Basic Math Lesson-02

Addition & Subtraction of Fractions:

If we add or subtract an integer from fractions, Least Common Multiple (L.C.M) is required. For instance:

¹⁄₂ + 5 = ?

You should know that every integer has a denominator (i.e. 5 can be written as ⁵⁄₁). So above thing can be written as:

¹⁄₂ + 5 = ¹⁄₂ + ⁵⁄₁

All of you know we have to take L.C.M of the values in denominators i.e. 2 and 1. As 2 is multiple of 1, so L.C.M will be 2

➩ ¹⁄₂ + ⁵⁄₁ = ⁽¹⁺¹⁰⁾⁄₂
➩             = ¹¹⁄₂ Answer!

Steps to add or subtract fractions:

Suppose, we need to find value of ¹⁄₁₂ + ¹⁄₁₆

For that purpose, we need L.C.M of 12 and 16.

(12, 16)
➩ 4 (3, 4)
➩ 4 × 3 × 4
➩ 48

Similarly,

⁵⁄₁₂ – ³⁄₁₆ = ^{(4 (5) – 3 (3))}⁄₄₈
➩ = ¹¹⁄₄₈ Answer!

You will not be asked such simple questions in exam, rather these are steps to solve complex questions. Be quick, but accurate!

Multiplication & Division of Fractions:

Multiplication:

When a fraction multiplies with another fraction, numerator of one fraction multiplies with numerator of the other fraction and denominator of one fraction multiplies with denominator of the other fraction. For instance:
⁵⁄₂ × ³⁄₄ = ^{(5 × 3)}⁄_{(2 × 4)}
➩           = ¹⁵⁄₈

Also, when a fraction is multiplied with an integer, the integer multiplies only with the numerator as follows:

Also, note that in case of multiplication of fractions, you can switch numerator with the other numerator as follows:

Or you may simply do this in single step and eliminate extra steps, as 2 in first denominator will cancel out 8 and 3 in second denominator will cancel out 15 as follows:

Also, I will advise you to always try to make division first instead of multiplication, because this will ease the calculation and simplify the expression or equation quickly. Mathematically,

Division:

When a fraction X is divided by another fraction Y, then you may simply multiply fraction X with inverse of fraction Y (i.e. ¹⁄_Y).
For instance, let’s say fraction X is ⁵⁄₂ and fraction Y is ³⁄₄:

Complex division:

It may be possible that two fractions can be written in denominator. Let’s understand how to solve such fractions:
For instance,

While simplifying such expressions, always start from the bottom and use L.C.M to solve it as follows:

BODMAS or PEMDAS Rules:

For purpose of arithmetic calculation, you have had studied BODMAS/PEMDAS rule in math in schooling era. This rule is important for beginners because any mistake in this mostly results to incorrect answer. So let’s study PEMDAS:

This rule is very famous and applied in every arithmetic calculations. Violation from this may result to wrong answer. Many people commit mistake in this rule while solving an arithmetic expression or equation.

According to PEMDAS rule:

• Every calculation must begin from the smallest parenthesis “( )” and then gradually move to the larger parenthesis “{ }”and finally largest parenthesis “[ ]”.
• Exponentials must be solved first before other arithmetic operations (i.e. ÷, ×, +, and –).
• Then do either multiplication or division as convenient. But keep in mind that when both multiplication and division are consecutively come, start solving from left to right and solve whichever comes first.
• Finally do either addition or subtraction as convenient.

Addition & Subtraction:

Adding big integers without using calculator:

In schooling era, you have learned to use folloing way to add 4594 and 5945:

 

Addition of Decimal:

Using similar way, you can add decimal to decimal as well as integers to decimal as follows:

 

Basic rules for addition:

Let’s consider a, b & c are numbers (i.e. number means that it can be positive, negative and decimal or integer):

• a + b = b + a. For instance, 5 + 3 = 3 + 5; the result will remain 8.
• a + (b + c) = (a + b) + c. For instance, 4 + (5 + 1) = (4 + 5) + 1; the result will remain 10.
• If a = b, then, a + c = b + c. In other words, if we add both sides of an equation with a number ‘c’ (i.e. number means that it can be positive, negative, decimal or integer), Left Hand Side (L.H.S) would remain equal to Right Hand Sight (R.H.S) of the equation. So we can add any number on both sides of equation. For instance, if x = 12, so x + 3 = 12 + 3.

Subtracting big integers without using calculator:

In school level, also, you have learned to use following way to subtract 4594 from 5945:

Important Note: In subtraction operation, when the upper digit is less than lower digit (as in above figure, upper digit 4 which is less than lower digit 9), we need to take 1 carry from next left upper digit (as in above figure, next left upper digit is 9, which is left of upper digit 4)..

Basic rules of Subtraction:

Let’s consider a, b & c are numbers:

• a – b = a + (– b) or you can write a – b = –b + a; therefore, a – b ≠ b – a.
• a – (b – c) = a – b + c; this is because –ve sign will multiply inside the parenthesis and will change signs of both b & c; Therefore, a – (b – c) ≠ (a – b) – c.
• (a – b) – c = a – b – c; this is because, if we open the parenthesis, nothing will change.
• If a = b, then, a – c = b – c. In other words, if we subtract both sides of an equation with a number, Left hand side would remain equal to Right hand sight of the equation. So we can subtract any equation on both sides by same number.

Addition & Subtractions combined:

At this point we need to learn how to deal with addition & subtraction at a time in an expression. For this purpose, let’s solve:
4 + 3 – 5 = ?

You can either do addition first or subtraction first. Both ways are correct as follows:
4 + 3 – 5 = (4 + 3) – 5 = 7 – 5
= 2

Or

4 + 3 – 5 = 4 + (3 – 5) = 4 + (– 2) = 4 – 2
= 2

 
Similarly, Let’s solve:
– 44 + 22 + 84 + 61 – 88 – 20 = ?

Always apply smart thinking and avoid doing stupid lengthy calculations as below:
– 44 + 22 + 84 + 61 – 88 – 20 = 22 + 84 + 61 – 44 – 88 – 20

By Taking –ve sign common = (22 + 84 + 61) – (44 + 88 + 20)
And, blaa blaa …

Such things will not enhance your problem solving skill rather you will remain stuck to such habit and hence unable to explore your brain. Remember that, Quantitative Reasoning is more about thinking smartly than just doing math.

Simply think in your mind the following:
– 44 + 22 + 84 + 61 – 88 – 20 = ?

You must make pairs of two such that their solutions are easily identified in mind. For instance, (22 & – 20) and (84 & – 88).
After rearranging in such a way,
– 20 + 22 + 84 – 88 – 44 + 61 = 2 – 4 + 17
                    = 15 Answer!

Many of you may think that it’s just a minor thing. But this minor thing will make significant difference in your level when you make habit of such things. As mentioned earlier, use your hand work as minimal as you can and your brain as maximum as you can while solving each step.

Multiplication:

We have learned how to add numbers. If we add same number many times, it’s called multiplication. For instance, if you add two 3’s, it means you have multiplied 3 with 2:

3 + 3
By taking 3 common,
3(1 + 1) = 3 × 2 = 6
 
Similarly, if you add five 3’s, it means you have multiplied 3 with 5. i.e.,
3 + 3 + 3 + 3 + 3
By taking 3 common,
3(1 + 1 + 1 + 1 + 1) = 3 × 5 = 15
 
Important Note: When we take common, it would replace with ‘1’ from all values from which the common value has taken..

Many signs indicate the relationship as multiplication. For instance,
3 × 5 = 3 (5) = (3) 5

All above expressions show multiplications. When a number is multiplied by zero, the result is zero.
3 × 0 = 0 + 0 + 0 = 0

Important Note: When we take common, it would replace with ‘1’ from all values from which the common value has taken..
 

Multiplication of negatives with other number:

When a negative number (i.e. either integer or decimal) multiplies with a positive number, it always results to a negative number. For instance,
(–3) × 5 = –15

Furthermore, when a negative number multiplies with another negative number, it always results to a positive number. For instance,
(–3) × (–5) = 15

Similarly, if more than two numbers involve, we can solve each pair of numbers one by one. For instance,
5 × (–2) × 4 × (–3) × (–5) = {5 × (–2)} × 4 × {(–3) × (–5)}
{–10} × [4 × {15}]
= {–10} × [60]
= – 600 Answer!
 

Multiplication of big numbers:

Let’s multiply 15 with 25.

➩             15 × 25 = 375 Answer!
 
Alternatively, you may do so by using below shortcut:
➩   15 × 25 = 15 × (20 + 5)
➩       = (15 × 20 + 15 × 5)
➩       = 300 + 75
➩       = 375 Answer!

Important Short ways without calculator: If you know the table of 15, you can split 25 into 20 + 5. After that, rather you multiply 15 with 25, you simply need to multiply 15 with 20 and then with 5. After this, you need to add the result. This split is necessary because you know that 15 × 20 = 300. Actually, you know 15 times 2 is 30, so 15 times 20 must be 300. Also, you know 15 × 5 = 75, so this calculation becomes easier and much quicker when you don’t have calculator in your exam.

Note that, when you multiply 15 with 20, it gives same result when you multiply 15 with (2 × 10). Mathematically,
15 × (2 × 10) = (15 × 2) × 10 {As we know, a × (b × c) = (a × b) × c}
= 30 × 10 = 300

Also, it is important to note that when you multiply 15 with 14, it gives same result when you multiply 15 with (10 + 4). Remember that, here, we will multiply 15 with both 10 and with 4, because there’s sign of addition (i.e. +) between 10 and 4. Mathematically,
15 × (10 + 4) = (15×10 + 15×4) {As we know, a × (b + c) = ab + ac}
= 150 + 60 = 210 Answer!

Similarly, When you multiply 15 with 8, it gives same result as you multiply 15 with (10 – 2). Mathematically,
15 × (10 – 2) = (15×10 – 15×2)                     {As we know, a × (b – c) = ab – ac}
= 150 – 30 = 120 Answer!
 

Multiplication of Decimals:

If we are asked to find the product of 0.25 & 0.15. First convert the decimal to fractions by using the following method:

Multiplication with negative numbers:

Now let’s discuss what happens when the negative integers are entertained in process of multiplication. For that purpose, we’ll discuss different aspects of negative integers and positive integers combined.

Multiplication of one positive and one negative number:
When a positive integer multiplies with a negative integer, it always results to a negative integer. i.e.,
4 × (–5) = –20
Similarly,
(–3) × 4 = –12
 
Multiplication of two negative numbers:
When a negative integer multiplies with another negative integer, it always results to a positive integer. i.e.,
(–3) × (–5) = 15

Similarly,
(–4) × (–4) = 16
 
Multiplication of one positive vs two negatives:
When a positive integer multiplies with two negative integers, it always results to a positive integer. i.e.,
3 × (–4) × (–5) = ?

To do this task, first we need to separately calculate result of the two negatives, which would definitely a positive as we learnt above, i.e.,
3 × (–4) × (–5) = 3 × {(–4) × (–5)} = 3 × {20} = 3 × 20

Now it’s a form where the two positive integers multiply, that result to another positive i.e., = 60
 
One negative vs two negatives:
When a negative integer multiplies with two negative integers, it always results to a negative integer. i.e.,
(–3) × (–4) × (–5) = ?

Firstly, we need to solve the two negative integers, that would result to a positive one i.e.,
(–3) × (–4) × (–5) = (–3) × {(–4) × (–5)} = (–3) × {20} = (–3) × 20

Now, it becomes a case where one positive and one negative multiplies, that results to a negative.
(–3) × 20 = –60
 
Multiplication of Negatives vs addition of negatives:
Few students who are new to arithmetic don’t even know the difference between multiplying two or more negative numbers and adding two or more negative numbers.

Before learning this, you should know that if the two numbers are adjacent to each other with parenthesis, this means that the two numbers are multiplying with each other. i.e.:

(–3)(–4) = (–3) × (–4)

On other hand, (–3) ± (–4) ≠ (–3) × (–4)

± operation means either with addition (i.e. +) or subtraction (i.e. –)

Also, remember: (–3)(–4) ≠ –3 – 4

Basic Rule of Multiplication:

Let’s consider a, b & c are numbers. Then,

• a × b = b × a
• a × (b × c) = (a × b) × c
• a × (–b) = – ab, in other words, if the two numbers ( one positive and other negative) multiply it results to a negative number, rather than positive.
• (–a) × (–b) = ab; in other words, if the two negative numbers multiply it results to a positive number, rather than negative.
• If a = b, then, a × c = b × c. In other words, if we multiply both sides of an equation with a number, Left hand side would remain equal to Right hand sight of the equation.

Division:

When we break a rod by making ONE cut at the middle, it break into two equal parts, we simply divide it into two, each of which resulted to half of the previous rod length. Diagrammatically,

We use it in mathematical form as ¹⁄₂ or 1 ÷ 2, stated as ‘1 is divided by 2’, or simply ‘1 by 2’or ‘1 upon 2’

5 ÷ 10 is the same as 1 ÷ 2, because it can further be simplified as,

Also, if we break the same rode by making two cuts of equal lengths, it breaks into three equal parts, we simply divide it into three, each of which resulted to one-third of the previous rod length.

From above diagram, note that the two successive cuts of equal lengths break the ruler/rod into three pieces. And each piece length is one-third of the original length of the ruler/rode. Similarly, three successive cuts of equal lengths break the ruler/rod into 4 pieces, making each piece length one-fourth of the original length of the ruler/rode. And four successive cuts break it into 5 pieces, making each piece length one-fifth of the original length of the ruler/rode. By following this patter, “How many cuts will be required to break a ruler or a rod into 19 pieces?” Well, you know the sequence (i.e. pattern). Number of pieces is always one greater than number of cuts. Thus, in order to break the ruler or a rod into 19 pieces, you need 18 successive cuts, making each piece length one upon eighteenth (i.e. ¹⁄₁₈) of the original length of the ruler/rode.

On other hand, the concept of reverse scenario is also important. For instance, “21 successive cuts to a ruler or a rod break the ruler/rod into how many pieces?” As we know, number of pieces is always one greater than number of cuts, so 21 successive cuts break the ruler/rod into 22 pieces, making each piece length one upon twenty two (i.e. ¹⁄₂₂) of the original length of the ruler/rode.

Similarly, 3 ÷ 15 can be simplified as 1 ÷ 5.

Important Note: When a number is divided by 0, the result can’t be find, i.e., undefined or infinity (∞).

Mathematically, ⁰⁄₀ = ¹⁄₀ = ²⁄₀ = ³⁄₀ = ……… = ∞
 

Division of fractions:

When, we divide ( ⁴⁄₅ ) by ( ²⁄₃ ), we should know, the point in mind that, If we multiply the value instead of dividing by the value, then the dividing value would be reversed.

 

Division of decimals:

Dividing decimals is as simple as dividing integers. You just need to convert decimal into integer by multiplying and dividing with respective power of 10. For instance,
2.5 ÷ 1.5 = ^2.5⁄_1.5

To change decimals into integer, we need to multiply numerator and denominator with 10, i.e.:

After simplification, we’ll get:
= ⁵⁄₃ Answer!
 
Similarly, let’s divide a complex decimal with an integer as follows:
1.024 ÷ 32 = ^1.024⁄₃₂

To change decimals into integers, let’s multiply numerator and denominator with 10³ (i.e. 1000) as follows:

After simplification, we’ll get:
= ³²⁄₁₀₀₀ Answer!       or       = 0.032 Answer!
 

Division of positive and negative combined:

Now let’s discuss what happened when the negative integers are entertained in process of division. For that purpose, we’ll discuss different aspects of negative integers and positive integers combined.

Division of positive and negative:
When a positive integer is divided by a negative integer, it always results to a negative number, i.e.,
5 ÷ (– 2) = ?

In order to solve it, we need to convert in form of fraction, containing a nominator and a denominator, i.e.,
⁵⁄_(-2)
 

Basic rules for division:

Let’s consider a, b & c are numbers:

 

Multiplication & Division Combined:

Let’s learn how to solve expressions containing both multiplication & division. Let’s solve the below expression:
²⁵⁄₄ × 12
Note that ²⁵⁄₄ as a whole multiplies with 12. It doesn’t mean that 12 will multiply with both 25 as well as 4. Rather, 12 will multiply with only numerator (i.e. 25). Basically, 12 means ¹²⁄₁; thus, when ²⁵⁄₄ multiplies with ¹²⁄₁, numerator multiplies with numerator and denominator multiplies with denominator as follows:

But, you are recommended to solve by following short way (i.e. division first):

So, we’ve learned that a dividing number can be written on any one side at a time below two numbers that are multiplying with each other. i.e., in above example, 25 & 12 are two numbers that are multiplying with each other. So in this case, 4 (which is dividing) can be written on either sides below the two numbers.

Let’s practice a bit more on multiplication & divisions. The more we practice, the more we become quick in learning. Try yourself at first, to simplify the following:
(64 ÷ 4) × (28 ÷ 8)
Solution:

In some instances, where multiplications and divisions are adjacent to each other, always start solving from left side and gradually move towards right side.

For instance, 4 ÷ 2 × 2 = ?

Incorrect way: 4 ÷ 8 × 2 = 4 ÷ 16 = ¹⁄₄
Correct way: 4 ÷ 8 × 2 = (4 ÷ 8) × 2 = ¹⁄₂ × 2 = 1               (First solving division, and then multiplication)

In other cases, we need following the BODMAS rule. For instance,
5 + 5 × 5

If we solve addition first rather than multiplication, the answer will be incorrect.
Incorrect: 5 + 5 × 5 = (5 + 5) × 5 = 10 × 5 = 50
Correct: 5 + 5 × 5 = 5 + (5 × 5) = 5 + 25 = 30

Similarly,
5 + 5 ÷ 5
Incorrect way: 5 + 5 ÷ 5 = 10 ÷ 5 = 2
Correct way: 5 + 5 ÷ 5 = 5 + 1 = 6
 

Distribution and Factorizing:

When an integer is multiplies two or more numbers inside parenthesis such that the numbers inside brackets are having arithmetic operation of addition or subtraction, the integer will multiply with both of the numbers that are inside brackets. For instance,

4 × (5 + 3)

Here, 4 is number which is multiplying with two numbers inside brackets. Here the values inside the brackets are having arithmetic operation of addition. Hence, 4 will multiply with both of the numbers inside brackets. i.e.:

= 4×5 + 4×3 = 20 + 12
= 32

Also, you may solve it by simply adding 5 and 3, then multiply the result with 4 as below:
= 4 × (8) = 32

4 × (5x + 3)

Here, we cannot add 5x and 3, so, we have to distribute 4 inside brackets, i.e. multiply 4 with both of the numbers inside brackets. After simplifying, we’ll get:

= 20x + 12
 
Similarly, 5 × (2x + 3y + 4z)
Incorrect way: 5 × (2x + 3y + 4z) = 5 × {(2 + 3 + 4) (x + y + z)} = 5 × {9(x + y + z)} = 45x + 45y + 45z
Correct way: 5 × (2x + 3y + 4z) = 10x + 15y + 20z

And similarly,   4 × (5x + 8y – 9z)
Incorrect way: 4 × (5x + 8y – 9z) = 4 × {(5 + 8 – 9) (x + y – z)} = 4 × (4) (x + y – z) = 16x + 16y – 16z
Correct way: 4 × (5x + 8y – 9z) = 20x + 32y – 36z
 
Note that the reverse case also true, i.e.:
20x + 32y – 36z = 4 × (5x + 8y – 9z)

And,

10x + 15y + 20z = 5 × (2x + 3y + 4z)

Impact of removing brackets on numbers inside brackets:

This concept is very important for arithmetic. Let’s understand this through examples:

Let’s assume we need to simplify the following expression:
4x + (2x + 5y – 8z)

In this expression, if we open the parenthesis (i.e. brackets), the signs of all numbers that are inside parenthesis will not change upon removing parenthesis, i.e.:
= 4x + 2x + 5y – 8z = 6x + 5y – 8z
 
But in other case, when there’s negative sign outside the parenthesis, the signs of all numbers that are inside parenthesis will change upon removing parenthesis, i.e.,
4x – (2x + 5y – 8z)

In the above expression, if we open the parenthesis, the signs will be changed such that positive sign will become negative, while negative sign will become positive. i.e.:
= 4x – 2x – 5y + 8z             (As 2x and 5y were positive, so become negative; and 8z was negative, so become positive)
= 2x – 5y + 8z

All Arithmetic Operations combined:

Let’s use all four arithmetic operations (i.e.: +, –, × and ÷): 6 – 3 ÷ 2 + 1 × 5 = ?
According to the rule, we have to use division & multiplication first, and then use addition & subtraction. i.e.:

And, if you need to convert the original expression (i.e. 6 – 3 ÷ 2 + 1 × 5 = ?) in decimal, solve in decimals from beginning as follows:
6 – 3 ÷ 2 + 1 × 5 = 6 – 1.5 + 5 = 4.5 + 5 = 9.5 Answer!

Note that both answers are same, i.e.: ¹⁹⁄₂ = 9.5
 
Now, let’s move ahead, by including parenthesis as well as exponents:
5 – 3(4 + 2³) + 4² ÷ 2²

By applying PEMDAS rule:
5 – 3(4 + 2³) + 4² ÷ 2² = 5 – 3(4 + 8) + 16 ÷ 4
                                  = 5 – 3(12) + 16 ÷ 4
                                  = 5 – 36 + 4
                                  = –27 Answer!
 

Similarly,
–3 × 12 + 4 ÷ 8 – (4 – 6) = ?

By applying PEMDAS rule:

 

Number line:

In a number-line, numbers are always increases from left to right. For instance, consider the number-line below:

Now, suppose we need to find the distance between 2 and 5 in a number-line. We know that the number 5 represents its distance from 0, also number 2 represents its distance from 0 in a number line as shown below:

Basically, the number 5 represents its distance from 0. Also, 2 represents its distance from 0. Every number in a number-line represents its distance from 0. In order to get distance between any two points on a number-line, you should take (higher number) – (lower number). So distance between 2 and 5 is: 5 – 2 = 3
 
Similarly, the distance between 5 and –3 can be calculated as follows:

Mathematically, the distance between 5 & –3 is:
5 – (–3) = 5 + 3 = 8

Note that, the distance between any two numbers is always (Larger number – Smaller number) (e.g. 5 – (–3), because 5 is larger, and –3 is smaller).

So far, we’ve learned number-line from one basic scenario. Let’s try to play with the common sense of your brain through this topic. This will help you improving some problem solving skill and improve your thinking power in broader way. 🙂
 

Example Scenario:

From the given number link above, if the marks are equally spaced (i.e. all adjacent points are equidistant), which of the following statement MUST be true?

I. a – b = 0
II. b – a < 0
III. a – b < 1 – (–1)         or         a – b < 2

A) I only
B) II only
C) III only
D) Both I & III
E) Both II & III

First, try yourself before going below for answer with explanation.
 

Solution:

First of all, given that marks on the number-line are equally spaced. This means that distance between a and 1 is same as distance between 0 and a. Also, this is same as distance between b and 0 as well as the distance between (–1) and b.

Let’s check statement I: a – b = 0

As you see that a and b are different points, so it is impossible that their difference is 0. So, statement (I) cannot be true.
 
According to statement II: b – a < 0

As you can see that b is less than a, so (smaller number – larger number) is always negative. Thus, statement II MUST be true.
 
According to statement III: a – b < 1 – (–1) or a – b < 2

As you can see that the distance between a and b is less than the distance between 1 and (–1), thus statement III is also MUST be true.

Hence, choice E is correct answer!
 
Note that we have used distance approach rather than plugin approach. Have you seen how quick and easy it is? You are damn sure on your answers! Let’s discuss another basic example scenario.
 

Example Scenario:

Which of the following statement MUST be true? (Indicate all such statements.)

I. y > x + b
II. y + b > x + a
III. a × b > y

A) I only
B) II only
C) III only
D) Both I & II
E) Both II & III

Solution:
First of all, pay attention to the words ‘MUST be true’ in question. Secondly, note that there’s no information given about the space between marks on the number-line in this question. So we are not sure whether all the marks are equally spaced. So in that scenario, remember that the marks may be equally spaced or not equally spaced. Therefore, you need to be careful!

So, let’s begin with first statement:
  I. y > x + b

From the given number-line, you can see clearly y > x (i.e. y is greater than x), because y is on the right side of x in a number-line. As we know, numbers increase when we move to right side of a number line.

Furthermore, b is a negative number. It is now irrelevant question to think How much is its value? You have enough information that b is negative. So, when a negative number (i.e. b) is added to a positive number (i.e. x), the result will always be less than the positive number (i.e. x). Because, you are adding a negative number to a positive number, this in-fact means you have reduced the positive number or subtracted some value from the positive number. Mathematically, the expression becomes x + b

Now, according to given question, y is greater than x, so y MUST always be greater than x + b. Because x + b is less than x, and x is less than y. Thus, x + b is less than y. Therefore, statement (I) MUST be true.
 
Now, let’s analyze statement II:
  II.   y + b > x + a

Again, here you may quickly see from the number-line that
y > x
Also, b > a

From the two inequalities, we can conclude that y + b > x + a

In other words, the sum of two bigger numbers (i.e. y and b) is always greater than the sum of two less numbers (i.e. x and a). This basic concept of inequality will also be discussed in detail and with advance level scenarios in advance level preparation later on after you complete the basics. Therefore, statement (II) MUST also be true.
 
Now, let’s analyze statement III:
  III.  a × b > y

Both a and b are negative, but the product of two negatives is always positive. So, the product of a and b could be greater than y. But this product can also be less than y or even it could be equal to y. Therefore, we do not have sufficient information whether the product of a and b is greater than y.

Therefore, statement (III) could be true, but is NOT a MUST be true statement.

Thus, Choice D is correct answer!

Constant vs Variable:

A constant is a number, whose value remain same. And, we know that a number can be an integer or a decimal. Constant is basically written in form of numbers (e.g. 0, 1, 2, 3, …). On other hand, a variable has more than one value (i.e. its value is not remained same in all conditions). Variable is basically written in form of alphabets (e.g. x, y, z, …). Most of the times, we see the combination of both variables and constant in an equation or expression. For instance,
x + y = 10

Here, 10 is a constant, while x and y are variables.

Note that, you can put infinite values for x and y, but the constant (i.e. 10) cannot change!
 

Expression vs Equation:

An expression (or sometimes called arithmetic expression) is a combinations of arithmetic operations (i.e. addition, subtraction, multiplication, division and exponential values etc.) without any sign of equality.

On other hand, an equation is combination of arithmetic operations for numbers that include a sign of equality (i.e. =). Below are some examples of expressions and equations:

An equation has a Left Hand Side (or L.H.S) and a Right Hand Side (or R.H.S). On other hand, an expression has no L.H.S or a R.H.S. For instance, let’s see equation 2x + y = 15. Here L.H.S of the equation is 2x + y and R.H.S of the equation is 15.

How to simplify a basic expression?

Simplifying an expression requires some rules of BODMAS/PEMDAS that you have already learned so far. Advance level expressions simplification will be discussed in advance level preparation plan. So, let’s learn how to solve an equation.
 

How to solve an equation?

Simple/Linear Equations: These equations have maximum 1 exponential power on a variable or variables (i.e. x). For instance,
a = 5                                         (Note that a¹ and a means same, i.e. in both cases a has 1 exponent)

And,
a + b = 15

Basically, there are two ways to solve simple equations:
    a)   Plugin
    b)   Solving simultaneously
 

a) Plugin value of one variable:

This method is mostly used by many people and is simple but a bit longer. Let’s discuss with some examples. For instance, consider:
        a = 5
And, a + b = 15

Just, put value of a = 5 in the above equation as below:
     (5) + b = 15                         (We simply plugin 5 in place of variable a in second equation, because a = 5)
5 + b = 15 –––––––––––– (Eq. 1)

For getting variable b on one side, we need to subtract 5 from Left Hand Side (L.H.S) of the equation. According to the subtraction rules as discussed above, we can subtract by a number on both sides of the equation. So, we need to subtract 5 on both sides of the equation, in order to get value of variable b. After subtracting both sides of the equation by 5, we’ll get:
➩   5 + b – 5 = 15 – 5
➩           b = 10

Thus, we have solved the two equations for value of a and b, i.e.: a = 5, and b = 10.
 

Similarly, let’s solve equations for value of a and b, if a + b = 15 and 2a + b = 24
 

Solution:

a + b = 15 –––––––––– (Eq. A)
2a + b = 24 –––––––––– (Eq. B)

From (Eq. A), subtracting both sides of (Equation A) by b, we’ll get:
➩         a = 15 – b
By putting this value of a (i.e. 15 – b) in (Eq. B), we’ll get:
➩ 2(15 – b) + b = 24

    ➩ 30 – 2b + b = 24
    ➩ – 2b + b = 24 – 30                           (By taking variable b on one side, and constant on other)
    ➩ –b = –6
    ➩ b = 6                           (By dividing both sides of the previous equation by –1)

Now, you can put this value of b = 6, in either (Eq. A) or (Eq. B), you will get value of a. You should put value of b in that equation which is easier. So, let’s put the value of b in (Eq. A), which is bit more easy and simple:
    ➩   a + 6 = 15
    ➩   a = 15 – 6
    ➩   a = 9

Thus, and we have successfully solved the given equations for a and b: i.e. a = 9 and b = 6.
 

b) Solving simultaneously:

Now, let’s learn a shorter and recommended method of solving the two different equations simultaneously that involve two variables:

Actually, we have subtracted (a + b = 15) from (2a + b = 24). So, the signs of all values of the first equation, which is placed at bottom (i.e. a + b = 15) will be changed as shown above.
 

Why we change signs of equation which is at the bottom?

To understand this, let’s learn how to subtract expression (5x + 4) from another expression (7x + 9)

7x + 9 – (5x + 4)

As we can see that the sign outside the brackets is negative, so we change the signs of all values inside the brackets, upon opening the brackets (i.e. upon removing the brackets). So,

= 7x + 9 – 5x – 4 = 2x + 5

Through simultaneously solving method:

Note that sign of the bottom expression has changed. In same way, while solving equations, we have to change the signs of bottom equation when we need to subtract the two equations.
 

Similarly, let’s learn another scenario and solve below equations for x and y:
x – y = 12 ––––––––––––– (Eq. 3)
5x + 6y = 38 –––––––––––– (Eq. 4)
 

Solution:

In this scenario, you can clearly see that sign of variable y in the above two equations are different (i.e. one is positive and other is negative). So, in such scenarios, always try to equate the coefficients* of y in both equations and then add the two equations so variable y must cancel out and we can easily find value of x.

*Coefficients are those numbers that multiply with a variable. For instance, in equation (5x + 4y = 32) coefficient of x is 5 and that of y is 4.
 
In order to equate the coefficients of y in both equations, we should multiply (Equation 3) with 6 on both sides of the equation, and multiply (Equation 4) with 1 as below:

If we multiply (Eq. 4) with 1, the equation will remain the same. Let’s multiply (Eq. 3) on both sides with 6.
➩   6(x – y) = 6 (12)
➩   6x – 6y = 72 ––––––––––––– (Eq. 5)

Now, by adding (Eq. 4) and (Eq. 5), we’ll get value of x as follows:

    ➩   11x = 110
    ➩       x = ¹¹⁰⁄₁₁ = 10

By putting this value of x in (Eq. 3) or (Eq. 4), (Eq. 3 is preferred being easier than Eq. 4), well get value of y as follows:
    ➩   (10) – y = 12
    ➩   10 – 12 = y
    ➩       y = –2

Thus, the equations have been solved for x and y, i.e.: x = 10, y = –2
 

Similarly, let’s solve the equations for m and n, if m + 3n = 12; and 3m + 5n = 48
First, try yourself, then go ahead to check your answer.
 

Solution:

m + 3n = 12 ––––––––––––– (Eq. X)
3m + 5n = 40 –––––––––––– (Eq. Y)

In order to find value of variable m, we need to equate coefficients of n for both equations. For that purpose, we should multiply (Eq. X) with 5 on both sides, and (Eq. Y) with 3 on both sides. So, let’s do this for (Eq. X) as below:

      5(m + 3n) = 5(12)
  ➩     5m + 15n = 60 –––––––––––– (Eq. Z)

Similarly, by multiplying (Eq. Y) with 3 on both sides:
        3(3m + 5n) = 3(40)
  ➩     9m + 15n = 120

Now, we have same coefficients of (Eq. Z) and the above equation. Both have positive sign with variable n, so we should subtract (Eq. Z) from the above equation as below:

  ➩     m = ⁶⁰⁄₄ = 15

By putting this value of m in (Equation X), we’ll get value of n as follows:

      15 + 3n = 12
  ➩     3n = 12 – 15
  ➩         3n = –3
  ➩         n = –1

Hence, values of m and n are 15 and –1 respectively.
 

Occasionally, the solving the question is not about simply math, rather it’s about thinking out of the box! Let’s learn with an example scenario:

Find the sum of m and n if: 5m + 3n = 12 and 3m + 5n = 44
 

Solution:

You can solve this question in same way as we have done previous questions, i.e. find values of m and n, and then find the sum. But there’s a shorter way if you think out of the box!

Think out of the box: Notice that, you just need to find the sum of m and n rather than its individual values. If you analyze it bit more, you will come to the point where you can answer it just in single step or two. The two equations have 5m and 3m (giving total of 8m upon addition); also both have 3n and 5n (giving total of 8n upon addition). Thus, you will have 8m + 8n on L.H.S upon adding two equations, and then by taking 8 common, you can find value of m + n as below:

8(m + n) = 56
m + n = ⁵⁶⁄₈ = 7 Answer!

Infinite solutions:

In few questions, you may get infinite solutions or the answer cannot be determined. In such questions, you need to select choice that states “Cannot be determined”. For instance, solve the following equations for x:
5x + 4y = 40
15x + 12y = 120
 

Solution:

It seems that the two equations are different. In fact these two equations are basically same. Actually, first equation is 5x + 4y = 40, and the second equation is three times the first equation. In other words, if we multiply first equation by three on both sides, we’ll get the second equation as follows:

5x + 4y = 40
3 (5x + 4y) = 3 (40)                     (Multiplying by 3 on both sides of previous equation)
➩ 15x + 12y = 120
Also, if we simplify second equation by taking 3 common on L.H.S (Left Hand Side), and then simplify, we’ll obtain the first equation as follows:

        15x + 12y = 120
    ➩   3(5x + 4y) = 120                   (By taking 3 common on L.H.S of previous equation)
    ➩   5x + 4y = 40                     (Dividing by three on both sides of the previous equation)

As the two equations are basically same, so we cannot solve it for x or y. Answer cannot be determine or infinite.
 

Quadratic Equations:

Those equations, which value of exponent of a variable ‘x or y’ always 2, are quadratic equations. For instance,
x² = 1,     and     x² – 10x + 24 = 0, are quadratic equations.

The general form of quadratic equation is:
ax² + bx + c = 0                   {Note that ax² ≠ (ax)², you will learn further about this in advance level study plan in this online course}

Where, a = Coefficient of x², b = Coefficient of x, and c = Constant (i.e. an integer)
 

Solving Quadratic Equations:

There are two ways to solve such equations:
    1.   By Factorization
    2.   Using Quadratic Formula
 

1. By Factorization:

This way is mostly useful when the coefficient of x² is 1, or can be converted to 1 by taking common etc. Take a look into the quadratic equation below, and let’s solve it for the value of x by using factorizing:
2x² – 20x = 48

First we need to make it like general form of quadratic equation i.e ax² + bx + c = 0
    ➩   2x² – 20x – 48 = 0                   (After subtracting both side by 48)
Now, we need to make the coefficient of x² as 1. So let’s take common 2 from left hand side of the equation:
    ➩   2(x² – 10x – 24) = 0

By dividing this equation by 2 on both sides, we’ll get:
    ➩   x² – 10x – 24 = 0

Now, we can solve this quadratic equation by factorizing. So let’s factorize the above equation.

Factorizing the quadratic equation involve careful breakdown of the coefficient of x such that the sum of the values (which were broken-down) must be equals to the coefficient of x including the sign (whether + or –). Furthermore, the product of the breakdown values must be equals to the value constant (i.e. c in general form of equation). So let’s factorize the quadratic equation: x² – 10x – 24 = 0             Or             x² + (– 10)x + (–24) = 0

The coefficient of x is –10. At this point, we need to think about two integers, whose sum should be –10 (i.e. equals to b or coefficient of x) and whose product should equals to –24 (i.e. equals to c or constant). After bit brainstorming, we can think of the integers that gives sum 10 and product 24 are either 6 & 4, and 12 & –2. Now, think the following possible combinations in your mind based on given quadratic equation:
x² – 10x – 24 = 0

(–6, –4) | It’s sum gives –10, but its product doesn’t give – 24. Eliminated!
(2, –12) | It’s sum gives –10, also its product gives – 24. That’s it!
(12, –2) | It’s product gives –24, but sum doesn’t give –10. Eliminated!
(6, –4) | It’s product gives –24, but its sum doesn’t give –10. Eliminated!

So, let’s split –10x into 2x and –12x as below:
x² – 10x – 24 = 0
    ➩   x² + 2x – 12x – 24 = 0
Note that we have broken-down –10 into 2 and –12, such that its sum should be equal to –10. And product should be –24. But why we have eliminated the other possible pairs? The reason is that, although the sum of the pairs gives back –10, but the product of these pairs does not give the constant –24. And in some cases, product gives –24, but sum doesn’t give –10. For instance, the sum of –6 and –4 gives –10, but the product of –6 and –4 gives +24, instead of –24. Therefore, we cannot –10 into –6 and –4. And in other pairs, we have similar issues. Hence, we have broken-down –10 into +2 and –12 so that the sum must give us back –10, while the product must give the constant –24.

Now, we have: x² + 2x – 12x – 24 = 0
At this stage, always write down two brackets such that one pair contain x2 and x and the other contain x and the constant as below:
(x² + 2x) – (12x + 24) = 0

To avoid such mistake, it’s better to do following steps at beginner level,

Factorization Steps:
x² – 10x – 24 = 0
    ➩   x² + 2x – 12x – 24 = 0
    ➩   (x² + 2x) + (–12x – 24) = 0

By taking –ve sign common (basically, –ve sign = –1)
    ➩   (x² + 2x) – (12x + 24) = 0
These were the steps that we skipped in previous calculations.

Now, by taking x common from first bracket and 12 common from second bracket as follows:
x(x + 2) – 12(x + 2) = 0

    ➩   (x + 2) (x – 12) = 0
Now, remember that when product of two values equals to 0, then at least one of the two values must be zero. For instance, if a × b = 0, then three possibilities are there a = 0, or b = 0, or both a and b are zero.

Similarly, the product of (x + 2) & (x – 12) gives zero implies that:
Either, x + 2 = 0 or x – 12 = 0
    ➩   x = –2 or x = 12

But, few quadratic equations need to be solved only through using quadratic formula, because factorization method is complex in such scenarios. For instance, consider the following equation: 3x² – 10x + 3 = 0

We cannot take 3 common from L.H.S (Left Hand Side) of the equation, and the coefficient of x² is 3. In such cases (i.e. where the coefficient of x² is some integer and that cannot be taken as common), we will not solve such questions through factorization, rather we’ll use quadratic formula to solve such questions.

Using Quadratic Formula:

If the general form of quadratic equation is considered, the quadratic formula is:

ax² + bx + c = 0 ––––––––––––– (Eq. Q)

For instance, let’s solve the same question as we have solved using factorization method previously.
x² – 10x – 24 = 0

After converting this into general form of quadratic equation (i.e. Eq. Q), we’ll get:
    ➩   (1)x² +(–10)x + (–24) = 0

By comparing this equation with general form of quadratic equation i.e. (Eq. Q), we’ll get:
a = 1, b = –10 and c = –24

Now, by putting these values of a, b and c in quadratic equation formula to find value of x as follows:

Note that these values are same as we have already find through factorization method.

Now, let’s solve the question which was hard to solve through factorization method as we discussed at end of the factorization topic. Below is that question, and solve it for x:
3x² – 10x + 3 = 0
    ➩   3x2 + (–10x) + 3 = 0
After comparing it’s coefficients with general form of quadratic equation (i.e. Eq. Q), we’ll get:
a = 3
b = –10
c = 3

Now, by using quadratic formula, well get:

Other Scenarios:
Sometimes, arithmetic questions regarding solving equations are difficult to solve for many people, despite these questions are quite easy. Such questions frequently tested in exams now-a-days, so let’s discuss a scenario and learn how to think and solve under such circumstances:
If 2x + 3y + 4z = 140, and 14x + 28z = 56. What is y?

First try yourself, then see below for solution:
 

Solution:

Notice that, first equations includes three variables (i.e. x, y and z), while the second equation includes two variables (i.e. x and z). While, we need to find value of y, the variable which is not present in second equation. That is the problem. A problem solving skill can help you how to deal with such problem.

After bit brainstorming, the solution of the problem relies on the value of 2x + 4z. If we sort out this, we can easily get value of 3y after putting value of 2x + 4z in first equation (i.e. 2x + 3y + 4z = 140). If we look at the second equation (which involves only two variables x and z) we can clearly see that can be determined if we take 7 common as follows:
14x + 28z = 56

By taking 7 common from L.H.S of the equation:
7(2x + 4z) = 56

By dividing above equation with 7 on both sides, we’ll get:
2x + 4z = ⁵⁶⁄₇
2x + 4z = 8
First equation (i.e. 2x + 3y + 4z = 140) can be rearranged to: 2x + 4z + 3y = 140

Now, as we have found value of 2x + 4z (i.e. 8), so let’s put 8 in place of 2x + 4z in above equation as follows:
2x + 4z + 3y = 140
8 + 3y = 140
3y = 140 – 8 = 132
y = ¹³²⁄₃ = 44 Answer!

Hence, problem solved!
 

Similarly, let’s discuss another scenario:
If 2x + y + 3z = 120,   x + 3y + z = 240,  and  3x + 2y + 2z = 300;   what is x + y + z?

First try yourself, then see below for solution:
 

Solution:

Note that all three equations include three variables (i.e. x, y and z), while we need to find the sum of these three variables (i.e. x + y + z). So, we must think of doing something with the given three equations, such that we can get value of x + y + z.

For this purpose, we see there are total 6x in three equations, total 6y in three equations, as well as 6z in three equations such that if we add three equations, we can take 6 common; thereby we can get x + y + z, as follows:

By taking 6 common on L.H.S of the equation, we’ll get:
6(x + y + z) = 660
x + y + z = ⁶⁶⁰⁄₆
x + y + z = 110 Answer!

Hence, problem solved!
 

Basic Algebraic Rules:

Suppose, x and y are numbers:

Some commonly used formulae:

(a + b) (a + b) = (a + b)²
a² + b² + 2ab = (a + b)²
(a – b) (a – b) = (a – b)²
a² + b² – 2ab = (a – b)²
(a + b) (a – b) = a² – b²

Other important Algebraic rules:

Any number raised to power 0 is always equals to 1. Even 0 raised to power 0 is also equals to 1. i.e.:
x⁰ = 1; and 0⁰ = 1
Also, remember that when two or more exponential expressions are adding or subtracting, try to take common to simplify it, i.e.:

Simplify the following: 5⁴ + 10²
Always breakdown the above into primes in order to take common, as below:
5⁴ + 10² = 5⁴ + (5×2)²
= 5⁴ + 5²×2²
= (5²×5²) + 5²×2²

We can see that 5² can be taken as common from the above expression, so by taking 5² common, we’ll get:
= 5² (5² + 2²)
= 25 (25 + 4)
= 25 (29)
= 25 (30 – 1)
= 750 – 25 = 725 Answer!
 

Alternate way (without taking common if you know large exponential value (e.g. 5⁴ or 25²):

5⁴ + 10² = (5²)² + 100
= 25² + 100
= 625 + 100 = 725 Answer!