# Inequalities – CA Foundation Maths Study Material

This Inequalities – CA Foundation Maths Study Material is designed strictly as per the latest syllabus and exam pattern.

## Inequalities – CA Foundation Maths Study Material

Previous Year Exam Questions

Question 1.
Graphs of Inequalities are drawn below: [1 Mark, Nov. 2006] L1: 5x + 3y = 30 L2: x + y = 9
L3: y = $$\frac{x}{2}$$
L1: y = $$\frac{x}{2}$$
The common region (shaded part) shown in the diagram refers to the inequalities :
(a) 5x + 3y ≤ 30
x + y ≤ 9
y ≤ $$\frac{1}{2}$$x
y ≤ x/2
x ≥ 0, y ≥ 0

(b) 5x + 3y ≥ 30
x + y ≤ 9
y ≥ x/3
y ≤ x/2
x ≥ 0, y ≥ 0

(c) 5x + 3y > 30
x + y ≥ 9
y ≤ x/3
y ≥ x/2
x ≥ 0, y ≥ 0.

(d) 5x + 3y > 30
x + y < 9 y ≥ 9 y ≤ x/2 x ≥ 0, y ≥ 0 Solution : (b) Tricks : Go by choices Take a point of the common region. Let the testing point is (5 ; 3) of the region. This point satisfies all given inequations of option (b). Question 2. If |x + $$\frac{1}{4}$$| > $$\frac{7}{4}$$, then:
(a) x < $$\frac{-3}{2}$$ or x > 2
(b) x < -2 or x > $$\frac{3}{2}$$
(c) -2 < x < $$\frac{3}{2}$$ (d) None of these [1 Mark, Nov. 2006] Answer: (b) |x + $$\frac{1}{4}$$| > $$\frac{7}{4}$$ Question 3.
If $$\left|\frac{3 x-4}{4}\right| \leq \frac{5}{12}$$, the solution set is:
(a) {x: $$\frac{19}{18}$$ ≤ x ≤ $$\frac{29}{18}$$}
(b) {x: $$\frac{7}{9}$$ ≤ x ≤ $$\frac{17}{9}$$}
(c) {x: $$\frac{-29}{18}$$ ≤ x ≤ $$\frac{-19}{18}$$}
(d) None of these [1 Mark, Feb. 2007]
(b) is correct. (b) is correct Question 4.
On solving the inequalities 6x + y ≥ 18; x + 4y ≥ 12; 2x + y ≥ 10, we get the following situation: [1 Mark, Feb. 2007]
(а) (0, 18), (12,0), (4, 2) & (7, 6)
(b) (3, 0), (0,3), (4, 2), & (7, 6)
(c) (5, 0), (0, 10), (4, 2) & (7, 6)
(d) (0,18), (12, 0), (4, 2), (0,0) and (7, 6)
(a) For 6x + y = 18

 x 3 2 y 0 6

Point are (3 ; 0); (2 ; 6)

For x + 4y = 12

 x 0 4 y 3 2

Point are (0 ; 3) ; (4 ; 2)
and 2x = y = 10

 x 0 4 y 4 2

Points are (3; 4) & (4; 2)
Solving eqn. 6x + y = 18 & 2x + y = 10
Subtracting 4x = 8 x = 2
Putting x = 2 in 2x + y = 10
we get
2 × 2 + y = 10
∴ y = 6
Point is (2; 6)
(a) is correct
Tricks : Go by choices
Point (0; 8) satisfy eqn. 6x + y = 18
Point (12 ; 0) satisfies eqn x + 4y = 12
Point (4 ; 2) satisfies eqns x + 4y = 12 and 2x + y = 10
Point (2 ; 6) satisfies eqns 6x + y = 18 and 2x + y = 10
(a) is Correct

Question 5.
A car manufacturing company manufactures cars of two types A and B. Model A requires 150 man-hours for assembling, 50 man-hours for painting and 10 man¬hours for checking and testing. Model B requires 60 man-hours for assembling, 40 man-hours for painting and 20 man-hours for checking and testing. There are available 30 thousand man-hours for assembling, 13 thousand man-hours for painting and 5 thousand man-hours for checking and testing. Express the above situation using linear inequalities. Let the company manufacture x units of type A model of car and y units of type B model of car.
Then, the inequalities are :
(a) 5x + 2y ≥ 1000; 5x + 4y ≥ 1300, x + 2y ≤ 500; x ≥ 0, y ≥ 0.
(b) 5x + 2y ≤ 1000, 5x + 4y ≤ 1300, x + 2y ≥ 500; x ≥ 0, y ≥ 0.
(c) 5x + 2y ≤ 1,000, 5x + 4y ≤ 1300, x + 2y ≤ 500; x ≥ 0, y ≥ 0.
(d) 5x + 2y = 1000, 5x + 4y ≥ 1300, x + 2y = 500 ; x ≥ 0, y ≥ 0.
Models

 Conditions A(x) B(y) Total Assembly 150 man hrs 60 man hrs 30,000 man hrs Painting 50 man hrs 40 man hrs 13,000 man hrs Checking & Testing 10 man hrs 20 man hrs 5,000 man hrs

Ineqns. are
[150x + 60y ≤ 30,000] ÷ 30 ⇒ 5x + 2y ≤ 1,000
[50x+40y ≤ 13000] ÷ 10 ⇒ 5x + 4y ≤ 1300
[10x + 20y ≤ 5000] ÷ 10 ⇒ x + 2y ≤ 500
x ≥ 0 & y ≥ 0
(b) is correct

Question 6.
The rules and regulations demand that the employer should employ not more than 5 experienced hands to 1 fresh one and this fact is represented by : (Taking experienced person as x and fresh person as y) [1 Mark, Aug. 2007]
(a) y ≥ $$\frac{x}{5}$$
(b) 5y ≤ x
(c) 5y ≥ x
(d) None.
(a) & (c)
1 Fresh with 5 experienced maximum employees.
y Fresh with 5y experienced maximum employees.
From Question
x ≤ 5y ⇒ 5y ≥ x, OR, y ≥ x/5
(a) & (c) are correct.

Question 7.
The shaded region represents: [1 Mark, Aug. 2007] (a) 3x + 2y ≤ 24, x + 2y ≥ 16, x + y ≤ 10x, x ≥ 0,y ≥ 0
(b) 3x+2y ≤ 24, x + 2y ≤ 16, x + y ≥ 10, x ≥ 0, y ≥ 0
(c) 3x + 2y ≤ 24, x + 2y ≤ 16, x + y ≤ 10, x ≥ 0, y ≥ 0
(d) None of these.
(c) Tricks : Go by choices
Take a point (1,1) satisfies all inequations of option (c)
(c) is Correct

Question 8.
The shaded region represents: [1 Mark, Nov. 2007] (a) 3x + 5y ≤ 15, 5x + 2y ≥ 10, x, y ≥ 0
(b) 3x + 5y ≤ 15, 5x + 2y ≤ 10, x, y ≥ 0
(c) 3x + 5y ≥ 15, 5x + 2y ≥ 10, x, y, ≥ 0
(d) None of these.
(b) Tricks : Go by choices
Let point is (1,1) of the common region Point (1,1) satisfies all ineqns. of option (b)
option (b) is correct. Question 9.
The shaded region represents : [1 Mark, June 2008] (a) x + y > 6, 2x – y > 0
(b) x + y < 6, 2x – y > 0
(c) x + y > 6,2x – y < 0 (d) None of these. Answer: (a) Tricks : Go by choices Point (7 ; 1) satisfies all conditions of option (a) (a) is correct Question 10. If a > 0 and b < 0, it followings that: [1 Mark, June 2008] (a) $$\frac{1}{a}>\frac{1}{b}$$
(b) $$\frac{1}{a}<\frac{1}{b}$$ (c) $$\frac{1}{a}=\frac{1}{b}$$ (d) None of these Answer: (a) a > 0 ⇒ $$\frac{1}{a}$$ > 0
b < 0 ⇒ $$\frac{1}{b}$$ < 0 ∴ $$\frac{1}{a}>\frac{1}{b}$$
∴ (a) is correct

Question 11.
The Linear relationship between two variables in an inequality :
(a) ax + by ≤ c
(b) ax. by ≤ c
(c) axy + by ≤ c
(d) ax + bxy ≤ c
Linear eqn is ax + by = c
∴ (a) option is correct

Question 12.
The solution of the inequality $$\frac{(5-2 x)}{3} \leq \frac{x}{6}$$ – 5 is: [1 Mark, June 2010]
(a) x ≥ 8
(b) x ≤ 8
(c) x = 8
(d) none of these
(a) is correct. or 10 – 4x ≤ x – 30
or 10 + 30 ≤ x + 4x
or 5x ≥ 40
or x ≥ 8
option (a) is correct

Question 13.
Solution space of inequalities 2x + y ≤ 10 and x – y ≤ 5: [1 Mark, June 2011]
(i) includes the origin.
(ii) includes the point (4,3) which one is correct? i
(a) Only (i) (b) Only (ii)
(c) both (i) and (ii)
(d) none of the above
(a) is correct ;
Tricks : Go by choices
(0, 0) satisfies both ineqns. but (4; 3) does not satisfy 1st
(a) is correct

Question 14.
On the average,experienced person does 5 units work while a fresh one 3 units work daily but the employer have to maintain the output of atleast 30 units of work per day. [1 Mark, Dec. 2011 & 12]
The situation can be expressed as.
(a) 5x + 3y ≤ 30
(b) 5x + 3y ≥ 30
(c) 5x + 3y = 30
(d) None of these
(b) Let No. of experienced persons = x and No. of Freshers = y
∴ 5x + 3y ≥ 30 Question 15.
Find the range of real of x satisfying the inequalities 3x – 2 > 7 and 4x – 13 > 15. [1 Mark, June 2012]
(a) x > 3
(b) x > 7
(c) x < 7
(d) x < 3 Answer: (b) is correct. 3x – 2 > 7 ⇒ 3x > 9 ∴ x > 3 ………..(1)
4x > 15 + 13 ⇒ 4x > 28 ∴ x >7 …………(2)
Clearly From (1) and (2); x > 7 satisfies both
(b) is correct.

Question 16.
The shaded region represents : [1 Mark, Feb. 2008] (a) x + y ≤ 5, x ≥ 2, y ≤ 1
(b) x + y ≤ 5, x ≥ 2, y ≥ 1
(c) x + y ≥ 5, x ≥ 2, y ≥ 1
(d) None of these
Tricks: Go by choices, option (b)

Question 17.
The union forbids the employer to employ less than 2 experienced person (x) to each fresh person (y),This situation can be expressed as: [1 Mark, June 2013]
(a) x ≤ y/2
(b) y ≤ x/2
(c) y ≥ x/2
(d) none
(b) is correct
No. of Fresh persons for x Experienced person = $$\frac{x}{2}$$
$$\frac{x}{2}$$ ≥ y (given) ∴ y ≤ $$\frac{x}{2}$$

Question 18.
The solution of the inequality
8x + 6 < 12x + 14 is
(a) (-2, 2)
(b) (-2, 0)
(c) (2, ∞)
(d) (-2, ∞)
(d) is correct
8x + 6 < 12x + 14
or – 8 < 4x
or -2 < x x > -2
∴ Soln. is (-2; ∞)

Question 19.
The graph of linear inequalities
7x + 9y < 63; x + y > 1;
0 ≤ x ≤ 6 and 0 ≤ y ≤ 6 has been given below (a) BCDB and DEFD
(b) Unbounded
(c) HFGH
(d) ABDFHKA
(d) Clearly common region is ABDFHKA.

Question 20.
Which of the following graph represents the in equality x + y ≤ 6 is [1 Mark, Dec. 2014] (d) None of these
(a) is correct. The graphical representation of x + y ≤ 6 is as follows : Question 21.
The graph of linear inequalities
x + y ≥ 5; x + y ≤ 5; 0 ≤ x ≤ 4 and 0 ≤ y ≥ 2 is given below: The common region of the inequalities will be: [1 Mark, Dec. 2014]
(a) OABCEO
(b) ECDE
(c) Line Segment DC
(d) Line Segment BC
(c)

Question 22.
The common region represented by the inequalities 2x + y ≥ 8, x + y ≥ 12, 3x + 2y ≤ 34 is: [1 Mark, June. 2015]
(a) Unbounded
(b) In feasible i
(c) Feasible and bounded
(d) Feasible and unbounded S
(C) is correct.  clearly It is Feasible and bounded.

Question 23.
By lines x + y = 6,2x – y = 2, the common region shown is the diagram refers to: [1 Mark, Dec. 2015]
(a) x + y ≥ 6, 2x – y ≤ 2, x ≥ 0, y ≥ 0
(b) x + y ≤ 6, 2x – y ≤ 2, x ≥ 0, y ≥ 0
(c) x + y ≤ 6, 2x – y > 2, x ≥ 0, y ≥ 0
(d) None of these
(b) is correct
Tricks: Go by choices
A point (1,1) (let) satisfies all inequations of (b). Question 24.
The common region of x + y ≤ 6; x + y ≥ 3, is for shown by shaded region: [1 Mark, June 2016] (d) None of these
(a) is correct.
Tricks : Go by choices.
Clearly a point of the common region of option (a) satisfy all given constraints x + y ≤ 6 & x + y ≥ 3.

Question 25.
The inequalities x1 + 2x2 < 5; x1 + x2 > 1; x1 > 0; x2 > 0 represents the region. [1 Mark, Dec. 2016] (a) is correct
Tricks : Go by choices.

Question 26.
A dietitian wishes to mix together two kinds of food so that the vitamin content of the mixture is atleast 9 units of vitamin A, 7 units of vitamin B, 10 units of vitamin C and 12 units of vitaminD. The vitamin content per kg. of each food is shown below: Assuming x kgs of food I is to be mixed with y kgs of food II the situation can be expressed as: [1 Mark, June 2017]
(a) 2x + y ≤ 9; x + y ≤ 7; x + 2y ≤ 10; 2x + 3y ≤ 12 ; x ≥ 0, y ≥ 0
(b) 2x + y ≥ 30; x + y ≤ 7; x + 2y ≥ 10; x + 3y ≥ 12; x ≥ 0; y ≥ 0
(c) 2x + y ≥ 9; x + y ≤ 7; x + y ≤ 10; x + 3y ≥ 12; x ≥ 0, y ≥ 0
(d) 2x + y ≥ 9; x + y ≥ 7; x + 2y ≥ 10; 2x + 3y ≥ 12; x ≥ 0; y ≥ 0
Atleast → Minimum
So, use > Sign here.
Constraints are:
2x + y ≥ 9;
x + y ≥ 1
x + 2y ≥ 1
2x + 3y ≥ 12
(d) is correct.

Question 27.
The shaped region represented by the inequalities
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x ≥ 0, y ≥ 0 (d) None
Tricks : Go by choices
Option (b) is correct.

Question 28.
In the following diagram, the region represented by the inequalities
x + 2y ≤ 10, x + y ≤ 6. x ≤ 4 & x ≥ 0, y ≥ 0 is: [1 Mark, June 2018] (c) ACD
(d) DEG
(a)
Tricks : Go by choices

Question 29.
The linear relationship between two variables in an inequality: [1 Mark, May 2018]
(a) ax + by ≤ c
(b) ax.by ≤ c
(c) axy + by ≤ c
(d) ax + bxy ≤ c
(a)
Standard form of Linear Eqn. is
ax + by = c.
So; ax + by ≤ c is a Linear Ineqn. Question 30.
On Solving the Inequalities 5x + y ≤ 100, x + y ≤ 60, x ≥ 0, y ≥ 0, we get the following situation: [1 Mark, Nov. 2018]
(a) (0, 0), (20, 0), (10, 50) & (0, 60)
(b) (0, 0), (60, 0), (10, 50) & (0, 60)
(c) (0, 0), (20, 0), (0,100) & (10, 50)
(d) None of these
(a)
Tricks : Go by choices

Question 31.
An employer recruits experienced (x) and fresh workmen (y) under the condition that he cannot employ more than 11 people, x and y can be related by the inequality: [1 Mark, June 2019]
(a) x + y ≠ 11 ;
(b) x + y ≤ 11, x ≥ 0, y ≥ 0
(c) x + y ≥ 11, x ≥ 0, y ≥ 0
(d) None of these
(b)
Clearly x + y ≤ 11.
and x ; y > 0.

Question 32.
The solution set of the inequations x + 2 > 0 and 2x – 6 > 0 is
(a) (- 2 , ∞);
(b) ( 3 , ∞)
(c) (- ∞ , – 2 )
(d) (-∞, – 3) [1 Mark, June 2019]
∵ x + 2 > 0 ⇒ x > -2
and 2x – 6 > 0 ⇒ x > 3
⇒ x = {-1; 0, 1, 2, 3, 4,………..} (1)
and 2x – 6 > 0 ⇒ x > 3
⇒ x = {4 ; 5 ; 6 ; 7 ;………} (2)
From (1) and (2); we get x = {4, 5, 6, ………}satisfies both conditions.
∴ Solution Set = (3; ∞)

Question 33.
The common region represented by the following inequalities
L1 = X1 + X2 < 4;
L2 = 2X1 – X2 > 6 (a) OABC;
(b) Outside of OAB
(c) ΔBCE
(d) ΔABE
Solution:
(d)
Clearly Common region is ΔABE.
(d) is correct.