Three tap together drain a complete tank in 5 hours. If first drains this tank in 10 hours alone, th

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24 Dec 2014 Time and Work

Three tap together drain a complete tank in 5 hours. If first drains this tank in 10 hours alone, then minimum hours taken by second tap will be (in any condition)

OPtion

1) 10
2) 8
3) 4
4) 15
5) 20
6) 2
7) 1
8) 40
9) 5
10)25

Solution

Let time taken by first, second and third tap are t1, t2, t3
Then according to given condition
1/t1 + 1/t2 + 1/t3 = 1/5 (i)
and
1/t1 = 1/10
so
1/t2 + 1/t3 = 1/5 - 1/10 = 1/10
for t3 = Infinite
t2(min) = 10
Option 1)

Correct Option: 1 Best Solution (14)

kumari jyoti   10 years AGO

tap 1 will empty half of the tank alone in 5 hours.
if tap 3 is assumed to be of nil efficiency then tap 2 will also empty half of the tank in 5 hrs.
so, it will alone empty the tank in 10 hrs

Like? Yes (3) | No   

Vinay Sipani    10 years AGO

In 1 hour,three taps can drain 1/5 th of tank
=> (1/x) + (1/y) + (1/z) = 1/5

Given x = 10
=> (1/y) + (1/z) = 1/10
=> 10y + 10z = yz
=> z = 10y / (y - 10)

y is minimum => z is maximum
so, (y - 10) = 0 => y = 10 hrs

In other words, for z not to be negative,(y - 10) is to be positive
=> y - 10 = 0 => y = 10 hrs

Like? Yes (2) | No   

Md.Sultan Khichi   10 years AGO

They all drain the tanx in 5 hrs.
First tab alone drain it in 10 hrs.
Second tab minimum time =[1/5-1/10]=1/10
Ans. Ob 1

Like? Yes (1) | No   

Ashwin Dinesh   10 years AGO

guessed!

Like? Yes (1) | No   

Rajat Rokade   10 years AGO

10 /2=5 so atleast 10 hours

Like? Yes (1) | No   

Rahul Rathod   10 years AGO

10 /2=5 so atleast 10 hours

Like? Yes (1) | No   

susanta   10 years AGO

3 tap done 1 hr=1/5
1st done in 1 hr=1/10
2nd+3rd done in 1 hr=1/5-1/10
=1/10
2nd+3rd done full work =10 hr
if 3rd tap is not working then 2nd must take minimum 10 hr

Like? Yes (1) | No   

SREENIVASA RAO PALLAPU   10 years AGO

T1 + T2 + T3 will complete 1/5 th of the total work in 1 hour.
T1 alone complete 1/10 th of the total work in 1 hour.
=> T2 + T3 will complete 1/10 th of the total work in 1 hour
i.e T2 + T3 =1/5 - 1/10 = 1/10
so in any condition T2 will take at least 10 hours to complete the work( either T3 is on/off).

Like? Yes (1) | No   

PAWAN KUMAR   10 years AGO

If there is no condition, then we could consider that the 3rd tap is not working (for minimum time by 2nd tap).
Therefore, 1/5 = 1/10 + 1/tap2 => 1/tap2 = 1/10
=> tap2 = 10 hours. Ans.

Like? Yes (1) | No   

priyanka   10 years AGO

Work done by (B+C) in 1 hr is (1/10)
B+C together takes 10 hrs
(1/B)+(1/C)=(1/10)
Therefore B>10

Like? Yes (1) | No   

kiran lata   10 years AGO

let A,B,C are three taps
A+B+C=5hrs
only A take=10 hrs
thus A drain half of tank in 2.5hrs.
B+c=2.5hrs
thus to complete drain a tank , B takes=4*2.5hrs=10hrs

Like? Yes (1) | No (1)  

guru   10 years AGO

1/5 = 1/10 + 1/x
1/x = 1/10
x=10

Like? Yes (1) | No (1)  

Pradeep Singh   10 years AGO

every tap has same efficiency so second tap also will take min 10 hours

Like? Yes (1) | No (1)  

Aparna Bhattacharya   10 years AGO

Three tap together drain a complete tank in 5 hours=1/5 hrs
first pipe drains this tank in 10 hours alone=1/10 hrs
so,then minimum hours taken by second or third tap will be
=( 1/5 - 1/10) = 1/10 or 10 hours.
Option : 1

Like? Yes (1) | No (2)