Important CAT Inequalities Questions (with Notes) PDF

Inequalities is one of the key topics in the CAT Quants Section. CAT Inequalities questions appear in the CAT and other MBA entrance exams every year. You can check out these  CAT Inequalities Questions from Previous years. In this article, we will look into some important CAT Inequalities Questions (with Notes) PDF. These are a good source for practice; If you want to practice these questions, you can download the CAT Inequalities Questions PDF below, which is completely Free.

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• CAT Inequalities Questions – Tip 1: Be thorough with all the basics of this topic. Understand the important Mod formulas well. If you’re starting the prep, firstly understand the CAT Algebra Syllabus; CAT Inequalities questions are not very tough, and hence must not be avoided.
• CAT Inequalities Questions – Tip 2: Based on our analysis of the previous year’s CAT questions,  this was the CAT Inequalities weightage: 2-3 questions were asked on this topic (in CAT 2021).
• Practice these CAT inequalities questions PDF. Learn all the major formulae from these concepts. You can check out the Important CAT Inequalities questions & Formulas PDF here.

Question 1: 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

1) 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 2: 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) $-8 \leq x \leq 8$

2) 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 3: 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) $p^{-1} + q^{-1} + r^{-1}$

3) 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 4: The number of integers n satisfying -n+2 ≥ 0 and 2n ≥ 4 is

a) 0

b) 1

c) 2

d) 3

4) 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

Question 5: Which of the following values of x do not satisfy the inequality $(x^2 – 3x + 2 > 0)$ at all?

a) $1\leq x \leq 2$

b) $-1\geq x \geq -2$

c) $0 \leq x \leq 2$

d) $0\geq x \geq -2$

5) Answer (A)

View Video Solution

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;  x>2

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

So the answer will be A.

Question 6: The number of positive integer valued pairs (x, y), satisfying 4x – 17 y = 1 and x < 1000 is:

a) 59

b) 57

c) 55

d) 58

6) Answer (A)

View Video Solution

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

a) -2/5

b) 2/17

c) 10/17

d) None of these

7) Answer (D)

View Video Solution

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 8: 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) 9:4

b) 81:4

c) 1:4

d) 25:4

8) Answer (D)

View Video Solution

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 = 2b  – 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 9: For how many integers n, will the inequality $(n – 5) (n – 10) – 3(n – 2)\leq0$ be satisfied?

9) Answer: 11

View Video Solution

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 + 1 = 11.

Question 10: 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) 1

b) 2

c) 3

d) 4

10) Answer (C)

View Video Solution

Solution:

Let the roots of the equation $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 11: The smallest integer $n$ such that $n^3-11n^2+32n-28>0$ is

11) Answer: 8

View Video Solution

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 factorize n^2-9n+14 as (n-2)(n-7).

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

$\Rightarrow$ $n^3-11n^2+32n-28>0$

$\Rightarrow$ $(n-2)^2(n-7)>0$

Therefore, we can say that n-7>0

Hence, n$_{min}$ = 8

Question 12: If x is a real number, then $\sqrt{\log_{e}{\frac{4x – x^2}{3}}}$ is a real number if and only if

a) $1 \leq x \leq 3$

b) $1 \leq x \leq 2$

c) $-1 \leq x \leq 3$

d) $-3 \leq x \leq 3$

12) Answer (A)

View Video Solution

Solution:

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

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

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

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

1=< x=< 3

Question 13: 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) 8

b) 9

c) 10

d) 12

13) Answer (B)

View Video Solution

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

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

.’. Option B is correct.

Question 14: The number of pairs of integers $(x,y)$ satisfying $x\geq y\geq-20$ and $2x+5y=99$

14) Answer: 17

View Video Solution

Solution:

We have 2x + 5y = 99 or $x=\frac{\left(99-5y\right)}{2}$

Now $x\ge\ y\ \ge\ -20$ ; So $\frac{\left(99-5y\right)}{2}\ge\ y\ ;\ 99\ge7y\ or\ y\le\ \approx\ 14$

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,

then (99 – 5y) must be even, which will happen when “y” is odd. However, there are only 17 odd values of “y” be -20 and 14.

Hence the number of possible values is 17.

Question 15: For real x, the maximum possible value of $\frac{x}{\sqrt{1+x^{4}}}$ is

a) $\frac{1}{2}$

b) $1$

c) $\frac{1}{\sqrt{3}}$

d) $\frac{1}{\sqrt{2}}$

15) Answer (D)

View Video Solution

Solution:

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

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 as $\frac{1}{\sqrt{\ 2}}$

Question 16: if x and y are positive real numbers satisfying $x+y=102$, then the minimum possible valus of $2601(1+\frac{1}{x})(1+\frac{1}{y})$ is

16) Answer: 2704

View Video Solution

Solution:

Now we have $2601\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=2601\left(\frac{xy+y+x+1}{xy}\right)$

Now we know that x+y=102. Substituting it in the above equation

$2601\left(\frac{xy+y+x+1}{xy}\right)=2601\left(\frac{103}{xy}+1\right)$

Maximum value of xy  can be found out by AM>= GM relationship

$\ \frac{\ x+y}{2}\ge\ \sqrt{xy}\ or\ \ \sqrt{\ xy}\le\ 51\ or\ xy\le\ 2601$

Hence the maximum value of “xy” is 2601. Substituting in the above equation we get

$2601\left(\ \frac{\ 103+2601}{2601}\right)=2704$

Question 17: The number of integers n that satisfy the inequalities $\mid n – 60 \mid < \mid n – 100 \mid < \mid n – 20 \mid$ is

a) 21

b) 19

c) 18

d) 20

17) Answer (B)

View Video Solution

Solution:

We have $\mid n – 60 \mid < \mid n – 100 \mid < \mid n – 20 \mid$.

Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line. This means that when the absolute difference from a number is larger, n would be further away from that number.

Example: The absolute difference of n and 60 is less than that of the absolute difference between n and 20. Hence, n cannot be $\le\ 40$, as then it would be closer to 20 than 60, and closer on the number line would indicate lesser value of absolute difference. Thus we have the condition that n>40.

The absolute difference of n and 100 is less than that of the absolute difference between n and 20. Hence, n cannot be $\le\ 60$, as then it would be closer to 20 than 100. Thus we have the condition that n>60.

The absolute difference of n and 60 is less than that of the absolute difference between n and 100. Hence, n cannot be $\ge80$, as then it would be closer to 100 than 60. Thus we have the condition that n<80.

The number which satisfies the conditions are 61, 62, 63, 64……79. Thus, a total of 19 numbers.

Alternatively

as per the given condition : $\mid n – 60 \mid < \mid n – 100 \mid < \mid n – 20 \mid$.

Dividing the range of n into 4 segments. (n < 20, 20<n<60, 60<n<100, n > 100 )

1) For n < 20.

|n-20| = 20-n, |n-60| = 60- n, |n-100| = 100-n

considering the inequality part :$\left|n-100\right|<\ \left|n\ -20\right|$

100 -n < 20 -n,

No value of n satisfies this condition.

2) For 20 < n < 60.

|n-20| = n-20, |n-60| = 60- n, |n-100| = 100-n.

60- n < 100 – n and 100 – n < n – 20

For 100 -n < n – 20.

120 < 2n and n > 60. But for the considered range n is less than 60.

3) For 60 < n < 100

|n-20| = n-20, |n-60| = n-60, |n-100| = 100-n

n-60 < 100-n and 100-n < n-20.

For the first part 2n < 160 and for the second part 120 < 2n.

n takes values from 61 …………….79.

A total of 19 values

4) For n > 100

|n-20| = n-20, |n-60| = n-60, |n-100| = n-100

n-60 < n – 100.

No value of n in the given range satisfies the given inequality.

Hence a total of 19 values satisfy the inequality.

Question 18: $f(x) = \frac{x^2 + 2x – 15}{x^2 – 7x – 18}$ is negative if and only if

a) -5 < x < -2 or 3 < x < 9

b) x < -5 or -2 < x < 3

c) -2 < x < 3 or x > 9

d) x < -5 or 3 < x < 9

18) Answer (A)

View Video Solution

Solution:

$f(x) = \frac{x^2 + 2x – 15}{x^2 – 7x – 18}$<0

$\frac{\left(x+5\right)\left(x-3\right)}{\left(x-9\right)\left(x+2\right)}<0$

We have four inflection points -5, -2, 3, and 9.

For x<-5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative. Hence, the overall expression will be positive. Similarly, when x>9, all four terms will be positive.

When x belongs to (-2,3), two terms are negative and two are positive. Hence, the overall expression is positive again.

We are left with the range (-5,-2) and (3,9) where the expression will be negative.

Question 19: For a real number x the condition $\mid3x-20\mid+\mid3x-40\mid=20$ necessarily holds if

a) $10<x<15$

b) $9<x<14$

c) $7<x<12$

d) $6<x<11$

19) Answer (C)

View Video Solution

Solution:

Case 1 : $x\ge\frac{40}{3}$
we get 3x-20 +3x-40 =20
6x=80
x=$\frac{80}{6}$=$\frac{40}{3}$=13.33
Case 2 :$\frac{20}{3}\le\ x<\frac{40}{3}\ $
we get 3x-20+40-3x =20
we get 20=20
So we get x$\in\ \left[\frac{20}{3},\frac{40}{3}\right]$
Case 3 $x<\frac{20}{3}$
we get 20-3x+40-3x =20
40=6x
x=$\frac{20}{3}$
but this is not possible
so we get from case 1,2 and 3
$\frac{20}{3}\le\ x\le\frac{40}{3}$
Now looking at options
we can say only option C satisfies for all x .
Hence 7<x<12.

Question 20: The number of distinct pairs of integers (m,n), satisfying $\mid1+mn\mid<\mid m+n\mid<5$ is:

20) Answer: 12

View Video Solution

Solution:

Let us break this up into 2 inequations [ Let us assume x as m and y as n ]

| 1 + mn | < | m + n |

| m + n | < 5

Looking at these expressions, we can clearly tell that the graphs will be symmetrical about the origin.

Let us try out with the first quadrant and extend the results to the other quadrants.

We will also consider the +X and +Y axes along with the quadrant.

So, the first inequality becomes,

1 + mn < m + n

1 + mn – m – n < 0

1 – m + mn – n < 0

(1-m) + n(m-1) < 0

(1-m)(1-n) < 0

(m – 1)(n – 1) < 0

Let us try to plot the graph.

If we consider only mn < 0, then we get

But, we have (m – 1)(n – 1) < 0, so we need to shift the graphs by one unit towards positive x and positive y.

So, we have,

But, we are only considering the first quadrant and the +X and +Y axes. Hence, if we extend, we get the following region.

So, if we look for only integer values, we get

(0,2), (0,3),…….

(0,-2), (0, -3),……

(2,0), (3,0), ……

(-2,0), (-3,0), …….

Now, let us consider the other inequation as well, in which |x + y| < 5

Since one of the values is always zero, the modulus of the other value is less than or equal to 4.

Hence, we get

(0,2), (0,3), (0,4)

(0,-2), (0, -3), (0, -4)

(2,0), (3,0), (4,0)

(-2,0), (-3,0), (-4,0)

Hence, a total of 12 values.

Question 21: 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

21) Answer: 6

View Video Solution

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.

Question 22: 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 $4m^2+2m+1$

22) 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 23: 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) $-2 \leq w \leq -0.5$

23) 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 24: 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

24) 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 25: 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

25) 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 26: 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

26) 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 27: $m$ 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

27) 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 28: 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

28) 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 29: 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

29) 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 30: 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

30) 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 31: 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

31) Answer (C)

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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 32: 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

32) Answer (C)

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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 33: 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

33) 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 34: 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

34) Answer (B)

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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 35: x, y and z are three positive integers such that x > y > z. Which of the following is closest to the product xyz?

a) (x-1)yz

b) x(y-1)z

c) xy(z-1)

d) x(y+1)z

35) Answer (A)

View Video Solution

Solution:

The expressions in the four options can be expanded as

xyz-yz; xyz-xz; xyz-xy and xyz+xz

The closest value to xyz would be xyz-yz, as yz is the least value among yz, xz and xy.

Option a) is the correct answer.

InstructionsFor these questions the following functions have been defined.

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

Question 36: Given that $x >y> z> 0$. Which of the following is necessarily true?

a) $la(x, y, z) < le(x, y, z)$

b) $ma(x, y, z) < la(x, y, z)$

c) $ma(x, y, z) < le(x, y, z)$

d) None of these

36) Answer (D)

View Video Solution

Solution:

Best approach to these type question remain assuming values and checking

Case – 1.x=8 ; y=7 ; z = 5

la (x,y,z) = 12

le (x,y,z) = 2

ma (x,y,z) = 7

Case -2: Let us try to find values for which la(x,y,z) and le(x,y,z) would be equal. In such a case, ma(x,y,z) would also be the same.

So max(x-y,y-z)= min(x+y, y+z)

As x>y>z>0, min(x+y, y+z) = y+z

So max(x-y, y-z) =y+z

Either x-y=y+z or y-z = y+z

So x=2y+z or z=0

But z cannot be 0 according to given condition.

So, x=2y+z

Let us assume y=2 and z=1

So x=5

la (x,y,z)= 3

le (x,y,z) = 3

ma (x,y,z)= 3

based on these two cases we can deduce that non of the given options holds true.

So the correct option to choose is D – None of these.

Question 37: What is the value of ma(10, 4, le((la10, 5, 3), 5, 3))?

a) 7

b) 6.5

c) 8

d) 7.5

37) Answer (B)

View Video Solution

Solution:

$Ma(10, 4, le((la10, 5, 3), 5, 3))$
Or $Ma(10, 4, le(8, 5, 3))$
Or $Ma (10,4,3)$
Or $\frac{1}{2} (6+7) = 6.5$

Question 38: 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)).

a) 5

b) 12

c) 9

d) 4

38) Answer (C)

View Video Solution

Solution:

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

So, these are some of the most important Questions on CAT inequalities. Practice these CAT inequalities questions PDF. Learn all the major formulae from these concepts. You can check out the Important CAT Inequalities questions & Formulas PDF here.