21.% of liquid B initially present in the vessel = 2 / (3 + 2) × 100 = 40%
% of liquid B finally present in the vessel = 4 / (1 + 4) × 100 = 80%
The second solution is liquid B which is being mixed and it has 100% liquid B.
80% of liquid B present in the resultant mixture may be taken as average percentage. So, using rule of alligation on liquid B per cent, we can write,
or 1 : 2
The ratio of liquid left in the vessel to liquid B being mixed = 1: 2
Since the quantity of liquid B being mixed is 20 liters, the quantity of liquid left in the vessel is 10 liters.
Therefore, the total quantity of liquid initially present in the vessel = 10 + 20 = 30 liters
Quantity of liquid A = 3 / (2 + 3) × 30 = 18 liters.
.02 =(2/100) x 100% =2%
Let N / 11 = 233
Then, N = 233 x 11 = 2563
? Missing digit is 5.
121012 = 12 x 10084 + 4
? remainder = 4
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
Subtract 20, 25, 30, 35, 40, 45 from successive numbers. So 0 is wrong.
NA
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