there: 1 - (cause it doesn't matter how many water there is to solve this question) - i) pipe 1 : it clears the water up in 1 hours, so the amount of water it cleans per 1 hours is 1 - ii) pipe 2 : it clears the water up in 2 hours, so the amount of water it cleans per 1 hours is $\frac{1}{2}$ - iii) if we add it to gather, then it will be $\frac{3}{2}$ - (cause one plus half) - iv) then we can solve it by $\frac{2}{3}$ hours or 40 minutes

Pipe 1 drains 1 tank per hour (1T/h).

Pipe 2 drains 1 tank per 2 hours or half a tank per hour (0.5T/h).

Together they drain 1 and a half tanks per hour (1.5T/h). It’s easier to change this too 3 tanks per 2 hours (3T/h) so we don’t have a decimal.

Then I divided by 3 to give me how long it would take for 1 tank to drain (2 hours = 120 minuets, 120/3 = 40 minuets.

All calculations together

1T/h + 0.5T/h = 1.5T/h x 2 = 3 tanks every 2 hours = 3 tanks every 120 minuets = 1 tank every 40 minuets

First, you change 1 hour into 60 minutes and 2 hours into 120 minutes. Next, you determine how much of the tank is emptied in 1 minute which is $\frac{1}{60}$ for the first pipe and $\frac{1}{120}$ for the second pipe. You add the two fractions together to see how much of the tank is emptied in 1 minute if both pipes are opened, which is $\frac{3}{120}$ or $\frac{1}{40}$ . If $\frac{1}{40}$ of the tank is emptied in one minute, then the entire tank will be emptied in 40 minutes.

40 minutes was the only choice offered between 30 minutes and 1 hour. Pipe one would empty tank in 1 hour. Two pipe ones would empty it in 30 minutes. Therefore 40 minutes was probably correct. Check to make sure: In 40 minutes pipe one would have emptied 40/60 or 2/3 of tank. Pipe two would have emptied 40/120 or 1/3 of tank. 1/3 + 2/3 = 1 whole tank.

Pipe 2 helps draining the tank when pipe 1 is open so it must drain faster than when only pipe 1 is open so the answer must be smaller than 1 hour. 30 min is 1/2 of 1 hour so it would require pipe 1 and 2 be the same so the answer must be 40 min.

You calculate the inverse ratio of pipe and time for each pipe alone. These inverse ratios must be the same since we are talking about the same tank. Therefore, P1=2P2. This means that the performance of both pipes is 3 times the performance of pipe 2 and 1/3 of 2 hours is 40 minutes.

There is a little problem in this question.

The rate of pipes depend on the pressure of water,so the answer isn't 40mins,certainly.

sorry for my poor English.

这题有个小问题。水管流速取决于水压，因此答案绝对不是40分钟。

为我糟糕的英语感到抱歉。

If pipes has a constant rate,and volume is 1,so the rate of pipe A is 1,B is 0.5,so the sum of rate is 1.5,time is 2/3, it means 40 mins.

如果水管是匀速的话，设容量为1，则甲水管速度为1，乙为0.5，合速度1.5，则时间2/3，即40分钟。

Let pipe 1 be P 1 and Pipe 2 be P 2 and Eff be the efficiency which is the time.

If Eff P 1 is 2\times Eff P 2 then it simply means we 2P 2 will make 1P 1.

Thus by using both pipes we can represent P 1 in terms of P 2 that is. P 1 = 2P 2.

Eff P 1 = 120mins therefore

Eff3 P 1 = 120/3 =40mins

• Pipe 1 drains 1 water tank per hour
• Pipe 2 drains $\frac{1}{2}$ a water tank per hour
• So both pipes would drain 1 $\frac{1}{2}$ water tanks per hour
• Doubling the time: Both pipes drain 3 water tanks per 2 hours
• Change from hours to minutes: both pipes drain 3 water tanks per 120 min
• Easy maths: So 1 water tank would drain in 120/3 = 40 minutes.

Let the volume be 1 gallon Volume emptied in 1 min when pipe 2 is closed= (1/60) gallons Volume emptied in 1 min when pipe 1 is closed= (1/120) gallons

Therefore, volume emptied in 1 min when both are open= (1/60)+(1/120)=(1/40) gallons

1 min:(1/40)=(x):1 => x=40 min

I kind of fudged this one based on the answers provided. If the faster pipe empties the entire tank in 1 hour, then together, they will empty the tank faster. So the final three options of 1 hour, 1.5 hours, and 3 hours are immediately out.

Then I thought about how much of the tank each of them could empty in half an hour. The faster pipe empties exactly half the tank, but the slower one doesn't finish emptying its half. So the 30 minute (1 half hour) option can't work either. The only choice left was 40 minutes.

While this didn't solve the problem directly, it would be sufficient to find the answer on an SAT exam or an exam for a course.

P1+P2=?

P1= 1h P2= 2h

1+1/2= 1h 2/2+1/2=1h 3/2=1h Entonces 2/3*1h =40 minutos

Here's an algebraic solution... (I'm really just adding this to practice my $\LaTeX$ .) I welcome any formatting comments: $\begin{aligned} v &= \text{volume} \\ t &= \text{time (in hours)} \\ p_{1}, p_{2} &= \text{drainage rates of pipes 1 \& 2} \end{aligned}$

If $\left. \begin{aligned} p_{1}t &=v\\ t&=1 \end{aligned}\right\} \Rightarrow p_{1}=v$

and $\left. \begin{aligned} p_{2}t &=v\\ t&=2 \end{aligned}\right\} \Rightarrow p_{2}=\frac{v}{2} \text{,}$

$\text{then what is } t \text{ when } (p_{1}+p_{2})t=v\text{?}$

$\begin{aligned} (p_{1}+p_{2})t&=v \\ (v+\frac{v}{2})t&=v \\ \frac{3}{2}vt&=v \\ \frac{3}{2}t&=1 \end{aligned}$

$t=\frac{2}{3} \textit{ or } 40 \text{ min}$

Let a represent Pipe 1 and b , Pipe 2.

For Pipe 1 to empty water -with Pipe 2 closed- it takes 60 minutes (1 hour). That is,
60 a = 1..... 1 ; where the result, 1, represents a whole (emptying the water tank).

For Pipe 2 to empty water -with Pipe 1 closed- it takes 120 minutes (2 hours). That is,

120 b = 1..... 2 ; where 1 here also represents a whole.

For both pipes to empty the water tank, at a time, t ,

t ( a + b ) = 1...... 3 ;

We'll now solve for t ,

From ... 1 , a = 1/60;

From ... 2 , b = 1/120.

Substituting these values for a and b in ... 3 ,

t ((1/60) + (1/120)) = 1

t ({2+1}/120) = 1

t (1/40) = 1

t = 40

Thus, the time, t, required by Pipe 1 and Pipe 2 to empty the tank is 40 minutes .

In 1 hour, Pipe 1 can empty 1 tank.

In 2 hours, Pipe 2 can empty 1 tank. So, in 1 hour, Pipe 2 can empty 1/2 tank.

In 1 hour, Pipes 1 and 2 together can empty (1 + 1/2) tank = 3/2 tank. So, In 2/3 hours, Pipes 1 and 2 together can empty 1 tank. 2/3 hours = 40 minutes.

This was my thinking:

Pipe 1 clears the tank in 1 hour, meaning in 30 minutes, the tank will be half-full, and in 15 minutes, $\frac{1}{4}$ of it will be full.

Pipe 2 clears the tank in 2 hours, meaning after an hour, the tank will be half full, after 30 min, $\frac{1}{4}$ full, and after 15 min, $\frac{1}{8}$ full.

So, if both pipes are open for 15 minutes, the tank will be drained $\frac{3}{8}$ of the way ( $\frac{1}{4} + \frac{1}{8} = \frac{3}{8}$ ). Divide that by $3$ and we get the amount that will be drained after 5 minutes: $\frac{1}{8}$ , given that both pipes are open. With that, we can now calculate the time it takes for both pipes to drain the tank by simply multiplying $8 * 5$ , indicating the eight $\frac{1}{8}$ s that is drained for every 5 minutes. Our final answer is 40 minutes.

I did it more through an estimation(math-free) that pointed me to the correct answer. Pipe 1 drains 50% in 30mins, Pipe 2 drains 50% in 1 hour.

So anything taking longer than 1 hour it's impossible as an answer since the tank drains in 1 hour with pipe 1 opened. Secondly, it would be impossible for the tank to drain completely in only 30 mins, since Pipe 1 drains 50% and Pipe 2 drains 25%.

So the draining would take somewhere between 30 mins and 1 hour. We have one answer that fits right there.

It seems complicated at first, but did this in less than 10 seconds. In real cases, if you want the exact values, you go math

In 2 hours, Pipe 1 can empty 2 tanks (1 each hour) and Pipe 2 can empty 1 tank (1 in 2 hours) Hence, combined, Pipe 1 and Pipe 2 can empty 3 tanks in 2 hours So they can empty 1 tank in $2/3$ hours = 40 mins

1 takes 1 hour to empty 2 tales 2 hours to empty

1 takes 2/3s of the "mathematical weight" (idk the terms)

What i did was:

1 (if there were two of them) would take 30 mins each 2 (if there were two of them) would take 60 mins each

Now the difference is 30 minutes and 60 minutes

Since 1 takes 2/3's the mathematical weight, what is 2/3's downwards (or 1/3 upwards from 30 - 60) from 60 to 30? ... 20 minutes, therefor 40 mins is the answer.

let volume of container be L.
in 2 hrs: hole (1) can empty 2L and hole(2) can empty L. So, total 3L can be emptied in 2 hrs . therefore L will be emptied in 2/3 hrs = 40min

Given the rates for each pipe, the answer must be in minutes, so we have two options available, 30 and 40 minutes. In 30 minutes the pipe 1 alone should drain at least half of the tank, and the pipe 2 wouldn't be able to drain the other half the tank in 30 minutes, so this answer is incorrect. The answer left is 40 minutes.

If pipe 2 takes 2 hours to drain, that is 120 minutes. If pipe 1 takes 1 hour to drain, that is 60 minutes. If we assume therefore that the volume of the cylinder is 120 litres, the flow rate from pipe 2 will be 60 ltr/ hr and pipe 1 will be 120 ltr / hr. By opening both pipes 1 and 2, the flow rate increases to 180 ltr / hr. 120 ltr divided by 180 ltr / hr equals 2/3 of an hour or 40 minutes.

Assume the tank contains 2 units of water. Pipe 1 then expels 2 units per hour. Pipe 2 then expels 1 unit per hour. Pipe 1 + Pipe 2 thus expel 3 units of water per hour.

$\frac{2 Units}{3 Units per Hour}$ = $\frac{2}{3}$ Hours = 40 minutes

Since pipe 1 works twice as fast as pipe 2, in the same amount of time pipe 1 will have drained twice the amount of water as pipe 2. In other words, pipe 1 will have drained 2/3 of the water, while pipe 2 while have drained 1/3. Therefore, the solution can be found by calculating either how long it will take pipe 1 to drain ⅔ of the water, or how long it will take pipe 2 to drain ⅓ of the water. Pipe 1: 1 container/ 60 min = ⅔ container/ 40 min. Pipe 2: 1 container/ 120 min = ⅓ container/ 40 min. So, it will take both pipes 40 min to drain the water.