Quantitative Aptitude
ALLIGATION MCQs
Let C.P. of 1 litre milk be Re. 1.
S.P. of 1 litre of mixture = Re.1, Gain = \(\frac{50}{3}\) %
So, C.P. of 1 litre of mixture = \(\left(100\times\frac{3}{350}\times1\right)=\frac{6}{7}\)
By the rule of alligation, we have:
C.P. of 1 litre of water = 0
C.P. of 1 litre of milk = Rs.1
Main Price = \(\frac{6}{7}\)
So, Ratio of water and milk = \(\frac{1}{7}:\frac{6}{7}=1:6.\)
By the rule of alligation:
Cost of 1 kg of 1st kind = 720 p
Cost of 1 kg of 2nd kind = 570 p
Main Price = 630 p
So, Required ratio = 60 : 90 = 2 : 3.
S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%.
C.P. of 1 kg of the mixture = Rs. \(\left(\frac{100}{110}\times68.20\right)=Rs. 62.\)
By the rule of alligation, we have:
Cost of 1 kg tea of 1st kind = Rs. 60
Cost of 1 kg tea of 2nd kind. = Rs. 65
Main Price = Rs, 62
So, Required ratio = 3 : 2.
Let the price of the mixed variety be Rs. x per kg.
By rule of alligation, we have
Cost of 1 kg of Type 1 rice = Rs.15
Cost of 1 kg of Type 2 rice = Rs. 20
Main Price = Rs. x
So, \(\left(\frac{20-x}{x-15}\right)=\frac{2}{3}\)
60 - 3x = 2x - 30
5x = 90
x = 18.
Let the quantity of the wine in the cask originally be x litres.
Then, quantity of wine left in cask after 4 operations = \(\left[x\left(1-\frac{8}{x}\right)^{4}\right]liters\)
So, \(\left(\frac{x\left(1-\frac{8}{x}\right)^{4}}{x}\right)=\frac{16}{81}\)
\({\left(1-\frac{8}{x}\right)^{4}}=\left(\frac{2}{3}\right)^{4}\)
\({\left(\frac{x-8}{x}\right)}=\frac{2}{3}\)
3x - 24 = 2x
x = 24
By the rule of alligation, we have:
Profit on 1st part = 8%
Profit on 2nd part = 18%
Main Profit = 14%
Ration of 1st and 2nd parts = 4 : 6 = 2 : 3
So, Quantity of 2nd kind = \({\left(\frac{3}{5}\times1000\right)}kg=600 kg.\)
According to the question,
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Acid}}\,\,\,\,\,\,{\text{Water}} \cr
& {\text{Vessel A}}\,\,{\text{4}}\,\,\,:\,\,\,\,{\text{3}} \cr
& {\text{Vessel B}}\,\,\,2\,\,\,\,{\text{:}}\,\,\,\,\,{\text{3}} \cr} $$
Now using alligation,