How to Prepare for Quantitative Aptitude for CAT (2014)

Block III: Arithmetic and Word-based Problems

...BACK TO SCHOOL

As you are already aware, this block consists of the following chapters:

Percentages,

Profit & Loss,

Interest,

Ratio, Proportion and Variation,

Time and Work,

Time, Speed and Distance

To put it very simply, the reason for these seemingly diverse chapters to be under one block of chapters is: Linear Equations

Yes, the solving of linear equations is the common thread that binds all the chapters in this block.

But before we start going through what a linear equation is, let us first understand the concept of a variable and it’s need in the context of solving mathematical expressions.

Let us start off with a small exercise first:

Think of a number.

Add 2 to it.

Double the number to get a new number.

Add half of this new number to itself.

Divide the no. by 3.

Take away the original number from it.

The number you now have is......... 2!!

How do I know this result?

The answer is pretty simple. Take a look. I am assuming that you had taken the initial number as 5 to show you what has happened in this entire process.

The above is a perfect illustration of what a variable is and how it operates.

In this entire process, it does not matter to me as to what number you have assumed. All I set up is a kind of a parallel world wherein the number in your mind is represented by the variable X in my mind.

By ensuring that the final value does not have an X in it, I have ensured that the answer is independent of the value you would have assumed. Thus, even if someone had assumed 7 as the original value, his values would go as: 7, 9, 18, 27, 9, 2.

What you need to understand is that in Mathematics, whenever we have to solve for the value of an unknown we represent that unknown by using some letter (like x, y, a etc.) These letters are then called as the variable representations of the unknown quantity.

Thus, for instance, if you come across a situation where a question says: The temperature of a city increases by 1°C on Tuesday from its value on Monday, you assume that if Monday’s temperature was t, then Tuesday’s temperature will be t + 1.

The opposite of a variable is a constant. Thus if it is said in the same problem that the temperature on Wednesday is 34° C, then 34 becomes a constant value in the context of the problem.

Thus although you do not have the actual value in your mind, you can still move ahead in the question by assuming a variable to represent the value of the unknowns. All problems in Mathematics ultimately take you to a point which will give the value of the unknown—which then becomes the answer to the question.

Hence, in case you are stuck in a problem in this block of chapters, it could be due to any one of the following three reasons:

Reason 1: You are stuck because you have either not used all the information given in the problem or have used them in the incorrect order.

In such a case go back to the problem and try to identify each statement and see whether you have utilized it or not. If you have already used all the information, you might be interested in knowing whether you have used the information given in the problem in the correct order. If you have tried both these options, you might want to explore the next reason for getting stuck.

Reason 2: You are stuck because even though you might have used all the information given in the problem, you have not utilized some of the information completely.

In such a case, you need to review each of the parts of the information given in the question and look at whether any additional details can be derived out of the same information. Very often, in Quants, you have situations wherein one sentence might have more than one connotation. If you have used that sentence only in one perspective, then using it in the other perspective will solve the question.

Reason 3: You are stuck because the problem does not have a solution.

In such a case, check the question once and if it is correct go back to Reasons 1 and 2. Your solution has to lie there.

My experience in training students tells me that the 1^st case is the most common reason for not being able to solve questions correctly. (more than 90% of the times) Hence, if you consider yourself to be weak at Maths, concentrate on the following process in this block of chapters.

THE LOGIC OF THE STANDARD STATEMENT

What I have been trying to tell the students is that most of the times, you will get stuck in a problem only when you are not able to interpret a statement in the problem. Hence, my advise to students (especially those who are weak in these chapters)—concentrate on developing your ability to decode the mathematical meaning of a sentence in a problem.

To do this, even in problems that you are able to solve (easily or with difficulty) go back into the language of the question and work out the mathematical reaction that you should have with each statement.

It might not be a bad idea to make a list of standard statements along with their mathematical reactions for each chapter in this block of chapters. You will realise that in almost no time, you will come to a situation where you will only rarely encounter new language.

Coming back to the issue of linear equations:

Linear equations are expressions about variables that might help us get the value of the variable if we can solve the equation.

Depending upon the number of variables in a problem, a linear equation might have one variable, two variables or even three or more variables. The only thing you should know is that in order to get the value of a variable, the number of equations needed is always equal to the number of variables. In other words, if you have more variables in a system of equations than the number of equations, you cannot solve for the individual values of the variables.

The basic mathematical principle goes like this:

For a system of equations to be solvable, the number of equations should be equal to the number of variables in the equations.

Thus for instance, if you have two variables, you need two equations to get the values of the two variables, while if you have three variables you will need three equations.

This situation is best exemplified by the situation where you might have the following equation. x + y = 7. If it is known that both x and y are natural numbers, it yields a set of possibilities for the values of x & y as follows: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). One of these possibilities has to be the answer.

In fact, it might be a good idea to think of all linear equation situations in this fashion. Hence, before you go ahead to read about the next equation, you should set up this set of possibilities based on the first equation.

Consider the following situation where a question yields a set of possibilities:

Four enemies A, B, C and D gather together for a picnic in a park with their wives. A’s wife consumes 5 times as many glasses of juice as A. B’s wife consumes 4 times as many glasses of juice as B. C’s wife consumes 3 times as many glasses of juice as C and D’s wife consumes 2 times as many glasses of juice as D. In total, the wives of the four enemies consume a total of 44 glasses of juice. If A consumes at least 5 glasses of juice while each of the other men have at least one glass, find the least number of drinks that could have been consumed by the 4 enemies together.

(1)9    (2) 12

(3)11   (4) 10

In the question above, we have 8 variables—A, B, C & D and a, b, c, d – the number of glasses consumed by the four men and the number of glasses consumed by the four wives.

Also, the question gives us five informations which can be summarized into 5 equations as follows.

a = 5A

b = 4B

c = 3C

d = 2D

and a + b + c + d = 44

Also, A>5.

Under this condition, you do not have enough information to get all values and hence you will get a set of possibilities.

Since the minimal value of A is 5, a can take the values 25, 30, 35 and 40 when A takes the values 5, 6, 7 and 8 respectively. Based on these, and on the realization that b has to be a multiple of 4, c a multiple of 3 and d a multiple of 2, the following possibilities emerge:

At A = 5

In this case the answer will be 10, since in the case of a=35, b=4, c=3 & d=2, the values for A,B,C and D will be respectively 7,1,1 and 1. This solution is the least number of drinks consumed by the 4 enemies together as in all the other possibilities the number of glasses is greater than 10.

Such utilisations of linear equations are very common in CAT and top level aptitude examinations.

The relationship between the decimal value and the percentage value of a ratio:

Every ratio has a percentage value and a decimal value and the difference between the two is just in the positioning of the decimal point.

Thus 2/4 can be represented as 0.5 in terms of its decimal value and can be represented by 50% in terms of its percentage value.

Pre-assessment Test

Directions for Questions 11 and 12: Answer the questions based on the following information.

An Indian company purchases components X and Y from UK and Germany, respectively. X and Y form 40% and 30% of the total production cost. Current gain is 25%. Due to change in the international exchange rate scenario, the cost of the German mark increased by 50% and that of UK pound increased by 25%. Due to tough competitive market conditions, the selling price cannot be increased beyond 10%.

Directions for Questions 20 and 21: The petrol consumption rate of a new model car ‘Palto’ depends on its speed and may be described by the graph below:

Directions for Questions 22 and 23: Answer the questions based on the following information:

There are five machines—A, B, C, D, and E situated on a straight line at distances of 10m, 20 m, 30 m, 40 m and 50m respectively from the origin of the line. A robot is stationed at the origin of the line. The robot serves the machines with raw material whenever a machine becomes idle. All the raw materials are located at the origin. The robot is in an idle state at the origin at the beginning of a day. As soon as one or more machines become idle, they send messages to the robot-station and the robot starts and serves all the machines from which it received messages. If a message is received at the station while the robot is away from it, the robot takes notice of the message only when it returns to the station. While moving, it serves the machines in the sequence in which they are encountered, and then returns to the origin. If any messages are pending at the station when it returns, it repeats the process again. Otherwise, it remains idle at the origin till the next message(s) is (are) received.

SCORE INTERPRETATION ALGORITHM FOR PRE-ASSESSMENT TEST OF BLOCK III

If You Scored: <7: (In Unlimited Time)

Step One

Go through the Block III Back to School Section carefully. Grasp each of the concepts explained in that part carefully. I would recommend that you go back to your Mathematics school books (ICSE/ CBSE) Class 8,9 & 10 if you feel you need it.

Step Two

Move into each of the chapters of Block III one by one.

When you do so, concentrate on clearly understanding each of the concepts explained in the chapter theory.

Step Three

Then move onto the LOD 1 exercises. These exercises will provide you with the first level of challenge. Try to solve each and every question provided under LOD 1. While doing so do not think about the time requirement. Once you finish solving LOD 1, revise the questions and their solution processes.

Step Four

Go to the first review test given at the end of the block and solve it. While doing so, first look at the score you get within the time limit mentioned. Then continue to solve the test further without a time limit and try to evaluate the improvement in your unlimited time score.

In case the growth in your score is not significant, go back to the theory of each chapter and review each of the LOD 1 questions for all the chapters.

Step Five

Move to LOD 2 and repeat the process that you followed in LOD 1 in each of the chapters. Concentrate on understanding each and every question and it’s underlying concept.

Step Six

Go to the second review test given at the end of the block and solve it. Again, while doing so measure your score within the provided time limit first and then continue to solve the test further without a time limit and try to evaluate the improvement that you have had in your score.

Step Seven

Move to LOD 3 only after you have solved and understood each of the questions in LOD 1 & LOD 2. Repeat the process that you followed in LOD 1 – going into each chapter one by one.

Step Eight

Go to the remaining review tests given at the end of the block and solve them. Again, while doing so measure your score within the provided time limit first and then continue to solve the test further without a time limit and try to evaluate the improvement that you have had in your score.

In case the growth in your score is not significant, go back to the theory of each chapter and review each of the LOD 1,2 & 3 questions for all the chapters.

If You Scored:7–15 (In Unlimited Time)

Although you are better than the person following the instructions above, obviously there is a lot of scope for the development of your score. You will need to work both on your concepts as well as speed. Initially emphasize more on the concept development aspect of your preparations, then move your emphasis onto speed development. The following process is recommended for you:

Step One

Go through the Block III Back to School Section carefully. Revise each of the concepts explained in that part. Going through your 8th, 9th and 10th standard books will be an optional exercise for you. It will be recommended in case you scored in single digits, while if your score is in two digits, I leave the choice to you.

Step Two

Move into each of the chapters of Block III one by one.

When you do so, concentrate on clearly understanding each of the concepts explained in the chapter theory.

Step Three

Then move onto the LOD 1 & LOD 2 exercises. These exercises will provide you with the first level of challenge. Try to solve each and every question provided under LOD 1 & 2. While doing so do not think about the time requirement. Once you finish solving LOD 1, revise the questions and their solution processes. Repeat the same process for LOD 2.

Step Four

Go to the first review test given at the end of the block and solve it. While doing so, first look at the score you get within the time limit mentioned. Then continue to solve the test further without a time limit and try to evaluate the improvement in your unlimited time score.

Step Five

Go to the second review test given at the end of the block and solve it. Again, while doing so measure your score within the provided time limit first and then continue to solve the test further without a time limit and try to evaluate the improvement that you have had in your score.

In case the growth in your score is not significant, go back to the theory of each chapter and review each of the LOD 1& LOD 2 questions for all the chapters.

Step Six

Move to LOD 3 only after you have solved and understood each of the questions in LOD 1 & LOD 2. Repeat the process that you followed in LOD 1 – going into each chapter one by one.

Step Seven

Go to the remaining review tests given at the end of the block and solve them. Again, while doing so measure your score within the provided time limit first and then continue to solve the test further without a time limit and try to evaluate the improvement that you have had in your score.

In case the growth in your score is not significant, go back to the theory of both the chapters and re solve all LODs of all the chapters. While doing so concentrate more on the LOD 2 & LOD 3 questions.

If You Scored 15+ (In Unlimited Time)

Obviously you are much better than the first two category of students. Hence unlike them, your focus should be on developing your speed by picking up the shorter processes explained in this book. Besides, you might also need to pick up concepts that might be hazy in your mind. The following process of development is recommended for you:

Step One

Quickly review the concepts given in the Block III Back to School Section. Only go deeper into a concept in case you find it new. Going back to school level books is not required for you.

Step Two

Move into each of the chapters of Block III one by one.

When you do so, concentrate on clearly understanding each of the concepts explained in the chapter theory.

Then move onto the LOD 1 & LOD 2 exercises. These exercises will provide you with the first level of challenge. Try to solve each and every question provided under LOD 1 & 2. While doing so try to work on faster processes for solving the same questions. Concentrate on how you could have solved the same question faster. Also try to think of how much time you took over the calculations.

Step Three

Go to the first review test given at the end of the block and solve it. While doing so, first look at the score you get within the time limit mentioned. Then continue to solve the test further without a time limit and try to evaluate the improvement in your unlimited time score.

Step Four

Go to the second review test given at the end of the block and solve it. Again, while doing so measure your score within the provided time limit first and then continue to solve the test further without a time limit and try to evaluate the improvement that you have had in your score.

Step Five

In case the growth in your score is not significant (esp. under time limits), review each of the LOD 1 & LOD 2 questions for all the chapters.

Step Six

Move to LOD 3 only after you have solved and understood each of the questions in LOD 1 & LOD 2. Repeat the process that you followed in LOD 1 – going into each chapter one by one.

Step Seven

Go to the remaining review tests given at the end of the block and solve them. Again, while doing so measure your score within the provided time limit first and then continue to solve the test further without a time limit and try to evaluate the improvement that you have had in your score.

Chapter 5. PERCENTAGES

INTRODUCTION

In my opinion, the chapter on Percentages forms the most important chapter (apart from Number Systems) in the syllabus of the CAT and the XLRI examination. The importance of ‘percentages’ is accentuated by the fact that there are a lot of questions related to the use of percentage in all chapters of commercial arithmetic (especially Profit and Loss, Ratio and Proportion, Time and Work, Time, Speed and Distance).

Besides, the calculation skills that you can develop while going through the chapter on percentages will help you in handling Data Interpretation (DI) calculations. A closer look at that topic will yield that at least 80% of the total calculations in any DI paper is constituted of calculations on additions and percentage.

BASIC DEFINITION AND UTILITY OF PERCENTAGE

Percent literally, means ‘for every 100’ and is derived from the French word ‘cent’, which is French for 100.

The basic utility of Percentage arises from the fact that it is one of the most powerful tools for comparison of numerical data and information. It is also one of the simplest tools for comparison of data.

In the context of business and economic performance, it is specifically useful for comparing data such as profits, growth rates, performance, magnitudes and so on.

Mathematical definition of percentage The concept of percentage mainly applies to ratios, and the percentage value of a ratio is arrived at by multiplying by 100 the decimal value of the ratio.

For example, a student scores 20 marks out of a maximum possible 30 marks. His marks can then be denoted as 20 out of 30 = (20/30) or (20/30) × 100% = 66.66%.

The process for getting this is perfectly illustrated through the unitary method:

Then the value of x × 30 = 20 × 100

x = (20/30) × 100 Æ the percentage equivalent of a ratio.

Now, let us consider a classic example of the application of percentage:

Example: Student A scores 20 marks in an examination out of 30 while another student B scores 40 marks out of 70. Who has performed better?

Solution: Just by considering the marks as 20 and 40, we do not a get clear picture of the actual performance of the two students. In order to get a clearer picture, we consider the percentage of marks.

Thus, A gets (20/30) × 100 = 66.66%

While B gets (40/70) × 100 = 57.14%

Now, it is clear that the performance of A is better.

Consider another example:

Example: Company A increases its sales by 1 crore rupees while company B increases its sales by 10 crore rupees. Which company has grown more?

Solution: Apparently, the answer to the question seems to be company B. The question cannot be answered since we don’t know the previous year’s sales figure (although on the face of it Company B seems to have grown more).

If we had further information saying that company A had a sales turnover of ` 1 crore in the previous year and company B had a sales turnover of ` 100 crore in the previous year, we can compare growth rates and say that it is company A that has grown by 100%. Hence, company A has a higher growth rate, even though in terms of absolute value increase of sales, company B has grown much more.

IMPORTANCE OF BASE/ DENOMINATOR FOR PERCENTAGE CALCULATIONS

Mathematically, the percentage value can only be calculated for ratios that, by definition, must have a denominator. Hence, one of the most critical aspects of the percentage is the denominator, which in other words is also called the base value of the percentage. No percentage calculation is possible without knowing the base to which the percentage is to be calculated.

Hence, whenever faced with the question ‘What is the percentage …?’ always try first to find out the answer to the question ‘Percentage to what base?’

CONCEPT OF PERCENTAGE CHANGE

Whenever the value of a measured quantity changes, the change can be captured through

(a)Absolute value change or

(b)Percentage change.

Both these measurements have their own advantages and disadvantages.

Absolute value change: It is the actual change in the measured quantity. For instance, if sales in year 1 is ` 2500 crore and the sales in year 2 is ` 2600 crore, then the absolute value of the change is ` 100 crore.

Percentage change: It is the percentage change got by the formula

Percentage change = × 100

= × 100 = 4%

As seen earlier, this often gives us a better picture of the effect of the change.

Example: The population of a city grew from 20 lakh to 22 lakh. Find the

(a)percentage change

(b)percentage change based on the final value of population

Solution:

(a)percentage change = (2/20) × 100 = 10%

(b)percentage change on the final value = (2/22) × 100 = 9.09%

Difference between the Percentage Point Change and the Percentage Change

The difference between the percentage point change and the percentage change is best illustrated through an example. Consider this:

The savings rate as a percentage of the GDP was 25% in the first year and 30% in the second year. Assume that there is no change in the GDP between the two years. Then:

Percentage point change in savings rate = 30% – 25% = 5 percentage points.

Percentage change in savings rate = × 100 = 25%.

PERCENTAGE RULE FOR CALCULATING PERCENTAGE VALUES THROUGH ADDITIONS

Illustrated below is a powerful method of calculating percentages. In my opinion, the ability to calculate percentage through this method depends on your ability to handle 2 digit additions. Unless you develop the skill to add 2 digit additions in your mind, you are always likely to face problems in calculating percentage through the method illus- trated below. In fact, trying this method without being strong at 2-digit additions/subtractions (including 2 digits after decimal point) would prove to be a disadvantage in your attempt at calculating percentages fast.

This process, essentially being a commonsense process, is best illustrated through a few examples:

Example: What is the percentage value of the ratio: 53/81?

Solution: The process involves removing all the 100%, 50%, 10%, 1%, 0.1% and so forth of the denominator from the numerator.

Thus, 53/81 can be rewritten as: (40.5 + 12.5)/81 = 40.5/81 + 12.5/81 = 50% + 12.5/81

= 50% + (8.1 + 4.4)/81 = 50% + 10% + 4.4/81

= 60% + 4.4/81

At this stage you know that the answer to the question lies between 60 – 70% (Since 4.4 is less than 10% of 81)

At this stage, you know that the answer to the calculation will be in the form: 6a.bcde ….

All you need to do is find out the value of the missing digits.

In order to do this, calculate the percentage value of 4.4/81 through the normal process of multiplying the numerator by 100.

Thus the % value of = =

[Note: Use the multiplication by 100, once you have the 10% range. This step reduces the decimal calculations.]

Thus = 5% with a remainder of 35

Our answer is now refined to 65.bcde. (1% Range)

Next, in order to find the next digit (first one after the decimal add a zero to the remainder;

Hence, the value of ‘b’ will be the quotient of

b Æ 350/81 = 4 Remainder 26

Answer: 65.4cde (0.1% Range)

c Æ 260/81 = 3 Remainder 17

Answer: 65.43 (0.01% Range)

and so forth.

The advantages of this process are two fold:

(1)You only calculate as long as you need to in order to eliminate the options. Thus, in case there was only a single option between 60 – 70% in the above question, you could have stopped your calculations right there.

(2)This process allows you to go through with the calculations as long as you need to.

However, remember what I had advised you right at the start: Strong Addition skills are a primary requirement for using this method properly.

To illustrate another example:

What is the percentage value of the ratio ?

223/72 Æ 300 – 310% Remainder 7

700/72 Æ 9. Hence 309 – 310%, Remainder 52

520/72 Æ 7. Hence, 309.7, Remainder 16

160/72 Æ 2. Hence, 309.72 Remainder 16

Hence, 309.7222 (2 recurs since we enter an infinite loop of 160/72 calculations).

In my view, percentage rule (as I call it) is one of the best ways to calculate percentages since it gives you the flexibility to calculate the percentage value up to as many digits after decimals as you are required to and at the same time allows you to stop the moment you attain the required accuracy range.

Effect of a Percent Change in the Numerator on a Ratio’s Value

The numerator has a direct relationship with the ratio, that is, if the numerator increases the ratio increases. The percentage increase in the ratio is the same as the percentage increase in the numerator, if the denominator is constant.

Thus, is exactly 10% more than . (in terms of percentage change)

Percentage Change Graphic and its Applications

In mathematics there are many situations where one is required to work with percentage changes. In such situations the following thought structure (Something I call Percentage Change Graphic) is a very useful tool:

What I call Percentage Change Graphic (PCG) is best illustrated through an example:

Suppose you have to increase the number 20 by 20%. Visualise this as follows:

The PCG has 6 major applications listed and explained below: PCG applied to:

1.Successive changes

2.Product change application

3.Product constancy application

4.A Æ BÆA application

5.Denominator change to Ratio Change application

6.Use of PCG to calculate Ratio Changes

Application 1: PCG Applied to Successive Changes

This is a very common situation in most questions.

Suppose you have to solve a question in which a number 30 has two successive percentage increases (20% and 10% respectively).

The situation is handled in the following way using PCG:

Illustration

A’s salary increases by 20% and then decreases by 20%. What is the net percentage change in A’s salary?

Solution:

Hence, A’s salary has gone down by 4%

Illustration

A trader gives successive discounts of 10%, 20% and 10% respectively. The percentage of the original cost price he will recover is:

Solution:

Hence the overall discount is 35.2% and the answer is 64.8%.

Illustration

A trader marks up the price of his goods by 20%, but to a particularly haggling customer he ends up giving a discount of 10% on the marked price. What is the percentage profit he makes?

Solution:

Hence, the percentage profit is 8%.

Application 2: PCG applied to Product Change

Suppose you have a product of two variables say 10 × 10.

If the first variable changes to 11 and the second variable changes to 12, what will be the percentage change in the product? [Note there is a 10% increase in one part of the product and a 20% increase in the other part.]

The formula given for this situation goes as: (a + b + ab/100)

Hence, Required % change = 10 + 20 +

(Where 10 and 20 are the respective percentage changes in the two parts of the product) (This is being taught as a shortcut at most institutes across the country currently.)

However, a much easier solution for this case can be visualized as:

. Hence, the final product shows a 32% increase.

Similarly suppose 10 × 10 × 10 becomes 11 × 12 × 13

In such a case the following PCG will be used:

Hence, the final product sees a 71.6 percent increase

(Since, the product changes from 100 to 171.6)

Also note that this process is very similar to the one used for calculating successive percentage change.

Application for DI:

Suppose you have two pie charts as follows:

If you are asked to calculate the percentage change in the sales revenue of scooters for the company from year one to year two, what would you do?

The formula for percentage change would give us:

i.e.

Obviously this calculation is easier said than done.

However, the Product change application of PCG allows us to execute this calculation with a lot of ease comparatively. Consider the following solution:

Product for year one is: 0.2347 × 17342.34

Product for year two is: 0.2655 × 19443.56

These can be approximated into:

234 × 173 and 265 × 194 respectively (Note that by moving into three digits we do not end up losing any accuracy. We have elaborated this point in the chapter on Ratio and Proportions.)

The overall percentage change depends on two individual percentage changes:

234 increases to 265: A % change of 31/234 = 13.2 % approx. This calculation has to be done using the percentage rule for calculating the percentage value of the ratio

173 increases to 194 – A percentage change of approximately 12%.

Thus PCG will give the answer as follows:

Hence, 26.76 % increase in the product’s value. (Note that the value on the calculator for the full calculation sans any approximations is 26.82 %, and given the fact that we have come extremely close to the answer—the method is good enough to solve the question with a reasonable degree of accuracy.)

Application 3 of PCG: Product Constancy Application

(Inverse proportionality)

Suppose you have a situation wherein the price of a commodity has gone up by 25%. In case you are required to keep the total expenditure on the commodity constant, you would obviously need to cut down on the consumption. By what percentage? Well, PCG gives you the answer as follows:

Hence, the percentage drop in consumption to offset the price increase is 20%.

I leave it to the student to discover the percentage drop required in the second part of the product if one part increases by 50 percent.

Table 5.1 gives you some standard values for this kind of a situation.

Application 4 of PCG: AÆBÆA.

Very often we are faced with a situation where we compare two numbers say A and B. In such cases, if we are given a relationship from A to B, then the reverse relationship can be determined by using PCG in much the same way as the product constancy use shown above.

Illustration

B’s salary is 25% more than A’s salary. By what percent is A’s salary less than B’s salary?

A drop of 25 on 125 gives a 20% drop.

Hence A’s salary is 20% less than B’s.

Application 5 of PCG Æ Effect of change in Denominator on the Value of the Ratio

The denominator has an inverse relationship with the value of a ratio.

Hence the process used for product constancy (and explained above) can be used for calculating percentage change in the denominator.

For instance, suppose you have to evaluate the difference between two ratios:

Ratio 1 : 10/20

Ratio 2 : 10/25

As is evident the denominator is increasing from 20 to 25 by 25%.

If we calculate the value of the two ratios we will get:

Ratio 1 = 0.5, Ratio 2 = 0.4.

% change between the two ratios = × 100 = 20% Drop

This value can be got through PCG as:

Hence, 20% drop.

Direct process for calculation

To find out the percentage change in the ratio due to a change in the denominator follow the following process:

In order to find the percentage change from 10/20 to 10/25, calculate the percentage change in the denominator in the reverse fashion.

i.e. The required percentage change from R₁ to R₂ will be given by calculating the percentage change in the denominators from 25 to 20 (i.e. in a reverse fashion) & not from 20 to 25.

Table 5.1 Product Constancy Table, Inverse Proportionality Table, A Æ B Æ A table, Ratio Change to Denominator table

Application 6: Use of PCG to Calculate Ratio Changes:

Under normal situations, you will be faced with ratios where both numerator and denominator change. The process to handle and calculate such changes is also quite convenient if you go through PCG.

Illustration

Calculate the percentage change between the Ratios.

Ratio 1 = 10/20 Ratio 2 = 15/25

The answer in this case is 0.5 Æ 0.6 (20% increase). However, in most cases calculating the values of the ratio will not be easy. The following PCG process can be used to get the answer:

When 10/20 changes to 15/25, the change occurs primarily due to two reasons:

(A)Change in the numerator (Numerator effect)

(B)Change in the denominator (Denominator effect)

By segregating the two effects and calculating the effect due to each separately, we can get the answer easily as follows:

Numerator Effect The numerator effect on the value of the ratio is the same as the change in the numerator. Hence, to calculate the numerator effect, just calculate the percentage change in the numerator:

In this case the numerator is clearly changing from 10 to 15 (i.e. a 50% increase.) This signifies that the numerator effect is also 50%.

Denominator Effect As we have just seen above, the effect of a percentage change in the denominator on the value of the ratio is seen by calculating the denominator’s percentage change in the reverse order.

In this case, the denominator is changing from 20 to 25. Hence the denominator effect will be seen by going reverse from 25 to 20 i.e. 20% drop.

With these two values, the overall percentage change in the Ratio is seen by:

This means that the ratio has increased by 20%.

I leave it to the student to practice such calculations with more complicated values for the ratios.

Implications for Data Interpretation

Percentage is perhaps one of the most critical links between QA and Data Interpretation.

In the chapter theory mentioned above, the Percentage Rule for Percentage Calculations and the PCG applied to product change and ratio change are the most critical.

As already shown, the use of PCG to calculate the percentage change in a product (as exhibited through the pie chart example above) as well as the use of PCG to calculate ratio changes are two extremely useful applications of the concepts of percentages into DI.

Applying Percentages for the special case of comparing two ratios to find the larger one.

Suppose you have two ratios to compare. Say R₁ = N₁/D₁ and R₂ = N₂/D₂

The first step is to find the ten percent ranges for each of these ratios. In case, they belong to different ranges of 10% (say R1 lies between 50-60 while R2 lies between 70 to 80), it becomes pretty simple to say which one will be higher.

In case, both of these values for percentage of the ratios belong to the same ten percent range, then we can use the following process

Step 1: Calculate the percentage change in the numerator

Step 2: Calculate the percentage change in the denominator.

There could be four cases in this situation, when we move from Ratio₁ to Ratio₂:

Case 1: Numerator is increasing while denominator is decreasing Æ obviously the net effect of the two changes will be an increase in the ratio. Hence, R₂ will be greater.

Case 2: Numerator is decreasing while denominator is increasing Æ obviously the net effect of the two changes will be a decrease in the ratio. Hence, R₁ will be greater.

It is only in the following cases that we need to look at the respective changes in the Numerator and denominator.

Case 3: Numerator and denominator are both increasing

Calculate the percentage value of the respective increases. If the numerator is increasing more than the denominator the ratio will go up. On the other hand, if the denominator is increasing more than the numerator, Ratio₂ will be smaller than Ratio₁. (Note: Compare in percentage values)

Case 4: Numerator and denominator are both decreasing Æ

Calculate the percentage value of the respective decreases. If the numerator is decreasing more than the denominator the ratio will go down. On the other hand, if the denominator is decreasing more than the numerator, Ratio₂ will be greater than Ratio₁.

FRACTION TO PERCENTAGE CONVERSION TABLE

The following percentage values appear repeatedly over the entire area where questions can be framed on the topic of percentage. Further, it would be of great help to you if you are able to recognize these values separately from values that do not appear in the Table 5.2.

Some Utilisations of the Table

• The values that appear in the table are all percentage values. These can be converted into decimals by just shifting the decimal point by two places to the left. Thus, 83.33% = 0.8333 in decimal value.
• A second learning from this table is in the process of division by any of the numbers such as 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 16, 24 and so on, students normally face problems in calculating the decimal values of these divisions. However, if one gets used to the decimal values that appear in the Table 5.2, calculation of decimals in divisions will become very simple. For instance, when an integer is divided by 7, the decimal values can only be .14, .28, .42, .57, .71, .85 or .00. (There are approximate values)
• This also means that the difference between two ratios like can be integral if and only if xis divisible by both 6 and 7.

This principle is very useful as an advanced short cut for option based solution of some questions. I leave it to the student to discover applications of this principle.

Table 5.2 Percentage Conversion Table

Formula for any cell = Column value × 100/Row value

Calculation of Multiplication by Numbers like 1.21, 0.83 and so on

In my opinion, the calculation of multiplication of any number by a number of the form 0.xy or of the form 1.ab should be viewed as a subtraction/addition situation and not as a multiplication situation. This can be explained as follows.

Example: Calculate 1.23 × 473.

Solution: If we try to calculate this by multiplying, we will end up going through a very time taking process, which will yield the final value at the end but nothing before that (i.e. you will have no clue about the answer’s range till you reach the end of the calculation).

Instead, one should view this multiplication as an addition of 23% to the original number. This means, the answer can be got by adding 23% of the number to itself.

Thus 473 × 1.23 = 473 + 23% of 473 = 473 + 94.6 + 3% of 473 = 567.6 + 14.19 = 581.79

(The percentage rule can be used to calculate the addition and get the answer.)

The similar process can be utilised for the calculation of multiplication by a number such as 0.87

(Answer can be got by subtracting 13% of the number from itself and this calculation can again be done by percentage rule.)

Hence, the student is advised to become thorough with the percentage rules. Percentage calculation & additions of 2 & 3 digit numbers.

WORKED-OUT PROBLEMS

Problem 5.1 A sells his goods 30% cheaper than B and 30% dearer than C. By what percentage is the cost of C’s goods cheaper than B’s goods.

Solution There are two alternative processes for solving this question:

1. Assume the price of C’s goods as p.:Then A’s goods are at 1.3 p and B’s goods are such that A’s goods are 30% cheaper than B’s goods. i.e. A’s goods are priced at 70% of B’s goods.

Hence, 1.3 p Æ 70

B’s price Æ 100

B’s price = 130 p/70 = 1.8571 p

Then, the percentage by which C’s price is cheaper than B’s price =

(1.8571 p – p) × 100/(1.8571 p) = 600/13= 46.15%

Learning task for student Could you answer the question: Why did we assume C’s price as a variable p and then work out the problem on its basis. What would happen if we assumed B’s price as p or if we assumed A’s price as p?

2. Instead of assuming the price of one of the three as p, assume the price as 100.

Let B = 100. Then A = 70, which is 30% more than C. Hence C = 23.07% less than A (from Table 4.1) = approx. 53.84. Hence answer is 46.15% approximately.

(This calculation can be done mentally if you are able to work through the calculations by the use of percentage rule. The students are advised to try to assume the value of 100 for each of the variables A, B and C and see what happens to the calculations involved in the problem. Since the value of 100 is assumed for a variable to minimise the requirements of calculations to solve the problems, we should ensure that the variable assumed as 100 should have the maximum calculations associated with it.)

Problem 5.2 The length and the breadth of a rectangle are changed by +20% and by –10% respectively. What is the percentage change in the area of the rectangle.

Solution The area of a rectangle is given by: length × breadth. If we represent these by:

Area = L × B = LB Æ then we will get the changed area as

Area_(NEW) = 1.2 L × 0.9 B = 1.08 LB

Hence, the change in area is 8% increase.

Problem 5.3 Due to a 25% price hike in the price of rice, a person is able to purchase 20 kg less of rice for ` 400. Find the initial price.

Solution Since price is rising by 25%, consumption has to decrease by 20%. But there is an actual reduction in the consumption by 20 kg. Thus, 20% decrease in consumption is equal to a 20 kg drop in consumption.

Hence, original consumption is: 100 kg of rice.

Money spent being ` 400, the original price of rice is ` 4 per kg.

(There, you see the benefit of internalising the product constancy table! It is left to the student to analyze why and how the product constancy table applies here.)

Problem 5.4 A’s salary is 20% lower than B’s salary, which is 15% lower than C’s salary. By how much percent is C’s salary more than A’s salary?

Solution The equation approach here would be

A = 0.8 B

B = 0.85 C

Then A = 0.8 × 0.85 C

A = 0.68 C (Use percentage change graphic to calculate the value of 0.68)

Thus, A’s salary is 68 % of C’s salary.

If A’s salary is 68, B’s salary is 100.

Using percentage change graphic

68100

Students are advised to refrain from using equations to solve questions of this nature. In fact, you can adopt the following process, which can be used while you are reading the problem, to get the result faster.

Assume one of the values as 100. (Remember, selection of the right variable that has to take the value of 100 may make a major difference to your solving time and effort required. The thumb rule for selecting the variable whose value is to be taken as 100 is based on three principal considerations:

Select as 100, the variable

1.With the maximum number of percentage calculations associated with it.

2.Select as 100 the variable with the most difficult calculation associated with it.

3.Select as 100 the variable at the start of the problem solving chain.

The student will have to develop his own judgment in applying these principles in specific cases.

Here I would take C as 100, getting B as 85 and A as 68.

Hence, the answer is (32 × 100/68).

Problem 5.5 The cost of manufacture of an article is made up of four components A, B, C and D which have a ratio of 3 : 4 : 5 : 6 respectively. If there are respective changes in the cost of +10%, –20%, –30% and +40%, then what would be the percentage change in the cost.

Solution Assume the cost components to be valued at 30, 40, 50 and 60 as you read the question. Then we can get changed costs by effecting the appropriate changes in each of the four components.

Thus we get the new cost as 33, 32, 35 and 84 respectively.

The original total cost was 180 the new one is 184. The percent change is 4/180 = 2.22%.

Problem 5.6 Harsh receives an inheritance of a certain amount from his grandfather. Of this he loses 32.5% in his effort to produce a film. From the balance, a taxi driver stole the sum of ` 1,00,000 that he used to keep in his pocket. Of the rest, he donated 20% to a charity. Further he purchases a flat in Ganga Apartment for ` 7.5 lakh. He then realises that he is left with only ` 2.5 lakh cash of his inheritance. What was the value of his inheritance?

Solution These sort of problems should either be solved through the reverse process or through options.

Reverse process for this problem He is left with ` 2.5 lakh after spending ` 7.5 lakh on the apartment.

Therefore, before the apartment purchase he has ` 10 lakh. But this is after the 20% reduction in his net value due to his donation to charity. Hence, he must have given ` 2.5 lakh to charity (20% decrease corresponds to a 25% increase). As such, he had 12.5 lakh before the charity. Further, he must have had ` 13.5 lakh before the taxi driver stole the sum. From 13.5 lakh you can reach the answer by trial and error trying whole number values. You will get that if he had 20 lakh and lost 32.5% of it he would be left with the required 13.5 lakh.

Hence, the answer is ` 20 lakh.

This process can be done mentally by: 2.5 + 7.5 = 10 lakh Æ +25% Æ 12.5 lakh Æ + 1 lakh Æ 13.5 lakh.

From this point move by trial and error. You should try to find the value of the inheritance, which on reduction by 32.5%, would leave 13.5 lakh. A little experience with numbers leaves you with ` 20 lakh as the answer. This process should be started as soon as you finish reading the first time.

Through options Suppose the options were:

(a) 25 lakh   (b) 22.5 lakh

(c) 20 lakh   (d) 18 lakh

Start with any of the middle options. Then keep performing the mathematical operation in the order given in the problem. The final value that he is left with should be ` 2.5 lakh. The option that gives this, will be the answer. If the final value yielded is higher than ` 2.5 lakh in this case, start with a value lower than the option checked. In case it is the opposite, start with the option higher than the one used.

As a thumb rule, start with the most convenient option—the middle one. This would lead us to start with ` 20 lakh here.

However, if we had started with ` 25 lakh the following would have occurred.

25 lakh –32.5% Æ 16.875 lakh –1 lakh Æ 15.875 lakh –20% – 7.5 lakh, should equal 2.5 lakh Æ (Prior to doing this calculation, you should see that there is no way the answer will yield a nice whole number like 2.5 lakh. Hence, you can abandon the process here and move to the next option)

Trying with 20 lakh, 20 –32.5% Æ 13.5 lakh –1ac. Æ 12.5 lakh –20% Æ 10 lakh – 7.5 lakh = 2.5 lakh Æ Required answer.

LEVEL OF DIFFICULTY (I)

LEVEL OF DIFFICULTY (II)

Directions for Questions 36 to 38: Read the following passage and answer the questions.

In a recent youth fete organised by Mindworkzz, the entry tickets were sold out according to the following scheme:

This offer could have been availed only when tickets were bought in a fixed lot according to the scheme and any additional ticket was available at its original price.

Questions 49 and 50: Study the following table and answer the questions that follow.

LEVEL OF DIFFICULTY (III)

Directions for Questions 13 to 16: Read the following and answer the questions that follow.

Two friends Shayam and Kailash own two versions of a car. Shayam owns the diesel version of the car, while Kailash owns the petrol version.

Kailash’s car gives an average that is 20% higher than Shayam’s (in terms of litres per kilometre). It is known that petrol costs 60% of its price higher than diesel.

Directions for Questions 17 to 23: Read the following and answer the questions that follow.

In the island of Hoola Boola Moola, the inhabitants have a strange process of calculating their average incomes and expenditures. According to an old legend prevalent on that island, the average monthly income had to be calculated on the basis of 14 months in a calendar year while the average monthly expenditure was to be calculated on the basis of 9 months per year. This would lead to people having an underestimation of their savings since there would be an underestimation of the income and an overestimation of the expenditure per month.

Directions for Questions 28 to 30: Read the following and answer the questions that follow the BSNL announced a cut in the STD rates on 27 December 2011. The new rates and slabs are given in the table below and are to be implemented from 14 January 2012.

Slab Details

Directions for Questions 31 to 35: Read the following caselet and answer the questions that follow.

The circulation of the Deccan Emerald newspaper is 3,73,000 copies, while its closest competitors are The Times of Hindustan and India’s Times, which sell 2,47,000 and 20% more than that respectively (rounded off to the higher thousand). All the newspapers cost ` 2 each. The hawker’s commissions offered by the three papers are 20%, 25% and 30% respectively (these commissions are calculated on the sale price of the newspaper). Also, it is known that newspapers earn primarily through sales and advertising.

Directions for Questions 38 to 40: Read the following and answer the questions that follow.

A Eurailexpress train has 2 AC I bogeys having 24 berths each, 3 AC II bogeys having 45 berths each, 2 AC III bogeys having 64 berths each and 12 3-tier bogeys having 64 berths each. There are no general bogeys in the train. If 200 euros is the cost of an AC 3-tier berth from London to Glasgow, answer the following questions:

Directions for Questions 45 to 48: In an economy the rate of savings has a relation to the investment in industry for that year and for the following three years and the investment in industry for that year has a relation to the total production output in the next 4 years.

Level of Difficulty (I)

Hints

Level of Difficulty (II)

1.Assume the initial value to be 100 and solve.

3.Total number of students = Full fee waver + 50% concession + No concession.

6.Assume initial value of price = 100.

Since, the price is a mutiplicative function, we have

100 × 1.1 × 1.2 × 0.8 × 1.25 × 1.5

Solve using percentage change graphic.

8.Salary ratio is 2.25 : 2.2.6666.

Hence 0.41666 = ` 40.

Then, 6.9166 = ` 664

9.Solve using options

13.Assume initial amount of gold to be 100.

Then he gives away: 50 + 25 + 12.5 = 87.5

But 87.5 = 1309000 kg

Hence, 100 = 149600

18.If initial income = 100, initial food expenditure = 25.

New income = 120

Since, food expenditure is constant at 25, the percentage of the new income = 20.833.

Percentage point change = 25 – 20.833 = 4.166

24.Solve using options.

25.Use standard formulae of percentage.

Alternatively, this problem can also be solved by assuming values for p, x, y and z. Then compare the options to see which one fits.

29.Solve using options.

31.100 × 1.125 × 0.92 = 103.5

34.Solve using options.

35.Reduction required = × 2568

36-38.To maximise the discount, tickets, need to be bought in two groups of 18 and 6 respectively. The maximum possible price per ticket will occur when tickets are bought in sets of 6.

39.Solve through options.

Checking for option (a) Æ Final voting is 200 for and 400 against.

Hence, the motion is rejected by 200 votes. This means that if none of the things had occurred then the motion would have been passed by 400 votes. i.e. 600 (for) and 200 (against) 1/3 of 600 were abducted and the opponents doubled their voted from 200 to 400.

Since all the values fit; the answer is (a).

40.Take 45% men = 25% women.

42.Since, runs scored = overs × run rate. If overs reduce by 25%, run rate will go up by 33.33%. Hence, Australia could have scored any number of runs.

47.Valid votes = 64%

The second placed candidate gets 20% votes.

Then the winner can get between 20.01% to 44% votes.

Level of Difficulty (III)

1.Assume initial raw material price to be 100. This means that the initial labour cost is 25. Hence the net cost is 125. Now, since there is a 15% increment in raw material cost and the labour cost has gone up to 30% of the raw material cost, it is clear that the new total expenditure is 115* 1.3 = 149.5. Reduce the cost to 125 by reducing the usage of raw materials used.

2.Assume that 50, 40 and 20 hours are available. There is no need to use 10% waste of time in this question.

4.Half as large again means 1.5 times (or an addition of 50%).

5.Assume values for M and x and solve through options.

6.A = 1.5 B , A – C = 180000 and B = 1.05 C. Solve to get A, B and C. Also, A + B + C = 90% of total voters on voting list. This will give you the answer.

Ideally solve this question through options.

7.Clock loses 0.5% of 168 hours in the first week and gains 0.25% of 168 hours in the second week. Hence, net loss is 0.25% of 168 hours.

8.Vawal uses 133.33 litres of petrol every month, while the price of petrol has gone up by ` 1.96. Hence, the increase in expenditure = 133.33 * 1.96 = ` 261 approximately.

11.Maximum revenue for the shopkeeper will occur when the minimum discount offer is used by the customer. This level is 4%.

12.This is the case of maximum discounts.

Hints for Questions 13–16

13.Average in liter per kilometre multiplied by the Cost of fuel in `/litre will give the required cost per kilometre.

14.Shyam’s car gives 20 km/litre means 0.05 litres per kilometre then Kailash’s car gives 0.06 litre/km. However, since we do not know the price of petrol or diesel we cannot find out the difference in the cost of travel.

15.This question is the opposite of question 13.

16.Cost of petrol is ` 25 per liter. Cost per kilometre for Shyam = 12.5 × 0.05

Also, cost per kilometre for Kailash = 25 × 0.06

Hints for Questions 17–23

Estimated average savings

= –

17.The value will depend on the values of annual expenditure which is not available.

18.Average estimated monthly expenditure is given for the island of Hoola Boola Moola and not for Mr. Boogle Woogle’s community.

19.Original estimated savings = 87 – 22 = 65 Moolahs.

New estimated savings = 1218/12 – 198/12 = 85.

24.0.72 × 1.15 × 0.68 × 1.11.

25.Solve through options: A 15% reduction on the correct answer will give a profit of 30.05%.

Option (a) is correct.

26.The last discount being 22,950, it means that the value prior to this 15% discount must have been 1,53,000 checking with options:

200,000 17,000 1,53,000. Hence option (c) is correct.

27.For 45% of the journey in city driving conditions, 54% of the fuel is consumed.

Hence, for the remaining 55% journey, 46% fuel is left.

Required increase in fuel efficiency

= × 100.

28.The maximum percentage reduction in peak rates is for the 200 – 500 category.

29.

33.Loss to be made up everyday = 373000(8 – 1.60)

= 6.4 × 373000.

No. of cc required to be sold =

34.Advertising rates have not been mentioned. Hence, we cannot solve the question.

36-40.The ticket cost are:

AC III Æ 100 (assume), AC – II Æ 120,

AC I Æ 190, 3 Tier Æ 47.5, General Æ 40.

Also, AC – II = 780 Euros for a London – Paris journey

36.(47.5 – 40) × 6.5 = 48.75

37.(100 + 120 + 190 + 47.5 + 40) × 6.5.

38.Maximum revenues on a return journey means 100% bookings both ways.

39. × 100
40. = = 30%

42.Original percentage difference = 30%

At 60% aperture opening the smaller gas will last = 173.33 hours.

Similarly, the larger gas will last = 160 hours.

Thus, the smaller gas lasts × 100 = 8.33% more than the larger gas.

Then, required answer = × 100 = 72.22%

44.The 5% point increase in savings rate will account for a 2.5% increase in investment in 2005–06 and a further 1.25% increase in investment in 2006–07.

Thus, Indian investment is 2006-07 = 2 million × 1.025 × 1.0125 similarly, calculate for Pakistan.

45.Use the same process as for the previous question.

46.Cannot be determined since we do not know the initial values of the production output.

47.Since there is a 2.5% increase in investment in 2005–06, there will be a 2.5% increase in manufacturing production is 2006–07.

Then, GDP growth rate = = 9.26%.

Solutions and Shortcuts

Level of Difficulty (I)

1.It can be clearly seen that 700% of 9 = 63 is the highest number.

2.0.25 N = 75 Æ = 300. Thus, 0.45 × 300 = 135.

3.20% of 50% of 75% of 70% = 20/100 × 50/100 × 75/100 × 70 = 0.2 × 0.5 × 0.75 × 70 = 5.25.

A quicker way to think here would be: 20% of 70 = 14 Æ 50% of 14 = 7 Æ 75% of 7 = 5.25

4.41 (3/17)% = 700/17%. As a fraction, the value = 700/(17 × 100) = 7/17.

5.The following PCG will give the answer:

Hence, the percentage reduction required is 28.56% (40/140)

6.100 Æ 140 (after a 40% increase) Æ 105. The reduction from 140 to 105 is 25% and hence, it means that he needs to reduce his consumption by 25%.

7.Assume Ram as 100. Shyam will be 133.33 and Bram will be 80

Thus, Bram’s goods are 40% cheaper than Shyam’s (53.33/133.33)

8.Total votes = 6000. Valid votes = 75% of 6000 = 4500. Bhiku gets 65% of 4500 votes and Mhatre gets 35% of 4500. Hence, Mhatre gets: 0.35 × 4500 = 1575 votes.

9.If the candidate has inadvertently increased his height by 25% the correction he would need to make to go back to his original height would be to reduce the stated height by 20%.

10.Let Arjit’s height be H. Then, H × 1.15 = 120 Æ H = 120/1.15 = 104.34.

11.Let the number be N. Then, 5N should be the correct outcome. But instead the value got is 0.2N. Change in value = 5N – 0.2N = 4.8N. The percentage change in the value = 4.8N × 100/5N = 96%.

12.The percentage difference would be given by thinking of the percentage change between two numbers: (x – 5) to (x + 5) [‘What he wanted to get’ to ‘what he got by mistake’].

The value of the percentage difference in this case depends on the value of x. Hence, this cannot be answered. Option (d) is correct.

13.65% of x = 13% of 2000 Æ 0.65 x = 260 Æ x = 400

14.From the first statement we get that out of 80 litres of the mixture, 20 litres must be milk. Since, we are adding water to this and keeping the milk constant, it is quite evident that 20 litres of milk should correspond to 20% of the total mixture. Thus, the amount in the total mixture must be 100, which means we need to add 20 litres of water to make 100 litres of the mixture.

15.(50/100) × (a/100) × (b) = (75/100) × (b/100) × (c) Æ 50a = 75c Æ c = 0.667a

16.

Hence, the required answer is 32%

17.The area of a triangle depends on the product base × height.

Since, the height increases by 40% and the area has to increase by 60% overall, the following PCG will give the answer.

The required answer will be 20/140 = 14.28%

18.The volume goes up by:

Hence, 98%

19. (AÆBÆA use of PCG)

\ Answer = 30/130 = 23.07%

20.100 75 120

We have assumed initial expenditure to be 100, in the above figure. Then the final expenditure is 120. The percentage change in consumption can be seen to be 45/75 × 100 = 60%

21.If the price of rice has fallen by 20% the quantity would be increased by 25% (if we keep the expenditure constant.)

This means that 20 kgs would increase by 25% to 25 kgs.

22.91 – 0.3N = N Æ 1.3 N = 91 Æ N = 70.

23.B + 60% of A = 175% of B Æ

60% of A = 75% of B.

i.e. 0.6A = 0.75B

A/B = 5/4

Apparently it seems that A is bigger, but if you consider A and B to be negative the opposite would be true.

Hence, option (d) is correct.

24.The winning candidate gets 56% of the votes cast and the losing candidate gets 44% of the votes cast. Thus, the gap between the two is 12% of the votes cast = 144 votes. Thus, the votes cast = 1200. Since, this is 80% of the number of voters on the voting list, the number of people on the voting list = 1200/0.8 = 1500.

25.100000 110000 121000

26.10000 Æ 11000 (after a 10% increase) Æ 10450 (after a 5% decrease) Æ 12540 (after a 20% increase)

27.His investments are 2000, 3000 and 5000 respectively. His dividends are: 200, 750 and 1000, which means total dividend = 1950.

28.Sushant 1080, hence Mohit = 1080 × 1.2 = 1296. Rajesh = 1296/0.9 = 1440. 1440 out of 2000 means a percentage of 72%.

29.30% students got a final score of 13. 10% students got a final score of 33 (inclusive of grace marks.) 35% students got a final score of 60

Hence, average score of the class

= = 37.6

30.Ram would spend 20%, 12% and 20.4% respectively on household expenditure, books and clothes. His savings would account for 100 – 20 – 12 – 20.4 = 47.6% of his income. Since the savings = 9520, we get 0.476 ×Income = 9520 Æ Income = 9520/0.476 = 20000

31.The only logic for this question is that Hans’ salary would be more than Bhaskar’s salary. Thus, only option (a) is possible for Hans’ salary.

32.By using options, you can easily see that option (a) satisfies.

2500 females means 3000 males.

Increase = 2500 × 0.2 + 3000 × 0.11 = 830

33.If Ashu’s salary =100, then Vicky’s salary = 175. Ashu’s new salary = 125, Vicky’s new salary = 175 × 1.4 = 245. Percentage difference between Vicky’s salary and Ashu’ salary now = 120 × 100/125 = 96%.

34.Let the second row contain 100 books. Then, the first row would contain 125 books and the third row would contain 75 books. The total number of books would be 100 + 125 + 75 = 300. But this number is given as 600 which means that the total number of books would be double the assumed value for each row. Thus, the first row would contain 125 × 2 = 250 books.

35.Since the only iron contained in the ore is 90% of 25%, the net iron percentage would be 22.5%. Thus, 60 kg should be 22.5% of the ore Æ 60/0.225 = 266.66

36.The data is insufficient since the number of matches to be played by India this year is not given. (You cannot assume that they will play 40 matches.)

37.100000 Æ 110000 (after 1 year) Æ 121000 (after 2 years) Æ 133100 (after 3 years and at the start of the fourth year).

38.Total people present = 700 + 500 + 800 = 2000.

Indians = 0.2 × 700 + 0.4 × 500 + 0.1 × 800 = 420 = 21% of the population. Thus, 79% of the people were not Indians.

39.Price of a cow after increase = 2400. Price of a calf after 30% increase = 1820. Cost of 12 cows and 24 calves = 12 × 2400 + 24 × 1820 = 72480

40.If we take Ram as 100, we will get Bobby as 125 and Chandilya as 83.33. This means Chandilya’s goods are priced at 2/3^rd Bobby’s and hence he sells his goods 33.33% cheaper than Bobby.

41.1 Bottle Æ 0.5x metres

? Bottles Æ 400 meters

Using unitary method, we get no. of bottles = 400/0.5x = 800/x Bottles.

42.(100 × 0.8 × 0.25)% = 80000 kg Æ 20% = 80000 kg. Thus, the total quantity of hematite mined = 400000.

43.The total cost for a year = 2,50,000 + 2% of 2,50,000 + 2000

= 2,55,000 + 2000 = 2,57,000

To get a return of 15% he must earn: 2,57,000 × 0.15 = 38,550 in twelve months.

Hence, the monthly rent should be 38550/12 = 3212.5.

44.The sales price of the first shirt is 8/7 × 42 = ` 48.

Hence, I am being offered a discount of ` 6 on a price of ` 48 – a 12.5% discount.

The sales price of the second shirt is 7/6 × 36 = ` 42.

Hence, I am being offered a discount of ` 6 on ` 42 – a 14.28% discount. Hence, the second shirt is a better bargain.

45.72% must have voted for Sonia Gandhi and 16% for Sushma Swaraj. Hence, 88 × 3 = 264.

46.The following Venn diagram would solve this problem:

We can clearly see from the above figure that 65% of the people passed both subjects. Since this value is given as 325, we get that the total number of students who appeared for the exam is 500.

47.Ravana spends 30% on house rent, 21% on children’s education and 24% on clothes. Thus, he spends 75% of his total salary. He thus saves 25% of his salary which is given as being equal to 2500. Thus, his salary is ` 10000.

48. (final sales figure)

Hence, the required price drop is 67.5/150 = 45% drop. Thus there is a drop of 250 × .45 = 112.5

49.A C % increase in income means the new income is A (1 + C/100) while a D% increase in expenditure means that the new expenditure would be X(1 + D/100). Thus, the new savings = A(1 + C/100) – X(1 + D/100)

50.In 2001, BMW = 15%, Maruti = 50% and hence Honda = 35%

Level of Difficulty (II)

1.

42.1875

Now, 42.1875 = ` 16,875

Hence

Also, year 2 donation is 56.25 × 400 = 22500

2.The following figure shows the percentage of failures:

From the figure it is clear that 60% of the people have failed in at least one subject, which means that 40% of the students would have passed in both subjects. This value is given as 880 people. Hence, there would be 880/0.4 = 2200 students who would appear in the examination.

3.The thought process would go like:

If we assume 100 students

This means that a total of 12 people are getting a fee waiver. (But this figure is given as 90.)

Hence, 1 corresponds to 7.5.

Now, number of students not getting a fee waiver

= 51 boys and 37 girls

50% concession Æ 25.5 boys and 18.5 girls (i.e. a total of 44.)

Hence, the required answer = 44 × 7.5 = 330

4.Solve using options. Checking for option (b), gives us:

200000 Æ 180000 Æ 171000 Æ 153900 Æ 146205

(by consecutively decreasing 200000 by 10% and 5% alternately)

5.Total characters in her report = 25 × 60 × 75.

Let the new no. of pages be n.

Then:

n × 55 × 90 = 25 × 60 × 75

n = 22.72

This means that her report would require 23 pages. A drop of 8% in terms of the pages.

6.The following percentage change thinking would give us the value of the percentage increase as 98%

7.The total raise of salary is 87.5% (That is what 15/8 means here).

Using the options and PCG, you get option (c) as the correct answer.

8.October : November : December = 9 : 8 : 10.666 since, he got ` 40 more in December than October, we can conclude that 1.666 = 40 Æ 1 = 24.

Thus, total Bonus for the three months is:

0.4 × 27.666 × 24 = 265.6

9.Solve through trial and error using the options. 12% (option a) is the only value that fits the situation.

10.9% increase is offset by 8.26% decrease. Hence, option (b) is correct.

11.The expenditure increase can be calculated using PCG as:

100 Æ 112.5 Æ 123.75.

A 23.75% increase.

12.Rajesh’s scores in each area is 65 and 82 respectively out of 100 each. Since, the exam is of a total of 250 marks (100+100+50) he needs a total of 195 marks in order to get his target of 78% overall. Thus, he should score 195-65-82 = 195-147=48 marks in Sociology which would mean 96%.

13.The total wealth given would be 50% + 25% (which is got by 50% of the remaining 50%) + 12.5% (which is got by 50% of the remaining 25%). Thus, the total wealth given by him would be equivalent to 87.5% of the total. Since, this is equal to 130900 kilograms of gold, the total gold would be:

130900 × 8/7 = 149600.

14.Population at the start = 100.

Population after 2 years = 100 × 1.08 × 1.01 × 1.08 × 1.01 = 108.984

Thus, the required percentage increase = 18.984%

15.After the migrations, 72.9% of the people would remain in the country. This would comprise females and males in the ratio of 1:2 (as given) and hence, the women’s population left would be 1/3^rd of 72.9% = 24.3% which is given as being equal to 364500. Thus, the total population would be

364500 × 100/24.3 = 1500000

16.24% of the total goes to urban Gujarat $72 m

\ 1% = $ 3 mn.

The required value for Rural AP

= 50% of 20% = 10%

Hence, required answer = $ 30 mn

17.In the previous question, the total FDI was $ 300 mn.

A growth of 20% this year means a total FDI of $360 mn.

The required answer is 12% of 10% of 360 mn

= 1.2% of 360 = $4.32 mn.

18.The income goes to 120. Food expenditure has to be maintained at 25. (i.e. 20.833%)

Hence, percentage point drop from 25 to 20.833 is 4.16%

19.Assume the initial surface area as 100 on each side. A total of 6 such surfaces would give a total surface area of 600. Two surface areas would be impacted by the combined effect of length and breadth, two would be affected by length and height and two would be affected by breadth and height. Thus, the respective surface areas would be (110.25 twice, 126 twice and 126 twice) Thus, new surface area = 220.5 + 504 = 724.5. A percentage increase of 20.75%. Option (d) is correct.

20.Option (c) fits the situation as if the ratio is 10:9, the value of B’s salary would first go up from 10 to 12 and then come down from 12 to 9 (after a 25% decrease). On the other hand, the value of A’s salary would go up from 9 to 11.25 and then come back to 9 (Note that a 25% increase followed by a 20% decrease gets one back to the starting value.)

21.Initial quantity of milk and water = 12 and 48 liters respectively. Since, this is already containing 20% milk, adding more milk to the mixture cannot make the mixture reach 15% milk. Hence, it is not possible.

22.On `100 he saves `6. On 115 he still saves `6. Thus, his expenditure goes up from 94 to 109- a percentage increase of 15 on 94 = 15.95%.

23.B = 100, A = 150, C = 100, D = 160. D is 160% of B. Note that this does not change if all the values are incremented by the same percentage value.

24.Think about this problem through alligation. Since, A spends 12% of his money and B spends 20% of his money and together they spend 15% of their money- we can conclude that the ratio of the money A had to the money B had would be 5:3. Hence, Total money with A = 5 × 1200/8 =750.

Money spent by A = 12% of 750 = 90.

Money left with A = 750 × 90 = 660.

25.Chanda would have spent 12% of Maya

Thus, her percentage of expenditure would be 0.12 M × 100/C = 12 M/C

26.Option (d) is correct and can be verified experimentally by using values for x, y, z and p.

27.The weekly change is equal to ` 1,68,000.

Hence, the daily collection will go up by 1,68,000/7 = 24,000.

28.The total population of the town can be taken as 9 + 8 + 3 = 20.

The number of literates would be:

80% of 9 + 70% of 8 + 90% of 3 = 7.2 + 5.6 + 2.7 = 15.5

15.5 out of 20 represents a 77.5% literacy rate.

29.Solve using options. 2/25 fits the requirement.

30.Let the exam be of 100 marks. A obtains 36 marks (10% or 1/10^th less than the pass marks) while B obtains 32 marks (11.11% or 1/9^th less than A). The sum of A and B’s marks are 36 + 32 = 68. To pass C can obtain 28 marks less than 68 – which is a percentage of 41(3/17)%. If C obtains 28% less marks than 68 or if C obtains 40% less marks than 68 he would still pass. Thus, option (d) is correct.

32.If Z = 100, X = 80 and Y = 72.

Thus, Y is less then X by 10%

33.Assume values of x% = 10% and the original price as 100, then the final price = K/100 = 99 Æ K = 9900.

(Note: After an increase of 10% followed by a decrease of 10% a price of 100 would become 99).

Put these values of x, and K in the options. The option that gives a value of 100 for the original price should be the correct answer.

Option (d) is correct.

34.The correct answer should satisfy the following condition: If ‘x’ is the increased salary

x × 0.8 × 0.1 = (x – 4800) × 0.8 × 0.12.

None of the first 3 options satisfies this.

In fact, solving for x we get x = 25800.

Option (d) is correct.

35.A sales tax of 7% on a price of 2568 would amount to a tax amount of 179.76. Since, the price is rounded off to the next higher integer, the tax would be rounded off to `180. This would also be the amount of discount (or reduction in price) that Reena is asking for.

36.The minimum price occurs at:

18 × 30 + 6 × 36 + 1 × 40

Hence, 796/25 = 31.84

37.36 × 24 + 40 × 1 = 904

Required answer = 904/25 = 36.16.

38.If the ticket lots are halved, the maximum discount will be available for 9 tickets (25%). A maximum number of 16 tickets can be bought in ` 532 as: 9 tickets for ` 30 each. 6 tickets for ` 32 each and 1 ticket for ` 40

39.Solve using options.

Checking for option (a) will go as: According to this option 400 people have voted against the motion. Hence, originally 200 people must have favoured the motion. (Since, there is a 100% increase in the opponents)

This means that 200 people who were for the motion initially went against it.

This leaves us with 400 people who were for the motion initially (after the abduction.)

1/3^rd of the original having been abducted, they should amount to half what is left.

This means that 600 (for) and 200 (against) were the original distribution of 800.

This option fits perfectly (given all the constraints) and hence is the correct answer.

40.1 man is married to 1 woman.

Hence, 45% of men = 25% of women.

i.e. 0.45 M = 0.25 W

Hence =

Women to men ratio of 9:5

Using alligation, the required answer is 32.14

41.The required weight of the bucket to the water when full is 3:2.

If both the weights (bucket and water) are integers, then the total weight must be a multiple of 5.

Only option (c) shows this characteristic.

42.We do not have sufficient information to solve the question.

43.A 25% hike in the price would result in a 20% drop in consumption (if we are keeping expenditure constant). Thus, the drop in what he can buy of 20kg is equivalent to 20% of the original consumption. Hence, the original consumption should be 100 kg and the new consumption should be 80 kg. The increased price of rice would be 400/80 = `5

44.Income of the salesman = 1200 + (1600 × x)

Where x is the number of ` 10000 sales he achieves over the initial ` 10000.

For 1200 + 1600 × x = 7600

We get x = 4.

This means that the sales value must be ` 50000.

45.A sales value of ` 9000 cannot be achieved.

46.This question is based on a product constancy situation. A 25% increment in the commission (How?? Note: When the commission goes up by 5 percentage points from 20 to 25, there is a 25% increment in the commission) would get offset by a 20% drop in the volume of the transaction.Option (d) is correct.

47.Out of a total of 100% votes; 80% voted. 16% were invalid and 20% went to the second placed candidate. This means that the maximum the winner can get is 44%. Options a, b and c are greater them 44% and hence cannot be correct. Hence, none of these.

48.At 12 noon, the watch would show the correct time (since till then the temperature range was below 40°C). The watch would gain 2% every hour between 12 and 4. An hour having 3600 seconds, it would gain 72 seconds in each of these hours. Thus, at 7 pm it would be 72 × 4 = 288 seconds ahead. The time exhibited would be 7: 04: 48.

49.Pepsi worth ` 16 would be containing 60 grams of vitamins.

50.Option (a) would cost: 6 + 7.5 = 13.5

Option (b) would cost: 12 + 3.2 = 15.2

Option (c) would cost: 4 + 3.2 + 2.8 = 10

Option (d) would cost: 12 + 2.8 = 14.8

Option (c) is the cheapest.

Level of Difficulty (III)

1.Let the initial price of raw materials be 100. The new cost of the same raw material would be 115.

The initial cost of labour would be 25 and the new cost would be 30% of 115 = 34.5

The total cost initially would be `125.

The total cost for the same usage of raw material would now be: 115+34.5= 149.5

This cost has to be reduced to 125. The percentage reduction will be given by 24.5/149.5 = 17 % approx.

2.Let the initial times allotted be: 50,40 and 20 hours. Then, the time used in each activity is:

20,12 and 4 hours. Thus, 36 hours out of 110 are used in all.

Hence, the answer is 36/110 = 32.72 %

3.The following structure would follow:

Passed all: 5%

Passed 4: 20% of 90% =18%

Passed 1: 25% of 90% = 22.5%

Passed 2: 24.5%

Passed None: 5%

Passed 3: Rest (100 – 5-18-22.5-24.5-5 = 25%)

But it is given that 300 people passed 3. Hence, 25% = 300.

Hence, 1200 students must have appeared in the test.

4.The third gallery making the capacity ‘half as large again’ means: an increase of 50%.

Further, it is given that : 4(first + third) = 12 (second)

In order to get to the correct answer, try to fit in the options into this situation.

(Note here that the question is asking you to find the capacity of the second gallery as a percentage of the first.)

If we assume option (a) as correct – 70% the following solution follows:

If second is 70, then first is 100 and first + second is 170. Then third will be 85 (50% of first + second).

Then the equation:

4 × (100 + 85) should be equal to 12 × 70

But this is not true.

Through trial and error, you can see that the third option fits correctly.

4 × (100 + 80) = 12 × 60.

Hence, it is the correct answer.

5.Let the initial percentage of salt be 10% in 100 liters of sea water in the flask.

10% of this is poured out (i.e. 10 liters are poured out) and the water heated so as to increase the percentage of salt in the beaker 5 times (we have assumed M as 5 here.)

This means that there will be 30% salt in the beaker. Since, the salt concentration is increased by only evaporating water, the amount of salt remains the same.

Initially the salt was 10% of 10 liters (= worth 1 liter). Hence, the water must have been worth 9 liters.

Now, since this amount of salt becomes worth 50% of the total solution, the amount of water left after evaporation would have been 1 liter and the total would be 2 liters.

When the 2 liters are mixed back again: The new concentration of salt in sea water would go up. In this specific case by alligation we would get the following alligation situation:

Mix 90 liters of 10% salted sea water with 2 liters of 50% salted sea water.

The result using alligation will be: [10 + 40/46] % concentration of salted sea water. The value of the increase percentage will be 400/46. (this will be the value of x)

Now, try to use the given options in order to match the fact that originally the flask contained 100 liters of sea water.

Use M = 5, x = 400/46,

Only option (b) matches the situation.

= 100

6.The only values that fit this situation are C 25%, B 30%, and A 45%. These are the percentage of votes polled. (Note: these values can be got either through trial and error or through solving c + c + 5 + 1.5 (c + 5) = 100%

Then, 20% is 18000 (the difference between A & C.)

Hence, 90000 people must have voted and 100000 people must have been on the voter’s list.

7.The net time lost over two weeks would be 0.25% of a week’s time (since in the first week the clock loses ½% and in the second week the clock gains ¼% on the true time.)

A week contains 168 hours. Hence, the clock loses 0.42 hours.i.e. 25.2 minutes or 25 minutes 12 seconds. Hence, the correct time would be 12:25:12.

8.Traveling for 2400 kms at 18 kmph, Vawal will use 133.33 liters of petrol every month. The increase in expenditure for Vawal will be 133.33 × 0.7 × 28 = ` 262 (approx).

9.The required answer will be given by: (7/107) × 2400 = 157 km

10.The original expenditure is 28* 133.333 = ` 3733.333

The new expenditure will be given by 28 × 1.07 × n/18 where n = the no. of kilometers to travel.

Since the new expenditure should increase by ` 200, its value has to be equal to ` 3933.333

This gives us n = 2363.15

Hence, the answer is e.

11.The shopkeeper would get the maximum revenue when everybody opts for a 4% resale of the right. In such a case, the revenue for the shopkeeper from each customer would be: 96% of 4000 = 4000 – 160 = 3840. hence, total revenue is 38400.

12.Similarly, the highest discount would be if everybody opts for the 15% discount. In such a case, the total discount would be: 600 × 10 = 6000.

13-16.Detailed solutions for 13–16 are given in the hints of LOD III.

17–23.The average income estimated would be: Annual Income/14 (Underestimated savings).

The average monthly expenditure would be: Annual expenditure/9 (Overestimated expenditure)

17–19 are explained in the hints of LOD III.

20.x/14 = 87. Hence, annual income = 1218.

New income = 1218/12 = 101.5

Change in estimated income due to the change in process of average calculation = 14.5/87 Æ 16.66% increase.

21.Estimated monthly income would go up, while the estimated monthly expenditure would go down. Hence, Savings (estimated) would increase.

22.Cannot be determined since the percentage change would depend on the actual values which are not available for this question.

23.The estimated monthly expenditure would change from: x/9 to x/11. Hence, percentage drop in the ratio will be 2/11 Æ 18.18%

24 to 29 are explained in the hints to LOD III.

31-34.The following table will give a clearer picture of the situation:

31.Reduction of = 30.32%

32.The percentage difference between the revenues is: (746–594) × 100/746 = 20.37

Hence, the required value is 30.32/20.37 = 1.488

33.The day’s cost of printing 373000 copies of Deccan Emerald is: 373000 × 8 = 2984000

Out of this, the paper recovers 596800. The remaining cost to be recovered would be: 2387200.

At ` 3000 per cc, 795.733 cc will have to be booked on any given day in order to obtain the cost. This represents 99.46% of the total value.

35.Times of Hindustan:

Total cost = 2,47,000 × 7.5 = 18,52,500

Net revenues from newspaper sales is 3,95,200

Cost to be covered through advertising = 18,52,500 – 3,95,200 = 14,57,300.

At an ad rate of `1800 per cc, they would have to sell 809.61 cc i.e.73.6%

Similar calculations for India’s Times will give 79.2%.

Hence, the percentage point difference = 5.6

36.If Ac 3^rd costs 100, Ac 2^nd would cost 120 and AC 1^st would cost 190. 3 Tier ticket would cost : 47.5 and general ticket would cost 40.

AC 2^nd Æ 780 = 120

Then the difference between 3 Tier and general ticket would be: 7.5 × 780 = 48.75

37.Total cost Æ 100 + 120 + 190 + 47.5 + 40 = 497.5

This gives (497.5/120) × 780 = 3233.75.

43.Hursh Sarma’s savings:

Required Ratio = 4800/900 = 5.333

48.Assume he has 1200 francs, 1200 DM and 1200 Liras. If he converts everything to francs, the result will be:

1200 DM will convert to 240 Euros which will convert to 960 francs. But 51 Moolas = 1 franc. Thus the value of 1200 DM in terms of Moolas goes up from 1200 × 36 = 43200 to 960 × 51 = 48960. This increase in value has occurred only because of the change of currency. Hence, he should convert all his DM into Francs. However, before concluding on this you also will need to consider the effect of Liras.

It is evident that 1200 DM will yield 240 Euros, which would yield 720 Liras (since 1 euro is 3 lira), which in turn would yield 720 × 70 = 5040 Moolas.

Thus, it is evident that by converting DM into Liras the increase in value is higher than that achieved by converting DM into Francs.

Similarly, converting Francs to Liras also increases the value of the Francs.

1200 × 51 becomes equivalent to 900 × 70.

Note: The thought process goes like this: 1200 Francs = 300 Euros (since 1 euro = 4 francs). Further 300 Euros equals 900 liras which equal 900 × 70 Moolas.

49.Cannot be determined since the conversion from dollar to Euro is not given, neither is the inter currency exchange rate between Lira, Francs and DMs.

50.Obviously, both a and c are required in order to answer this question.