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# CAT Inequalities Questions PDF [Most Important]

Inequalities form an important part of CAT Algebra. You can expect close to 2-3 questions in the 22-question format of the paper. Inequalities is often one of the most feared topics among the candidates. In this article, we will look into some Inequalities Questions for the CAT Exam. If you want to practice these questions, you can download the PDF, which is completely Free.
Inequalities-based questions appear in the CAT test almost every year. Inequalities are one of the simplest topics to solve and score on the CAT; therefore questions with inequalities should not be avoided.

• CAT Inequalities – Tip 1: The concept of Inequalities questions covered in the CAT Algebra in the Quantitative Aptitude portion of the CAT Exam are: Range of Inequalities, Modulus functions, Possible solutions, and so on.
• CAT Inequalities – Tip 2: The subject also covers linear and quadratic equations, root finding, polynomials, functions, and other topics.
• CAT Inequalities – Tip 3: For CAT, inequalities are considered slightly challenging in inequalities topic. A strong foundation in this topic will aid a student in answering questions about Functions as well. Every year 2-3 questions are asked from Inequalities in the CAT Exam.

Question 1:Â Let a, b, c, d be four integers such that a+b+c+d = 4m+1 where m is a positive integer. Given m, which one of the following is necessarily true?

a)Â The minimum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2-2m+1$

b)Â The minimum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2+2m+1$

c)Â The maximum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2-2m+1$

d)Â The maximum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2+2m+1$

Solution:

Taking lowest possible positive value of m i.e. 1 . Such that a+b+c+d=5 , so atleast one of them must be grater than 1 ,

take a=b=c=1 and d=2

we get $a^2 + b^2 + c^2 + d^2 = 7$ which is equal to $4m^2+2m+1$ for other values it is greater than $4m^2+2m+1$ . so option B

Question 2:Â Given that $-1 \leq v \leq 1, -2 \leq u \leq -0.5$ and $-2 \leq z \leq -0.5$ and $w = vz /u$ , then which of the following is necessarily true?

a)Â $-0.5 \leq w \leq 2$

b)Â $-4 \leq w \leq 4$

c)Â $-4 \leq w \leq 2$

d)Â $-2 \leq w \leq -0.5$

Solution:

We know $w = vz /u$ so taking max value of u and min value of v and z to get min value of w which is -4.

Similarly taking min value of u and max value of v and z to get max value of w which is 4

Take v = 1, z = -2 and u = -0.5, we get w = 4

Take v = -1, z = -2 and Â u = -0.5, we get w = -4

Question 3:Â If x, y, z are distinct positive real numbers the $(x^2(y+z) + y^2(x+z) + z^2(x+y))/xyz$ would always be

a)Â Less than 6

b)Â greater than 8

c)Â greater than 6

d)Â Less than 8

Solution:

For the given expression value of x,y,z are distinct positive integers . So the value of expression will always be greater than value when all the 3 variables are equal . substitute x=y=z we get minimum value of 6 .

$(x^2(y+z) + y^2(x+z) + z^2(x+y))/xyz$ = x/z + x/y + y/z + y/x + z/y + z/x

Applying AM greater than or equal to GM, we get minimum sum = 6

Question 4:Â The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x <= y is:

a)Â 7

b)Â 13

c)Â 14

d)Â 18

e)Â 20

Solution:

y = 38 => x = 1

y = 36 => x = 2

y = 14 => x = 13

y = 12 => x = 14 => Cases from here are not valid as x > y.

Hence, there are 13 solutions.

Question 5:Â What values of x satisfy $x^{2/3} + x^{1/3} – 2 <= 0$?

a)Â $-8 \leq x \leq 1$

b)Â $-1 \leq x \leq 8$

c)Â $1 \leq x \leq 8$

d)Â $1 \leq x \leq 18$

e)Â $-8 \leq x \leq 8$

Solution:

Try to solve this type of questions using the options.

Subsitute 0 first => We ger -2 <=0, which is correct. Hence, 0 must be in the solution set.

Substitute 8 => 4 + 2 – 2 <=0 => 6 <= 0, which is false. Hence, 8 must not be in the solution set.

=> Option 1 is the answer.

Question 6:Â If x > 5 and y < -1, then which of the following statements is true?

a)Â (x + 4y) > 1

b)Â x > -4y

c)Â -4x < 5y

d)Â None of these

Solution:

Substitute x=6 and y=-6 ,

x+4y = -18

x = 6, -4y = 24

-4x = -24, 5y = -30

SoÂ none of the options out of a,b or c satisfies .

Question 7:Â x and y are real numbers satisfying the conditions 2 < x < 3 and – 8 < y < -7. Which of the following expressions will have the least value?

a)Â $x^2y$

b)Â $xy^2$

c)Â $5xy$

d)Â None of these

Solution:

$xy^2$ will have it’s least value when y=-7 and x=2 and equals 98.
So $xy^2>98$

$x^2y$ will have it’s least value when y=-8 and x=3 and equals -72.
So, $x^2y > -72$

$5xy$ will have it’s least value when y=-8 and x=3 and equals -120
So, $5xy > -120$

So, of the three expressions, the least possible value is that of 5xy

Checkout: CAT Free Practice Questions and Videos
Question 8:Â $m$ is the smallest positive integer such that for any integer $n \geq m$, the quantity $n^3 – 7n^2 + 11n – 5$ is positive. What is the value of $m$?

a)Â 4

b)Â 5

c)Â 8

d)Â None of these

Solution:

$n^3 – 7n^2 + 11n – 5 = (n-1)(n^2 – 6n +5) = (n-1)(n-1)(n-5)$
This is positive for n > 5
So, m = 6

Question 9:Â If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1+b)(1+c)(1+d)?

a)Â 4

b)Â 1

c)Â 16

d)Â 18

Solution:

Since the product is constant,

(a+b+c+d)/4 >= $(abcd)^{1/4}$

We know that abcd = 1.

Therefore, a+b+c+d >= 4

$(a+1)(b+1)(c+1)(d+1)$
= $1+a+b+c+d+ ab + ac + ad + bc + bd + cd+ abc+ bcd+ cda+ dab+abcd$
We know that $abcd = 1$
Therefore, $a = 1/bcd, b = 1/acd, c = 1/bda$ and $d = 1/abc$
Also, $cd = 1/ab, bd = 1/ac, bc = 1/ad$
The expression can be clubbed together as $1 + abcd + (a + 1/a) + (b+1/b) + (c+1/c) + (d+1/d) + (ab + 1/ab) + (ac+ 1/ac) + (ad + 1/ad)$
For any positive real number $x$, $x + 1/x \geq 2$
Therefore, the least value that $(a+1/a), (b+1/b)….(ad+1/ad)$ can take is 2.
$(a+1)(b+1)(c+1)(d+1) \geq 1 + 1 + 2 + 2 + 2+ 2 + 2 + 2 + 2$
=> $(a+1)(b+1)(c+1)(d+1) \geq 16$
The least value that the given expression can take is 16. Therefore, optionÂ C is the right answer.

Question 10:Â Let x and y be two positive numbers such that $x + y = 1.$
Then the minimum value of $(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$ is

a)Â 12

b)Â 20

c)Â 12.5

d)Â 13.3

Solution:

Approach 1:

The given expression is symmetric in x and y and the limiting condition (x+y=1) is also symmetric in x and y.

=>This means that the expression attains the minimum value when x = y
x=y=1/2
So, the value = $(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$ =Â $(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$ =12.5

Approach 2:

$(x+1/x)^2$ +Â $(y+1/y)^2$ =Â $(x+1/x+y+1/y)^2$ – $2*(x+1/x)(y+1/y)$

Let x+1/x and y+1/y be two terms. ThusÂ (x+1/x+y+1/y)/2 would be their Arithmetic Mean(AM) and $\sqrt{(x+1/x)(y+1/y)}$ would be their Geometric Mean (GM).

Therefore, we can express the above equation asÂ $(x+1/x)^2$ +Â $(y+1/y)^2$ = $4AM^2$ – $GM^2$. As AM >= GM, the minimum value of expression would be attained when AM = GM.

When AM = GM, both terms are equal. That is x+1/x = yÂ +1/y.

Substituting y=1-x we get

x+1/x = (1-x) + 1/(1-x)

On solving we get 2x-1 = (2x-1)/ x(1-x)

So either 2x-1 = 0 or x(1-x) = 1

x(1-x) = x * y

As x and y are positive numbers whose sum = 1, 0<= x, yÂ <=1. Hence, their product cannot be 1.

Thus, 2x-1 = 0 or x=1/2

=> y = 1/2

So, the value = $(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$ =Â $(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$ = 12.5

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