*Tuesday, March 26th, 2019*

We usually deal with a lot of equations in the Quant Section equating RHS and LHS. In inequalities, we do have LHS and RHS but these are not equal, any of the following case is possible:

LHS >= RHS ; LHS > RHS ; LHS < RHS ; LHS <= RHS

- We can add or subtract same number from both sides with no change in the truth of the inequality. If a > b, then a+k > b+k e.g. If 8 > 6 then 8 + 2 > 6 + 2 and if 9 < 5 then 9 – 3 < 5 – 3
- We can multiply or divide both sides with the same number, however the sign will depend as follows:
- It will not change the sign of the inequality if the number is positive. If a>b, then ak > bk ; k > 0 e.g. 8 > 6 => 8 Ã— 4 > 6 Ã— 4
- It will not change the sign of the inequality if the number is negative. If a > b, then ak < bk; k< 0 e.g. 9 > 5 => 9 Ã— (-3) < 5 Ã— (-3)

Inequalities can be of different types like linear inequalities, quadratic inequalities. Let us discuss the methods to solve these inequalities one by one.

Linear inequalities are the ones in which the highest power of the variable is 1. These are the simple inequalities to solve, which need the knowledge of the basic algebraic rules to solve them.

Quadratic inequalities are ones in which the highest power of the variable is 2.

Now to solve such inequalities use the following steps:

- Make right hand side of the inequality equal to zero by transposing the terms (if any) from left hand side.
- Factorize the expression on the left-hand side and make sure that the coefficient of ‘x’ is positive in each factor.
- Equate to zero both the factors to get the critical points. Plot these points on the number line.
- Now, there will be three regions on the number line and the rightmost region will give you positive inequalities, middle one will give you negative inequalities and the leftmost part will again give you positive inequalities.

Similar approach can be followed in cubic inequalities and the higher ones.

Some Basic Rules and Things to Remember in case of Inequalities are:

- Arithmetic Mean > = Geometric Mean (G.M.) for any set of +ve numbers
- Modulus functions, i.e. the magnitude of the variable & not the sign,Â |4| = 4, |-4| = 4
- Maxima-Minima concepts dealing with products and sum

So, we have discussed the basic concepts and procedures to solves the basic inequalities problems, now let us try solve a few of the problems.

Below are the problems that we will put our hands on:

**Ques:** Find the largest integer x that satisfies: (x+2) (x-1)^{2}(x-4) (x+2)^{2}(x-3) < 0

A. 4

B. 3

C. -2

D. -3

**Â ****Sol: **

In the given equation, 2^{nd} and 3^{rd} term are squares, so these can be neglected as will always be +ve

So, we are left with (x+2) (x-4) (x-3) < 0

Substituting the individual terms to 0 we get x=-2, 4 & 3

We get the range of x as (-âˆž,-2) U (3,4), so the largest integer satisfying the given equation becomes -3

Hence option C is the answer

**Ques:** Solve the inequality: x^{3}Â â€“ 5x^{2}Â + 8x â€“ 4 > 0.

**Â ****Sol:**

Let m, n, o be the roots of this cubic equation

m + n + o = 5

mn + no + om = 8

mno = 4

This happens when m = 1, n = 2 and o = 2

We can also use polynomial remainder theorem

The sum of the coefficients = 0

=> P(1) = 0

=> (x – 1) is a factor of the equation.

We can now find the other two by dividing the polynomial by (x – 1) and then factorizing the resulting quadratic equation.

(x – 1) (x – 2) (x – 2) > 0

If x is less than 1, the equation is â€“ve

If x is between 1 and 2, the equation is +ve

If x is greater than 2, the equation is +ve

Hence, the solution is as follows: 1 < x < 2 OR x > 2

The answer is (1, 2) âˆª (2, âˆž)

**Ques:** How many positive integer values can x take that satisfy the inequality (x – 8) (x – 10) (x – 12)…….(x – 100) < 0?

A. 25

B. 30

C. 35

D. 40

**Â ****Sol: **

When x = 8, 10, 12, ….100 equation becomes 0.

When x = 101, 102 or beyond, all the terms are positive, so the equation becomes +ve.

So, the numbers remaining are 1, 2, 3, …7 and then odd numbers from 8 to 99.

If x =1,

All the terms become -ve. There are totally 47 terms in this list, so product of these will be negative.

Also, x = 1 works, giving other values of x =2, 3, 4, 5, 6, and 7.

When x = 9, there is one positive terms and 46 negative terms. So, the product will be positive.

When x = 11, 2 terms are +ve and rest 45 are -ve, giving the -ve product.

When x = 13, 3 terms are +ve and rest 44 are -ve terms, giving the +ve product.

Now, alternate odd numbers need to be counted, starting from 11.

So, the numbers that will work for this inequality are 1, 2, 3, 4, 5, 6, 7…and then 11, 15, 19, 23, 27, 31,….. and so on.

Regarding the last term 99, there are 46 positive terms and 1 negative term.

We need to know how many terms are there in the list 11, 15, 19,….99. These can be written as

4 * 2 + 3

4 * 3 + 3

4 * 4 + 3

…

4 * 24 + 3

A set of 23 terms. So, total number of values = 23 + 7 = 30, a +ve integer

Choice B is the correct answer.

**Â ****Ques:** The sum of three distinct natural numbers is 25. What is the maximum value of their product?

**Sol:**

Let the three natural numbers be a, b and c. We know that m + n + o = 25.

Let us substitute some values and see where this is headed

m = 1, n = 2, o = 22, Product = 44

m = 2, n = 3, o = 20, product = 120

m = 5, n = 6, o = 14, product = 420

The numbers need to be as close to each other as possible. This property is an extension of the AM-GM Inequality.

25/3âˆ¼8,Â So, we should choose a, b, and c close enough to 8.

8 + 8 + 9 = 25. But the numbers need to be distinct. 8 * 7 * 10 come closest. The maximum product would be 560.

Hence the answer is The maximum product is 560.

**Ques:** Consider integers m, n such that â€“ 3 < m < 4, â€“ 8 < n < 7, what is the maximum possible value of m^{2}Â + mn + n^{2}?

A. 60

B. 67

C. 93

D. 84

**Sol:**

Trial and error is the one of the best approach for this type of question.

m^{2}Â and n^{2}Â are both positive and depend on |m| and |n|.

If m, n are large negative or large positive numbers, m^{2}Â and n^{2}Â will be high.

mn will be positive if m, n have the same sign, and negative if they have opposite signs.

So, for m^{2}Â + mn + n^{2}Â to be maximum, so both m & n need to be positive or both negative.

Let us try two scenarios

m = â€“ 2, n = â€“ 7: m^{2}Â + mn + n^{2}Â = 4 + 14 + 49 = 67

m = 3, n = 6: m^{2}Â + mn + n^{2}Â = 9 + 15 + 36 = 60

The equation will be most sensitive to the highest power.

The equation will be more sensitive to the term with the greater value.

In this question, we have a tradeâ€“off between higher value for m^{2}Â and n^{2}. For n^{2}, the choice is between 6^{2}Â and (â€“7)^{2}. This impact will overshadow the choice for p (where we are choosing between â€“2 and 3).

So, the maximum value for the expression would be 67.

Choice B is the correct answer.

**Ques:** For how many integer values does the following inequality hold good? (x + 2) (x + 4) (x + 6)……..(x + 100) < 0?

A. 25

B. 50

C. 49

D. 47

**Sol:**

(x + 2) (x + 4) (x + 6) ……..(x + 100) < 0

Now, the above expression will be zero for x = â€“2, â€“4, â€“6, â€“ 8â€¦..â€“100.

For x > â€“ 2 all the terms will be positive, hence the product will be positive.

For x < â€“ 100, all the terms will be negative and since there are 50 terms (even number), the product will be positive.

Now, if x = â€“ 99, the term x + 100 would be positive, everything else would be negative, so the expression would have 49 negative terms and one positive term. So the product would be negative.

Overall the expression will be negative if there are exactly 49 negative terms, or exactly 47 negative terms, or exactly 45 termsâ€¦. Or so on, up to exactly one negative term.

Exactly 49 negative terms= x = â€“ 99

Exactly 47 negative terms= x= â€“ 95

Exactly 45 negative terms= x = â€“ 91

……………

Exactly 1 negative term=. x = â€“ 3

So, x can take values {â€“3, â€“7, â€“11, â€“15, â€“19â€¦. â€“99}. We need to compute how many terms are there in this list.

In other words, how many terms are there in the list {3, 7, 11, ….99}. Now, these terms are separated by 4, so we can write each term as multiple of 4 + â€˜some constantâ€™.

Or 3=0 * 4 + 3

7=1 * 4 + 3

11=2 * 4 + 3

………………….

99=24 * 4 + 3

We go from 0 * 4 + 3 to 24 * 4 + 3, a total of 25 terms.

Hence the answer is 25

Choice A is the correct answer.

**Ques:** Consider three distinct positive integers a, b, c all less than 100. If |a – b| + |b – c| = |c â€“ a|, what is the maximum value possible for b?

A. 98

B. 99

C. 50

D. 100

**Sol:**

|q â€“ p| is the distance between p and q on the number line and |p â€“q| is the same as |q â€“p|.

We have |a -b| + |b -c| = |c â€“ a| i.e. the sum of some two of the distances is equal to the third.

So, the point b is in between a and c. We can have a or c to be 99 and b to be 98, since all the three have to be distinct.

Maximum value b can take is 98.

Choice A is the correct answer.

**Ques:** x^{4}Â â€“ 4x^{3}Â + ax^{2}Â â€“ bx + 1 = 0 has positive real roots. What is the maximum possible value of a + b?

A. 20

B. 12

C. 8

D. 10

**Â ****Sol:**

If we have an equation of the form ax^{4}Â + bx^{3}Â + cx^{2}Â + dx + e = 0 with Roots p, q, r and s.

We can say sum of the roots, m + n + o + p =Â âˆ’b/a

Sum of the products taken two at a time, mn + mo + mp + no + np + op = c/a

Sum of the products taken three at a time, mno + mnp + mop + nop =Â âˆ’d/a

Product of the roots, mnop = e/a

So, if m, n, o, p were roots of the given equation

m + n + o + p = 4,

mnop = 1

This means Arithmetic mean of m, n, o, p = 1 and geometric mean of m, n, o, p = 1.

m, n, o, p are positive real numbers. AM = GM. This gives all 4 numbers are equal.

Or, this expression is (x-1)^{4}

a = mn + mo + mp + no + np + op = 6

-(-b) = mno + mnp + mop + nop = 4

a = 6, b = 4. a + b = 10

Hence the answer is 10

Choice D is the correct answer.

All questions from CAT Exam Logical Reasoning

Logical Reasoning – Set 1: A high security research lab requires the researchers to set a pass key sequence based on the scan of the five fingers of their left hands.

Logical Reasoning – Set 2: Eight friends: Ajit, Byomkesh, Gargi, Jayanta, Kikira, Manik, Prodosh and Tapesh are going to Delhi from Kolkata

Logical Reasoning – Set 3: In an 8 X 8 chessboard a queen placed anywhere can attack another piece if the piece is present in the same row

Logical Reasoning – Set 4: A tea taster was assigned to rate teas from six different locations â€“ Munnar, Wayanad, Ooty, Darjeeling, Assam and Himachal.

Logical Reasoning – Set 5: Four cars need to travel from Akala (A) to Bakala (B). Two routes are available, one via Mamur (M) and the other via Nanur (N).

Logical Reasoning – Set 6: A new airlines company is planning to start operations in a country.

Logical Reasoning – Set 7: In a square layout of size 5m Ã— 5m, 25 equal sized square platforms of different heights are built.

Logical Reasoning – Set 8: There are 21 employees working in a division, out of whom 10 are special-skilled employees (SE) and the remaining are regular skilled employees (RE).

Logical Reasoning – Set 9: Healthy Bites is a fast food joint serving three items: burgers, fries and ice cream.

You can also see:Â Coded Inequality â€“ Tips and Tricks to Solve Questions in Logical Reasoning

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