In what time would a cistern be filled by three pipes whose diameters

Question

Competition Mode Question

In what time would a cistern be filled by three pipes whose diameters are 1 cm , 4/3 cm, 2 cm, running together, when the largest alone will fill it in 61 minutes, the amount of water flowing in by each pipe being proportional to square of its diameter ??

A. 18 minutes
B. 20 minutes
C. 40 minutes
D. 36 minutes
E. 72 minutes.

A. 18 minutes
B. 20 minutes
C. 40 minutes
D. 36 minutes
E. 72 minutes.

Archit3110 · Accepted Answer

from given info
dia === flow rate
1=====1
4/3====16/9
2======4
so largest dia 2 cm ; flow rate = 4 cm2/min vol of tank ; 4*61 ; 244 cm^3
since all three pipes are open together so total combined flow rate ; 1+16/9+4 = 61/9
time taken ; 61/9*x = 244
x= 36 mins
IMO D

In what time would a cistern be filled by three pipes whose diameters are 1 cm , 4/3 cm, 2 cm, running together, when the largest alone will fill it in 61 minutes, the amount of water flowing in by each pipe being proportional to square of its diameter ??

A. 18 minutes
B. 20 minutes
C. 40 minutes
D. 36 minutes
E. 72 minutes.

A. 18 minutes
B. 20 minutes
C. 40 minutes
D. 36 minutes
E. 72 minutes.

eakabuah · Answer

Diameter of Tap A = 1cm, diameter of Tap B=4/3 cm, and diameter of Tap C=2cm. The largest tap, Tap C fills the cistern in 61 minutes.
We are informed that the amount of water flowing into the cistern is directly proportional to the square of the diameter of the tap. Hence the time to fill the cistern is inversely proportional to the diameter of the tap.

hence using proportionality, we can determine the time for Taps A and B, given time taken for Tap C=61minutes.
Time for A, ta = 61*4/1 = 144 minutes
Time for B, tb= 61*4*9/16 = 549/4 minutes

Hence the time taken when all three taps work together = ta*tb*tc/(ta*tb + ta*tc + tb*tc)
=1,205,604/56,745.25
=36minutes

Hence the answer is D.

joohwangie · Answer

In what time would a cistern be filled by three pipes whose diameters are 1 cm , 4/3 cm, 2 cm, running together, when the largest alone will fill it in 61 minutes, the amount of water flowing in by each pipe being proportional to square of its diameter ??

A. 18 minutes
B. 20 minutes
C. 40 minutes
D. 36 minutes
E. 72 minutes.

A. 18 minutes
B. 20 minutes
C. 40 minutes
D. 36 minutes
E. 72 minutes.

the 2 extra pipes together is little bigger than the diameter of the largest one so it should be at least half faster, therefore, less than 30 minutes.
I would say A

exc4libur · Answer

(d_1)^2*(time_1)=(d_2)^2*(time_2)…time_1=61 
 (2)^2*(61)=(1)^2*(time_2)…4(61)=(time_2)…time_2=244 
 (2)^2*(61)=(4/3)^2*(time_2)…4(61)(9)/16=(time_3)…time_3=549/4

rate_n=\frac{1.cistern}{time_n}

\frac{1.cistern}{add.rates}=time.to.fill 
 time.to.fill=\frac{1}{1/61+1/244+1/549/4}=\frac{1}{5/244+4/549}=\frac{1}{5/(4*61)+4/(9*61)}=\frac{61}{5/4+4/9}=36

Answer (D)

lacktutor · Answer

The rate of the amount of water flowing in by each pipe= 1 : 16/9 : 4

The capacity of cistern =x+(16/9)x+4x=(61/9)x
-->The largest one is 4x. It takes 61 minutes to fill the cistern alone.

All pipes started and finished at the same time.
So, finding the time which the largest one spent is enough to answer the question.
 
(61/9)x --> 61 minutes
4x ------> a minutes

--> a= (4x*61):(61/9)x= 36 minutes

The answer is D.

effatara · Answer

"The flow of water flowing in through each pipe is directly proportional to the square of the diameter of that pipe" which means that:
d^2*k=f (where 'd' is the pipe dia, 'k' is a constant and 'f' is the quantity of water flowing in per unit of time).
Let's assume the capacity of the cistern is 122 liters. So the largest pipe (2 cm dia) will fill @ 122/61=2 liters/minute. Then:
(2^2)*k=2....> k=1/2

The values of 'f' for the other two pipes can thus be determined as follows:
(4/3)cm dia: (4/3)^2*(1/2)=8/9 liters/minute
1 cm dia: 1^2*(1/2)=1/2 liters/minute

So when all 3 pipes work together, total inflow of water into the cistern per minute is 2+(8/9)+(1/2)=61/18 liters.
Time needed to fill 122 liters = 122/(61/18) = (122*18)/61 = 2*18 = 36 minutes.

ANS: D

rudi0setyawan · Answer

the amount of water flowing in by each pipe being proportional to the square of its diameter

water debit of three:
= 1^2 + (4/3)^2 + 2^2
= 1 + 16/9 + 5
= 61/9 cm^2

since 4 cm^2 needs 61 minutes, so 
4 * 61 = 61/9 * T
T = 4*9
T = 36 minutes

PSKhore · Answer

Rate is directly proportional to the square of the diameter and inversely proportional to time.

time = k/square(diameter)

from pipe 3, we can calculate k

61 = k/4
k = 61X4 min
Likewise, calculate for pipe1 and pipe 2

t1 = 61X4/1 = 244

t2 = 61X4/(16/9) = 549/4

Total time:

(1/244) + (1/61) + (4/549) = 1/61(1/4 + 1 + 4/9)

eventually 61 cancels out and you are left with 1/36
so 36 minutes

PSKhore · Answer

Rate is directly proportional to the square of the diameter and inversely proportional to time.

time = k/square(diameter)

from pipe 3, we can calculate k

61 = k/4
k = 61X4 min
Likewise, calculate for pipe1 and pipe 2

t1 = 61X4/1 = 244

t2 = 61X4/(16/9) = 549/4

Total time:

(1/244) + (1/61) + (4/549) = 1/61(1/4 + 1 + 4/9)

eventually 61 cancels out and you are left with 1/36
so 36 minutes

In what time would a cistern be filled by three pipes whose diameters

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

In what time would a cistern be filled by three pipes whose diameters

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards