Some Basic concepts
Mixture: Aggregate of two or more than two type of quantities gives us a mixture.
For eg: 1 liter of a mixture contains 250 ml water and 750 ml milk. That means 1/4 of the mixture is water and 3/4 of the mixture are milk. In other words, 25% of the mixture is water and 75% of the mixture is milk.
Mean-Price: The cost price of a unit quantity of the mixture is called the mean price.
Alligation: It is a method of solving arithmetic problems related to mixtures of ingredients. This rule enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of the desired price.
The Alligation method is applied for percentage value, ratio, rate, prices, speed etc and not for absolute values. That is, whenever per cent, per hour, per kg, per km etc., are being compared, we can use Alligation.
Rule of Alligation
1. If two ingredients are mixed, then
- i) (Quantity of cheaper)/(Quantity of dearer) = (C.P. of dearer) - (Mean Price)/ (Mean Price) - (C.P. of cheaper)
(Cheaper Quantity) : (Dearer Quantity) = (d - m) : (m - c)Example: 300 gm of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% in the solution?
Solution: The existing solution has 40% sugar, and sugar is to be mixed; so the other solution has 100% sugar. So, by alligation method;
The two mixture should be added in the ratio 5:1
Therefore, required sugar = 300/5*1 = 60gm
Trick: Quantity of sugar added = (required % value-present % value)/(100 - required% value)
Solution: 300(50-40)/100-50 = 60gms
Example: In what ration must water be mixed with milk to gain 20% by selling the mixture at cost price?
Solution: Let C.P. of milk be Re. 1 per liter
Then, S.P. of 1 liter of mixture = Re.1
Gain obtained = 20%.
C.P. of 1 liter of mixture = Rs. [(100/120)*1] = Rs. 5/6
By the rule of alligation, We have:
The Ratio of water and milk = 1/6 : 5/6
Example: How many kgs. of wheat costing Rs.8 per kg must be mixed with 86 kg of rice costing Rs. 6.40 per kg so that 20% gain may be obtained by Belling the mixture at Rs. 7.20 per kg?
Solution: S.P. of 1 kg mixture = Rs. 7.20, Gain = 20%.
C.P. of 1 kg mixture = Rs.[(100/120)*7.20] = Rs. 6.
By the rule of alligation, we have:
Wheat of 1st kind: Wheat of 2nd kind = 60 : 200 = 3:10.
Let x kg of wheat of 1st kind be mixed with 36 kg of wheat of 2nd kind.
Then, 3 : 10 = x : 36 or 10x = 3 * 36 or x = 10.8 kg.
2. Suppose 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 = [x(1-y/x)n]units.
Example: A Container contains 80 kg. of milk. Form this container, 8kg. of milk was taken out and replaced by water. This process was further repeated two times. How much milk is now contained by the container?
Solution: Amount of milk left = 80[(1-(8/80)3)]kg
= 58.34kg.
3.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 be added = Q liters
Quantity of ingredient in the given mixture = x% of P = x/100*P
Percentage of the ingredient in the final mixture = Quantity of the 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]
The quantity of ingredient to added Q = P * (y-x/100-y)
4. 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 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 be added further be w liters.
Therefore, in new mixture:
Quantity of milk = x * a/(a+b) liters ➝ Eq... (i)
Quantity of water = [x * b/(a+b)] + w liters ➝ Eq... (ii)
➝ c/d = Eq... (i)/ Eq... (ii)
Quantity of water to be added further, w=x* (ad-bc)/c(a+d)
5. 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 n = Pn * x/(Pn + Qn).... and so on
Similarly,
Quantity of Q in vessel 1 = Q1 * x/(P1 + Q1)
Quantity of Q in vessel n = Qn * x/ (Pn + Qn).... and so on
Therefore, when content of all these vessels is mixed in one large vessel, then
Quantity of P/Quantity of Q = Sum of quantities of P in different vessels / Sum of quantities of Q in different vessels Quantity of P / Quantity of Q
= [P1/(P1+Q1) +......Pn/(Pn + Qn)] / [Q1/(P1 + Q1)/ +..... Qn/(Pn + Qn)]
Some frequently asked questions on mixtures and Alligations
# You Will be given the Quantity of ingredients and their price. You will be asked to find the mean price of the resultant mixture.
# You will be given a desired Quantity/Price of a mixture and also the price of ingredients. You have t find out in what ratio/Percentage of ingredients should be mixed to get the desired Quantity.
# Find the resultant quantity when you concentrate/dilute the mixture.
# Find the ratio in which 2 Quantities should be mixed so that the resultant mixture can be sold mixture can be sold to gain x% profit.
Related Article:-
No comments:
Post a Comment