# CAT Inequalities Questions PDF [Most Important]

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InequalitiesÂ 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 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.

Question 1:Â For real x, the maximum possible value of $\frac{x}{\sqrt{1+x^{4}}}$ is

a)Â $\frac{1}{2}$

b)Â $1$

c)Â $\frac{1}{\sqrt{3}}$

d)Â $\frac{1}{\sqrt{2}}$

Solution:

NowÂ $\frac{x}{\sqrt{\ 1+x^4}}=\ \frac{\ 1}{\sqrt{\ \ \frac{\ 1+x^4}{x^2}}}=\frac{1}{\sqrt{\ \frac{1}{x^2}+x^2}}$

Applying A.M>= G.M.

$\frac{\left(\frac{1}{x^2}+x^2\right)}{2}\ge\ 1\ or\ \ \frac{1}{x^2}+x^2\ge\ 2$ Substituting we get the maximum possible value of the equation asÂ $\frac{1}{\sqrt{\ 2}}$

Question 2: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 3: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 4: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 $p^{-1} + q^{-1} + r^{-1}$

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.

Question 5: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

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

Question 6:Â Which of the following values of x do not satisfy the inequality $(x^2 – 3x + 2 > 0)$ at all?

a)Â $1\leq x \leq 2$

b)Â $-1\geq x \geq -2$

c)Â $0 \leq x \leq 2$

d)Â $0\geq x \geq -2$

Solution:

After solving given equation, we will have inequality resolved to:

(x-1)(x-2)>0

Or we can say range of x will be as follows:

x<1; Â x>2

Hence, option A has a set of values which don’t lie in the possible range of x.

So theÂ answer will be A.

Question 7:Â The number of positive integer valued pairs (x, y), satisfying 4x – 17 y = 1 and x < 1000 is:

a)Â 59

b)Â 57

c)Â 55

d)Â 58

Solution:

y = $\frac{4x-1}{17}$

The integral values of x for which y is an integer are 13, 30, 47,……

The values are in the form 17n + 13, where $n \geq 0$

17n + 13 < 1000

=> 17n < 987

=> n < 58.05

=> n can take values from 0 to 58 => Number of values = 59

Question 8:Â If | r – 6 | = 11 and | 2q – 12 | = 8, what is the minimum possible value of q / r?

a)Â -2/5

b)Â 2/17

c)Â 10/17

d)Â None of these

Solution:

| r-6 | = 11 => r = -5 or 17

| 2q – 12 | = 8 => q = 10 or 2

So, the minimum possible value of q/r = 10/(-5) = -2

Question 9:Â If a and b are integers of opposite signs such that $(a + 3)^{2} : b^{2} = 9 : 1$ and $(a -1)^{2}:(b – 1)^{2} = 4:1$, then the ratio $a^{2} : b^{2}$ is

a)Â 9:4

b)Â 81:4

c)Â 1:4

d)Â 25:4

Solution:

Since the square root can be positive or negative we will get two cases for each of the equation.

For the first one,

a + 3 = 3b .. i

a + 3 = -3b … ii

For the second one,

a – 1 = 2(b -1) … iii

a – 1 = 2 (1 – b) … iv

we have to solve i and iii, i and iv, ii and iii, ii and iv.

Solving i and iii,

a + 3 = 3b and a = 2bÂ  – 1, solving, we get a = 3 and b = 2, which is not what we want.

Solving i and iv

a + 3 = 3b and a = 3 – 2b, solving, we get b = 1.2, which is not possible.

Solving ii and iii

a + 3 = -3b and a = 2b – 1, solving, we get b = 0.4, which is not possible.

Solving ii and iv,

a + 3 = -3b and a = 3 – 2b, solving, we get a = 15 and b = -6 which is what we want.

Thus, $\frac{a^2}{b^2} = \frac{25}{4}$

Question 10:Â For how many integers n, will the inequality $(n – 5) (n – 10) – 3(n – 2)\leq0$ be satisfied?

Solution:

$(n – 5) (n – 10) – 3(n – 2)\leq0$
=> $n^2 – 15n + 50 – 3n + 6 \leq 0$
=> $n^2 – 18n + 56 \leq 0$
=> $(n – 4)(n – 14) \leq 0$
=> Thus, n can take values from 4 to 14. Hence, the required number of values are 14 – 4 +Â 1 = 11.

Question 11:Â The minimum possible value of the sum of the squares of the roots of the equation $x^2+(a+3)x-(a+5)=0$ is

a)Â 1

b)Â 2

c)Â 3

d)Â 4

Solution:

Let the roots of the equationÂ $x^2+(a+3)x-(a+5)=0$ be equal to $p,q$

Hence, $p+q = -(a+3)$ and $p \times q = -(a+5)$

Therefore, $p^2+q^2 = a^2+6a+9+2a+10 = a^2+8a+19 = (a+4)^2+3$

As $(a+4)^2$ is always non negative, the least value of the sum of squares is 3

Question 12:Â The smallest integer $n$ such that $n^3-11n^2+32n-28>0$ is

Solution:

We can see that at n = 2, $n^3-11n^2+32n-28=0$ i.e. (n-2) is a factor of $n^3-11n^2+32n-28$

$\dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$

We can further factorizeÂ n^2-9n+14 as (n-2)(n-7).

$n^3-11n^2+32n-28=(n-2)^2(n-7)$

$\Rightarrow$Â $n^3-11n^2+32n-28>0$

$\Rightarrow$Â $(n-2)^2(n-7)>0$

Therefore, we can say that n-7>0

Hence, n$_{min}$ = 8

Question 13:Â If x is a real number, then $\sqrt{\log_{e}{\frac{4x – x^2}{3}}}$ is a real number if and only if

a)Â $1 \leq x \leq 3$

b)Â $1 \leq x \leq 2$

c)Â $-1 \leq x \leq 3$

d)Â $-3 \leq x \leq 3$

Solution:

$\sqrt{\log_{e}{\frac{4x – x^2}{3}}}$ will be real ifÂ $\log_e\ \frac{\ 4x-x^2}{3}\ \ge\ 0$

$\frac{\ 4x-x^2}{3}\ >=\ 1$

$\ 4x-x^2-3\ >=\ 0$

$\ x^2-4x+3\ =<\ 0$

1=< x=< 3

Question 14:Â Among 100 students, $x_1$ have birthdays in January, $X_2$ have birthdays in February, and so on. If $x_0=max(x_1,x_2,….,x_{12})$, then the smallest possible value of $x_0$ is

a)Â 8

b)Â 9

c)Â 10

d)Â 12

Solution:

$x_0=max(x_1,x_2,….,x_{12})$

$x_0$ will be minimum if x1,x2…x12 are close to each other

100/12=8.33

.’.Â max$(x_1,x_2,….,x_{12})$ will be minimum ifÂ $(x_1,x_2,….,x_{12})$=(9,9,9,9,8,8,8,8,8,8,8,8,)

.’. Option B is correct.

Question 15:Â The number of pairs of integers $(x,y)$ satisfying $x\geq y\geq-20$ and $2x+5y=99$

We have 2x + 5y = 99 orÂ $x=\frac{\left(99-5y\right)}{2}$
Now $x\ge\ y\ \ge\ -20$ ;Â SoÂ $\frac{\left(99-5y\right)}{2}\ge\ y\ ;\ 99\ge7y\ or\ y\le\ \approx\ 14$
So $-20\le y\le14$. Now for this range of “y”, we have to find all the integral values of “x”. As the coefficient of “x” is 2,