Tap A can fill a tank in 4 hours whereas tap B and tap C can empty it in respectively 12 hours and 1

M4maths Previous Puzzles

15 May 2015 Work and Time

Tap A can fill a tank in 4 hours whereas tap B and tap C can empty it in respectively 12 hours and 16 hours separately. All the 3 tap are opened with tank empty. After 2 hours tap B is closed, after how much time of closing of B, tank will be filled completely.

OPtion

1) 4 hrs 13 min 20 sec
2) 3 hrs 54 min 18 sec
3) 4 hrs 15 min 30 sec
4) 3 hrs 46 min 40 sec
5) 5 hrs 2 min 20 sec
6) 5 hrs 10 min 40 sec
7) 6 hrs 12 min
8) 5 hrs 40 min
9) 2 hrs 58 min 20 sec
10)5 hrs 10 min 20 sec

Solution

When all the 3 tap are opened the work of 1 hrs
= 1/4 - 1/12 - 1/16
= 12-4-3/ 48 = 5/48
Work of 2 hrs = 5/48 * 2 = 5/24 part
remaining work = 1- 5/24 = 19/24 part
After closing tap B,
the work of remaining 2 taps
per hr = 1/4 - 1/16 = 3/16
So extra hours required = (19/24)/3/16
= 19*16/ 24*3 = 38/9 hrs
Therefore
4 hrs & 2/9 hrs
4 hrs & (2*60)/9 min
4 hrs & 40/3 min
4 hrs & 13min & 1/3 min
Therefore
4 hrs & 13min & 20 sec
Option 1)

Correct Option: 1 Best Solution (24)

RATAN KUMAR BARMAN   10 years AGO

A can fill 1/4th of the tank in 1 hour
B can empty 1/12th of the tank in 1 hour
C can empty 1/16th of the tank in 1 hour
after opening all the pipes for 2 hour, the filled portion of the tank is 2(1/4-1/12-1/16)= 10/48
remaining portion =(1-10/48)= 38/48
now after closing B, A & C together can fill {1/4-1/16)= 3/16 in 1 hour.
3/16 ...... in 3600 seconds
so 38/48.... in (16/3*38/48*3600) seconds= 15200 seconds = 4 hrs 13 min 20 sec

Like? Yes (7) | No   

Surit Datta   10 years AGO

After calculating we get the correct answer.
So option 1
that's it.

Like? Yes (3) | No   

aman aditya   10 years AGO

let capacity of tank=48 liter (taking L.C.M)
A can fill 48/4=12 l in an hr
B can empty 48/12= 4 l in an hr
C can empty 48/16= 3 l in an hr
In first two hr =2*(12-(4+3))=10 l of water is there
water to be filled= 48-10=38 l
now B is closed so total transaction in one hr=12-3=9 l
so time=(38/9)*60=4 hr 13 min 20 sec

Like? Yes (2) | No   

MRITUNJAY   10 years AGO

resuitant of tap A B C=1/4-1/12-1/16=5/48
workdone in 2 hours=10/48
remaining work=1-10/48=38/48
workdone by tap A and c together=1/4-1/16=3/16
so time required =38/48*16/3=38/9=4 hr 13 min 20 sec

Like? Yes (2) | No   

ravi singh   10 years AGO

Answer is 4hr 13min 20 sec

Tap A filled tank in 1 hr= 1/4

Tap B empty tank in 1 hr= 1/12

Tap C empty tank in 1 hr= 1/16

Let, when tap B is closed,
X more hr will be tanken to filled the tank completely.
So,

(X+2)/4 - 2/12 - (X+2)/16 = 1

Then, X = 76/18 hrs
Which is equal to 4hr 13min 20 sec.

Like? Yes (2) | No   

b.subbarao   10 years AGO

as per the data part of tank will be filled in one hour=1/4-1/12-1/16=5/48
tank part filled after 2 hours=5/48*2=10/48
the remaining part of the tank=1-10/48=38/48
after 2 hours b tap stopped so the reining A&B are continued
A&B taps will fill the tank in one hour=1/4-1/16=3/16
hence the remaining tank filled by A&B in th duration of=38/48/(3/16)=38/48*16/3=38/9 hours=4 rs.13min.20sec.
so option 1 is correct answer

Like? Yes (2) | No   

kishore   10 years AGO

filling means +ve emptying means -ve
A B C LCM
4 12 16 48
12 4 3
12-(4+3) = 5
for first two hours 10 parts work has been done.
so remaining 48-10 = 38 parts of work will be done by only A and C so 48/9 = 4hours 13min 20 sec

Like? Yes (2) | No   

vivek   10 years AGO

work done by a,b,c in 1 hr= (1/4-1/12-1/16)=5/48
so, work done by them in 2 hr = 2*5/48=10/48
now remaining work will be 1-10/48=38/48
also, a and c can work together. so, work done by a and c in 1 hr =(1/4-1/16)=9/48
so, remaining unit of work = 38 unit
so required time = 38/9 hr
=4 hr13 min20sec

Like? Yes (2) | No   

Shiva Teja   10 years AGO

Part filled when Taps A,B and C are opened for 2 hours
=(1/4-1/12-1/16)*2=5/24

Remaining part to be filled =1-5/24=19/24

After closing B Time required to fill he tank completely
=(19/24)/(1/4-1/16)=38/9 hours
=4 hours 13 minutes 20 seconds.

Like? Yes (1) | No   

atmakuri jagadish   10 years AGO

In 2 hours,
Tap a can fill 1/2
tap b can empty 1/6
tap c can empty 1/8
So after 2 hours, tank will be filled by (1/2) - (1/6) - (1/8) = 5/24
Tank remaining to be filled is 19/24
If x hours is taken for tank to be filled
then tap a will fill x/4 of the tank
tap c will empty x/16
Therefore , (x/4) - (x/16) = 19/24
Solving this we get x as 38/9
So this can be written as 4 hours + ((2/9)*60) minutes
this is nothing but 4 hours + 13 minutes + ((3/9)*60) seconds
This can be further written as 4 hours + 13 minutes + 20 seconds
So option 1

Like? Yes (1) | No   

Karthick Raja   10 years AGO

Pipes fill in 1 hr
pipe A = (1/4)
pipe B = (1/12)
pipe C = (1/16)
Tank fill when all the pipes open in 2 hrs

(2/4)-(2/12)-(2/16)=(5/24)

Tank fill in 2 hrs = 1-(5/24) = (19/24)

after 2 hrs pipe B is closed
so
(x/4)-(x/16)=(19/24)

Taking LCM
(4x-x)/16 = (19/24)

x= (19*16)/(24*3)
x=4.22
convert 0.22 in minutes and seconds
0.22*6=1.322=13min 20seconds

4hours 13 mins 20 seconds

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PAWAN KUMAR   10 years AGO

2*(1/A -{1/B+1/C}) = 10/48
Now, 10/48 + t*(1/A-1/C) = 1
=> t = 38/9 hrs = 4hr 13min 20sec.

Like? Yes (1) | No   

Roshni Murali   10 years AGO

work done in 2 hours=2*(1/4-(1/12+1/16))=5/24
remaining work=1-(5/24)=19/24
work done by pipe a and c in 1 hour=1/4-1/16=3/16
time taken by pipe a and c work together=16/3
time taken by a and c to complete remaining work=(19/24)*(16/3)=4 hours 13 mins 20 seconds

Like? Yes (1) | No   

poojasaisri   10 years AGO

lcm of 4,12,16 is 48
total work=>
48=(2*(12-4-3))+(x*(12-3))
=>x=38/9
=4.222
=4 hrs 13 min 20 sec

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vibhu raj gupta   10 years AGO

after calculation get time is 4hr 13min 20sec

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sai   10 years AGO

1 hour = 3/16 part
for 19/24 = takes 4 hours 40/3
4 hour 13 min 20 sec

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Afroz Baig   10 years AGO

Part filled by three taps in 2 hours
= 2(1/41/12-1/16) =5 /24 Part unfilled = 19/28 Remaining work is done by A & C in ((1/4 -1/16)^-1 ) 16/3 hours .Total time taken to fill the tank after tap B is closed is
19/24 *16/3 =38/9 hours =4hrs 13 min 20sec

Like? Yes (1) | No   

divya prakash singh   10 years AGO

the tap A fills x/240 l in 1 min and tap B empties x/720 l in 1 min and tap C empties x/960 in 1 min.if all taps are open then after 2 hrs tank will be 5x/24 l full.therefore still 19x/24 l litre tank remains empty.after 2 hrs tap B is closed so the rate will be x/320 per min.therefore 19x/24 l will be filled in 760/3 mins which is 253.33 mins which is approximately 4 hrs 13 mins and 20 secs

Like? Yes (1) | No   

Meenakshi   10 years AGO

A=1/4(fill the tank)
B=1/12(empty the tank)
C=1/16(empty the tank)
Part filled in 2hrs=2*[(1/4)-(1/12)-(1/16)]
=5/24
Remaining part=1-5/24=19/24
Part filled in 1 hr=(1/4)-(1/16)=3/16
1hr=3/16part
xhr=19/24part
x=38/9=4 hrs 13mins 20 sec

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Ashwin Dinesh   10 years AGO

x(1/4 - 1/16) = 19/24
So x = 4.22 = 4hrs 13mins 20secs
which is option 1.

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Mukesh   10 years AGO

=>2(1/4 - 1/12 -1/16)+ t(1/4 - 1/16)=1
=>t=38/9 hours = 4h 13 min 20 sec

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sourav pal   10 years AGO

Considering the unit work here
part filled in 2 hours=2*[(1/4) - (1/12) -(1/16)] = 5/24

==== remaining part = 1-5/24 = 19/24

net part filled in 1 hour will be(with tank B closed) =[(1/4) - (1/16)]=3/16

=== 3/16 part == 1hour

==== 19/24 part == (16/3)*(19/24) = 38/9 = 4.22 hrs

Like? Yes (1) | No   

Mohit   10 years AGO

a fill 25% in one hour
b empty 8.33% per hour
c empty 6.25% per hour so it takes 4 hours 13 minutes 20 second to fill the tank after closing of tap B after 2 hours.

Like? Yes (1) | No (1)  

SAGNIK DATTA   10 years AGO

let total tank be 48 units so A=12 units/hr B=12 C=16
In 2 hrs quantity filled = 10 remaining 38/9 =4.22 hours

Like? Yes (1) | No (1)