Alligation & Mixture- 2 - Free MCQ Practice Test with solutions, UPSC

The buying price is :
100/110 × 68.20 = 62
Let the amount of tea worth rs 60 be x and the amount of tea worth rs 65 be y
The total amount is : x + y
60x + 65y = 62(x + y)
60x + 65y = 62x + 62y
65y - 62y = 62x - 60x
3y = 2x
3/2 = x / y
3 : 2

Let initially vessel have 8 l of liquid and x lof this liquid be replaced with water then quantity of water in new

Let's assume cost of water is 0,
And the amount of water in mixture be x litres, and amount of milk in mixture be y litres,
So as per question the mixture would be sold a 8 Rs/litre
=> Cost of water per litre times amount of water in mixture of 1 litre + Cost of milk per litre times amount of milk in mixture of 1 litre must be equal to required cost of 1 litre mixture times 1 litre(total mixture quantity)
Mathematically, 0*x + 12*y = 8*1
=> 12 y = 8
=> y=2/3 litre
Now since total mixture is 1 litre,
So x = 1-y
x = 1-(2/3) = 1/3 litre
So ratio of water to milk in mixture, to get cost of required mixture to 1 litre = x/y = (1/3)/(2/3)
=> Ratio of water to milk in mixture is 1:2
 

Let the sugar sold at 8% gain = x
Therefore, Sugar sold at 18% gain = (1000 - x)
Let CP of sugar = Rs.y per kg
Total CP = Rs.100y
Therefore,
(108/100 * xy) + 118/100(1000-x)y
114/100*1000y
= 108xy + 118000y - 118xy = 114000y
10x = 4000
x = 400
Quantity sold at 18% profit
= 1000 - 400 = 600 kg

Let C.P. of 1 litre milk be Re. 1
Then, S.P. of 1 litre of mixture = Re. 1, Gain = 25%.
C.P. of 1 litre mixture = Re.(100/125 * 1) = 4/5
By the rule of alligation, we have

Ratio of milk to water = 4/5 : 1/5
=> 4:5
Hence, percentage of water in the mixture = 1/5 x 100%    
= 20%

Let C.P. of 1 litre milk be Rs. 1
S.P. of 1 litre of mixture = Rs. 1, Gain = 
∴ C.P. of 1 litre of mixture = 
BY the rule of alligation, we have:

∴ Ration of water and milk =