Numbers

Ratio and proportion are one the most basic and easy to understand concepts for several MBA entrance exams particularly the CAT exam. This doesn’t mean that it should be overlooked, skipped, or not paid adequate attention to. Questions from ratio and proportion are usually tested along with other concepts such as triangles, allegation, and mixtures, therefore to excel in the other concepts the knowledge of ratio and proportion is vital. The basic principle of ratio and proportion is used when one compares 2 data sets of the same type. One way to do this is to find out the difference (a-b). Another method of comparison could be by division or finding out ratios. (For example, a/b also written as a: b).

For example, the cost of a book is Rs. 250 and the cost of a pen is Rs. 50. If we compare the two numbers using the difference method, the difference is Rs. 200. If we compare by division or ratios, they would be in the ratio:     250/50 = 5/1. The cost of the book is five times that of the pen.

Below, the details about the topic of ratio and proportion for different MBA competitive exams is given:-

CAT

Number System
YearNo of QuestionsGood attemptDifficulty

 

2019

 

Slot 11Difficult
Slot 231Difficult

 

2018

 

Slot 144Easy
Slot 242 to 3Moderate

 

2017

 

Slot 143Easy
Slot 243Easy

XAT

 

Number System
YearNo of QuestionsGood attemptDifficulty

 

2020

 

 

2

 

 

1

 

 

Moderate

 

 

2019

 

 

3

 

 

2

 

 

Moderate

 

 

2018

 

 

2

 

 

1

 

 

Easy – Moderate

 

SNAP

 

Number System
YearNo of QuestionsGood attemptDifficulty
2019
3
2
Easy-Moderate
2018
4
2
Moderate
2017
5
2 to 3
Moderate

List of Concepts within the Topic

The list of concepts that are covered in the Number System chapter is as follows: –

  • Concept of the ratio which can be defined as the ratio of the quantity A to the quantity B is a relation that tells us what multiple or fraction the quantity A is of the quantity B
  • Comparison of ratios
  • The concept of proportion can be defined as when 2 quantities are equal, the 4 quantities composing them are said to be proportional
  • Operations on ratios
  • Operations on proportions
  • Direct and inverse proportions- when the amount of a certain entity increases when the amount of another entity increases, they are said to be in direct proportional to each other. And are depicted as ‘x∝y’ and if x increase, as y decreases, then they are inversely proportional to each other ‘x∝1/y’. 
 

Details on This Page

Some Questions from Previous Papers

Some of the important CAT questions on ratio and proportion that appeared in the previous year papers are: –

1. Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is- 5: 6, 3: 4, 2 : 3, 4: 5 (CAT 2019 slot 1)

 

Let side of equilateral triangle ABC = 3a  So side of hexagon = a  Area of triangle = (root3 /4 )*(3a)^2 = 9a^2 *(root3/4)      Area of hexagon with side a = 6*(root3 /4 )*(a)^2
Ratio of areas of hexagon to that of triangle = { 6*(root3 /4 )*(a)^2 } / { 9a^2 *(root3/4)} = 6/9 = 2/3                                Thus required ratio = 2 : 3

2. A person invested a total amount of Rs 15 lakh. A part of it was invested in a fixed deposit earning 6% annual interest, and the remaining amount was invested in two other deposits in the ratio 2: 1, earning annual interest at the rates of 4% and 3%, respectively. If the total annual interest income is Rs 76000 then the amount (in Rs lakh) invested in the fixed deposit was-

(CAT 2019 slot 1)

 

Final interest at the end of the year = 76000,

Interest rate = 76000/1500000 *100 = 76/15 = 5 1/15

Combine interest rate from amount deposited at 4% and 3% interest rate = (4*2 + 3)/(2+1) =11/3 %

Let the amount ratio of amount deposited in fixed deposit and other deposit be x:y so

x/y = ( 76/15 -11/3 )/(6 – 76/15)

x/y = (76-55)/(90 -76) = 19:14

 

amount deposited in 6% = 19/33 *15 =8.63 lakh = 9 lakh

 

3. The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157 : 3, then the sum of the two numbers is- 50, 85, 95, 58 (CAT 2019 slot 2)

 

Let the number be a and b.

So ab = 616

Given, (a^3 –b^3)/(a-b)^3 = 157/3

(a-b)*(a^2 + b^2 +ab)/(a-b)*(a^2 + b^2 -2ab) = 157/3

Let a^2 + b^2 =k

So (k+ab)/(k-2ab) = 157/3

3k + 3ab = 157k – 314ab

154k = 117ab = 317*616

k = 317*616/154 = 1268

As we know (a+b)^2 = (a^2 + b^2) + 2ab = k + 2ab = 1268 + 2*616

(a+b)^2 =2500 = 50^2

So a+b = 50

 

4.  Amala, Bina, and Gouri invest money in the ratio 3: 4: 5 in fixed deposits having respective annual interest rates in the ratio 6: 5: 4. What is their total interest income (in Rs) after a year, if Bina’s interest income exceeds Amala’s by Rs 250? 7000, 6000, 6350, 7250

(CAT 2019 slot 2)

 

Ratio of their interest at the end of a year = 3*6 : 4*5 : 5*4 = 18 : 20:20

As per question 20x -18x = 250

2x = 250

X = 125

So total interest = (18+20+20)*x = 58*125 = 7250

 

5. If the rectangular faces of brick have their diagonals in the ratio 3: 2 3–√3: 15−−√15, then the ratio of the length of the shortest edge of the brick to that of its longest edge is- 1 : 3–√3, 2: 5–√5, 2–√2 :3–√3, 3–√3: 2 (CAT 2019 slot 1)

 

Let the size of bricks are l*b*h such that l > b> h

As we know diagonals = (l^2 + b^2 )^1/2 , (l^2 + h^2 )^1/2 , and (h^2 + b^2 )^1/2

Thus ratio of squares of diagonals = (l^2 + b^2 ) : (l^2 + h^2) : (h^2 + b^2 ) = (√15)^2 : (2√3)^2 : 3^2

Or (l^2 + b^2 ) : (l^2 + h^2) : (h^2 + b^2 ) = 15 : 12 : 9 = ( 9 + 6) : (9 + 3) : (3 + 6)

By comparing we can say l^2 = 9, h^2 = 3 and b^2 = 6

So l = 3 and h = √3

Required ratio of h/l = √3/3 = 1: √3

 

How to prepare for the topic

Level 1 

The student should first and foremost understand the basics of ratio and proportion and practice conversions from ratios to fraction form to percentage such as 0.25 is 25% which is in turn ¼. The CAT exam tests the skills and understanding of the aspirant as much as speed and mental calculation. The aspirant needs to practice calculations without calculators and as during the actual exam, using the calculator will be inconvenient and time consuming. If calculator is needed for complex calculations, do it in a timed manner and on the computer. At the present level, NCERT books from classes 10th to 12th can be referred to for brushing up the basics and practicing entry-level calculations.     

Level 2

At this level, more focus should be on higher level problems and solving them accurately and without looking at the solutions. The speed of solving the questions should not be focused on at this stage. An Excel sheet can be maintained to monitor the progress in the questions by the student. Any questions which are deemed to be difficult and can’t be solved in the first go should be recorded and revised after a week, if the difficulty is faced again, then the concept should be revised. For the intermediate level, study matter from a coaching center can be referred to as it has graded material with sections divided into different levels of difficulty.

Level 3

At the next level, the aspirant needs to focus on the speed on solving the questions and also solving them correctly and with accuracy without looking at solutions. Mocks should be given at this level. Several mocks should be given in an environment which is similar to the actual exam- without disturbance and in a timed manner. A complete analysis needs to be carried out. Section- wise tests will have questions of varying difficulty, attempting questions of the same difficulty level is not conducive as the real exam is a mixture of questions of different difficulty levels. Taking up a mock test series from a coaching institute will be beneficial. The art of selecting which questions to attempt and which to leave un- attempted needs to be practiced now, also which topics should be attempted and which to leave is also identified right now.

 

Inspiring Education… Assuring Success!!
Ⓒ 2020 – All Rights Are Reserved