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A stationary engine has enough fuel to move 12 hours when its tank is 4/5 full, how much
hours will it run when the tank is 1/3 full
Read Solution (Total 14)
-
- when engine is (4/5) full,it moves for 12 hr
so, when tank is full, it will move = 12/(4/5) = 15 hr
So, when it is 1/3 full, it will move = 15*(1/3)= 5 hours - 10 years agoHelpfull: Yes(46) No(0)
- Ans: 5hrs
fuel and time is directly propotional
4/5fuel=12 hrs
1fuel=12*5/4=15hrs
1/3 fuel= 15/3= 5hrs
- 10 years agoHelpfull: Yes(6) No(0)
- 12----->4/5
x------->1/3
12/x = (4/5)/(1/3) =>12*(1/3)=x*(4/5)
=>x=5hrs - 10 years agoHelpfull: Yes(3) No(0)
- ans is 5hr..
if 4/5 takes ---->12hr
1/3 takes----->
12*(1/3)/4/5=5 - 10 years agoHelpfull: Yes(2) No(0)
- when engine is (4/5) full,it moves for 12 hr
so, when tank is full, it will move = 12/(4/5) = 15 hr
So, when it is 1/3 full, it will move = 15*(1/3)= 5 hours - 10 years agoHelpfull: Yes(2) No(0)
- Ans. 5 hours
4/5 full tank has fuel to move engine for 12 hours
full tank has fuel to move engine for 12*5/4 hours
1/3 full tank has fuel to move engine for 12*5*1/(4*3) hours = 5 hours - 10 years agoHelpfull: Yes(0) No(0)
- 12 hours in 4/5th of tank , so the engine will run for 15 hours in full tank.
for 1/3rd of tank, the engine will run for 5 hours. - 10 years agoHelpfull: Yes(0) No(0)
- when tank is 4/5 full, engine goes for 12 hrs
1 full, engine goes for 12*5/4 hrs
1/3 full, engine goes for 12*5/4*1/3=5hrs - 10 years agoHelpfull: Yes(0) No(0)
- 5 hours is correct
- 10 years agoHelpfull: Yes(0) No(0)
- it will take 5 hours
- 10 years agoHelpfull: Yes(0) No(0)
- 4/5 ---->12 hrs
1/3----->x hrs
x*4/5=1/3*12
i.e x=5 hrs
- 10 years agoHelpfull: Yes(0) No(0)
- ans is 5 hours
solve it by unitary method. - 10 years agoHelpfull: Yes(0) No(0)
- 4/5:1/3::12:x
12*x=12*5
then x=5 - 10 years agoHelpfull: Yes(0) No(0)
- 5 hrs is the answer
- 9 years agoHelpfull: Yes(0) No(0)
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