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]

L_{1}: 5x + 3y = 30 L_{2}: x + y = 9

L_{3}: y = \(\frac{x}{2}\)

L_{1}: 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]

Answer:

(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)

Answer:

(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.

Answer:

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.

Answer:

(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.

Answer:

(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.

Answer:

(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

Answer:

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

Answer:

(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

Answer:

(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

Answer:

(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

Answer:

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

Answer:

(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, ∞)

Answer:

(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

Answer:

(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

Answer:

(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

Answer:

(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

Answer:

(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

Answer:

(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

Answer:

(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 x_{1} + 2x_{2} < 5; x_{1} + x_{2} > 1; x_{1} > 0; x_{2} > 0 represents the region. [1 Mark, Dec. 2016]

Answer:

(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

Answer:

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

Answer:

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]

(a) OADGO

(b) ADC

(c) ACD

(d) DEG

Answer:

(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

Answer:

(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

Answer:

(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

Answer:

(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]

Answer:

∵ 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

L_{1} = X_{1} + X_{2} < 4;

L_{2} = 2X_{1} – X_{2} > 6

(a) OABC;

(b) Outside of OAB

(c) ΔBCE

(d) ΔABE

Solution:

(d)

Clearly Common region is ΔABE.

(d) is correct.