GMAT Club Olympics 2024 (Day 8): Pipe A, working alone at its constant

Question

Pipe A, working alone at its constant rate, starts filling an empty pool and then stops. Subsequently, Pipe B, working alone at its constant rate, completes the task and fills the remaining part of the pool. If the entire filling process took 7 hours, and Pipe A pumped in three times as much water as Pipe B, how many hours will it take Pipe B, working alone at its constant rate, to fill the entire pool on its own?

(1) After 4 hours, half of the pool was filled.

(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.

(1) After 4 hours, half of the pool was filled.

(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Bunuel · Accepted Answer

GMAT Club Official Explanation:

Pipe A, working alone at its constant rate, starts filling an empty pool and then stops. Subsequently, Pipe B, working alone at its constant rate, completes the task and fills the remaining part of the pool. If the entire filling process took 7 hours, and Pipe A pumped in three times as much water as Pipe B, how many hours will it take Pipe B, working alone at its constant rate, to fill the entire pool on its own?

(1) After 4 hours, half of the pool was filled.

(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.

Pipe A pumping in three times as much water as Pipe B implies that Pipe A pumped in 75% of the pool's capacity, while Pipe B pumped in the remaining 25% of the pool's capacity.

Assuming Pipe A takes x hours to fill the entire pool and Pipe B takes y hours to fill the entire pool, Pipe A would require 3x/4 hours to fill 75% of the pool, and Pipe B would require y/4 hours to fill the remaining 25% of the pool. Hence, we are given that 3x/4 + y/4 = 7, which simplifies to 3x + y = 28 and we need to find the value of y.

(1) After 4 hours, half of the pool was filled.

Given that initially Pipe A filled 75% of the pool, it must have also filled initial 50% of the pool by itself. Hence, this statement implies that Pipe A filled half of the pool in 4 hours. Thus, Pipe A would need 8 hours to fill the entire pool, making x equal to 8. Substituting x = 8 into 3x + y = 28 gives y = 4. Sufficient.

(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.

This implies that the combined rate of Pipe A and Pipe B is three times that of Pipe A alone. Thus, 1/x + 1/y = 3*1/x. Simplifying this gives x = 2y. Substituting x = 2y into 3x + y = 28 gives y = 4. Sufficient.

Answer: D.

Kinshook · Answer

Given: Pipe A, working alone at its constant rate, starts filling an empty pool and then stops. Subsequently, Pipe B, working alone at its constant rate, completes the task and fills the remaining part of the pool.

Asked: If the entire filling process took 7 hours, and Pipe A pumped in three times as much water as Pipe B, how many hours will it take Pipe B, working alone at its constant rate, to fill the entire pool on its own?

Let us assume that it takes Pipe A and Pipe B working alone to fill the empty pool in a & b hours respectively

Let us also assume that Pipe A worked for x hours working alone.

x/a + (7-x)/b = 1
(x/a) : (7-x)/b = 3 : 1 
x/a  = 3/4; x = 3a/4
(7-x)/b = 1/4; (7-3a/4) = b/4; b = 28-3a

b = 28 - 3a = ?

(1) After 4 hours, half of the pool was filled.
Since Pipe A filled 3/4 of the pool, entire half of the pool was filled by Piple A working alone.
4/a = 1/2; a = 8;
b = 28 - 3a = 28 - 24 = 4 hours
It will take Pipe B, working alone at its constant rate, to fill the entire pool on its own in 4 hours
SUFFICIENT

(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.
(1/a + 1/b) = 3/a 
2/a = 1/b
a = 2b
b = 28 - 3a = 28 - 6b
7b = 28; b = 4 hours
It will take Pipe B, working alone at its constant rate, to fill the entire pool on its own in 4 hours
SUFFICIENT

IMO D

divi22 · Answer

Let pool volume be P.
Let the volumetric flow rate of A be Va.
Let the volumetric flow rate of B be Vb.
Let the time taken by A until it stops = ta.
Let the time taken by B after A stops until tank fill = tb.

According to the question, 
1. Va*ta = 3*Vb*tb

=> Va*ta = 3P/4 & Vb*tb = P/4 -----(3)

2. ta + tb = 7

From statement 1,
half of the pool is filled in 4 hrs => only A is working for 4 hrs to fill half of the pool because, according to question, A fills 3P/4.
=> Va*4 = P/2
=> Va = P/8
Using equation 3, ta = 6hrs => tb = 1hr => Vb = P/4
=> the time that only B will take to fill entire pool = P/Vb = P/(P/4) = 4hrs

So, statement 1 alone is sufficient to answer the question.

From Statement 2,
=> P/(Va+Vb) = (1/3)*(P/Va)
=> 2Va = Vb

From equation 1,
=> Va*ta =3*2*Va*tb
=>ta = 6tb

From equation2,
=> 6tb + tb = 7
=> tb = 1hr
=> Vb = P/4 (from equation 3)

=> the time that only B will take to fill entire pool = P/Vb = P/(P/4) = 4hrs

So, statement 2 alone is sufficient to answer the question.

The answer is D.

Fido10 · Answer

(1) After 4 hours, half of the pool was filled.

Since pipe A Pipe A pumped in three times as much water as Pipe B then Pipe A pumped 3V/4 and Pipe B pumped V/4, where V is the whole volume of the tank.
So : V/2 --> 4(h)
 3V/4-->x=3V/4 * 4 *2/V=  6(h) 
So A filled its part of (3V/4) in 6 hours, and B filled its part of (V/4) in 1 hour,  we can then conclude that B can fill the whole thank in 4h hours , Sufficient

(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.

let's condider TA = time requierd for A to fill its part of 3V/4 and TB = time required for B to fill its part of V/4
then :  TA+TB=7hours  (*)
so, 3V/4 -->TA
 V-->  4TA/3

A and B combined can do the job in a time that is one-third that of A which is 1/3 * 4TA/3 = 4TA/9

Both we have : (3/4)/TA + (1/4)/TB=9/4TA
=>3/TA+1/TB=9/TA
=>TA=6TB
 using (*) we conclude that TB=1h, then the rate of B to fill the whol tank is 1*4 =4 hours , Sufficient,

Correct answer is D

AthulSasi · Answer

Let  TA  be the time taken by pipe A to fill the pool alone. 
Let  Ta  be the time taken by pipe A in the given scenario. 
Let  TB  be the time taken by pipe B to fill the pool alone. 
Let  Tb  be the time taken by pipe b in the given scenario.

Given 1
Pipe A pumped 3 times as much as pipe B. 
Since A and B filled the pool completly, pipe A filled  \frac{3}{4}  of the pool and pipe B filled  \frac{1}{4}  of the pool.

i.e
Ta =  \frac{3}{4}* TA

Tb =  \frac{1}{4}* TB

Given 2
Ta = Tb = 7 hours.

Statement 1 
(1) After 4 hours, half of the pool was filled.
Since we know A filled 3/4 of the pool and B started only after A stopped, we can deduce that only A was working in the first 4 hours when the pool was filled upto half level. 
.
From here we get.
TA = 4*2 = 8 hours. 
i.e Pipe A working alone will take 8 hours to fill the pool.
Ta =  \frac{3}{4}* TA 
= 6 hours.

Tb = 7 - 6 = 1 hour.
 TB = 4* Tb =  4 * 1 = 4 hours.  
Hence sufficient.

Statement 2 
(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.

\frac{1}{TA}+\frac{1}{TA}=3 * \frac{1}{TA} 
From this
TA = 2 TB.

We know Ta + Tb = 7
 2 TB * \frac{3}{4}+ \frac{1}{4}* TB = 7 
6 TB + TB = 28
 TB = 4 hours.
Hence sufficient.

correct asnwer D

Abhijeet24 · Answer

Let x hours pipe A was opened and y hours pipe b was opened. and also suppose a and b be number of hours required by pipe A and pipe B to fill the pool working alone. 
x + y = 7.....................................(1)
and x/a + y/b = 1 ........................(2)
Since, Pipe A pumped in three times as much water as Pipe B.
x/a = 3/4 and y/b = 1/4 ...............(3)
3/4 th pool is filled with pipe A and 1/4 by pipe B.

Statement 1:  4/a = 1/2
a = 8 hrs
and now from (3), x = 6 and using (1),  y = 1.
b = 4 hrs.
Statement 1: Sufficient

Statement 2: 
1/a + 1/b = 3/a
1/b = 2/a..................................(4)
Putting this value in (2), x/a + 2y/a = 1
x + 2y = a
14 - x = a    ( from (1) )
14 = a + x............................(5)
putting this value in (3)
x/(14 - x) = 3/4
4x = 42 - 3x
7x = 42
x = 6 and then now putting this value in suitable equations to get the below value.
a= 8, y = 1 and then b = 4 
Statement 2: Sufficient

So, Option (D) is correct. EACH statement ALONE is sufficient to answer the question asked.

Purnank · Answer

Given -  Pipe A, working alone at its constant rate,  starts filling an empty pool and then stops.  Subsequently, Pipe B, working alone at its constant rate,  completes the task and fills the remaining part of the pool.  If the entire filling process took 7 hours, and Pipe A pumped in three times as much water as Pipe B ,
To find - how many hours will it take Pipe B, working alone at its constant rate, to fill the entire pool on its own?

Total time = 7Hrs
Pipe A pumped in  three times  as much water as Pipe B
Means Pipe A contributed 75% of filling of pool and Pipe B remaining 25%.

1st - After 4 hours, half of the pool was filled.
As we know pipe A started first and it contributed 75% means when tank was half filled it was due to work of pipe A.
Hence from here we can calculate time required to fill 75% as 4Hrs did 50% then 6Hrs did 75%
and we know total time was 7Hrs.
Hence for remaining 25% pipe B took = 1Hr
From comparison, If pipe B contributed for full 100% of the pool working alone then it would have taken 4Hrs to fill the entire pool on its own.
  Sufficient.

2nd - Working together  at their respective constant rates , Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.
Lets assume rate of pipe A as a units/hr and for B as b units/hr. And time required for A to fill pool completely by working alone as Ta
Therefore, lets form equation based on given statement,
 \frac{Ta}{3} * (a+b) = a*Ta 
on further simplification,  b=2a 
Plus we know from given data,
Pipe A pumped in three times as much water as Pipe B,
Means, lets assume pipe A was on for t hrs. Then,
 a*t = 3*b*(7-t)  Put b=2a in it
 a*t = 6*a*(7-t)  cancel out a
 t = 6*(7-t) 
 t = 6*7 - 6t 
 t=6  Means Pipe B worked for  = 7-t = 7-6 = 1Hr  and contributed 25% of pool.
From comparison, If pipe B contributed for full 100% of the pool working alone then it would have taken 4Hrs to fill the entire pool on its own.
  Sufficient.

Answer is D.

BGbogoss · Answer

Pipe A, working alone at its constant rate, starts filling an empty pool and then stops. Subsequently, Pipe B, working alone at its constant rate, completes the task and fills the remaining part of the pool. If the entire filling process took 7 hours, and Pipe A pumped in three times as much water as Pipe B, how many hours will it take Pipe B, working alone at its constant rate, to fill the entire pool on its own?

(1) After 4 hours, half of the pool was filled.

(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.

(1) After 4 hours, half of the pool was filled.
This means that A takes 4*2*3/4=6 hours to fill the 3/4 of the pool (Pipe A pumped in three times as much water as Pipe B)
Thus Pipe B takes 7-6=1hour to fill the 1/4 of the pool. Meaning it would take 4 hours to fill the pool alone.
Statement 1 alone is sufficient.

(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.
This statement also means that when pipe A fills the third of the pool, Pipe B fills the remaining two thirds of the Pool.

Suppose the volume of the pool is V and the rates of pipes A and B are rA and rB and tB the time it takes Pipe B, working alone at its constant rate, to fill the entire pool on its own

We have rB=2*rA

And 7hours = (3/4)*V/rA + (1/4)*V/rB
= (6/4)*V/rB + (1/4)*V/rB
= (7/4)*V/rB

=> 4Hours = tB

Statement 2 alone is sufficient.

Answer D

xhym · Answer

We know that the entire process took 7 hours so:  t_a+t_b=7

Also, if V is the volume of the pool, then Pipe A filled  \frac{3V}{4}  and pipe B filled  \frac{V}{4} .

What we're looking for is how long it will take pipe B to fill the entire pool on its own, which is  \frac{1}{B} .

Let's look at each statement with this in mind:

Statement 1

We know that A filled 3/4 of the pool on it's own which means that since the pool was only half filled in 4 hours, A still had 1/4 of the pool to fill by then. That means that we can find the rate for A:

4A = \frac{1}{2}  ->  A = \frac{1}{8}  (so 1/A=8 meaning that A would fill the entire pool in 8 hours).

That means that to fill the other 1/4 of the pool, it would take A 2 more hours. That leaves 1 hour for B to fill the remaining 1/4 of the pool (7 total - 6 for A = 1 for B), so we can calculate the rate for B.

1*B = \frac{1}{4}  ->  B = \frac{1}{4}  (So 1/B = 4 meaning that B would fill the entire pool in 4 hours).

-> Statement 1 is sufficient.

Statement 2

The total amount done by pipe A is  A*t_a , and for pipe B it is  B*t_B . Since pipe A pumped three times as much water as pipe B, we know that:

A*t_a = 3*B*t_b  but we know that  t_a+t_b=7  so  A*t_a = 3*B*(7-t_a)

Furthermore, from the statement, we know that:

A+B=3A

B=2A

Now, we can plug that in:

A*t_a = 3*B*(7-t_a)

A*t_a = 3*2A*(7-t_a)

t_a = 42-6*t_a

7*t_a = 42

t_a = 6  so  t_b = 1

B worked for 1 hour and filled 1/4 of the pool so:

1*B = \frac{1}{4}  ->  B = \frac{1}{4}  (So 1/B = 4 meaning that B would fill the entire pool in 4 hours).

-> Statement 2 is sufficient.

Therefore, the answer is  D. EACH statement ALONE is sufficient to answer the question asked.

Milkyway_28 · Answer

Let us sort through the information provided in the question.

h1 = how many hours A pumped
h2 = how many hours B pumped
3x = how much water A pumped
x = how much water B pumped
4x = Total water in pool
h1 + h2 = 7

"Pipe A pumped in three times as much water as Pipe B"

A*h1 = 3x
B*h2 = x
h1 + h2 = 7

Now onto the statements

(1) "After 4 hours, half of the pool was filled."

We know that A pumped 3/4 of the pool before B started. If it took pump A 4 hours to fill half the pool then it follows that it would take A 6 hours to fill 3/4 of the pool. Meaning, it took pump B 1 hour to fill 1/4 of the pool. Therefore it would take Pump B 4 hours to fill the pool.

Statement 1 is sufficient.

(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.

For this equation let us have h3 represent the time it takes Pipe A to fill the pool by itself. With this, we can create the following equations.

A*h3 = 4x

\frac{{(A+B)*h3}}{3} = 4x

\frac{{(A+B)*h3}}{3} = A*h3

We can cancel the h3's and get

3A = A + B or 2A = B

Now we can plug this into our original equations for A and B

A*h1 = 3x
2A*h2 = x

Now we can substitute to get

A*h1 = 6A*h2 or h1 = 6*h2

Let us also not forget that h1 + h2 = 7

Therefore h1 = 6 and h2 = 1

This means that B fills x in 1 hour. The pool is 4x and therefore it takes B 4 hours to fill the entire pool.

Statement 2 is also sufficient, as is Statement 1

Answer is D

Suboopc · Answer

Volume filled by pipe A = 3V
Volume filled by pipe B = V
 \frac{3V}{R_A} + \frac{V}{R_B} = 7

Statement 1:
4 hrs half of the pool was filled.
Meaning, pipe A rate,  R_A = \frac{2V}{4} = \frac{V}{2}

\frac{3V}{R_A} + \frac{V}{R_B} = 7

R_B = V

Time for pipe B to fill on its own = 4V/V = 4 hrs
Sufficient
 
Statement 2:
 \frac{4V}{R_A + R_B} = \frac{4V}{R_A} * \frac{1}{3}

R_B = 2*R_A

\frac{V}{R_B} + \frac{3V}{{\frac{R_B}{2}}} = 7

R_B = V

Sufficient

D

smile2 · Answer

Let's say the overall volume is 1, then A has filled 3/4 and B has filled 1/4. 
Let x be the time taken by A to fill 3/4 volume. 
Let  R_a  and  R_b  be the rates of A and B respectively.

R_a = \frac{\frac{3}{4}}{x}= \frac{3}{4x}

R_b = \frac{\frac{1}{4}}{(7-x)}= \frac{1}{(28-4x) } 
To compute no. of hours it will take Pipe B, working alone at its constant rate, to fill the entire pool on its own, we need to find out value of x.
So this question is now modified to can you calculate the value of x?

From statement 1, we can compute the value of x
 \frac{1}{2/4}= \frac{3}{4x}

x = 6 hours
So time taken by B alone is 4 hours

From statement 2, we know that 
 \frac{3}{4x }+ \frac{1}{{28-4x}} =\frac{9}{4x}

\frac{1}{28-4x}= \frac{6}{4x} = \frac{3}{2x}

2x = 28*3 - 12x

14x = 28*3

x = 6 
So time taken by B alone is 4 hours

Therfore, each statement alone is sufficient. Hence the right ans choice is D.

DG1989 · Answer

Pipe A, working alone at its constant rate, starts filling an empty pool and then stops. Subsequently, Pipe B, working alone at its constant rate, completes the task and fills the remaining part of the pool. If the entire filling process took 7 hours, and Pipe A pumped in three times as much water as Pipe B, how many hours will it take Pipe B, working alone at its constant rate, to fill the entire pool on its own?

(1) After 4 hours, half of the pool was filled.

(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.

(1) After 4 hours, half of the pool was filled.
(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.

Solution:  Let's assume
R1 as the rate of Pipe A
R2 as the rate of Pipe B
T1​ be the time Pipe A works
T2 be the time Pipe B works

Given that,
T1 + T2 = 7  -------- (1)
R1 * T1 = 3 * R2 * T2 -------- (2)

We need to find  \frac{1}{X}​  when B​ is working alone.

Statement 1:   After 4 hours, half of the pool was filled
Since, A pumps 3 times the volume as compared to B, A fills  \frac{3}{4}  of the pool.
In this case, A worked for 4 hours, filled half of the pool, then worked for some time to fill  \frac{1}{4}  of the pool during the next 3 hours, and stopped.
Pipe B worked for the remaining time and filled  \frac{1}{4}  of the pool.

During 4 hours,
R1 * T1 =  \frac{1}{2}  of the pool

R1 * 4 =  \frac{1}{2}  of the pool

R1 =  \frac{1}{8}  of the pool

Thus Pipe A takes 8 hours to fill the entire pool.
Hence, it will take 6 hours to fill  \frac{3}{4}  of the pool.
As T1 + T2 = 7
T2 = 7 - 6
T2 = 1
Time for which B worked and filled  \frac{1}{4}  of the pool = 1 hour
Thus the rate of Pipe B when working alone = 4 hours 
  SUFFICIENT

Statement 2:   Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.
This means,  \frac{1}{R1}  +  \frac{1}{R2}  =  \frac{3}{R1} 
 \frac{2}{R1}  =  \frac{1}{R2}

\frac{R2}{R1}  =  \frac{1}{2}

From (2)
R1 * T1 = 3R2 * T2
 \frac{R2}{R1}  =  \frac{T1}{3T2}

\frac{T1}{3T2}  =  \frac{1}{2}

T1 =  \frac{3T2}{2}

From (1)
T1 + T2 = 7
 \frac{3T2}{2}  + T2 = 7

\frac{5T2}{2}  = 7

by solving the above equation, we can determine the values of T1, T2, R1, and eventually R2.
  SUFFICIENT

The correct answer is Option D

DhanyaAbhirami · Answer

We have a relation between the work done by them : A did 3 times work as B
Let work done = 4 units
A did 3 units
B did 1 unit
(1) 
A did 2 units in 4 hours
A did 1 unit in 2 hour
B did 1 unit in 1 hour
To do 4 units of work, B takes 4 hours
SUFFICIENT
(2) 
A and B working together has same speed as 3 A's working together
B has same speed as 2A's
B did 1 unit, took time T
A did 1 units, took time 2T
A did 3 units, took time 6T
T + 6T = 7 hours
T = 1 hour
To do 4 units of work, B takes 4 hours
SUFFICIENT
Ans is D

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