TCS
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Numerical Ability
Geometry
In a vessel, there are 10 litres of alcohol. An operation is defined as taking out five litres of what is present in the vessel and adding 10 litres of pure water to it. What is the ratio of alcohol to water after two operations?
a) 1 : 5
b) 2 : 3
c) 1 : 6
d) 3 : 2
Read Solution (Total 18)
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- answer is a i.e 1:5
here vessel initially contains 10 litres of alcohol and 0 litres of water
in first operation 5 ltrs of alcohol is taken out and 10 ltrs of water is added.
so after first operation 5 ltrs of alcohol and 10 ltrs of water are there in the vessel
in the second operation 5 ltrs are again taken out
but as now alcohol and water are in the ratio of 1:2
or alcohol is one third of total quantity and water is two thirds of total quantity
so amount of alcohol taken = one third of total amount taken
=5/3 ltrs
and amount of alcohol taken = two thirds of total amount taken = 10/3 ltrs
now quantity of alcohol remaining = 5-5/3 ltrs= 10/3 ltrs
and amount of water remaining = 10-10/3 ltrs = 20/3 ltrs
after that 10 ltrs of water is added
so amount of water after that = 20/3+10 = 50/3 ltrs
so ratio of alcohol to water = (10/3)/(50/3) = 1:5
- 11 years agoHelpfull: Yes(111) No(6)
- Final concentration = Initial concentration (1−replacement quantity/final volume)
Final concentration = =1×(1−10/15)=13
Final concentration = 13×(1−10/20)=16
So ratio of alcohol : water = 1 : 5 - 10 years agoHelpfull: Yes(6) No(14)
- Final concentration = Initial concentration (1−replacement quantity/final volume)
Final concentration = =1×(1−10/15)=1/3
Final concentration = 13×(1−10/20)=1/6
so,after 2 operations concentration is 1/6.hence,final volume should be 6 unit.
So ratio of alcohol : water = 1 : 5 as (1+5=6)
- 9 years agoHelpfull: Yes(5) No(6)
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- : Final concentration = Initial concentration(1−replacement quantityFinal volume)
Final concentration = 1×(1−10/15)=13
Final concentration = 13×(1−10/20)=16
So ratio of alcohol : water = 1 : 5
- 10 years agoHelpfull: Yes(1) No(12)
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