Quantitative Aptitude

Ratio and Proportion, Averages, Mixtures and Alligation

Importance in CAT

• Ratio and Proportion, Averages and Mixtures and Alligation are broadly categorized under Arithmetic

• Its importance in CAT is relatively less (2 to 4%) but questions related to these topics are more frequent in regional entrance tests

Key Concepts • Ratio and Proportion Ratios (Scaling and comparison of ratios) Proportion Continued proportion (geometric mean)

• Averages Averages Arithmetic, geometric and harmonic mean Weighted average Median and mode

• Mixtures and Allligation Alligation cross Alligation line Successive replacements

Types of Questions frequently asked

• Ratio and Proportion Work and Remuneration (Partnerships) Questions based on compenendo-dividendo-invertendo-alternendo Miscellaneous questions based on ratios and proportion

• Averages, Mixtures and Alligation Questions based on Mean, Median and Mode Questions based on Weighted averages Questions based on application of Averages and Mixtures and

Alligation

Properties - Ratios and Proportion

If a × d > b × c then . This property is used to compare ratios.

If the given simultaneous equations are: p1x + q1y + r1z = 0 and p2x + q2y + r2z = 0

then, x : y : z = q1r2 − q2r1 : r1p2 − r2p1 : p1q2 − p2q1

Properties - Ratios and Proportion If a < b, then If a > b, then

If a : b :: c : d or a/b = c/d, then

Properties - Ratios and Proportion

Examples – Ratio and Proportion• Asha, Altheda and Amata had a total of Rs. 2750 with them. They

decided to divide this money between themselves such that 1/4th of Asha’s share was equal to 1/5th of Altheda’s share, which in turn was equal to half of Amata’s share. How much money did Amata receive?

Solution: Assume that the money received by Asha, Altheda and Amata is x, y, and z. Hence,

∴ y = 5x/4 and z = x/2 Thus, the ratio of Asha, Altheda and Amata becomes 1 : 5/4 : 1/2; which is equivalent to 4 : 5 : 2. Hence, Amata received 2/11 × 2750 = Rs. 500

Examples – Ratio and Proportion• Let a, b, c, d and e be integers such that a = 6b = 12c, and 2b = 9d = 12e.

Then what is the value of Solution: The given data states that a = 6b = 12c … (1) and 2b = 9d = 12e … (2) Divide equation (1) by 12 (l.c.m of 1, 6 and 12) to get ∴ a : b : c = 12 : 2 : 1 Divide equation (2) by 36 (l.c.m of 2, 9 and 12) to get ∴ b : d : e = 18 : 4 : 3 b is common in both the given ratios so we will find out the l.c.m. of and 9 which is 18. Thus a : b : c can be written as 108:18:9 ∴ a : b : c : d : e = 108 : 18 : 9 : 4 : 3 ∴ a = 108k; b = 18k; c = 9k; d = 4k and e = 3k where k is an integer Thus

Examples – Ratio and Proportion• Three numbers are in continued proportion. Their mean proportional is

10 and the sum of the other two is 29. Find the numbers.

Solution: Let a, b and c be the numbers which are in continued proportion. Thus ⇒ b2 = ac Then, b = 10 b2 = ac = 100 and a + c = 29

∴a = 25 and c = 4 or a = 4 and c = 25

Examples – Ratio and Proportion• Solve the following equation: Solution: Using the Componendo and Dividendo law, we get Taking square roots on both the sides, we get

7x + 5 = 49x – 35 or 7x + 5 = –49x + 35 42x = 40 or 56x = 30 x = 20/21 or x = 15/28

Examples – Ratio and Proportion

Solution:

We will solve the given problem using the rule: Add the first two ratios and subtract the third ratio from it to get Similarly, add the second and third ratio and subtract the first ratio from it to get Similarly, add the first and third ratio and subtract the second ratio from it to get

Each of the above obtained equations are equal to each other

Examples – Ratio and Proportion• Amber Chew opened a departmental store at Great India Palace in

Noida by investing Rs. 20 million. After a few months, her brother Sheesh Chew joined the business and invested Rs. 30 million. At the end of the year, the profit was shared in the ratio of 3 : 2. After how many months did Amber’s brother join the business?

Solution: Amber Chew invested Rs. 20 million for a total period of 12 months. Assume that Sheesh Chew joined the business after x months. Then, Sheesh Chew invested Rs. 30 million, but only for a period of (12 – x) months. Amber and Sheesh shared profits in the ratio 3 : 2. ∴ x = 20/3 = 6.67 months Thus, Amber’s brother joins the business after 6 months (and 20 days).

Averages – definitions and formulae• Average of a set of given data is a form of proportions.• It is usually the middle value of a set of data and is also known

as mean.

• Average (Geometric mean) = root of product of all the terms given in a set of data.

• As an example harmonic mean is used to find average speed

when same distance is travelled with different speeds.• Relationship between means: Arithmetic Mean > Geometric Mean > Harmonic Mean

Averages – definitions and formulae• Weighted Average: Weight stand for relative importance given to

different values or parameters. Here, w1, w2, … wn are the individual weights.• Many problems of weighted averages which involve mixtures are

also solved using the method of Alligation.• Median: Middle value of a group of numbers in

ascending/descending order. Median of a set having n numbers = • Mode: The most frequent number in a set of numbers.

Examples - Averages• After every test, Rajiv calculates his cumulative average. QT and OB

were his last two tests. 83 marks in QT increased his average by 2. 75 marks in OB further increased his average by 1. Reasoning is the next test, if he gets 51 in Reasoning, his average will be _____.

Solution: Let the average marks of Rajiv and the number of tests he gave before giving QT be a and n respectively. ∵ 83 marks in QT increased his average by 2 … (i) 75 marks in OB further increased his average by 1 … (ii) Solving equation (i) and (ii), we get the value of n = 10 and a = 61 Hence, a × n = 610 He gets 51 in his next test Reasoning

Examples - Averages• The average weight of a school football team (consisting of 22 members)

decreases by half a kilogram if the goalie is not included. What is the goalie’s weight, if the average weight of the team initially was 60 kg?

Solution: It is given to us that the average weight of the team consisting of 22 players is 60 kg. Thus the total weight of all the players = 60 × 22 = 1320 Assume that the goalie’s weight is x. The average weight of the team excluding the goalie is 59.5 Alternatively, we can use the direct formula to calculate the goalie’s age New Average

Examples - Averages• Ankush is planning to take a housing loan from a bank. He is confused

about taking a floating or fixed rate of interest. The fixed rate of interest has remained constant at 12% for the last 3 years. The floating rate of interest was 12%, 15% and 14% in the last 3 years. Based on this data, should Ankush opt for floating or fixed rate of interest?

Solution: The fixed rate of interest remained constant at 12% for the last three years. The floating rate of interest was 12%, 15% and 14% Geometric mean is used to calculate average interest rates. ∴ Average rate of interest for 3 years = (1.12 × 1.15 × 1.14)1/3 = 1.136 The average floating rate of interest for the three years is 13.6%. Hence, based on this data, Ankush should opt for fixed rate of interest.

Examples - Averages• A final year MBA student gets 50% in the exam and 80% in the

assignments. If the exam should count for 70% of the final result and the assignment for 30%, what will be the final score of the student, if professors decide to use weighted harmonic mean to uneven performances?

Solution: Let W1 and W2 be the weights assigned to the exam and assignment

respectively. ∴ W1 = 70% and W2 = 30%

Let X1 and X2 be percentage marks obtained by the student in the exam

and assignment respectively. ∴ X1 = 50% and X2 = 80%

Examples - Averages

• Find the median and mode of the set of the numbers 1, 2, 3, 7, 5, 9, 2, 12, 2, 13, 4, 10, 2, 8, 2, 6?

Solution: The numbers can be arranged in an ascending order as follows. 1, 2, 2, 2, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13 There are 16 numbers present in this set. Thus the median of the given set is the arithmetic mean of the 8th and 9th numbers. ∴ Median = (4 + 5)/2 = 4.5

The number ‘2’ occurs maximum number of times. Hence, the mode is 2.

Mixtures And Alligation – Concepts and Applications

• Problems related to weighted averages can be solved using the method of Alligation

• Concepts : Cross Alligation Line Alligation

Successive replacement• Rule of Alligation: , where w1 and w2 are weights of

given parameters and x is the weighted average.• This method is used for problems related to: -- Mixtures -- distribution of a parameter in two different groups• Successive Replacement: If a vessel of x litres of A and y litres of A is

removed and replaced by water and the process is followed n times, then the amount of A remaining in the solution after n replacements

Examples – Mixtures and Alligation• 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 will solve the given problem first by using the weighted average method and then by using Alligation Cross and Alligation Line. Let the two solutions be solution A and solution B. Solution A contains petrol and diesel in the ratio 2 : 3 which implies that 2/5th portion of the solution is petrol. Similarly 3/10th portion of solution B is petrol. Assume that A and B are mixed in the ratio a : b such that 7/20th portion of the final mixture should be petrol. The formula of weighted average = On solving the above obtained equation, we will get a = b. Hence A and B should be mixed in 1 : 1 ratio.

Examples – Mixtures and Alligation• Let us solve the same problem by using Alligation Cross method. Amount of petrol in solution A = 2/5 = 8/20 Amount of petrol in solution B = 3/10 = 6/20 Amount of petrol required in the final solution = 7/20

It is clearly visible that equal weights of both the solutions will be needed to obtain a mixture which has the desired amount of petrol. This is consistent with the result that we had obtained using the weighted average method.

Examples – Mixtures and Alligation• Let us solve the previous example using Alligation Line method now. Amount of petrol in solution A = 2/5 = 8/20 Amount of petrol in solution B = 3/10 = 6/20 Amount of petrol required in the final solution = 7/20 The weighted average rule can be represented as Alligation Line as follows

2/5 7/20 3/10

a b Now by using the Alligation rule which states that , we get Thus we get the same answer by each of the three methods.

Examples – Mixtures and Alligation• 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? Use Alligation Cross method to solve the problem

Solution: Let the required average be x. By using Alligation Cross method, we get

By using the Alligation rule, we get ∴ 50 − x = 2x – 88 ∴ 3x = 138 ∴ x = 46 Hence, the average of the new class will be 46 marks.

Examples – Mixtures and Alligation• A man in charge of maintaining the pH level of a swimming pool is given

two bottles 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? Use Alligation Line method to solve this problem.

Solution: 74 65 52

w1 w2

It is given that concentration in first bottle is 74%, concentration in the second bottle is 52% and the concentration of chlorine in the mixture is 65% Using the Alligation rule, we get Thus the amount of chlorine in second bottle =

Example – Successive Replacements

• 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: Let x be the capacity of the vessel and y be the amount of petrol replaced by kerosene.

It is given to us that n is 2 and ratio in which petrol and kerosene are present after 2 iterations is 49:32.

Let the quantity of petrol after 2 iterations be 49m and the quantity of kerosene be 32m

Example – Successive Replacements

∴ 7x = 9x – 90

∴ x = 45 litres

∴ Capacity of the vessel (x) = 45 litres Thus the capacity of the vessel is 45 litres