Mixtures
(Alligation)
The problems on this topic generally involve
the theory of ratio & proportion and at times basics of
percentages and profit & loss.
Types of Mixture
There are two types of mixtures: (i)
Simple, (ii) Compound
Simple Mixture: When two different
ingredients are mixed together,
it is known as a simple mixture.
Compound Mixture: When two or more simple mixtures are mixed together to form another
mixture, it is known as a compound mixture.
Alligation Rule
This rule enables
us to find the proportion in which two or more ingredients at the given price must be
mixed to produce a mixture at
a given price.
Statement: The rule states
that when different quantities of different costs (values or strengths) are mixed
together to produce
a mixture of mean cost (value or strength), the ratio of the quantities are
inversely proportional to the differences in their costs from the mean cost (value or strength).
The CP of the
item that is cheaper is given by CP cheaper. The CP of the item that is
costlier (dearer) is given by CP dearer. The C.P. of unit quantity of
the final mixture is called the MEAN PRICE and is given by CP mean price•
Quantitycheaper =CPdearer — CPmean price
Quantltydearer CPmean price — CPcheaper
Applications of the Rule
(i)
The Alligation Rule is used to find the mean value of a
mixture when the prices
of two or more ingredients, which are mixed together and the proportion in which they are
mixed are given.
(ii)
It is also used to find the proportion in which the
ingredients at
given prices must be mixed to produce a mixture at a given price.
E31.In what
ratio should tea @ 35 per kg be mixed with tea @ 27 per kg so that mixture may
cost Rs.30 per kg?
Sol. Quantity of cheaper = 35 - 30 = 5
Quantity of dearer 30-27 3
Hence the two should be
mixed in the ratio 3 : 5. The above ratio calculated is 3 : 5 and not 5 : 3, i.e., tea @ 35
per kg to have 3 parts and tea @
27 per kg to have 5 parts. This may be checked mentally by the following simple rule :
"If mean
price is closer to cheaper C.P, then quantity of cheaper will be more and vice versa."
E32.In what ratio should two different types of mixtures containing milk and water
in the ratio of 5 : 1 and 2 : 1 respectively be mixed to obtain a final mixture
containing milk and water in the ratio
of 3 : 1?
Sol. Strength
of milk in mixture 1 = 5/6 (because there are 5 parts
of milk for every 1 part of water. Hence there are 5 parts
of milk for a total of 6 parts of the mixture). Strength
of milk in mixture 2 = 2/3 (because there are 2 parts
of milk for every 1 part of water. Hence there are 2 parts
of milk for a total of 3 parts of the mixture). Strength
of milk in the final mixture required = 3/4 (because there
are 3 parts of milk for every 1 part of water. Hence there
are 3 parts of milk for a total of 4 parts of the mixture). As we
can see, the mixture 1 is stronger in milk as compared to
mixture 2, hence we will refer to mixture 1 as the dearer mixture,
mixture 2 as the cheaper mixture and the final mixture as the mean
mixture.
5 - 3
Quantity of cheaper = 6 4 = 1
Quantity of dearer 3 - 2
4 3
Hence, both the types of mixtures will have to be mixed in
the
ratio of 1 : 1.
Two mixtures of same ingredients mixed (Compound mixture)
Mixture 1 has ingredients (A and
B) in a : b. quantity of ingredient A = a
quantity of ingredient B b
Mixture 2 has same
ingredients (A and B) in x : y. quantity
of ingredientA = x
quantity of ingredientB y
Now 'M' units of Mixture 1 and
'N units of Mixture 2 are mixed to form a
resultant mixture with ingredients (A and B) in the ratio qA
: qB .
Case I. When qA and qB are
to be found out. By alligation rule in the
resultant mixture,
Quantity of ingredient A =
qA =
M ( a/a+b ) + N (x/x+y)
Quantity of ingredient B
qB M
(b/a+b) + N (y/x+y)
Then, amount of ingredient A in the resultant mixture.
= qA x
(M +N)
(qA
+ qB)
Case II. When M and N are to be
found out (i.e. amount of mixtures)
Consider the
quantity of any one ingredient (say, A) in all the three mixtures, i.e. quantity of ingredient A in
In mix. 1 = a / a+b , i.e. a out
of (a+b).
In
Mix 2 = x/x+y , in the resultant
mix. = qA / qA + qB
By alligation rule we get, in the resultant mixture, Quantity
of 1st mix. = Q 1= (x/ x+y) – (qA/qA + qB)
Quantity of 2nd mix.
Q2 (qA/qA + qB)
– (a/a+b)
Then, amound of 1st mixture in
the resultant mixture
= Q1 x(M +N)
(Q1 +
Q2)
Amount of second mix. In the resultant mix.
= Q2 x(M +N)
(Q1
+ Q2)
E33. In two alloys, the ratio of
zinc to tin is 5 : 2 and 3 : 4. If 7 kg of the first alloy and 21 kg of the
second Amount of 2nd mixture in the resultant mixture alloy are mixed together to form a new alloy, then what will be
the ratio of zinc and tin in the new
alloy?
Sol. Quantity of X = Quantity of Zinc.
Quantity of Y = Quantity of Tin.
M=7kg N=
21 kg
a=5 b=2
x=3 y=4
Putting
these values in the formula, we get
Quantity of zinc = 7 x (5/5+2) + 21 x (3/3+4) = 14 = 1: 1
Quantity of tin 7 x ( 2/5+2) + 21 x (4/3+4) 14
Removal and Replacement by Equal Quantity
If a vessel contains
"a" litres of petrol, and if "b" litres be withdrawn and replaced by kerosene, then if
"b" litres of the mixture be withdrawn and replaced by kerosene, and the operation repeated 'n' times in all, then
Petrol left in vessel after nth operation = (a - b)n
Initial quantity of Petrol in vessel a
Note that
in the denominator the term "Initial quantity of Petrol in vessel" is equal to the
"Volume of mixture in the vessel after the nth operation."
E34.A
tea merchant buys two kinds of tea, the price of the first kind being twice
that of the second. He sells the mixture at
Rs.14/kg thereby making a profit of
40%• If the ratio of the first to second
kind of tea in the mixture is 2 : 3, then find the
cost price of each
kind of tea.
Sol. The cost of mixture = 14x100 = Rs.10/kg.
140
Ratio in which the cheaper and dearer is
mixed = 3:2.
Let the price of cheaper tea be Rs.x/kg and
dearer tea be Rs.2x/kg.
Applying the alligation rule, we get
3 = 2x -10
2 10 - x
30-3x=4x-20
.•. 7x=50
.•. x=71/7
Cost of cheaper tea = Rs. 71/7 and cost of dearer tea= Rs. 14 2/7.
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