Title: Tips on Solving Questions Based on Mixtures & Alligations
12 TIPS on cracking Aptitude Questions on Mixtures
and Alligations
2Tip 1 Volume of a substance in the final
solution Sum of volumes of the substance in the
ingredients
- Question How many liters of pure alcohol must be
added to a 100 liter solution that is 20
alcohol in order to produce a solution that is
25 alcohol? - Solution
- Here, let us add X liters of 100 alcohol to 100
liters of 20 alcohol to produce 25 alcohol.
Then, - Volume of alcohol in the original 100 liter
solution 20 x 100 liters 20 liters - Volume of alcohol in the pure alcohol solution
X liters - Total volume of alcohol in the final solution
25 x (100 X) liters - Volume of alcohol in the final solution Sum of
the volumes of alcohol in the two liquids being
mixed - 0.25 x (100 X) liters 20 X liters
- X 20 / 3 liters
3Tip 2 From a solution of volume V containing x
percent of a substance, if y liters are replaced
by water, the new percentage of the substance in
the solution x (V-y)/ V
Question 8 liters are drawn from a cask full of
pure wine and is then filled with water. This
operation is performed three more times. The
ratio of the quantity of wine now left in cask to
that of water is 16 65. How much wine did the
cask hold originally? Solution Let Vw be the
volume of wine in the cask originally. Let Vt be
the total volume of liquid in the cask. Let X be
the percentage of wine in the cask originally. X
original volume of wine/ total volume of liquid
in the cask Vw / Vt When 8 liters are drawn
out, the volume of wine is reduced by 8 X liters
while the total volume of liquid remains
unchanged as it is re-filled with water. Let X
be the new percentage of wine in the cask after
this operation X (original volume of wine
8 X) / total volume of liquid in the cask Vw
8 (Vw/ Vt) / Vt X (Vt 8) / Vt After 4
such replacement operations, X X (Vt 8)/
Vt 4 From the problem, X 16/ (16 65)
16/ 81 Also, since originally the cask was full
of pure wine, X 1 (Vt 8)/ Vt 4 16/
81 gt Vt 24 liters
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