An empty reservoir has two taps for water inlet – one at the top of th

Question

An empty reservoir has two taps for water inlet – one at the top of the reservoir and the other at the bottom. One way to fill the empty reservoir completely is to turn on the top tap only for 6 hours and then to turn on the bottom tap only for 4 hours. If the time taken by the top tap alone to fill the empty reservoir is 1.5 times the time taken by the bottom tap alone to fill the empty reservoir, how much time does it take to fill the empty reservoir if both the taps are turned on together?

A) 5 hours 20 minutes
B )5 hours 12 minutes
C) 5 hours
D) 4 hours 48 minutes
E) 4 hours 40 minutes

A) 5 hours 20 minutes
B )5 hours 12 minutes
C) 5 hours
D) 4 hours 48 minutes
E) 4 hours 40 minutes

Debo1988 · Accepted Answer

x - rate of top tap
y - rate of bottom tap
Total work to be completed = 6x + 4y
Also, y = 3x/2 (since given time taken by Top is 1.5 times that taken by Bottom and since rate and time are inversely related)
So, total work to be completed = 12x

Combined rate of Top and Bottom = x + 3x/2 = 5x/2

Time taken = Total work / Combined rate = (12x) /(5x/2) = 24/5 hrs = 4 hrs 48 mins

GMATinsight · Answer

If the time taken by the top tap alone to fill the empty reservoir is 1.5 times the time taken by the bottom tap alone to fill the empty reservoir

i.e. Bottom tap takes t hours time 
then top tap takes 1.5t hours

Top tap took 6 hours and bottom tap took 4 hours

but incidentally 6 = 1.5*4

i.e. both Taps must have filled half of teh tank independently

i.e. Independent Time taken by Top Tap to fill the reservoir = 2*6 = 12 hours
i.e. Independent Time taken by Bottom Tap to fill the reservoir = 2*4 = 8 hours

Time taken by both taps together = 1/  = 24/  = 24/5 = 4 hours 48 minutes

Answer: Option D

newyork2012 · Answer

Top=t , Bottom=b

Vb = 1.5 Vt
4*Vb=6*Vt

So to fill up the tank would take t alone 12 hours , b alone 8 hours

Together 1/12 + 1/8 = 20/96 ..so it will take t&b together 96/20 = 24/5=4hours ,48 min to finish

Posted from my mobile device

fireagablast · Answer

w=r*t
Both taps used at different times:
1=r1t1 + r2t2
1=6r1 + 4r2 --> t1=6, t2=4
1=6r1 + 4(1.5r1) -->r2=1.5r1
1=6r1+6r1
1=12r1
r1=1/12

both taps used at the same time:
1=(r1+r2)t
1=(r1+1.5r1)t
1=(1/12 * 1.5/12)t
t=24/5

24/5 = 4 *4/5 = 4 hours and 4/5*60min = 4 hours and 48min

sume3tss · Answer

could you please explain how one gets that both Taps must have filled half of the tank independently

effatara · Answer

Let the flow rate (capacity) of the Bottom Tap (BT) be 2 liters per hour. Then the capacity of the Top Tap (TP) is 3 liters per hour. TT and BT fills in 6*2 L and 4*3 L in 6 hrs and 4 hrs respectively which fills up the reservoir. Thus reservoir capacity is 12 +12 = 24 L. The combined capacity of the two taps is 5 L/hr. If 'x' hrs is the time taken by the two taps together to fill up the reservoir then:
5x=24...> x=24/5=4 hrs 48 minutes. ANS: D

Sutra · Answer

Let, the bottom tap takes x hours where the top one takes 1.5x or3x/2 hours.
ATQ.
12/3x+4/x=1
Hence we get x=8
So the top tap takes 12 hours and the bottom takes 8 hour and they take to fill the empty reservoir in
24/5 hours or 4 hours 48 min

Posted from my mobile device

prabsahi · Answer

Can someone please explain me this problem..
A more dependable and clear approach..

plaverbach · Answer

T= Rate Top
B= Rate Bottom

1.5T=B 
(x2) 3T=2B
(x2) 6T=4B

(+) 6T +4B=1  => (Work=Rate.Time)
(-)6T=4B
(=) 4B=1-4B

B= \frac{1}{8} 
T= \frac{1}{12}

B+T= \frac{5}{24}  ==> You can always sum rates

Work=Rate.Time ==> W=1 because you are trying to fill 1 reservoir
1= \frac{5}{24} *T

T= \frac{24}{5} =4 \frac{4}{5} =4hours+ \frac{4}{5} *60= 4h48

ndj5y2 · Answer

This question could definitely be worded better..

PSKhore · Answer

6/(1.5x) + 4/x = 1
x= 1/8

Total time:
1/8 + 1/12 = 5/24

=4.8 hours

An empty reservoir has two taps for water inlet – one at the top of th

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