MCQ Solutions - Inequalities - Quantitative Aptitude for CA Foundation
Q1: The owner of a new restaurant is designing its floor plan. His goal is to have a seating capacity of at least 410 customers. The small tables can seat 4 customers, while the large tables can seat 8 customers.
- x = the number of small tables
- y = the number of large tables
Select the inequality in standard form that describes this situation:
- 4x + 8y ≤ 410
- 4x + 8y > 410
- 4x + 8y ≥ 410
- 4x + 8y < 410
Solution: Each small table seats 4 and each large table seats 8. The restaurant wants at least 410 seats, meaning the total seating cannot be less than 410. “At least” → greater than or equal to (≥).
Q2: In order to earn some extra money, Nate is mowing lawns. He charges Rs.14 to mow a small lawn and Rs.29 to mow a large lawn. He would like to make at least Rs.100.
- x = the number of small lawns
- y = the number of large lawns
Select the inequality in standard form that describes this situation:
- 14x + 29y ≥ 100
- 29x + 14y ≤ 100
- 29x + 14y ≥ 100
- 14x + 29y ≤ 100
Solution: Nate wants to earn at least Rs.100, so his total earnings must be greater than or equal to 100, not less. Therefore the inequality is ≥ 100.
Q3: Ella is purchasing tickets at the county fair. A ride on the Ferris wheel requires 1 ticket and a ride on the roller coaster requires 2 tickets. She has enough money to purchase a maximum of 44 tickets.
- x = the number of Ferris wheel rides
- y = the number of roller coaster rides
Select the inequality in standard form that describes this situation:
- 2 + x + 1 + y ≤ 44
- 2x + y ≤ 44
- 1 + x + 2 + y ≤ 44
- x + 2y ≤ 44
Solution: Ferris wheel = 1 ticket per ride, roller coaster = 2 tickets per ride. She can buy a maximum of 44 tickets, meaning the total must be less than or equal to (≤) 44.
Q4: For tax purposes, a large corporation wants to spend more than Rs.10,300,000 on charitable works. The company is told that, in a developing country, it can build a hospital for Rs.2,380,000 and a school for Rs.13,500.
- x = the number of hospitals built
- y = the number of schools built
Select the inequality in standard form that describes this situation:
- 2,380,000x + 13,500y > 10,300,000
- 2,380,000x + 13,500y ≥ 10,300,000
- 13,500x + 2,380,000y ≥ 10,300,000
- 13,500x + 2,380,000y > 10,300,000
Solution: The company wants to spend more than 10,300,000 on charitable work. “More than” means a strict inequality (>), not ≥.
Q5: Which of the graphs best suits the inequality y > 1?
a) GRAPH 1
b) GRAPH 2
c) GRAPH 3
d) GRAPH 4
Ans: b) GRAPH 2
Solution: For y > 1, the line y = 1 is dashed (since equality is not included) and the region is above the line. Graph 2 correctly shows this shading.
Q6: Which of the graphs best suits the inequality x - y > -1?
a) GRAPH 1
b) GRAPH 2
c) GRAPH 3
d) GRAPH 4
Ans: a) GRAPH 4
Solution: x − y > –1 can be rewritten as y < x + 1. This represents the region below a dashed upward-sloping line. Graph 4 matches this.
Q7: Choose the ordered pair that is not the solution of the inequality whose graph is shown.
Graph: Line y = x - 2, shaded above including line
a) (2, -4)
b) (4, -2)
c) (4, -4)
d) (-2, 2)
Ans: c) (–2, 2)
Solution: This point lies outside the shaded solution region shown in the graph. All other points fall within the shaded area.
Q8: Identify the inequality which has the ordered pair (2, -3) as a solution.
a) 5x + 3y ≤ 0
b) 5x - 3y < 0
c) 5x - 3y ≤ 0
d) 5x + 3y ≥ 0
Ans: d) 5x + 3y ≥ 0
Solution: Substituting (x, y) = (2, –3):
5(2) + 3(–3) = 10 – 9 = 1, and 1 ≥ 0 is true.
This is the only inequality satisfied by the point.
Q9: Choose the inequality which has the ordered pair (0, -43) as a solution.
a) 3x + 2y ≥ 0
b) (4, -2)
c) 2x + 3y > 0
d) 2x - 3y > 0
Ans: d) 2x - 3y > 0
Solution: Substituting (0, –43):
2(0) – 3(–43) = 129, and 129 > 0 is true.
All other options give false statements.
Q10: Solution set of an _______ can be represented on a number line.
a) In equation
b) equation
c) Either (a) or (b)
d) neither (a) or (b)
Ans: d) neither (a) or (b) [inequality]
Solution: A single-variable inequality (like x > 3 or x ≤ 7) is graphed on a number line. Equations give points but inequalities represent intervals.
Q11: The inequalities x ≥ 0, y ≥ 0 indicates –
a) First quadrant
b) Second quadrant
c) Third Quadrant
d) Fourth Quadrant
Ans: a) First quadrant
Solution: x ≥ 0 means points to the right of the y-axis.
y ≥ 0 means points above the x-axis.
Together they represent the first quadrant.
Q12: 4x > -16 implies –
a) x greater than equal to -4
b) x less than -4
c) x greater than -4
d) x less than equal to -4
Ans: c) x greater than -4
Solution: 4x > –16 → divide both sides by 4 (positive), sign stays the same → x > –4.
Q13: Mr. A plans to invest up to Rs. 30000 in two stocks X and Y. Stock X(x) is priced at Rs. 175 and Stock Y(y) at Rs. 95/share. This can be shown by
a) 175x + 95y ≤ 30000
b) 175x + 95y > 30000
c) 175x + 95y = 30000
d) None of these
Ans: a) 175x + 95y ≤ 30000
Solution: He plans to invest up to Rs.30000, meaning the total cost must be less than or equal to the limit.
Q14: Vitamin A and B are found in food F1 and F2... The constraints are
a) 20 X1 + 60 X2 ≤ 80 , 30 X1 + 40 X2 ≤ 100, X1 ≤ 0; X2 ≤ 0
b) 20 X1 + 60 X2 ≥ 80 , 30 X1 + 40 X2 ≥ 100, X1 ≥ 0; X2 ≤ 0
c) 20 X1 + 60 X2 ≥ 80 , 30 X1 + 40 X2 ≥ 100, X1 ≥ 0; X2 ≥ 0
d) 20 X1 + 60 X2 ≤ 80 , 30 X1 + 40 X2 ≥ 100, X1 ≤ 0; X2 ≥ 0
Ans: c)
Solution: Daily vitamin requirements are minimum values, so the total vitamins from foods F1 and F2 must be greater than or equal to the required amounts.
Q15: Suppose a man needs a minimum of 50 units of carbohydrates, 40 units of proteins... The mathematical inequalities are:
a) 4 X1 + X2 ≥ 50 , 5 X1 + 3 X2 ≤ 40, X1 ≥ 0; X2 ≤ 0
b) 4 X1 + X2 ≤ 50 , 5 X1 + 3 X2 ≥ 40, X1 ≤ 0; X2 ≥ 0
c) 4 X1 + X2 ≥ 50 , 5 X1 + 3 X2 ≥ 40, X1 ≥ 0; X2 ≥ 0
d) 4 X1 + X2 ≤ 50 , 5 X1 + 3 X2 ≤ 40, X1 ≤ 0; X2 ≤ 0
Ans: c)
Solution: The man needs at least 50 carbs and at least 40 proteins. Minimum requirements always use ≥ inequalities.
Q16: A man makes two types of furniture: chairs and tables... Constraints:
a) x + y ≤ 12, 5x + 2y ≥ 50, x ≤ 0 ; y ≥ 0
b) x + y ≥ 12, 5x + 2y ≤ 50, x ≥ 0 ; y ≤ 0
c) x + y ≤ 12, 5x + 2y ≤ 50, x ≥ 0 ; y ≥ 0
d) x + y ≥ 12, 5x + 2y ≥ 50, x ≤ 0 ; y ≤ 0
Ans: c)
Solution: Each machine has limited working hours. The time used must not exceed available time, so inequalities are ≤. Non-negative production gives x, y ≥ 0.
Q17: Graph to express the inequalities x + y ≤ 9 is –
a) 
b) 
c) 
d) None of these
Ans: a)
Solution: The inequality x + y ≤ 9 shades the region on or below the line x + y = 9. Graph (a) shows this exactly.
Q18: The inequalities X1 ≥ 0, X2 ≥ 0, are represented by one of the graphs shown below
a) First quadrant
b) Third quadrant
c) Second quadrant
d) Fourth quadrant
Ans: a)
Solution: x₁ ≥ 0 and x₂ ≥ 0 represent the first quadrant, and Graph (a) shades that region.
Q19: The inequality – X1 + 2 X2 ≤ 0 is indicated on the graph as :
a) [Above line]
b) [Below line]
c) [Right]
d) None of these
Ans: b)
Solution: –x₁ + 2x₂ ≤ 0 → 2x₂ ≤ x₁ → x₂ ≤ x₁/2.
This shades the region below the line with slope 1/2, correctly shown in Graph (b).
Q20: A firm plans to purchase hens (x) for its canteen. There cannot be more than 20 hens. This can be represented by
a) x < 20
b) x = 20
c) x > 20
d) none of these
Ans: d) none of these (x ≤ 20)
Solution: "Cannot be more than 20 hens" means x ≤ 20. No option shows ≤ 20, so “None of these” is correct.
Q21: A scooter company manufactures scooters of two models A and B... The constraints can be formulated by
a) 15x + 6y ≤300, x + 2y ≤ 50, x + 4y ≤ 120, x ≥ 0 ; y ≥ 0
b) 15x + 6y ≤300, x + 2y ≥ 50, x + 4y ≥ 120, x ≤ 0 ; y ≤ 0
c) 15x + 6y ≥ 300, x + 2y ≤ 50, x + 4y ≥ 120, x ≥ 0 ; y ≤ 0
d) 15x + 6y ≤300, x + 2y ≤ 50, x + 4y ≤ 120, x ≤ 0 ; y ≤ 0
Ans: a)
Solution: Each machine has limited hours, so the time spent must be ≤ available time. Production cannot be negative.
Q22: A company produces two types of leather belts... Constraints:
a) x + 2y ≥ 1000, x + y ≥ 800, x ≥ 400; y ≤ 700, x ≥ 0, y ≥ 0
b) x + 2y ≤ 1000, x + y ≤ 800, x ≤ 400; y ≤ 700, x ≥ 0, y ≥ 0
c) x + 2y ≥ 1000, x + y ≤ 800, x ≥ 400; y ≥ 700, x ≥ 0, y ≥ 0
d) x + 2y ≤ 1000, x + y ≥ 800
Ans: b)
Solution: Time, leather supply, and buckle limits impose maximum values → all inequalities use ≤, with variables non-negative.
Q23: The common region indicated on the graph is expressed by the set of five inequalities.
a) L1: x1 ≥0; L2: x2 ≥0; L3: x1+x2 ≤ 1; L4: x1-x2 ≥ 1; L5: x1 + 2x2 ≤ 0
b) L1: x1 ≥0; L2: x2 ≥0; L3: x1+x2 ≥ 1; L4: x1-x2 ≥ 1; L5: x1 + 2x2 ≤ 0
c) L1: x1 ≤ 0; L2: x2 ≤0; L3: x1+x2 ≤ 1; L4: x1-x2 ≥ 1; L5: x1 + 2x2 ≤ 0
d) None of these
Ans: a)
Solution: The shaded region matches all five inequalities: first-quadrant conditions plus three boundary lines with correct shading direction.
Q24: if m < n and a < b, then --
a) m - a < n – b
b) ma < nb
c) m/a < n/b
d) m + a < n + b
Ans: d) m + a < n + b
Solution: If m < n and a < b, adding gives (m + a) < (n + b). Addition preserves inequality direction. Other statements are not always true.
Q25: 5x < 20 implies
a) x < 4
b) x less than equal to 4
c) x > 4
d) x greater than equal to 4
Ans: a) x < 4
Solution: 5x < 20 → divide both sides by 5 → x < 4.
Q26: The rules and regulations demand that the employer should employ not more than 5 experienced hands to 1 fresh one and this fact can be expressed as –
a) y ≥ x/5
b) 5y ≥ x
c) both (a) and (b)
d) 5y ≤ x
Ans: b) 5y ≥ x
Solution: “Not more than 5 experienced per 1 fresh worker” → x ≤ 5y.
Rewriting: y ≥ x/5 or 5y ≥ x. Both are equivalent.
Q27: In a class of boys (x) and girls (y) the maximum seating capacity is 360. this can be shown by –
a) x + y ≤ 360
b) x + y ≠ 360
c) x + y ≥ 360
d) None of these
Ans: a) x + y ≤ 360
Solution: Maximum capacity is 360, meaning the number of students must not exceed 360.
Q28: if xy > 1 and z < 0, which of the following statement must be true?
I) x > z II) xyz < -1 III) xy/z < 1/z
a) I only
b) II only
c) III only
d) II and III only
Ans: c) III only
Solution: Dividing by negative z reverses inequality for III.
Q29: A firm makes two types of products: Type A and Type B... The constraints:
a) 3X1 + 3X2 ≤ 36, 5X1 + 2X2 ≤ 50, 2X1 + 6X2 ≤ 60, X1 ≥ 0, X2 ≥ 0
b) 3X1 + 3X2 ≥ 36, 5X1 + 2X2 ≥ 50, 2X1 + 6X2 ≥ 60, X1 ≥ 0, X2 ≥ 0
c) 3X1 + 3X2 ≤ 36, 5X1 + 2X2 ≥ 50, 2X1 + 6X2 ≤ 60, X1 ≥ 0, X2 ≤ 0
d) 3X1 + 3X2 ≥ 36, 5X1 + 2X2 ≤ 50, 2X1 + 6X2 ≥ 60, X1 ≤ 0, X2 ≤ 0
Ans: a)
Solution: Each product uses machine time, and the total time used cannot exceed the available hours. So all inequalities are ≤ with non-negative production.
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FAQs on MCQ Solutions - Inequalities - Quantitative Aptitude for CA Foundation
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