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Application of Quantitative Ability Concepts in Logical Reasoning Puzzles and Data Interpretation Sets

Monday, March 8th, 2021

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In the recent CAT papers, we have seen questions in Logical Reasoning and Data Interpretation section based on the concepts from Quantitative Ability. These questions can also be solved with simple logic but if solved with the Quantitative Ability concepts then solving becomes more convenient. Topics such as Mixture-Allegation, Percentages, Linear equations, numbers etc. are used to frame questions. In this post, we will look at some of the Quantitative Ability based applications in this section of CAT. We will not discuss the Quantitative Ability concepts in this post, rather we will implement them in the problems.

MIXTURE-ALLEGATION BASED QUESTIONS

Consider the following question:

The table gives the breakup of the percentages of fat, carbohydrates, gluten, and protein in three four kinds of rice – A, B, C, and D. The cost of each kind of rice per kg is mentioned in the table. Two or more of the rice varieties can be mixed to produce the desired ratios of fats, carbohydrates, gluten, and protein.           

RICE VARIETY	% FAT	% CARBOHYDRATE	% GLUTEN	% PROTEIN	COST/KG
A	35	15	25	10	150
B	20	20	25	15	125
C	75	30	30	35	100
D	60	25	20	20	200

Question 1: What is the least cost per Kg of a variety of rice which contains at least 50% fat?

Approach: Now, in order to get 50% fat, we need to mix two varieties of rice. One of the varieties would contain less than 50% and the other more than 50%. Mixing any two, if both of them contain more than 50% or less than 50% will not produce the desired variety. Looking from the table, A and B have less than 50% and C and D have more than 50%. We need to mix one of A and B with one among C and D. Since cost per kg is to be minimum, mixing C with B will give the least cost as these are the cheapest of the two. Using allegations,

We get the ratio of mixing as 6:5. Hence, the cost will be  = Rs111.36. (You may check for other ratios as well, the combination for the two cheapest will give the least cost per kg)

Question 2: If a variety, which contains at least 25% each of fat, carbohydrate, gluten, and protein, is to be prepared, then what would be the least cost per kg of that variety?

Approach: Our focus is again on least cost per kg, so we will mix the two cheapest varieties. We can acknowledge the fact that while mixing any two varieties, the resultant cost would always be between the prices of the two varieties being mixed. So, we mix varieties B and C to obtain the desired ratio. We need to check for the minimum ratio required for all the four components.

Gluten will always be more than (or equal to) 25% as variety B already has 25%

From the above, we will take the maximum ratio out of these minimum ratios. When we say maximum, we always take the maximum ratio with respect to the higher percentage variety. For instance, in the case of fat percentage, the minimum ratio in which B and C are mixed is 10:1 and since C has more percentage fat, mixing any more quantity of C will increase the fat percentage but adding more B will reduce the fat percentage in the resultant variety. Hence, if the ratio is 1:1, then C is more than 1/11^th of the resultant (C:B ratio is more than 1:10), which means that the resultant fat percentage is more than 25 percent. Hence, we take the maximum ratio as 1:1. The cost per kg then will be (100 + 125)/2 = Rs 112.5.

To solve such questions, in-depth knowledge of the concepts is required. Other methods to solve this question would be to check each possible combination without the use of allegations, which would be time-consuming and less convenient. Therefore, using a concept of allegations greatly helps in solving this kind of questions.

LINEAR EQUATION BASED QUESTIONS

Let us discuss this type with an example:

A shopkeeper offers certain discounts for different price range of garments – for every garment less than ₹ 2000 and more than ₹ 500, he offers ₹ 100 discount, for every garment at least ₹ 2000 and less than 5000 he offers ₹ 250 discount and for all garments more than and equal to ₹ 5000 he offers ₹ 500 discount.In particular day, four persons bought a certain number of garments and discount each received is given table:

Person	Total amount spent (without discount)	Discount
Amar	2,50,000	5000
Boman	45,000	2200
Chirag	46,000	3250
Danish	1,05,000	2500

 

Question 1: What is the minimum number of clothes Boman could have bought?

Approach: We can write the discount as the sum of the three types of discounts multiplied by the number of garments of each type – say ‘a’ garments in ₹ 100 discount category, ‘b’ garments in ₹ 250 discount category and ‘c’ garments in ₹ 500 discount category. So the equation will be:

Total discount = a*100 + b*250 + c*500

In order to find the minimum number of garments bought by Boman (the total number of garments bought is (a+b+c), we have to maximize the largest value in the equation. In addition, we need to check if the total amount spent is satisfied by the values obtained. For Boman, the discount is ₹2200, the maximum value of ‘c’ can be 4 in this case. 4*500 gives ₹2000 and the discount is₹ 2200, therefore the remaining ₹200 will be satisfied by either ‘a’ or ‘b’. For the smallest possible value of ‘b’ (b=1), the total of the equation will be₹ 2250. Thus, we need to take ‘a’ = 2 to satisfy the equation. Hence, the minimum number of garments bought by Boman is 6 (c+a). We can also see that for 4 garments of minimum ₹ 5000, the minimum sum of prices will be₹ 20,000 and total amount spent is ₹45,000, therefore c = 4 is a very much possible situation in our case.

QUESTION 2: What is the maximum number of garments that Amar could have bought?

APPROACH: From the equation:

Total discount = a*100 + b*250 + c*500

If we want to maximize a+b+c, we need to maximize the portion of ‘a’ in above equation. From the table, we see that Amar got ₹ 5000 as discount. If we take b and c as 0, a will be 50. The maximum amount of garment for category ‘a’ is 2000. For 50 garments, the maximum cost can be approximately 1,00,000 but as we can see from the table, the total amount spent is ₹ 2,50,000. We need to consider adding garments from higher range as well. If we reduce the value of ‘a’ by 1, then the discount covered will be 1900 and we still need to cover ₹ 100 from either ‘b’ category garments or ‘c’ category garments. But we cannot get ₹ 100 from any one of ‘b’ or ‘c’, hence we need to bring down the value of ‘a’ to 45, which will leave the remaining discount as 500. Since ‘c’ type garments are more than (or equal to) ₹ 5000, a single garment can cover the entire remaining total amount spent after we have maximized the value of ‘a’. Thus, the maximum number of garments Amar could have bought is 45+1 (a+c).

QUESTION 3: If Chirag and Danish bought the same number of garments, then what can be the possible value of the number of garments?

a.)32

b.)6

c.)20

d.)None of these

APPROACH: We need to make a+b+c for both Chirag and Danish equal. The best approach is to find out the minimum and a maximum number of garments each of Chirag and Danish could have bought. Applying the logic used in above questions, the number of garments for Chirag will range from 7 to 31 and the number of garments for Danish will range from 5 to 21. We have used the concept used in above two questions to get the range of values. From the two range of values, the common range is 7 to 21. So, any value from 7 to 21 is possible. Hence, option (c) is correct.

This set involves the knowledge of the formation of the basic linear equation and maximizing and minimizing the sum of three variables under given conditions.

PERCENTAGE BASED QUESTIONS

A scientist invented the lie-detector machine. The machine can detect 70% lie of every person but there is a bug in the machine, which shows 30% of the truth as a lie. In order to test the machine, different people are tested against it. The number of lies detected by the machine for each person tested is given below. Also the ratio of lie: truth statements for each person is mentioned.

PERSON	LIE DETECTED	LIE: TRUTH
Pawan	141	5:4
Sahil	138	1:2
Tarun	120	A:6

 

QUESTION: How many total statements did Pawan make?

Approach: in this type of question, we need to make a percentage equation.

Let a total number of statements made by Pawan be ‘n’. The lie: truth ratio for Pawan is 5:4. Let us say a number of lie statements are 5x and truth statements are 4x. The total number of statements will comprise of the sum of truth and lie statements as any statement can be either a truth or a lie.

n = 5x + 4x = 9x

We are also given that the machine detects 70% of the lie and it also considers 30% of the truth as lie. Hence, the total number of lie detected will be

(0.7)*5x + (0.3)*4x = 141

Solving this equation, we can calculate the value of ‘x’. Therefore, x = 30. Hence, the total number of statements made by Pawan is 9x = 270.

QUESTION: If Sahil and Tarun made the same number of statements, then what is the value of A ?

Approach: Following the method used in the last question, the number of statements made by Sahil is 300. Since Sahil and Tarun made the same number of statements, Tarun also made 300 statements. Hence, B = 300. Total number of statements is the sum of lie and truth statements,

300 = A*x + 6*x

Also, the number of lie statements detected by the machine is 138.

(0.7)*Ax + 0.3*6x = 138.

Solving the above two equations for A we get, A = 4 and x = 30.

In this post, we saw how some of the concepts of Quantitative Ability section used in the Logical Reasoning & Data Interpretation section of CAT. The three types discussed are the most common ones and the majority of questions are based on these. With this section getting tougher and trickier with the language, we might see more involvement of the concepts from Quantitative Ability to increase the toughness of questions. A sound knowledge of the concepts is very much required to solve such questions.

Other posts related to Logical Reasoning and Data Interpretation

Games and Tournaments for CAT Exam Logical Reasoning – Part 1
Games and Tournaments for CAT Exam Logical Reasoning – Part 2
Games and Tournaments for CAT Exam Logical Reasoning – Part 3
CAT Concepts – Introduction to Data Sufficiency
Basics of Data Interpretation – Approximation
How to Solve Number Series Questions in Logical Reasoning
How to prepare for LRDI
How to prepare for Logical Reasoning
How to Prepare for Data Interpretation
Boats and Streams – Tips Tricks and Questions for CAT Exam

CAT Questions related to Logical Reasoning and Data Interpretation

All questions from CAT Exam Logical Reasoning and Data Interpretation
Logical Reasoning – Set 1: A high security research lab requires the researchers to set a pass key sequence based on the scan of the five fingers of their left hands.
Logical Reasoning – Set 2: There are 21 employees working in a division, out of whom 10 are special-skilled employees (SE) and the remaining are regular skilled employees (RE).
Logical Reasoning – Set 3: Eight friends: Ajit, Byomkesh, Gargi, Jayanta, Kikira, Manik, Prodosh and Tapesh are going to Delhi from Kolkata by a flight operated by Cheap Air.
Logical Reasoning – Set 4: tea taster was assigned to rate teas from six different locations – Munnar, Wayanad, Ooty, Darjeeling, Assam and Himachal.
Logical Reasoning – Set 5: Four cars need to travel from Akala (A) to Bakala (B). Two routes are available, one via Mamur (M) and the other via Nanur (N).
Data Interpretation – Set 1: Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET).
Data Interpretation – Set 2: An old woman had the following assets: (a) Rs. 70 lakh in bank deposits
Data Interpretation – Set 3: Simple Happiness index (SHI) of a country is computed on the basis of three parameters: social support (S), freedom to life choices (F) and corruption perception (C).
Data Interpretation – Set 4: Funky Pizzaria was required to supply pizzas to three different parties.
Data Interpretation – Set 5: There were seven elective courses – El to E7 – running in a specific term in a college. Each of the 300 students enrolled had chosen just one elective from among these seven.

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