154E-ALLEGATIONS AND MIXTURE

Short Solutions:

It is the rule that is used to determine the mean value of the mixture when the prices of the individual items being mixed together and the proportion in which they are being mixed are given. Here, the value of the mixture is always higher than the lowest value and lower than the higher value of the items being mixed.

According to the Rule of Alligation:

Quantity of cheaper = Price of dearer - Mean price

Quantity of dearer   = Mean price - Price of cheaper

It can be also expressed as,

CP of 1 unit of cheap item(x)				CP of 1 unit of dearer item(y)
	↖		↗
		Mean Price(m)
	↙		↘
(y-m)				(m-x)

therefore,  (Cheaper quantity : Dearer quantity) = (y – m) : (m – x)
Where, mean price (m) is the cost price of a unit quantity of the mixture.
Also, if a container contains x units of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid is [x(1 - y/x)ⁿ] unit.

Example 1: How many kilograms of rice costing Rs 18 per kg must be mixed with 30 kg of rice costing Rs 14 per kg, so that the resultant mixture cost Rs 15 per kg.

Solution. Applying the rule of alligation, we have

CP of 1 unit of cheap item(x) = 14				CP of 1 unit of dearer item(y) = 18
	↖		↗
		Mean Price(m) = 15
	↙		↘
(y-m) = 18-15 = 3				(m-x)= 15-14= 1

therefore  = Quantity of cheaper rice = 3,   Quantity of dearer rice          1        

If cheaper rice is 3 kg, dearer rice is 1 kg.

If cheaper rice is 30 kg, dearer rice = (30 x 1) kg = 10 kg

                                                       3        

Example 2: In what proportion must a person mix rice Rs 12.00 per kg and Rs 14.40 per kg so as to make a mixture worth Rs 12.60 per kg?

Solution.  By the alligation rule,

CP of 1 unit of cheap item(x) = 12.00				CP of 1 unit of dearer item(y) = 14.40
	↖		↗
		Mean Price(m) = 12.60
	↙		↘
(y-m) = 14.40-12.60 =1.80				(m-x)= 1260-12.00 = 0.60

Quantity of cheaper rice = 180 = 3, Quantity of dearer rice = 60 = 1

therefore, He must mix rice in the ratio 3 : 1.

Example 3: In what proportion must water be mixed with milk to gain 20% by selling it at cost price?

Solution. Let CP of milk = Rs. 1 per litre

Therefore, SP of 1 L of mixture = Rs. 1, Profit = 20%

Therefore, CP of 1 L mixture = 1/120 x 100 = Rs. 5/6

CP of 1 litre water (x) = 0				CP of 1 litre of milk (y) = 1
	↖		↗
		Mean Price(m) = 5/6
	↙		↘
(y-m) = 1-5/6 =1/6				(m-x)= 5/6- 0= 5/6

Therefore, Required ratio = 1 : 5 = 1 : 5

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Question 7

If a mixture of alcohol and water in the proportion of 5:1 is sold at Rs.100 and the seller gains a profit of Rs.25, find the price of pure alcohol.
a) 80
b) 125
c) 90
d) 100
None of above

Question 8

Vessel ‘A’ contains 150 litres of mixture of milk and water in the ratio of 3:1. Vessel ‘B’ contains 200 litres of the mixture in the ratio of 5:2. If the mixture from both the vessels are poured into vessel ‘C’, find the ratio of the resulting mixture.
a) 15 : 7
b) 8 : 3
c) 15 : 2
d) 143 : 53
None of above

Question 9

A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. None of above

Question 10

An ore contains 5% of aluminium. To get 52 Kg aluminium, the quantity of the ore required is
A.520 kg
B.1040kg
C.780kg
D.1560kg
E.260kg

Question 11

A bottle is full of dettol. One third of it is taken out and then an equal amount is poured into the bottle to fill it. This operation is done four times. Find the final ratio of dettol and water in the bottle
A.49:65
B.65:16
C.9:16
D.16:65
E.9:25

Question 12

A DISHONEST MILKMAN PROFESS TO SELL HIS MILK AT COST PRICE BUT HE MIXES IT WITH WATER AND THEREBY GAIN 25%. THE PRECENTAGE OF WATER IN THE MIXTURE IS
A. 20
B. 25
C. 16
D. 65
E. 10