# CAT Inequalities Questions Download PDF [Most Important]

**Inequalities**Â is an important topic in the **CAT** **Quant** (Algebra) section. If you find these questions a bit tough, make sure you solve more CAT Inequalities questions. Learn all the important **formulas** and tricks on **how to answer questions**** on**** Inequalities**. You can check out these CAT Inequalities questions from the **CAT Previous year papers**. Practice a good number of sums in the CAT **Inequalities** so that you can answer these questions with ease in the exam. In this post, we will look into some important Inequalities Questions for CAT quants. These are a good source of practice for CAT 2022 preparation; If you want to practice these questions, you can download this Important **CAT Inequalities Questions** and answers **PDF** along with the video solutions below, which is completely Free.

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**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 $4m^2+2m+1$

**1)Â AnswerÂ (B)**

**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 $-2 \leq w \leq -0.5$

**2)Â AnswerÂ (B)**

**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

**3)Â AnswerÂ (C)**

**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

**4)Â AnswerÂ (B)**

**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 $-8 \leq x \leq 8$

**5)Â AnswerÂ (A)**

**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

**6)Â AnswerÂ (D)**

**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

**7)Â AnswerÂ (C)**

**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

**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

**8)Â AnswerÂ (D)**

**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

**9)Â AnswerÂ (C)**

**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

**10)Â AnswerÂ (C)**

**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

**Question 11:Â **If x>2 and y>-1,then which of the following statements is necessarily true?

a)Â xy>-2

b)Â -x<2y

c)Â xy<-2

d)Â -x>2y

**11)Â AnswerÂ (B)**

**Solution:**

This kind of questions must be solved using the counter example method.

x = 100 and y = -1/2 rules out option a)

x = 3 and y = 0 rules out options c) and d)

Option b) is correct.

**Question 12:Â **If x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have?

a)Â 5/3

b)Â $\sqrt{19}$

c)Â 13/3

d)Â None of these

**12)Â AnswerÂ (C)**

**Solution:**

The given equations are Â x + y + z = 5 — (1) , xy + yz + zx = 3 — (2)

xy + yz + zx = 3

x(y + z) + yz = 3

=> x ( 5 -x ) +y ( 5 – x – y) = 3

=> $ – y^2 – y (5 -x) – x ^2 + 5x = 3$

=> $ y^2 + y (x-5) + ( x ^2 – 5x +3) Â = 0 $

The above equation should have real roots for y, => Determinant >= 0

=>$b^2-4ac>0$

=> $ ( x – 5)^2 – 4(x ^2 – 5x +3 ) \geq 0$

=> Â $ 3x^2 -10x – 13 \leq 0$

=> Â $ -1 \leq x \leq \frac{13}{3}$

Hence maximum value x can take is $\frac{13}{3}$, and the corresponding values for y,z are $\frac{1}{3},\frac{1}{3}$

**Question 13:Â **If $x^2 + 5y^2 + z^2 = 2y(2x+z)$, then which of the following statements is(are) necessarily true?

A. x = 2y B. x = 2z C. 2x = z

a)Â Only A

b)Â B and C

c)Â A and B

d)Â None of these

**13)Â AnswerÂ (C)**

**Solution:**

The equation is not satisfied for only x = 2y.

Using statements B and C, i.e., x = 2z and 2x = z, we see that the equation is not satisfied.

Using statements A and B, i.e., x = 2y and x = 2z, i.e., z = y = x/2, the equation is satisfied.

Option c) is the correct answer.

**Question 14:Â **If u, v, w and m are natural numbers such that $u^m + v^m = w^m$, then which one of the following is true?

a)Â m >= min(u, v, w)

b)Â m >= max(u, v, w)

c)Â m < min(u, v, w)

d)Â None of these

**14)Â AnswerÂ (D)**

**Solution:**

Substitute value of u = v = 2, w = 4 and m = 1. Here the condition holds and options A and B are false. Hence, we can eliminate options A and B.

Substitute u = v = 1, w=2 and m= 1. Here m=min(u, v, w). Hence, option C also does not hold. Hence, we can eliminate option C.

Option d) is the correct answer.

**Question 15:Â **If pqr = 1, the value of the expression $1/(1+p+q^{-1}) + 1/(1+q+r^{-1}) + 1/(1+r+p^{-1})$

a)Â p+q+r

b)Â 1/(p+q+r)

c)Â 1

d)Â $p^{-1} + q^{-1} + r^{-1}$

**15)Â AnswerÂ (C)**

**Solution:**

Let p = q = r = 1

So, the value of the expression becomes 1/3 + 1/3 + 1/3 = 1

If we substitute these values, options a), b) and d) do not satisfy.

Option c) is the answer.

**Question 16:Â **From any two numbers $x$ and $y$, we define $x* y = x + 0.5y – xy$ . Suppose that both $x$ and $y$ are greater than 0.5. Then

$x* x < y* y$ if

a)Â 1 > x > y

b)Â x > 1 > y

c)Â 1 > y > x

d)Â y > 1 > x

**16)Â AnswerÂ (B)**

**Solution:**

$x*x < y*y$

or $x + 0.5x – x^2 < y + 0.5y – y^2$

$y^2 – x^2 + 1.5x – 1.5y < 0$

$(y – x)(y + x) – 1.5 (y – x) < 0$

$(y – x)(y + x -1.5) < 0$

$(x – y)(1.5 – (x + y)) < 0$

Now there will be two possibilities

$x < y$ and $(x + y) < 1.5$ ………..(i)

or $x > y$ and $(x + y) > 1.5$ …………(ii)

Among all options only option B satisfies (ii).

Hence, option B is the correct answer.

**Question 17:Â **The number of integers n satisfying -n+2Â â‰ĄÂ 0 and 2n â‰ĄÂ 4 is

a)Â 0

b)Â 1

c)Â 2

d)Â 3

**17)Â AnswerÂ (B)**

**Solution:**

-n+2 >= 0

or n<=2

and 2n>=4

or n>=2

So we can take only one value of n i.e. 2