Latest Microsoft Aptitude Question SOLUTION: There are three taps A,B and C.they can fill the tank in 10,20,and 25 hrs respectively.At first all of them are opened simultaneously.Then after 2 hrs tap C is closed and A and B

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There are three taps A,B and C.they can fill the tank in 10,20,and 25 hrs respectively.At first all of them are opened simultaneously.Then after 2 hrs
tap C is closed and A and B are kept running.After the 4rth hour tap B is closed.The remaining work is done by tap A alone.Find the percentage of the work done by Tap A by itself?

Asked In Microsoft tauseef arbab (9 years ago)

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Narasapu Hari prasad  (9 years ago)

Ans is 72%
percentage of work done by c in 2 days is 2/25*100=8%
percentage of work done by b in four days is 4/20*100=20%
so percentage of work done by A alone is 100-8-20=72%

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shivam srivastava  (9 years ago)

Ans is 72%
percentage of work done by c in 2 days is 2/25*100=8%
percentage of work done by b in four days is (4+2)/20*100=30%
so percentage of work done by A alone is 100-8-30=62%

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BHASKAR MANDAL  (9 years ago)

a percentage work = 100/10 = 10%
b =100/20 = 5% and c= 100/25=4%
all of them opened for 2hrs,and then a and b for another 2hrs (given after 4th hrs) then a done alone
so we can write
=> (10+5+4)*2 + (10+5)*2 + work percentage by a alone = 100
=>38+30+work by a alone = 100
=> work done by a alone = 100-68 = 32

so 32 is the answer.

Is this solution Helpfull? Yes (3) | No (13)  

amar sakpal  (9 years ago)

Ans is 72%
We have to calculate entire work done by A
Ans given by bhaskar calculate work of A after 4 hr(i.e 32%).. it do not consider before 4hr (which will be 40%)
Hence final ans will be 72%

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surbhi dubey  (9 years ago)

ans=32
work done by a,b and c in 2 hours=19/50
work done by a and b in another 2 hr=3/10
19/50+3/10+x=1
x=16/50*100%
32

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mohan  (5 years ago)

2% only
B'coz supposed 100 liter water present in tank
2 hours worked by all three =38
4 hours work by a and b= 60
And rest by only a means(100-98)=2
Then works has by itself a = 2%

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Partha Ghosh  (9 years ago)

right answer is 32%

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tirupathi reddy  (9 years ago)

Work done by A is 62%
1st method:
For 2 hours three taps can fill
= 2(1/10+1/20+1/25)
= 38/100

For 4 hours A and B worked
4(1/10+1/20)
= 60/100(or 3/5)

Remaining cap of tank is
1-(38/100)-(60/100)
= 2/100

2/100 part must be filled by A itself.

Finally A percentage of work is
= 2(1/10)+4(1/10)+2/100
= 62/100(62%)

Second method:
Lonely
Tap A fill tank in 10h
Tap B fill tank in 20h
Tap C fill tank in 25h

Total cap of tank is 100 units( LCM of 10, 20, 25)
For 1 hour
A can fill 10 units(100/10),
B can do 5 units(100/20),
C can do 4 units(100/25).

Firstly 3 taps runs for 2 hours
Completed units are 2(10+5+4) = 38 units

We have 62 units remaining
For the second time A and B work for 4 hours
Completed units are 4(10+5)
60 units of work is completed .

Finally A alone work for 2 units(100-38-60)
1 hour-----10 u
0.2 hours--2 units

Total work by A is (10(2+4)+2)/100
62/100(62%)

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kiritisaranam  (7 years ago)

actually mr tirupathi u have considered for the second equation as they have worked for 4 hrs but actually a and b combinely worked for 2 hrs only so your equation is wrong

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mohan  (5 years ago)

Ans will be 2%
Supposed take are is 100%
Fill by a,b,c in 2 hour=38%
Fill by a,b in 4 hours=60%
Fill by only itself by a= 100% - 98%
= 2% ans

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Raushan Kumar Paswan  (2 years ago)

A's one day work = 1/10
B's one day work = 1/20
C's one day work = 1/25
(A+B+C)'s one day work = (1/10+1/20+1/25) =19/100
(A+B)'s one day work = (1/10 +1/20) = 3/20

work is done by A+B+C in 2 hours is 19/50
work is done by A+B in 4 hours is 3/5
the remaining work is (1-(19/50+3/5)) = 1/50

in 6 hours, A do ((1/10)*6) works i.e 3/5 works.
total work is done by A by itself = (1/50)+(3/5)=31/50
%age of the work done by tap A by itself is ((31/50)/1)*100 i.e 62%.

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