CAT Inequalities Questions & Solutions

A sample of real CAT Inequalities past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 59 Inequalities questions in all — sign up free to practise them timed.

Q1.CAT 2025
Let $3\leq x\leq6$ and $\left[x^{2}\right] =\left[x\right]^{2}$ , where $[x]$ is the greatest integer not exceeding $x$ . If set $S$ represents all feasible values of $x$ , then a possible subset of $S$ is
- $\left(3,\sqrt{10}\right)\cup \left[5,\sqrt{26}\right)\cup \left\{6\right\}$
- $\left[3,\sqrt{10}\right] \cup \left[5,\sqrt{26}\right]$
- $\left[3,\sqrt{10}\right]\cup \left[4,\sqrt{17}\right]\cup \left\{6\right\}$
- $\left(4,\sqrt{18}\right)\cup\left[5,\sqrt{27}\right)\cup\left\{6\right\}$
Show solution
For n=3,4,5 and $x\in[n,n+1)$ we have $\lfloor x\rfloor=n$ , so the equation

$\lfloor x^2\rfloor=\lfloor x\rfloor^2=n^2$

$x^2\in[n^2,n^2+1)$ , i.e. $x\in[n,\sqrt{n^2+1})$

Thus for $3\le x\le6$

$S=[3,\sqrt{10})\ \cup\ [4,\sqrt{17})\ \cup\ [5,\sqrt{26})\ \cup{6}$

Option B and C have $\sqrt{10}$ included, which is not part of the original set. And Option D has $\sqrt{18}$ . So, it is not possible.

Option A is the answer.
Q2.CAT 2025
If a,b,c and d are integers such that their sum is 46, then the minimum possible value of $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}$ is
Answer: 2
Show solution
Given expression: $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}$

The given expression has just the sum of squares of the terms. So, the minimum value is either zero or positive.

If we can make all the values equal, we can get zero. But since all the values are integers and the sum 46 is not divisible by 4, we can't make everything equal.

So, the nearest four values are 12, 11, 11, 12.

With this, the minimum value is $(12-11)^2+(12-11)^2+(12-12)^2 = 2$
Q3.CAT 2023
If $x$ and $y$ are real numbers such that $x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$ , then the value $x - 2y$ is
- 0
- 1
- -1
- 2
Show solution
Given, $x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$

=> $x^2+4xy+4y^2+\left(x-2y-1\right)^2=0$

=> $\left(x+2y\right)^2+\left(x-2y-1\right)^2=0$

For the L.H.S. of the equation to be 0, each of the square terms should be 0 (as squares cannot be negative)

=> x - 2y - 1 = 0 => x - 2y = 1
Q4.CAT 2023
The population of a town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021, and increased by x% from the year 2021 to 2022, where x and y are two natural numbers. If population in 2022 was greater than the population in 2020 and the difference between x and y is 10, then the lowest possible population of the town in 2021 was
- 72000
- 74000
- 73000
- 75000
Show solution
It is given that the population of the town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021 and increased by x% from the year 2021 to 2022, where x and y are two natural numbers.

Hence, the population in 2021 is $100000\left(\ \frac{\ 100-y}{100}\right)$ .

The population in 2022 is $100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)$

It is also given that the population in 2022 was greater than the population in 2020 and the difference between x and y is 10.

Hence,

$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)>\ 100000$ , and (x-y) = 10

=> $100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 110+y}{100}\right)>\ 100000$

=> $\ \frac{\ 100-y}{100}\left(\ \frac{\ 110+y}{100}\right)>\ 1$

To get the minimum possible value of 2021, we need to increase the value of y as much as possible.

Hence, $\left(\ \ 100-y\right)\left\{\left(\ \ 100+y\right)+10\right\}>\ 10000$

=> $10000-y^2+1000-10y>\ 10000$

=> $y^2+10y<1000$

=> $y^2+10y+25<1025$

=> $\left(y+5\right)^2=1024\ <\ 1025$

=> $\left(y+5\right)^2=32^2$

=> $y=27$

Hence, the population in 2021 is 100000*(100-27) = 73000

The correct option is C
Q5.CAT 2022
The number of distinct integer values of n satisfying $\frac{4-\log_{2}n}{3-\log_{4}n} < 0$ , is
Answer: 47
Show solution
Let $\ \log_2n=y$

$\ \ \dfrac{\ 4-y}{3-\dfrac{y}{2}}<0$

$\ \ \left(4-y\right)\left(3-\dfrac{y}{2}\right)<0$

$\ \ \left(4-y\right)\left(6-y\right)<0$

$\ \ \left(y-4\right)\left(y-6\right)<0$

$4 < y < 6$

$4<\log_2n<6$

$2^4 < n < 2^6$

$16 < n < 64$

n can take values from 17 to 63(inclusive).

The number of n values possible = 47
Q6.CAT 2021
If n is a positive integer such that $(\sqrt[7]{10})(\sqrt[7]{10})^{2}...(\sqrt[7]{10})^{n}>999$ , then the smallest value of n is
Answer: 6
Show solution
$(\sqrt[7]{10})(\sqrt[7]{10})^{2}...(\sqrt[7]{10})^{n}>999$

$(\sqrt[7]{10})^{1+2+...+n}>999$

$10^{\frac{1+2+...+n}{7}}>999$

For minimum value of n,

$\frac{1+2+...+n}{7}=3$

1 + 2 + ... + n = 21

We can see that if n = 6, 1 + 2 + 3 + ... + 6 = 21.
Q7.CAT 2019
If x is a real number, then $\sqrt{\log_{e}{\dfrac{4x - x^2}{3}}}$ is a real number if and only if
- $1 \leq x \leq 3$
- $1 \leq x \leq 2$
- $-1 \leq x \leq 3$
- $-3 \leq x \leq 3$
Show solution
$\sqrt{\log_{e}{\dfrac{4x - x^2}{3}}}$ will be real if $\log_e\ \dfrac{\ 4x-x^2}{3}\ \ge\ 0$

$\dfrac{\ 4x-x^2}{3}\ \ge\ 1$

$\ 4x-x^2-3\ \ge\ 0$

$\ x^2-4x+3\ \le\ 0$

$1\le\ x\le\ 3$
Q8.CAT 2006
The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x <= y is:
- 7
- 13
- 14
- 18
- 20
Show 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.
Q9.CAT 2003
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
- Less than 6
- greater than 8
- greater than 6
- Less than 8
Show 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
Q10.CAT 2001
If x > 5 and y < -1, then which of the following statements is true?
- (x + 4y) > 1
- x > -4y
- -4x < 5y
- None of these
Show 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 .
Q11.CAT 2000
If x>2 and y>-1,then which of the following statements is necessarily true?
- xy>-2
- -x<2y
- xy<-2
- -x>2y
Show 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.
Passage
For these questions the following functions have been defined.

$la(x, y, z) = min (x+y, y+z)$
$le(x, y, z) = max(x -y, y-z)$
$ma (x, y, z) = \frac{1}{2} (le (x, y, z) + la (x, y, z))$
Q12.CAT 1997
For x=15, y=10 and z=9 , find the value of le(x, min(y, x-z), le(9, 8, ma(x, y, z)).
- 5
- 12
- 9
- 4
Show solution
Given expression can be reduced to
le(15, min(10,15-9) , le(9,8,12))
Or le(15,6,1) = 9

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CAT Inequalities Questions & Solutions

Q1.CAT 2025

Let $3\leq x\leq6$ and $\left[x^{2}\right] =\left[x\right]^{2}$ , where $[x]$ is the greatest integer not exceeding $x$ . If set $S$ represents all feasible values of $x$ , then a possible subset of $S$ is

$\left(3,\sqrt{10}\right)\cup \left[5,\sqrt{26}\right)\cup \left\{6\right\}$
$\left[3,\sqrt{10}\right] \cup \left[5,\sqrt{26}\right]$
$\left[3,\sqrt{10}\right]\cup \left[4,\sqrt{17}\right]\cup \left\{6\right\}$
$\left(4,\sqrt{18}\right)\cup\left[5,\sqrt{27}\right)\cup\left\{6\right\}$

Show solution

For n=3,4,5 and $x\in[n,n+1)$ we have $\lfloor x\rfloor=n$ , so the equation

$\lfloor x^2\rfloor=\lfloor x\rfloor^2=n^2$

$x^2\in[n^2,n^2+1)$ , i.e. $x\in[n,\sqrt{n^2+1})$

Thus for $3\le x\le6$

$S=[3,\sqrt{10})\ \cup\ [4,\sqrt{17})\ \cup\ [5,\sqrt{26})\ \cup{6}$

Option B and C have $\sqrt{10}$ included, which is not part of the original set. And Option D has $\sqrt{18}$ . So, it is not possible.

Option A is the answer.

Q2.CAT 2025

If a,b,c and d are integers such that their sum is 46, then the minimum possible value of $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}$ is

Answer: 2

Show solution

Given expression: $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}$

The given expression has just the sum of squares of the terms. So, the minimum value is either zero or positive.

If we can make all the values equal, we can get zero. But since all the values are integers and the sum 46 is not divisible by 4, we can't make everything equal.

So, the nearest four values are 12, 11, 11, 12.

With this, the minimum value is $(12-11)^2+(12-11)^2+(12-12)^2 = 2$

Q3.CAT 2023

If $x$ and $y$ are real numbers such that $x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$ , then the value $x - 2y$ is

0
1
-1
2

Show solution

Given, $x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$

=> $x^2+4xy+4y^2+\left(x-2y-1\right)^2=0$

=> $\left(x+2y\right)^2+\left(x-2y-1\right)^2=0$

For the L.H.S. of the equation to be 0, each of the square terms should be 0 (as squares cannot be negative)

=> x - 2y - 1 = 0 => x - 2y = 1

Q4.CAT 2023

The population of a town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021, and increased by x% from the year 2021 to 2022, where x and y are two natural numbers. If population in 2022 was greater than the population in 2020 and the difference between x and y is 10, then the lowest possible population of the town in 2021 was

72000
74000
73000
75000

Show solution

It is given that the population of the town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021 and increased by x% from the year 2021 to 2022, where x and y are two natural numbers.

Hence, the population in 2021 is $100000\left(\ \frac{\ 100-y}{100}\right)$ .

The population in 2022 is $100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)$

It is also given that the population in 2022 was greater than the population in 2020 and the difference between x and y is 10.

Hence,

$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)>\ 100000$ , and (x-y) = 10

=> $100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 110+y}{100}\right)>\ 100000$

=> $\ \frac{\ 100-y}{100}\left(\ \frac{\ 110+y}{100}\right)>\ 1$

To get the minimum possible value of 2021, we need to increase the value of y as much as possible.

Hence, $\left(\ \ 100-y\right)\left\{\left(\ \ 100+y\right)+10\right\}>\ 10000$

=> $10000-y^2+1000-10y>\ 10000$

=> $y^2+10y<1000$

=> $y^2+10y+25<1025$

=> $\left(y+5\right)^2=1024\ <\ 1025$

=> $\left(y+5\right)^2=32^2$

=> $y=27$

Hence, the population in 2021 is 100000*(100-27) = 73000

The correct option is C

Q5.CAT 2022

The number of distinct integer values of n satisfying $\frac{4-\log_{2}n}{3-\log_{4}n} < 0$ , is

Answer: 47

Show solution

Let $\ \log_2n=y$

$\ \ \dfrac{\ 4-y}{3-\dfrac{y}{2}}<0$

$\ \ \left(4-y\right)\left(3-\dfrac{y}{2}\right)<0$

$\ \ \left(4-y\right)\left(6-y\right)<0$

$\ \ \left(y-4\right)\left(y-6\right)<0$

$4 < y < 6$

$4<\log_2n<6$

$2^4 < n < 2^6$

$16 < n < 64$

n can take values from 17 to 63(inclusive).

The number of n values possible = 47

Q6.CAT 2021

If n is a positive integer such that $(\sqrt[7]{10})(\sqrt[7]{10})^{2}...(\sqrt[7]{10})^{n}>999$ , then the smallest value of n is

Answer: 6

Show solution

$(\sqrt[7]{10})(\sqrt[7]{10})^{2}...(\sqrt[7]{10})^{n}>999$

$(\sqrt[7]{10})^{1+2+...+n}>999$

$10^{\frac{1+2+...+n}{7}}>999$

For minimum value of n,

$\frac{1+2+...+n}{7}=3$

1 + 2 + ... + n = 21

We can see that if n = 6, 1 + 2 + 3 + ... + 6 = 21.

Q7.CAT 2019

If x is a real number, then $\sqrt{\log_{e}{\dfrac{4x - x^2}{3}}}$ is a real number if and only if

$1 \leq x \leq 3$
$1 \leq x \leq 2$
$-1 \leq x \leq 3$
$-3 \leq x \leq 3$

Show solution

$\sqrt{\log_{e}{\dfrac{4x - x^2}{3}}}$ will be real if $\log_e\ \dfrac{\ 4x-x^2}{3}\ \ge\ 0$

$\dfrac{\ 4x-x^2}{3}\ \ge\ 1$

$\ 4x-x^2-3\ \ge\ 0$

$\ x^2-4x+3\ \le\ 0$

$1\le\ x\le\ 3$

Q8.CAT 2006

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

Show 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.

Q9.CAT 2003

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

Less than 6
greater than 8
greater than 6
Less than 8

Show 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

Q10.CAT 2001

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

(x + 4y) > 1
x > -4y
-4x < 5y
None of these

Show 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 .

Q11.CAT 2000

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

xy>-2
-x<2y
xy<-2
-x>2y

Show 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.

Passage

For these questions the following functions have been defined.

la(x, y, z) = min (x+y, y+z)

le(x, y, z) = max(x -y, y-z)

ma (x, y, z) = \frac{1}{2} (le (x, y, z) + la (x, y, z))

Q12.CAT 1997

For x=15, y=10 and z=9 , find the value of le(x, min(y, x-z), le(9, 8, ma(x, y, z)).

Show solution

Given expression can be reduced to
le(15, min(10,15-9) , le(9,8,12))
Or le(15,6,1) = 9

47+ more Inequalities questions inside

Get the full CAT Inequalities set with timed practice, bookmarks and analysis — free to start.

Start practising free

CAT Inequalities previous year questions with solutions — AthenaPrep