CAT Inequalities Questions Download PDF [Most Important]

Inequalities 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.

Download Inequalities Questions for CAT

Join CAT 2023 Online Coaching

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

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

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

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

1) Answer (B)

View Video Solution

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: 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) $-0.5 \leq w \leq 2$

b) $-4 \leq w \leq 4$

c) $-4 \leq w \leq 2$

d) $-2 \leq w \leq -0.5$

2) Answer (B)

View Video Solution

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  u = -0.5, we get w = -4

Question 3: 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) Less than 6

b) greater than 8

c) greater than 6

d) Less than 8

3) Answer (C)

View Video Solution

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

a) 7

b) 13

c) 14

d) 18

e) 20

4) Answer (B)

View Video Solution

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

a) $-8 \leq x \leq 1$

b) $-1 \leq x \leq 8$

c) $1 \leq x \leq 8$

d) $1 \leq x \leq 18$

e) $-8 \leq x \leq 8$

5) Answer (A)

View Video Solution

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

a) (x + 4y) > 1

b) x > -4y

c) -4x < 5y

d) None of these

6) Answer (D)

View Video Solution

Solution:

Substitute x=6 and y=-6 ,

x+4y = -18

x = 6, -4y = 24

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

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

Question 7: 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) $x^2y$

b) $xy^2$

c) $5xy$

d) None of these

7) Answer (C)

View Video Solution

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: $m$ 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) 4

b) 5

c) 8

d) None of these

8) Answer (D)

View Video Solution

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: 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) 4

b) 1

c) 16

d) 18

9) Answer (C)

View Video Solution

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, option C is the right answer.

Question 10: 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) 12

b) 20

c) 12.5

d) 13.3

10) Answer (C)

View Video Solution

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 = $(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$ = $(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$ =12.5

Approach 2:

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

Let x+1/x and y+1/y be two terms. Thus (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 as $(x+1/x)^2$ + $(y+1/y)^2$ = $4AM^2$ – $GM^2$. 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 = y +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, y <=1. Hence, their product cannot be 1.

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

=> y = 1/2

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

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

a) xy>-2

b) -x<2y

c) xy<-2

d) -x>2y

11) Answer (B)

View Video Solution

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: 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) 5/3

b) $\sqrt{19}$

c) 13/3

d) None of these

12) Answer (C)

View Video Solution

Solution:

The given equations are  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)  = 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$
=>  $ 3x^2 -10x – 13 \leq 0$
=>  $ -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: 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) Only A

b) B and C

c) A and B

d) None of these

13) Answer (C)

View Video Solution

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: 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) m >= min(u, v, w)

b) m >= max(u, v, w)

c) m < min(u, v, w)

d) None of these

14) Answer (D)

View Video Solution

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

a) p+q+r

b) 1/(p+q+r)

c) 1

d) $p^{-1} + q^{-1} + r^{-1}$

15) Answer (C)

View Video Solution

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: 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) 1 > x > y

b) x > 1 > y

c) 1 > y > x

d) y > 1 > x

16) Answer (B)

View Video Solution

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: The number of integers n satisfying -n+2 ≥ 0 and 2n ≥ 4 is

a) 0

b) 1

c) 2

d) 3

17) Answer (B)

View Video Solution

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

Download CAT 2023 Study Plan PDF