# Arithmetic

#### Simple Interest & Compound Interest - 1

Mixtures And Alligations : 1

Mixtures And Alligations: Source: Indian recipe

Mixtures and Alligations is one of the most dreaded of all the all topics in Arithmetic Section. It is the continuation of the topic Averages with some new concepts being added into the list. Mixtures as the word suggests is the blending of two or more solutions with different compositions in order to obtain a new solution with the composition different from the initial solutions. To understand it in depth, let’s first understand the concept of weighted averages.

Weighted Average:

Let us say we have three sections A, B and C and each section appeared for a mathematics test. The following table gives the average marks and number of students in each of these sections.

 Section A Section B Section C Average marks 60 70 50 Number of students 40 20 30

Now, if all the sections are combined together, then what is the average marks of all the students combined?

This average value is called as the weighted average (Aw)

Aw = (Total marks obtained by all students)/(Total number of students)

Aw = 60*40+70*20+50*3040+20+30 = 58.88

This expression could also be written as:

Aw =60* 4090+70*2090+50*3090

(40/90) is the ratio of students in section A to the total number of students, similarly 20/90 is the ratio of students in section B to the total number of students and 30/90 is the ratio of students in section C, to the total number of students.

From this we can conclude that if we know the ratios, then also we can find the weighted average.

Thus, to find the weighted average, one needs to know the individual averages and the ratio of number of elements.

Let’s say we have two sets P and Q with average of set P being A1 and average of set Q being B1 and there being n1 and n2 elements in both the sets respectively. Thus the ratio of the number of elements is equal to the ratio of difference of difference of average of the second minus the weighted average to the difference of the weighted average and the average of the first set. Example 1:

A variety of sugar with Rs 20/kg is mixed with another sugar with Rs 50/kg to obtain a mixture with Rs 32 /kg. In what ratio must the two mixtures be mixed together?

Solution: Here A1 = Rs 20/kg, A2 = Rs 50/kg and Aw = Rs 32/kg

So, Thus, n1 :n2 = 18/12 = 3:2

Example 2: Two liquids A and B are in the ratio 5: 1 in container 1 and 1 : 3 in container 2. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1? (CAT 1996)

1. 2: 3         2. 4: 3         3.3: 2         4. 3: 4

Solution:  Here A1 is the average of the first liquid or the concentration of liquid A in the first container which is 5/6  and the concentration of liquid A in the second container is ¼ (out of the total). Also the weighted average is ½ Thus, the given ratio is 3:4 or the given contents in the two mixtures must be added in the ratio of 3:4.

So, option (4) is correct.

Example 3: How many litres of petrol must be added to a 24 litre mixture of petrol and kerosene which is in the ratio of 3:5, so that the resultant mixture is 40% petrol?

Solution:  The initial concentration of petrol = 3/8 and the final concentration of petrol = 40% = 2/5 .

As we are adding pure petrol, its concentration = 100% or 1 So, the two mixtures must be mixed in the ratio of (3/5)/(1/40) = 24:1

Initial quantity of mixture of petrol and kerosene = 24 litres

Pure petrol to be added = (1/24) * 24 = 1 litre

Example 4: A shopkeeper buys two varieties of rice, one at Rs 24/kg and the other at Rs 30/kg, mixes them and sells the mixture at Rs 31.2/kg making a profit of 20% on the transaction. In what ratio must the two varieties be mixed?

Solution: The shopkeeper sells the mixture at Rs 31.2/kg making a profit of 20%.

So, C.P. of the mixture = 31.2/1.2 = Rs 26/kg.

This means that on mixing the two varieties, the average value of the mixture obtained = Rs 26. Thus, the ratio in which they must be mixed is 4:2 or 2:1.

Example 5: A man buys spirit at Rs. 60 per litre, adds water to it and then sells it at Rs. 75 per litre. What is the ratio of spirit to water if his profit in the deal is 37.5%?

1. 9 : 1         2. 10 : 1         3. 11 : 1         4. None of these

Solution: Given that the man sells the mixture of spirit and water at Rs 75/litre and gains 37.5% on the transaction.

So, Effective C.P. (after mixing) = 75/1.375 = 54.54/litre or 600/11

Cost of water = Rs 0

Cost of spirit = Rs 60 So, ratio of spirit to water is (600/11)/(60/11) = 10:1

Thus, option (2) is correct.

Example 6: A 32L mixture of methanol and ethanol contains 40% ethanol. How much quantity of the mixture must be taken out and replaced by pure methanol, so that now the percentage of ethanol in the mixture becomes 25%?

Solution: As the quantity of mixture taken out is replaced by pure methanol, this can be viewed as mixing the remaining solution (of initial concentration) with pure methanol (0%)  to obtain 25% mixture (of ethanol) So, the quantity of initial mixture left: Pure methanol added = 25:15 = 5:3

Initial quantity = Methanol Added + Qty left

32 = 3x + 5x  or x = 4

Quantity of pure methanol to be added = 12 L

Example 7: Source:springfield

In a class there are a total of 54 students. Each boy contributed Rs 6 and each contributed Rs 8 for buying a farewell gift for their favourite teacher. The total amount collected was Rs 384. How many girls and boys are there in the class?

Solution: Total students = 54, So, B + G = 54

Average amount contributed by each boy = Rs 6 while that by girls = Rs 8

The weighted average = Total money/Total students = 384/54 = 64/9 Thus, ratio of boys to girls = 4:5

So total number of boys = (4/9)*54 = 24

Total number of girls = (5/9)*54 = 30

Example 8 : In a farm house there are only ducks and cows where each duck has two legs and each cow has 4 legs. The total numbers of heads was 150 while the total number of legs was 510. How many ducks and cows are there in the farm house?

Solution: The number of heads =150, so there are a total of 150 animals.

Each duck has 2 legs while each cow has 4 legs.

Average number of legs = Total number of legs/Total animals

= 510/150 = 17/5 Thus, Ducks : Cows =  3:7

Number of Cows = (7/10) *150 = 105

Number of Ducks = (3/10) * 150 = 45

Example 9: The population of a town is 42000. The population of males increases by 20% while the population of females increases by 34%. If the new population becomes 53,760, then what was the number of males initially?

Solution: The population of the town increases from 42000 to 53,760.

% Increase = 53760-4200042000 * 100 = 28%

Assume males to be a set whose population increases by 20%, mixed with another set of females whose population increases by 34%. Ratio of males : females = 6:8 = 3:4

Thus, initial number of males = (3/7)* 42000 = 18,000

Initial number of females = (4/7) * 42000 = 24,000

Example 10: Girdhari Lal buys two tables for a total cost of 32,000 and sells the first table at a profit of 20% and the other at a profit of 12%. The total amount received by Girdhari Lal on selling these tables is 36,800. What is the cost price of both these tables.

Solution: First method:

Let the C.P. of 1st table = Rs x and the C.P. of the other table = 32,000 –x

S.P. of 1st table = 1.2*x

S.P. of 2nd table = 1.12*(32,000-x)

Total S.P. = 36800

1.2x + 1.12(32,000-x) = 36800

0.08x = 960

Or x = 12,000

C.P. of the two tables would be 12,000 and 20,000.

Another Method: By alligations.

The average profit % on the entire transaction =( 36800-3200032000*100)=15%

Applying the method of alligations, we get : Thus, the ratio of C.P. of both the tables is 3:5

So, C.P. of 1st table = (3/8) * 32,000 = 12,000

C.P of 2nd table = (5/8)*32,000 = 20,000

Replacement:

In these type of problems, a part of the solution is constantly being replaced by another solution and one has to identify the quantities of original constituents left or the ratio of the constituents left. So let’s try to understand it with an example.

Let’s say we have a 10L solution of milk and water with milk being 80% and the rest being water. This means total amount of milk = 8L and water being 2L.

Now, if we remove 2 litres of the solution, then the amount of milk in these 2 litres will be 2*0.8 = 1.6 L and water removed = 2*0.2 = 0.4 L

Amount of milk left = 8 – 1.6 = 6.4 L

Amount of water left = 2 -0.4 = 1.6L

As we removed 2L from 10 L solution, i.e. 20% of the solution was removed and 80% left, we can also say that 20% of the components were removed and 80% were left.

Amount of milk left = 8(1 – 0.2)  = 8( 1 – 2/10)

Amount of water left = 2( 1-0.2) = 2(1 – 2/10)

Let’s say these 2 L which were removed were replaced by 2L of water. Thus, the total amount was again brought back to 10L.

Now again, 2 L of the new solution were removed and replaced by water.

Amount of milk left = 8(1 – 2/10)(1 -2/10) = 8(1 -2/10)2  = 5.12 L

Water left = 10 – 5.12 = 4.88

In general we can say that if such an operation is performed n times, the final amount of milk left = 8(1 -2/10)n

Here 8 is the initial amount of milk present, 2 is the amount of solution being replaced and 10 is the total volume of the solution in the container.

So, if a container of V volume is filled with a pure substance and is replaced by x litres of some other substance and this replacement is performed n times, then the total quantity left of the initial component is calculated by the formula:

Final amount left of the pure substance = Vf = V {1- x/V}n

Example 11: A cask of 20L is full with kerosene. 4L of kerosene are taken out and replaced with petrol. Again 4L of the solution are taken out and replaced by petrol and this process is repeated three times. What is the quantity f kerosene left in the cask and also the quantity of petrol in the cask?

Solution: This problem is the direct application of the formula learnt

Final quantity of kerosene = Vi {1 – x/Vi }n

Final quantity of kerosene = 20{ 1 – 4/20}3   = 20{ 1 – 4/20}3   =10.24

Final quantity of petrol = 20 – 10.24 = 9.76

Example 12: A cask contains 15 L of spirit. It is first replaced by 3 L of water. Then by 5 L of water and finally by 6L of water. What is the final quantity of spirit and water in the cask?

Solution:  Here the initial quantity of spirit = 15L

On first replacement, quantity of spirit taken out = 3L, i.e. 3/15 =1/5 as a fraction.

So quantity of spirit left 15 * (1 – 1/5)

On second replacement, quantity removed = 5L,i.e. 1/3 as a fraction

Quantity of spirit left = 15*(1 -1/5)(1-1/3)

On third replacement, quantity of spirit taken out = 6L, i.e. 2/5 as a fraction

Final quantity of spirit left = 15*(1-1/5)(1-1/3)(1-2/5) = 15* (4/5)*(2/3)*(3/5)

= 4.8L

Final quantity of water  = 15 – 4.8 = 10.2L

Example 13: A beaker of volume V litres is full of milk. 21 litres of milk is removed and replaced by water. Again 21 L of the solution is removed and replaced by water. If the ratio of milk to water is 16:33, then what is the value of V?

Solution:  Vf = Vi ( 1- x/Vi)n

Or VfVi=(1-xVi)n

The final ratio of milk to water = 16:33  or the ratio of final amount of milk to initial amount of milk = 16:49

Thus, Vf/Vi = 16/49

Here x = 21 and n = 2

16/49 = (1 – 21/V)2

4/7 = 1 – 21/V

21/V = 3/7   or V = 49 L