A tank can be filled by one tap 20 minute and by another 25 minute. Both the taps are kept open for

aman aditya   9 years AGO

let tank be of 100 liter(taking L.C.M of 20 and 25)
now first tank fill it in 20 min. it fill 5 lit/min
second tank fill it in 25 min . it fill 4 lit/min .
if both are open for 5 min=(5+4)*5=45 liter is filled.
so 100-45=55 liter is left
55/5=11 min.

Like? Yes (19) | No (2)  

sanjeev kumar   9 years AGO

1st and 2nd tap filled in first min=1/20+1/25=9/100
in 5 min=(9/100)*5=9/20
remaining space=1-9/20=11/20
11/20 filled by first tap=(11/20)*20=11 min.

Like? Yes (14) | No (3)  

Narendra Gopi   9 years AGO

work done by A in one day= 1/20
work done by B in one day=1/25
Both together work in a day is 1/20+1/25=9/100
5 days work is=5*9/100=45/100
remaining work=1-45/100=55/100
B is turn off A can complete the remaining work in 55/100*20=11days

Like? Yes (9) | No (2)  

santosh kumar patel   9 years AGO

Part filled in 5 minut = 5*(1/20+1/25)
= 9/20
Remaining part = 1-9/20 = 11/20

so, now 1/20 : 11/20 = 1 : X
X = 11

Like? Yes (7) | No (3)  

Dinni   9 years AGO

Both taps open =5 ( 1/20 +1/25 ) =9/20
Remaining =1-9/20 = 11/20.
Second tap is turned off.
First tap will fill 1/20 in 1 min
So 11/20 in 11min.

Like? Yes (3) | No   

satyam kumar   9 years AGO

5(A+B)+XA=1
A=1/20;
B=1/25;
A+B=9/100;
5*9/100+x*1/20=1;
SOLVING WE GET X=11;
so 11 hours more....total hours is 11+5=16.

Like? Yes (3) | No   

Altaf Hussain   9 years AGO

Let the capacity of the tank be 100.
Time taken by 1st tap is 20min, Rate=5
Time taken by 2nd tap is 25min, Rate=4
They are kept open for 5min i.e., 9*5=45, remaining 100-45=55 is filled by 1st tap in 55/5=11min

Like? Yes (3) | No (1)  

vinusha   9 years AGO

5 (1/20 + 1/25) =9/20 for 5min
Remaining 11/20 in 11 min

Like? Yes (2) | No   

Akansha   9 years AGO

Both taps are opened for 5min = 5 (1/20+1/25)=9/20
And second tap is turned off
Remaining portion= 1-9/20 =11/20
First tap will fill remaining in 11min.

Like? Yes (2) | No   

AK Garg   9 years AGO

let us assume work is 100 units
then tap 1 do 5 unit work and tap 2 do 4 unit work
so in first 5 mints they did 45 unit work
then remaing work is 55 unit so it is done by tap 1 in more 11 mins

Like? Yes (1) | No   

Soma Naresh   9 years AGO

1 min work of two pipes are 1/20, 1/25
5 min work by both pipes
=(5/20)+(5/25)=9/20
after 5 min second tap is closed
so first tap has to work to complete the work
so remaining work=1-(9/20)=11/20
tap one takes 11 mins to fill the tank

Like? Yes (1) | No   

avanthika   9 years AGO

Both taps open = 5 (1/20 + 1/25) =9/20
Remaining = 11/20 and second tap turned off
First tap fill 11/20th part in 11 minutes.

Like? Yes (2) | No (1)  

sagar kumar mishra   9 years AGO

1/20+1/25=9/100
total time taken to complete =100/9m
in 5 minute 45/100 part of wrk will be completed
therefore remaining part 55/100 will be completed by 1st tap =(55/100)*20=11m

Like? Yes (1) | No   

ramya   9 years AGO

Let 2 taps be A and B.
In 5 min A&B fill = 5(1/20 + 1/25) = 9/20 tank
Tap B is turned off. So,
The remaining time taken to fill the tank by tap A will be 1 - 9/20 = 11/20 tank
Therefore, The time taken by tap A to fill the remaining tank is: 11minutes

Like? Yes (1) | No   

Ashu Kumar   9 years AGO

LET EXTRA TIME TO FILL THE TANK IS X.
SO ( X+5/20)+5/25=1
SO X=11 MINUTE
SO CORRECT OPTION IS 8

Like? Yes (1) | No   

SOMA NARESH   9 years AGO

1 min work of two pipes are 1/20, 1/25
5 min work by both pipes
=(5/20)+(5/25)=9/20
after 5 min second tap is closed
so first tap has to work to complete the work
so remaining work=1-(9/20)=11/20
tap one takes 11 mins to fill the tank

Like? Yes (1) | No   

NITESH KUMAR SINGH   9 years AGO

when both the tank is open then time taken to fill the tank is 100/9.
so when both the tank is open for 5 minutes they can fill only 9/20 part of the tank so 11 /20 part of the tank is filled with after 11 minutes

Like? Yes (1) | No   

Ajay Kumar Gupta   9 years AGO

in one min both tank can fill tank (1/20+1/25)=9/100
in 5 min-tank will be 9/100*5=9/20 part
then 2nd is off.
by 1st tap-1/20 work in a minute.
so 11/20 lewft work in 11 min
so 11/20+9/20=1
so tank will full in 11 minutes.

Like? Yes (1) | No   

nishi   9 years AGO

1 min work of both taps =1/20 + 1/25 = 9/100
5 min work of both =9/100 * 5 =9/20
remaining work = 11/20
now 1/20 * x = 11/20
x=11

Like? Yes (1) | No   

vibha   9 years AGO

(1/20+1/25)*5 +(1/20)*x=1
solving x=11
as 1st tap work for x more min so he does x/20 more than 2nd

Like? Yes (1) | No   

S R Y Farhan   9 years AGO

1 work mean 100 percent work complete ----------------------------------------------eq 1
tap 1 => 1 min ----- 120 work complete
tap 2 => 1 min ----- 125 work complate
on doing 5 min together work ----> work completed = 5 *(120 + 125 ) = 920 .
work left to complete = 1 - 920 = 11/20.
now tap 1 requires 1120 work to complete in ______ time .
by using eq 1
time needed to complete laft work is = 1120 * 20 = 11 min.

Like? Yes (1) | No   

Harmeet kaur gurudatta   9 years AGO

1st tap fill in 20 min
2nd tap fill in 25 min
total capacity of tank=100
ie;1st fill 5 unit/min
2nd fill 4unit/min
if both open for 9 min they fill 45 unit (9*5)
remaining unit=100-45=55
now 2nd tap closes.1st tap fill in 11 min(55/5)
and=11 min

Like? Yes (1) | No   

velmala gangadhar   9 years AGO

let tank containing K lts of liquid
A tank capacity(pressure) is K/20 lts/min
B tank capacity(pressure) is K/25 lts/min
First A,B taps are kept open for 5 min then tank filled by
(K/20+k/25)*5 ==9K/20 part of tank
to fill remaining 11K/20 part of tank ,A tap will take 11 min...

Like? Yes (1) | No   

V.NIVETHA   9 years AGO

in five hours 9/20 of the tank will be filled so remaining 20/11 of the tank will be filled by tap a alone in 11 min

Like? Yes (1) | No   

vimal tiwari   9 years AGO

total time both will take to fill = 9/100
after 5 min = 45/100 = 9/20

remaing 11/20
first will fill it in 11 min

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pratik avinash   9 years AGO

1/20 + 1/25 = 9/100
work done by both in 5 min. = 5 * 9/100 = 9/20
r.w = 11/20
therefore remaining work done by A is 11 min.

Like? Yes (1) | No   

sarfarosh khan   9 years AGO

work done by A+B=45/100
remaining work is 55/100
first alone can do the work in 20 min
55/100 work done by firs is 55/100*20=8

Like? Yes (1) | No   

Ankit Kumar   9 years AGO

Let tank is of 100 L
Then speed of !st tap= 100/20 L/min=5 L/min
and speed of 2nd tap=100/25=4 l/min
Now both the tap are kept open for 5 min,then
total amount fill by both the tap =5*5 L+ 4*5 L=45 L
remaining =55 L which is to be filled by only 1st tap
hence time required to fill 55 L by 1st tap =55 L/5 L/min
=11 min

Like? Yes (1) | No   

vivek   9 years AGO

in 5 minutes work done by both pipes together is ((9/100)*5)=9/20
so required work will be 1-9/20= 11/20
work done by 1st pipe in 1 min= 1 unit(to fill a 20 unit tank)
then, time taken to fill 11 unit = 11 min.

Like? Yes (1) | No   

yamini shrivastava   9 years AGO

let the volume of the tank=v
tap 1 takes 20 mins to fill it and tap2 takes 25 mints to fill it up.
so the vol filled by tap1 in 1 min=v/20
tap2=v/25
togather the vol filled by them in 1 min =v/20+v/25=9v/100
so in 5 min the vol filled by both=9v/20
remaining vol =v-9v/20=11v/20.
v vol ifs filled by tap1 in 20 mins
11v/20 will be filled by tap1=11mins

Like? Yes (1) | No   

Kuheli   9 years AGO

1st tap
in 20 mint tank can filled 100 percent
in 5 mint tank can filled 25 percent
2nd tap
25 mints take to fill 100 percent
5 mints takes to fill 20 percent
so total in 5 mints tank can be filled 25 + 20 = 45 percent
remaining 55 percent can be filled by (100/20)*55
= 11minutes .

Like? Yes (1) | No   

Abhishek Anand   9 years AGO

As A takes 20 min to fill up so it is 100/20 = 5% efficient in doing work.
similarly for B 100/25 = 4%
Both works for 5 min so (5x5 + 4x5)= 45 % work is done
Rest 55% of work is to be done by tap A
so tap A will take 55/5=11 minutes more !

Like? Yes (1) | No   

GAUTAM CHAURASIA   9 years AGO

In 1 min part of tank filled = (1/20 + 1/25 ) = 9/100
In 5 Min = 9/20
Part of tank left unfilled = 1-9/20 = 11/20.
Time taken by Tap 1 to fill it = 11/20 * 20/1 = 11 min
Hence 11 More min will be requried

Like? Yes (1) | No   

Nizamuddin S   9 years AGO

When both taps are opened, in 1 minute the tank will be filled (1/20)+(1/25) part of tank that is 9/100 of the tank will be filled in 1 min .For 5 minutes 45/100 part of the tank will be filled and now second tap is turned off then the remaining 55/100 part of the tank will be filled in 11 minutes.So answer is 11.

Like? Yes (1) | No   

KajriSingh   9 years AGO

tank fill by two taps= 5(1/20+1/25)
remaining tank is 11/20
a fill a tank in=11/20*20/1=11

Like? Yes (1) | No   

Tarun Gupta   9 years AGO

Tank Filled by tap A in 1 min = 1/20
Tank Filled by tap B in 1 min = 1/25
Part filled in 5 min = 5(1/20 + 1/25) = 9/20
Remaining part = 1-9/20 = 11/20
Since Tap B is turn off
Then,
1/20:11/20 :: 1:x
x = 11/20 * 20 * 1
=> x = 11 min

Like? Yes (1) | No   

Archie   9 years AGO

1st tap in 20mins
2nd tap in 25mins
Let capacity of tank =LCM(20,25)=100 unit
1st tap's 1min work=5unit
2nd tap's 1min work=4unit
(1st+2nd tap's 1min work)=9unit
-----> Together they can fill 45unit in 5mins
Unit of tank left to be filled=55units
55units can be filled by 1st tap in 55units/(5unit/min)
=11mins
Hence 11mins more will be required to completely fill the tank .

Like? Yes (1) | No   

jayaram   9 years AGO

part filled in 5 minutes=5{(1/20)+(1/25)}=9/20
part remained=11/20;
time taken to fill remain part=(11/20)*20*1
time taken to fill the tank=11minutes

Like? Yes (1) | No   

Devendra Marghade   9 years AGO

5 min. filling by both the taps=5*[(1/20)+(1/25)]=9/20
Remaining tank=11/20, so time taken by first tank to fill remaining=(11/20)/(1/20)=11 min.
Option 8)

Like? Yes (1) | No (1)  

sidhu   9 years AGO

if tank size is100l 1st tank fillid in 20m. its 1minut to fill 5li.
2nd tank filled in 25min.its 1minut time to fill 4li,
both are open 5min then if fill 5*5+5*4=45; 100-45=55
then aftr 5m second is trnde of then complity filled time 55/5=11
ans=11

Like? Yes (1) | No (1)