Allegation and Mixture Quiz-1

The percentage of alcohol in the new mixture would be:

(2*20+3*80)/5

=280/5

=56%

The ratio of alcohol to water in the new mixture would be 56:44=14:11

Assume the capacity of the two containers is 88 litres each. When we mix 88 litres of the first and 88 litres of the second,the amount of spirit would be:

88*1/8 + 88*3/11=11+24=35 litres

Consequently the amount of water would be 2*88-35=176-35=141 litres.

Required ratio=35:141

 

We will consider 10 parts of liquid A and 4 parts of liquid B with each part a unit volume. For ease of understanding, the unit can even be taken as a liter. Unit can be any because all values, given and desired, are in percentages that are basically ratios of two values (so that units cancel out).

With this understanding, we will calculate the amount of water in each of 10 parts or 10 liters of liquid A and 4 parts or 4 liters of liquid B.

 

Let the adulterated fat = 0 Rs.

Butter = Rs 1

C.P of mixture = 1x 100 /125 = 4/5 Rs



=1:4 of adulterated fat = 1/5x 100 = 20%

The three vessels are of equal capacity (which we assume to be full since it is a PS question. I would prefer clearer direction though). The milk:water ratio is given for each as 4:3, 2:1 and 3:2. The reason we cannot just add the milk parts (4+2+3) and water parts (3+1+2) together is that they represent different fractions of the whole volume i.e. in the first vessel, milk is 4 parts out of a total of 7 parts while in the second vessel, milk is 2 parts out of a total of 3 parts etc. If we make the total number of parts in each ratio equal, then we can just add the 'milk parts' together and all the 'water parts' together.
Ratios:
4:3 (Total 7 parts)
2:1 (Total 3 parts)
3:2 (Total 5 parts)
How can we make the total number of parts equal in the 3 cases? By taking the LCM.
4*15 : 3*15 = 60:45 (Total parts 105)
2*35 : 1*35 = 70:35 (Total parts 105)
3*21 : 2*21 = 63:42 (Total parts 105)
When we mix the three solutions, the total number of parts of milk is 60+70+63 = 193
the total number of parts of water = 45+35+42 = 122
The required ratio = 193:122

Vessel A = kerosene = 3/11

Vessel B = 1/6

Vessel C = 1/5



Ratio = 11:24

Part taken out from vessel B =25/35*35=24L

                  C   :     A

                 first            1    :      2

               second        3    :      1

                  third          2   :       1



Ratio = 1:4

So 4 times second alloy should be mixed with first to get desired ratio