mix and allig

Contents

1. INTRODUCTION
2. USING WEIGHTED AVERAGES
3. RULE OF ALLIGATION
   1. ALLIGATION CROSS
   2. ALLIGATION LINE
4. SUCCESSIVE REPLACEMENT

Mixtures and Alligation

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INTRODUCTION

Two or more items are mixed together to form a mixture. Mixture problems involve finding the average attribute (cost, concentration, percentage, etc.) of the resulting mixture. Sometimes different liquids are mixed together to get a desired concentration of the mixture. Sometimes two or more items of different costs are mixed together. The concept of simple averages can be used to solve problems when two or more items are mixed in the same quantity. The concept of weighted averages is used to solve problems when different weights of two or more items are mixed.

USING WEIGHTED AVERAGES

Let’s look at an example to understand how weighted averages are used to solve mixture problems. If a milkman mixes 2 litres of milk costing Rs. 6 per litre and 1 litre of milk costing Rs. 9 per litre, then the per litre cost of the mixture can be calculated using weighted averages.

Here the attribute is the cost and the weight is the quantity of milk.

Note: If equal weights of the milks are mixed, the price of the mixture will be Rs. 7.50 per litre. This can be calculated by finding the simple average.

IMPORTANT:

• If 2 litres of milk costing Rs. 6 per litre and 1 litre of milk costing Rs. 9 per litre are mixed together, the price of the resultant mixture is less than Rs. 7.50. This is because the quantity of milk costing Rs. 6 per litre is more than the quantity of milk costing Rs. 9 per litre. Hence, the price of the mixture will be closer to the price of the milk with more quantity. In such a case we use the concept of weighted average to find the value of the attribute of the resultant mixture.

• The formula for weighted averages where two items are mixed can be represented as follows:

where x is the weighted average

and are the attributes (for example, cost, marks of students etc.) and

and are the weights (for example, weight in kilos/litres, number of students etc.)

REMEMBER:

• The attributes should all be of the same unit. Similarly, the weights should also be of the same unit.

Example 1:

Three varieties of premium basmati rice costing Rs. 60 per kg, Rs. 70 per kg and Rs. 90 per kg are mixed together in the ratio of 2 : 4 : 1 respectively. Find the cost of the resultant mixture. Solution: Cost of resultant mixture

IMPORTANT:

• In the above example, if the ratio of the weights of rice is changed to 2 : 5 : 1 or for that matter to 2 : n : 1 for any non negative value of n, the answer will remain same. This is because the weighted average is the same as the attribute (price) of the second variety of rice.

• When n items are mixed, the formula can be written as follows:

REMEMBER:

• Attribute and weight are a function of the items that are mixed. For example, if two acids have to be mixed, the attribute can be the concentration of acids and the weight can be the volumes of acids. If the marks of two groups of students are mixed, average marks of the two groups will be the attribute and the number of students in the two groups will become the weight.

Example 2:

A shopkeeper purchased 4 quintal tea at the rate of Rs. 110 per kg, 2 quintal tea at the rate of Rs. 140 per kg and 4 quintal tea at the rate of Rs. 120 per kg. He mixed the three varieties of tea. At what selling price should he sell the final mixture of tea to get a profit of 20%? Solution: Cost of resultant mixture Selling price of the mixture = cost price + 20% profit = 1.2 120 = Rs. 144 per kg

REMEMBER:

• The weighted average can be calculated even if the ratio of the weights is given instead of the absolute value of the weights.

Example 3:

A goldsmith mixes two types of alloys. He takes 6 kg of the first alloy containing gold and silver in the ratio 3 : 2 and 18 kg of the second alloy containing gold and silver in the ratio 2 : 3. What is the ratio of gold and silver in the final alloy mixture? Solution: Here the attribute is the average proportion of gold in the alloy mixture and weight is the quantity of the alloys. The proportion of gold in the first alloy is 3/5 and the proportion of gold in the second alloy is 2/5. The ratio of the weights of the two alloys is 6 : 18 = 1 : 3. 9 of 20 parts are gold. Hence, the other 11 of 20 parts are silver. Ratio of gold to silver in the final alloy mixture = 9 : 11

RULE OF ALLIGATION

The basic concept of alligation is the same as that of mixtures, i.e. weighted average. Alligation helps in finding the ratio in which two weights have to be mixed to get a given consistency of the mixture.

Example 4:

How much sugar costing Rs. 6 per kg must be mixed with 30 kg of sugar costing Rs. 9 per kg, so that the resultant mixture costs Rs. 7 per kg? Solution: Here the attribute is the cost of sugar per kg. w1 and w2 are the weights of varieties of sugar that cost Rs. 6 per kg and Rs. 9 per kg respectively. Here, x1 = 6 x2 = 9 w2 = 30 Using the formula for mixtures, we get, Cost of resultant mixture

IMPORTANT:

• We know that the average attribute is closer to the attribute of the item having more weight. Here, in the above example, the difference between the first attribute and the average attribute is 1 and the difference between the second attribute and the average attribute is 2; thus, the ratio of the weights of the first and the second item should be 2 : 1 i.e. inversely proportional to the ratio of the difference of attributes from average attribute. Hence, the weight of the first variety of sugar will be 60 kg. This is the basic idea behind the concept of alligation.

• Let us modify the formula of mixtures (for 2 items) to get the generalised formula for alligation:

In words, the alligation rule can be written as:

“If two items with different attributes are mixed together to get a resultant mixture with an average attribute, the ratio of the weights of the two items mixed will be inversely proportional to the deviation of attributes of these two items from the average attribute of the resultant mixture.”

In the relation given above, if x₁ < x < x₂, (x₂ x) is the deviation of the second attribute from the average attribute and (x x₁) is the deviation of the first attribute from the average attribute.

If we use the alligation rule to solve Example 4, we get,





Hence, w₁ = 2 w₂ = 2 30 = 60 kg.

REMEMBER:

• The alligation rule is only applicable when two items are mixed together.

PICTORIAL REPRESENTATION OF ALLIGATION RULE

ALLIGATION CROSS

The alligation rule can be represented and applied in a pictorial way (alligation cross) as follows:

We draw the above representation such that x₁ < x₂. Consequently, according to the principles of averages, x₁ < x < x₂.

Example 4 can be solved using the alligation cross as follows:

CASE 1: When x₁, x₂ and x are given, and the ratio of the weights w₁ : w₂ is asked OR when x₁, x₂, x and either of w₁ or w₂ are given, and the other is asked. (Example 4 was of Case 1)

Example 5:

32 litres of milk and water solution contains 84% milk. How much water should be added to this solution to reduce its concentration to 64%? Solution: Concentration of milk in the first solution = 84% Concentration of milk in the second solution = 0% [pure water contains 0% milk] Using the alligation cross, Ratio of the two solutions = 64 : 20 = 32 : 10 Since the weight of the first solution is 32 litres, hence the weight of second solution (pure water) is 10 litres.

Example 6:

Two solutions contain petrol and diesel in the ratio 2 : 3 and 3 : 7. In what ratio should the two solutions be mixed so that the ratio of the petrol and the diesel in the final mixture is 7 : 13? Solution: We can consider the proportions of petrol or diesel in the solutions to find the ratio in which the solutions are mixed. Let us take into consideration the proportion of petrol in the petrol-diesel solution. Proportion of petrol in the first solution = 2/5 = 8/20 Proportion of petrol in the second solution = 3/10 = 6/20 Proportion of petrol in the final mixture = 7/20 Using the alligation cross, Ratio of weights of two solutions = 1 : 1

Example 7:

When Charlie visited a chocolate factory, he stole from a tin-can some chocolate milkshake that had a 75% concentration of chocolate. He replaced this amount with a cheaper brand of milkshake that had only 25% chocolate concentration. Later, during their routine quality analysis, the authorities realized that the tin-can in question had only 40% chocolate concentration. What percentage of chocolate milkshake did Charlie steal from the tin-can? Solution: Consider the contents of the tin-can after the replacement. If we take the entire contents of the tin-can to be 1, and x to be the fraction of the stolen milkshake (which is equal to the fraction of 15% milkshake in the tin-can); then (1 – x) will be the fraction of the 75% milkshake still present in the tin-can. Using the alligation cross, 3x = 7 – 7x x = 7/10 = 0.7 Hence, Charlie stole 70% of the milkshake present in the can.

Example 8:

Sumit works as a state contractor for PWD and supplies bitumen mix for road construction. He has two varieties of bitumen, one at Rs. 42 per kg and the other at Rs. 25 per kg. How many kg of the first variety must Sumit mix with 25 kg of the second variety, so that he may, on selling the mixture at Rs. 40 per kg, gain 25% on the outlay? [IIFT 2007] (1) 30(2) 20 (3) 25(4) None of these Solution: The man wants to sell the mixture at Rs. 40/kg and make a profit of 25%. Hence, we have, Hence, option 4.

CASE 2: When x₁, x₂, w₁, and w₂ are given, and the average attribute x asked.

Example 9:

A class of 15 students got an average of 50 marks in an exam, while another class of 30 students got an average of 44 marks in the same exam. If all the students are combined into one class, then what will be the average marks of that class in the exam? Solution: 50 x = 2x – 88 3x = 138 x = 46 Hence, the average of the new class will be 46 marks.

CASE 3: When x₂ (or x₁), w₁, w₂ and x are given, and x₁ (or x₂) is asked.

Example 10:

When 7 litres of milk at Rs. 30 per litre is mixed with 3 litres of another brand of milk, the resultant mixture costs Rs. 23 per litre. What is the cost (per litre) of the 3-litre milk brand? Solution: 69 – 3x1 = 49 x1 = 20/3 = 6.67 Hence, the 3-litre brand of milk costs Rs. 6.67 per litre.

Conclusions: It is apparent that this method is only useful for case 1. There is no point in using Alligation cross for cases 2 and 3, since after the diagram is drawn it just results in the writing of the alligation rule equation. Hence, for cases 2 and 3 we will use another method – the Alligation Line method.

ALLIGATION LINE

Re-reading the note at the end of Example 4 will shed some light on the following method. We read there that the ratio of the weights of two items is inversely proportional to the ratio of the difference of their attributes from the average attribute.

The alligation rule can be represented and applied in the Alligation Line method as follows:

(Here again, we consider x₁ < x <; x₂)

So, according to what we read above,

Thus,

w₁ corresponds to (x₂ – x)… (i)

w₂ corresponds to (x – x₁)… (ii)

and

(w₁ + w₂) corresponds to (x₂ – x₁)… (iii)

Now let’s again solve Examples 8 and 9 using the Alligation Line method.

Example 9:

Alternate method of solution: A class of 15 students got an average of 50 marks in an exam, while another class of 30 students got an average of 44 marks in the same exam. If all the students are combined into one class, then what will be the average marks of that class in the exam? Solution: From Rule (iii), we have, (30 + 15 = 45) corresponds to (50 – 44 = 6) From Rule (i), we have, 30 corresponds to (50 – x) 50 – x = 4 x = 46 Hence, the average of the new class was 46 marks.

Example 10 :

Alternate method of solution: When 7 litres of milk at Rs. 30 per litre is mixed with 3 litres of another brand of milk, the resultant mixture costs Rs. 23 per litre. What is the cost (per litre) of the 3-litre milk brand? Solution: From Rule (i), we have, 3 corresponds to (30 – 23 = 7) From Rule (ii), we have, 7 corresponds to (23 – x1) 69 – 3x1 = 49 x1 = 20/3 = 6.67 Hence, the 3-litre brand of milk costs Rs. 6.67 per litre.

Example 11:

An artist bought two cans of equal capacity filled with paint to their brims. The first can contained a mixture of red and green paint in the ratio 5 : 3, while the second can contained a mixture of green and red paint in the ratio 9 : 7. If he mixes the two paints together in a new larger can, then what will be the ratio of the two paints (red : green) in that can? Solution: The ratios of red to green paint in the first and second can are 5 : 3 and 7 : 9 respectively. Now, let us take into consideration the proportion of red paint in the mixture: Proportion of red paint in the first mixture = 5/8 Proportion of red paint in the second mixture = 7/16 Using the alligation line, From Rule (iii), we have, From Rule (i), we have, Proportion of Red paint in the new mixture is 17/32. So, proportion of Green paint in the new mixture is 1 – 17/32 = 15/32 Hence, ratio of red to green paint in the new mixture is 17 : 15.

Example 12:

A man in charge of maintaining the pH level of a swimming pool is given two bottles (not necessarily of the same capacity) of chlorine; the first with a 74% concentration and the second with a 52% concentration. When he mixes the two liquids together, he gets 66 ml of a solution that has 65% chlorine concentration. What was the quantity of chlorine in the second bottle? Solution: Using the alligation line, From Rule (iii), we have, (w1 + w2) corresponds to 74 – 52 = 22 From Rule (i), we have, w1 also corresponds to 65 – 52 = 13 Now, (w1 + w2) is given to be 66 ml. Hence, quantity of chlorine in the second bottle, w2 = 66 – 39 = 27 ml

SUCCESSIVE REPLACEMENT

Let us take an example to understand the concept of successive replacement.

If a vessel has x litres of milk and y litres of milk is removed from it and replaced with y litres of water, the milk left in the vessel after the replacement is (x y) litres and the total milk-water solution available in the vessel is again x litres (x y + y). So after the replacement, the proportion of milk in the milk-water solution will be equal to:

For example, if 10 litres of milk is removed from 100 litres of milk and replaced with 10 litres of water, the proportion of milk in the resultant milk-water solution will be (100 10)/100 or 90/100 or 0.9.

The percentage of milk in the resultant milk-water solution will be 0.9 100 or 90%.

If the process is repeated, i.e. if 10 litres of this 90% milk-water solution (which contains 90 litres of milk and 10 litres of water) is removed (9 litres milk and 1 litre water will be there in the 10 litres milk-water solution removed) and replaced with 10 litres of water, the proportion of milk left in the vessel after the second iteration will be (90 9)/100 or 0.81 or 81% (which is 90% of 90%).

This can be generalized as,

where x is the original quantity,

y is the quantity that is replaced, and

n is the number of times the replacement process is repeated.

Example 13:

A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he has sold. What is the current proportion of water to milk? [CAT 2004] (1) 2 : 3(2) 1 : 2 (3) 1 : 3(4) 3 : 4 Solution: The quantity of milk and water is as shown in the table. Current proportion of water and milk is 40 : 60 = 2 : 3 Hence, option 1.

Example 14:

9 litres out of 90 litres milk are replaced with an equal quantity of water. Again 9 litres of the resultant milk water solution are replaced with an equal quantity of water. Find the final concentration of milk in the solution after the second iteration. Solution:

Example 15:

A petrol tank at a filling station has a capacity of 400 litres. The attendant sells 40 litres of petrol from the tank to one customer and then replenishes it with kerosene oil. This process is repeated with six customers. What quantity of pure petrol will the seventh customer get when he purchases 40 litres of petrol? [IIFT 2009] (1)20.50 litres(2)21.25 litres (3)24.75 litres(4)22.40 litres Solution: where x is the original quantity, y is the quantity that is replaced, and n is the number of times the replacement process is repeated. At the end of 6 replacements, each litre contains The customer who purchases 40 litres of petrol gets 40 0.53 21.2 litres of pure petrol. Hence, option 2.

Example 16:

A vessel is completely filled with petrol. 10 litres is drawn from this vessel and replaced with kerosene. 10 litres of the petrol kerosene mixture is again drawn from the vessel and replaced with kerosene. After the second iteration, the ratio of petrol and kerosene in the vessel is 49 : 32. Find the capacity of the vessel. Solution: 7x = 9x – 90 x = 45 litres Capacity of the vessel = Initial quantity of the petrol = x = 45 litres

Example 17:

A vessel is completely filled with a milk and water solution. The capacity of the vessel is 42 litres. 6 litres of this solution is replaced with pure water. The new concentration of milk in the milk-water solution is 30%. What was the concentration of milk in the original solution? Solution: Let the original concentration of milk be x%. The concentration of milk after the replacement is 30%. x = 35%

Example 18:

A dishonest employee steals 500 litres of pure 100% wine from a tank at the vineyard he works for and replaces the same with water. His friend notices him and he too yields to temptation and steals the same amount, again replacing it with water. A few days later, a routine check showed that the wine was not pure and that the ratio of wine to water in the tank was 121 : 23. The furious administration ordered that the wine-mixture in the tank be thrown away, and asked the manager to find out how much wine was initially in the tank. What was the manager’s reply (assuming he answered correctly)? Solution: x = 6000 litres Hence, the manager reported that 6000 litres of wine was initially in the tank.