# Inequalities and Modulus for CAT questions

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Inequalities and Modulus for CAT:

Question 1:

How many integers between -10 and 10 (excluding both) satisfy the inequality $$x^3-5x^2+8x-4 > 0$$

a) 7
b) 8
c) 9
d) 10

Question 2:

How many integer values of y satisfy the inequality $$(y+3)(y+6)……(y+150)<0$$?

a) 48
b) 46
c) 52
d) 50

Question 3:

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What is the sum of all the integral values of ‘x’ that satisfy the inequality 0 < x(x+1)(x+3)(x+4) < 190?

a) -9
b) -10
c) -11
d) -12

Question 4:

A badminton player before playing a major tournament had a win ratio of 2:3. Win ratio is the ratio of the number of matches won to the total number of matches played by a player. The player won 4 out of 5 matches played in the major tournament and his win ratio after the tournament is more than 69:100. What could be the maximum total number of matches played by the player including the matches played in the tournament?

a) 7
b) 14
c) 21
d) 26

Question 5:

If $$p$$ and $$q$$ are integers such that $$-5 < p < 2$$ and $$-8 < q < 9$$. Then find the maximum value of $$p^2+pq+q^2$$ ?

a) 93
b) 73
c) 85
d) 81

Question 6:

Find the largest integral value of x which satisfies the inequality (x – 4)(|x| – 5) < 0.

a) 5
b) -6
c) -5
d) -8

Question 7:

How many negative integral values does ‘x’ take:
||x – |x – 2| + 3| – 4| < 3

a) None
b) None
c) None
d) None

Question 8:

Suppose A and B represent natural numbers less than 10 and the number ‘9874A67B’ is divisible by both 9 and 11. What is the product of A and B?

a) 36
b) 63
c) 40
d) None of these

Question 9:

For how many positive integral values of x is the inequality |4-3x|<-1 true?

a) 0
b) 1
c) 2
d) 3

Question 10:

What is the solution set of the inequality ||x|-3| > 4?

a) $$(-\infty, -7) \bigcup (7, \infty)$$
b) $$(-\infty, -7)$$
c) $$(7, \infty)$$
d) None of these

Inequalities and Modulus Answers & Solutions:

1) Answer (A)

The given cubic equation is a simple cubic with integral roots.
On factorizing we get,
$$(x-1)(x-2)(x-2) > 0$$
Thus, the solution set is $$1 < x < 2$$ or $$x > 2$$
Thus only the integers 3, 4, 5, 6, 7, 8 and 9 satisfy the given inequality.

2) Answer (D)

Here, the given expression $$(y+3)(y+6)……(y+150)$$ is positive if y>-3 as all the terms are positive.
The expression is positive for $$y<-150$$.
The expression is 0 for y = -3,-6,-9 .,……
When y = -149,-148, exactly 49 terms are negative and the expression is negative.
When y = -143,-142, exactly 47 terms are negative and the expression is negative ………………
When y =-4,-5, exactly one term is negative.
When y = -149,-148,-143,1-42………………………,-5,-4 the value of the expression is negative.
1st,3rd and 5th are in AP.
Hence, -149 = -5+(n-1)(-6)
n = 25.
Similarly, 2nd,4th and 6th are also in AP and 25 terms more.
Hence, there are 50 terms in total.

3) Answer (B)

The value of the expression x(x+1)(x+3)(x+4) is positive for x > 0, negative for -1 < x < 0, positive for -3 < x < -1, negative for -4 < x < -3 and positive for x < -4
For x = -5, the value is -5*-4*-2*-1 = 40
For x = -6, the value is -6*-5*-3*-2 = 180
For x = -7, the value is -7*-6*-4*-3 = 504
For x = -2, the value is -2*-1*1*2 = 4
For x = 1, the value is 1*2*4*5 = 40
For x = 2, the value is 2*3*5*6 = 180
For x = 3, the value is 3*4*6*7 = 504
So, the values of ‘x’ which satisfy the inequality are -6, -5, -2, 1 and 2
The sum of the values is -10

4) Answer (D)

Let the number of matches played by the player before the tournament be 3n. The number of matches won by him be 2n so that his win ratio is 2:3.
After the tournament,
$$\frac{2n+4}{3n+5}>\frac{69}{100}$$
=> 200n+400 > 207n+345
=> 7n < 55
=> The max value of n is 7.
The maximum number of matches played by the player before the tournament is 3n which is 21.
The maximum number of matches played by the player after the tournament is 21+5 = 26.

5) Answer (A)

The maximum value of $$p^2+pq+q^2$$ occurs at the maximum negative value of both $$p$$ and $$q$$ or maximum positive value of both $$p$$ and $$q$$
Since, $$p^2$$ and $$q^2$$ terms would be maximised for extreme positive or extreme negative value of the given interger, but to maximise the term $$pq$$ we would need a positive value which can be obtained if and only if both $$p$$ and $$q$$ are both negative or both positive
For , p = -4 and q = -7 => $$p^2+pq+q^2$$ = 93
For , p = 1 and q = 8 => $$p^2+pq+q^2$$ = 73
Hence, Correct Option is A

6) Answer (B)

The two cases that satisfy the given inequality are:
Case 1: x – 4 > 0 and |x| – 5 < 0
x > 4 and -5 < x < 5
thus range of x is (4,5) – (1)
Case 2: x – 4 < 0 and |x| – 5 > 0
x < 4 and -$$\alpha$$ < x < -5 and 5 < x < $$\alpha$$
thus range of x is ($$\alpha$$ < x < -5) – (2)
From case (1) and (2) we can see that the largest integral value of x is -6.

7) Answer (2)

||x – |x – 2| + 3| – 4| < 3
=> -3 < |x – |x – 2| + 3| – 4 < 3
=> 1 < |x – |x – 2| + 3| < 7
If x >=2, the inequality becomes: 1 < | x – x + 2 + 3 | < 7
=> 1 < 5 < 7, which is true. So, for all x >=2, the inequality holds good
For x < 2, the equation becomes: 1 < | x + x – 2 + 3 | < 7
=> 1 < |2x + 1| < 7
For x >= -1/2, the inequality becomes 1 < 2x + 1 < 7 => 0 < 2x < 6 => 0 < x < 3
So, the intersection is 0 < x < 2
For x < -1/2, the inequality becomes 1 < -2x – 1 < 7 => 2x < -2 and 2x > -8
=> x < -1 and x > – 4
So, the intersection is -4 < x < -1
Let us check for the boundary cases: if x = -4, the expression doesn’t hold good
For x = -1 also, the expression doesn’t hold good. So, x takes 2 negative integral values.

8) Answer (A)

Divisibility rule of 9: 9+8+7+4+6+7+A+B is divisible by 9 (so should equal 45 or 54).
Divisibility rule of 11: 9+7+A+7 = 8+4+6+B or 9+7+A+7 = 8+4+6+B + 11.
Solving the above equations, only possible solution is A=9 and B=4. So, answer is a) 36

9) Answer (A)

The modulus function can never take negative values. Thus for no value of x is the inequality true.

10) Answer (A)

Case 1:
For $$-3 \leq x \leq 3$$, ||x|-3| = 3-|x|
=> 3-|x|>4 => |x| < -1. This is not true for any value of x.
Case 2:
For x<-3 and x>3, ||x|-3| = |x|-3
=> |x|-3>4 => |x| > 7 => $$x \epsilon (-\infty, -7) \bigcup (7, \infty)$$
Thus the solution set is as given in Option A.