GHAINT QUANT CONCEPTS - | MIXTURE & ALLIGATION | SWAG

Important formulas and shortcuts for Mixtures and Alligations 

1) Rule Of Alligation

Given , Quantity of cheaper ingredient = qc,
Cost price of cheaper ingredient = pc,
Quantity of dearer or costlier ingredient = qd,
Cost price of costlier or dearer ingredient = pd.
Consider, mean price of mixture as pm and quantity of mixture as qm.
We know, qm = qc + qd
Then we get,
(qc * pc + qd * pd) = qm * pm = (qc + qd) * pm
→ qc ( pm – pc) = qd (pd – pc)
→ qc / qd = (pd – pc) / ( pm – pc)

2) Quantity of ingredient to be added to increase the content of ingredient in the mixture to y%

If P liters of a mixture contains x% ingredient in it. Find the quantity of ingredient to be added to
increase the content of ingredient in the mixture to y%.
Let the quantity of ingredient to be added = Q liters
Quantity of ingredient in the given mixture = x% of P = x/100 * P
Percentage of ingredient in the final mixture = Quantity of ingredient in final mixture / Total quantity of
final mixture.
Quantity of ingredient in final mixture = [x/100 * P] + Q = [ P*x + 100 * Q] / 100
Total quantity of final mixture = P + Q
→ y/100 = [[ P*x + 100 * Q] / 100]/[P + Q]
→ y[P + Q] = [P*x + 100 * Q]

3) If n different vessels of equal size are filled with the mixture of P and Q

If n different vessels of equal size are filled with the mixture of P and Q in the ratio p1 : q1, p2 : q2, ……,
pn : qn and content of all these vessels are mixed in one large vessel, then
Let x liters be the volume of each vessel,
Quantity of P in vessel 1 = p1 * x / (p1 + q1)
Quantity of P in vessel 2 = p2 * x / (p2 + q2)
Quantity of P in vessel n = pn * x / (pn + qn)... and so on
Similarly,
Quantity of Q in vessel 1 = q1 * x / (p1 + q1)
Quantity of Q in vessel 2 = q2 * x / (p2 + q2)
Quantity of Q in vessel n = qn * x / (pn + qn)... and so on
Therefore, when content of all these vessels are mixed in one large vessel, then

4) If n different vessels of sizes x1, x2, …, xn are filled with the mixture of 

P and Q

 

If n different vessels of sizes x1, x2, …, xn are filled with the mixture of P and Q in the ratio p1 : q1, p2 : q2,……, pn : qn and content of all these vessels are mixed in one large vessel, then
Quantity of P in vessel 1 = p1 * x1/(p1 + q1)
Quantity of P in vessel 2 = p2 * x2/(p2 + q2)
Quantity of P in vessel n = pn * xn/(pn + qn)... and so on
Similarly,
Quantity of Q in vessel 1 = q1 * x1/(p1 + q1)
Quantity of Q in vessel 2 = q2 * x2/(p2 + q2)
Quantity of Q in vessel n = qn * xn/(pn + qn)
Therefore, when content of all these vessels are mixed in one large vessel

5) Quantity of ingredient to be added to change the ratio of ingredients in a mixture

In a mixture of x liters, the ratio of milk and water is a : b. If the this ratio is to be c : d, then the quantity
of water to be further added is:
In original mixture
Quantity of milk = x * a/(a + b) liters
Quantity of water = x * b/(a + b) liters
Let quantity of water to be added further be w litres.
Therefor in new mixture:
Quantity of milk = x * a/(a + b) liters → Equation(1)
Quantity of water = [x * b/(a + b) ] + w liters → Equation (2)
→ c / d = Equation (1) / Equation (2)

EXAMPLES

1. In what portion must rice at Rs. 3.10 per kg be mixed with rice at Rs. 3.60 per kg, so that the
mixture be worth Rs. 3.25 a kg? 

 Cheaper(310)--------------------------Dearer(360paise)
-----------------------mean(325)------------------------
-360-325=35-------------------------------325-310=15
Therefore, by rule of allegation :-
Quantity of cheaper/quantity of dearer = 35/15 = 7/3
Hence, required ratio = 7:3

2. Two types of oils having the rates of Rs. 4/kg and Rs.5/kg respectively are mixed in order to
produce a mixture having the rate of Rs. 4.60/kg. What should be the amount of the second type of oil if the amount of the first type of oil in the mixture is 40 kg? 

Cheaper(400)-------------------------Dearer(500)
---------------------Mean(460)-----------------------
-----40----------------------------------------60-------
Hence, by the rule of allegation they must be mixed in the ratio 40:60
if quantity of first type is 40kg , the quantity of second type will be 60 kg

3. How many kilograms of sugar worth Rs. 3.60 per kg should be mixed with 8 kg of sugar worth Rs.4.20 per kg , such that by selling the mixture at Rs. 4.40 per kg there maybe a gain of 10%? 

Gain = 10%
SP = Rs. 4.40/kg = 110% of CP
CP or Mean price = Rs. 4/kg
Now, Acc to allegation method:-
Cheaper(360)---------------Dearer(420)
------------------Mean(400)----------------
-----20------------------------------40-------
-----1-------------------------------2-----------
Ratio = 1:2
Required Quantity = 1:2 : : X:8
or X = 4kg

4. A mixture of 125 gallons of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture?

By applying the rule of allegation :-
20%-----------100%
---------25%----------
75-------------------5
Required ratio = 75 :5 or 15:1
Required amount of water = 1*125/15 = 8.33 gals
Direct formula :-
Water to be added Q = Initial mixture* (y-x)/100-y
y = required concentration
x= initial concentration
Q= 125*(25-20)/100-25 = 125*5/75 = 125/15
or Q = 25/3 = 8.33 gals

5. Milk and water are mixed in a vessel A in the ratio 5 : 3 and in vessel B in ratio 9 : 7 . In what ratio should quantities be taken from the two vessels so as to form a mixture in which milk and water will be in the proportion of 7 : 5 ?

In vessel A, milk = 5 / 5 + 3 = 5 / 8 of the weight of mixture
In vessel B milk = 9 / 9 + 7 = 9 / 16 of the weight of mixture
Now, we have to form a mixture in which milk be 7 / 12 of the weight of the mixture.
Now, acc to rule of allegation:-
5/8-------------------9/16
-----------7/12------------
1/48-------------------1/24
Required ratio = 1/48:1/24 = 1:2

6.A milk vendor has 2 cans of milk .The first contains 25% water and the rest milk. The second
contains 50% water. How much milk should he mix from each of the container so as to get 12 litres of milk such that the ratio of water to milk is 3:5?

Milk in 1st can = 75% = 3/4
Milk in 2nd can = 50% = 1/2
Milk in new Can = 5/8
Acc to allegation rule :-
3/4----------------------1/2
------------5/8---------------
1/8------------------------1/8 or
1-----------------------------1
Milk from each can = 1:1
Total milk = 2
Milk from Can 1 = 1*12/2 = 6 liters
Milk from can 2 = 1*12/2 = 6 liters

7.If 4 different vessels of equal size are filled with mixtures of A and B in the ratio 5:3, 2:3, 3:5, 7:9 and content of all these vessels are mixed in one large vessel. Then what is the ratio of A and B in the new vessel. 

If n different vessels of equal size are filled with the mixture of P and Q in the ratio p1 : q1, p2 : q2, ……,
pn : qn and content of all these vessels are mixed in one large vessel, then

Therefore, from above formula we have :-
A/B = (5/8)+ (2/5)+(3/8)+(7/16)/ (3/8)+(3/5)+(5/8)+(9/16)
A/B = (1/80)[50+32+30+35] / (1/80)[30+48+50+45]
A/B = 147/173
hence, ratio in the new vessel = 147:173

8. In what ratio must a person mix three kinds of wheat costing him Rs 1.20, Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? 

we can do it using allegation method twice first finding the ratio between kind A and kind B and then
between kind B and Kind C or kind A and Kind C .
Kind A cost = 120 paise
Kind B cost = 144 paise
Kind C cost = 174 paise
Mean cost = 141 paise
Now, Ratio between A and B
120-------------------144
-----------141-------------
3------------------------21 or
1-------------------------7
Therefore ratio between A and B = 1:7
Now, Case II we will find ratio between B and CI
144----------------------174
-------------141--------------
33--------------------------3
11--------------------------1
Hence, Ratio between B and C = 11:1
Now , A:B = 1:7 and
B:C = 11:1
A:B = 1:7 or 11:77
B:C = 11:1 or 77:7
Hence, A:B:C = 11:77:7

9. A man buys cows for Rs. 1350 and sells one so as to lose 6% and the other so as to gain 7.5% and on the whole he neither gains nor loses. How much does each cow cost?   

Using allegation method :-
6-----------------------------7.5
---------------0------------------
7.5------------------------------6
5---------------------------------4
Cost of I and II are in the ratio = 5:4
Therefore, Cost of I cow = 5*1350/9 = 750
Cost of II cow = 4*1350/9 = 600

10. There are 65 students in a class, 39 rupees are distributed among them so that each boy gets 80p and each girl gets 30p. Find the number of boys and girls respectively in the class.  

3900/65 = 60

Boys---------------------Girls
80-------------------------30
-----------60-----------------
30-----------------------20
3--------------------------2
65*3/5 = 39 boys
65*2/5 = 26 girls

THANK YOU :)