Quantitative Aptitude - SPLessons

Inequality Problems

Chapter 43

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Inequality Problems

shape Description

An algebraic inequality is a statement that one arithmetical expression is more than (or less than) another arithmetical expression. If every variable in the equation are raised to the first power, the inequality is said to be a linear inequality.

By decreasing it to a simpler inequality whose answer is clear, inequality is solved. The answer is not unique; it is inequation, since a numerous number of qualities may fulfill the inequality.


shape Methods

An inequality is expressed as any mathematical statement that uses one of the following symbols or inequalities.


To understand the concept of inequalities, we must keep in mind that equality means equal and inequality means there are five possibilities between terms/objects. Let us understand these types of possibilities through a table.


Signs Meaning
> Greater than
< Less than
Greater than or Equal to
Less than or Equal to
= Equal to


Let us consider two variables X and Y.


Signs Meaning
X > Y X Greater than Y
X < Y X is Less than Y
X ≥ Y X Greater than or Equal to Y
X ≤ Y X is Less than or Equal to Y
X = Y X is Equal to Y


Rules of Inequalities:
The relation between two inequalities can be established if they have a common term. For example,


a) A > B, B > C : By looking at this, we can easily define the relation that A > B > C which means A > C or C < A.

b) A < B, B < C : By looking at this, we can easily define that A < B C or C < A.

c) A ≥ B, B ≥ C : This implies that A ≥ B ≥ C, which means A ≥ C or C ≤ A.


A relation cannot be defined if they don’t have a common term. For example,

a) A > B, C > B : This implies that we cannot define a specific relation between A and C as both of them are greater than B.


b) B > A, D < B : This implies that a relation cannot be established between A and D as both are lesser than B.


c) A ≥ B, B ≤ C : This implies that a relation cannot be established between A and C as both are greater than or equal to B.


Solving Inequalities:-

Determine the values of the variable that make the inequality true. The methodology used to settle an inequality is like that used to solve a condition/equation. That is, by utilizing fundamental operations, attempt to identify the variable on one side of the inequality.


There are two basic rules for solving inequalities and are similar to the rules for solving equations. They are namely:

(i) When the same constant is added to (or subtracted from) both sides of an inequality, the direction of inequality is preserved, and the new inequality is equivalent to the original.


(ii) When both sides of the inequality are multiplied (or divided) by the same constant, the direction of inequality is preserved if the constant is positive, but reversed if the constant is negative. In either case the new inequality is equivalent to the original.


Example 1:

Statements: a) A > B b) B > C

Conclusions: a) A > C b) C > A

Solution:

    On combining both the statements, we get: A > B > C

    So, we can easily say that the conclusion a) follows i.e. A > C


Example 2:

Statements : a) A > B b) B < C

Conclusions : a) A > C b) A < C

Solution:

    Here, we can see that both A and C are greater than B but we cannot derive any relation between A and C.

    So the answer to the above questions will be “nothing can be concluded”.


Example 3:

Statement: A > B > C < D≥E

Conclusion: a) A > D b) D > B

Solution: Here, nothing can be concluded because there is no definite relation between A & D and D & B. we cannot say that which one is greater, equal or lesser.

shape Formulae

1. If a > b and k is positive constant then,


  • a + k > b + k

  • a – k > b – k

  • a x k > b x k

  • \(\frac{a}{k}\) > \(\frac{b}{k}\)


2. If a > b and k is negative constant then


  • a + k > b + k

  • a – k > b – k

  • a x k < b x k

  • \(\frac{a}{k}\) < \(\frac{b}{k}\)


3. a > b and b > c, then a > c.

4. a < b and b < c, then a < c.

5. If a > b and c > d, then a + c > b + d.

6. If a > b and c < d, then a – c > b – d.

shape Samples

1. Solve the inequalities \(x^2\) -3\(x\) < 0

Solution:

    Given equation is \(x^2\) -3\(x\) < 0

    Factorize the expression ie. \(x\)(\(x\) – 3) < 0

    For the product of two expressions to be less than (negative), one must be positive and the other must be negative.

    So, case(i) is \(x\) < 0 and \(x\) – 3 > 0

    case(ii) is \(x\) > 0 and \(x\) – 3 < 0

    In the first case \(x\) < 0 and \(x\) > 3

    Here, it is impossible because \(x\) cannot be less than 0 and greater than 3 at the same time.

    In the second case \(x\) > 0 and \(x\) < 3 which can be re-written as 0 < \(x\) < 3.


2. Which is greater \((100)^{50}\) or \((102)^{49}\)?

Solution:

    Given that

    \((100)^{50}\) and \((102)^{49}\)

    If a > b where a and b are positive, then \(\frac{a}{b}\) > 1.

    So, to check which number is greater, take the ratio of the two numbers

    \(\frac{(102)^{49}}{(100)^{50}}\) = \(\frac{1}{102}\) x \(\frac{(102)^{50}}{(100)^{50}}\)

    = \((\frac{102}{100})^{50}\) x \(\frac{1}{102}\)

    = \((1 + \frac{102}{100})^{50}\) x \(\frac{1}{102}\)

    = \((1 + \frac{1}{50})^{50}\) x \(\frac{1}{102}\) (since 2 ≤ (1 + \(\frac{1}{x}\)) < 3)

    = \(\frac{(102)^{49}}{(100)^{50}}\) < \(\frac{3}{102}\)

    i.e. \(\frac{(102)^{49}}{(100)^{50}}\) < 1

    Therefore, \((102)^{49}\) < \((100)^{50}\)


3. Solve 2\(x^2\) – 12\(x\) ≤ 0

Solution:

    Given that

    2\(x^2\) – 12\(x\) ≤ 0

    Divide the expression with 2 on both sides i.e.

    \(x^2\) – 6\(x\) ≤ 0

    \(x\)(\(x\) – 6) ≤ 0

    (\(x\) – 0)(\(x\) – 6) ≤ 0

    Therefore, solution would be \(x\)  ε [0,6] or 0 ≤ \(x\) ≤ 6.


4. Quantitative comparison of
\(x\) + 1 < 5 and \(y\) – 1 < 5


Quantity A Quantity B
\(y\) \(x\)


A. The quantity on the left is greater.
B. The quantity on the right is greater.
C. Both are equal.
D. The relationship cannot be determined without further information.

Solution:

    Given \(x\) + 1 < 5 and \(y\) – 1 < 5

    Column B: \(x\) + 1 < 5

    \(x\) + 1 – 1 < 5 – 1

    \(x\) < 4

    Column A: \(y\) – 1 < 5

    \(y\) – 1 + 1 < 5 + 1

    \(y\) < 6

    Now, even though 6 is greater than 4, relationship is not determined because \(x\) can be any number less than 4 and \(y\) can be any number less than 6.

    So, \(y\) could be 2 and \(x\) could be 3 or 2 and \(x\) could be -5 etc.
    Hence option D is correct.


5. Quantitative comparison of
5\(x\) + 2 \(y\) > 3


Quantity A Quantity B
9 15\(x\) + 6\(y\)


A. The quantity on the left is greater.
B. The quantity on the right is greater.
C. Both are equal.
D. The relationship cannot be determined without further information.

Solution:

    Given 5\(x\) + 2 \(y\) > 3

    By taking three times of 5\(x\) + 2 \(y\), 15\(x\) + 6\(y\) is obtained.

    By taking 3 times the given inequality i.e.

    5\(x\) + 2 \(y\) > 3

    (3) 5\(x\) + 2 \(y\) > (3) 3

    15\(x\) + 6\(y\) > 9

    15\(x\) + 6\(y\) is greater than 9.

    Therefore, option A is correct.