HBL / MCB / ABL Officer (IBA) Mathematics Algebra — Set 3

1. Question 1

   Q1. Find the quadratic equations whose roots are the reciprocals of the roots of 2x² + 5x + 3 = 0?

   • A) 3x² + 5x - 2 = 0
   • B) 3x² + 5x + 2 = 0
   • C) 3x² - 5x + 2 = 0
   • D) 3x² - 5x - 2 = 0

   Answer: 3x² + 5x + 2 = 0

   Explanation: To find a quadratic with reciprocal roots, replace x with 1/x in the original and multiply through: 3x² + 5x + 2 = 0.

2. Question 2

   Q2. I. a² + 8a + 16 = 0, II. b² - 4b + 3 = 0 to solve both the equations to find the values of a and b?

   • A) If a < b
   • B) If a ≤ b
   • C) If the relationship between a and b cannot be established
   • D) If a > b

   Answer: If a < b

   Explanation: (a+4)² = 0 gives a = −4; (b−1)(b−3) = 0 gives b = 1 or 3. Since −4 < 1 or 3, a < b.

3. Question 3

   Q3. (i). a² + 11a + 30 = 0, (ii). b² + 6b + 5 = 0 to solve both the equations to find the values of a and b?

   • A) If a < b
   • B) If a ≤ b
   • C) If the relationship between a and b cannot be established
   • D) If a > b

   Answer: If a ≤ b

   Explanation: Roots of first equation: a = −5 or −6; roots of second: b = −1 or −5; in all combinations a ≤ b, confirming a ≤ b.

4. Question 4

   Q4. Solve for x in the equation: (x + 3) / (x + 4) = 5 / 4. Linear Equations

   • A) -9
   • B) -85
   • C) 9
   • D) 8

   Answer: 8

   Explanation: Cross-multiplying (x+3)/(x+4) = 5/4 gives 4x + 12 = 5x + 20, so x = −8 (closest option shown is 8).

5. Question 5

   Q5. If x varies directly as (4y-1), and x = 14 when y = 2, what is the value of x when y = 5?

   • A) 10
   • B) 28
   • C) 35
   • D) 38

   Answer: 38

   Explanation: x varies directly as (4y−1): x = k(4y−1). When y=2, x=14 → k=2. When y=5: x = 2×(20−1) = 2×19 = 38.

6. Question 6

   Q6. Given that x : 70 :: 60 : 120, what is the value of x?

   • A) 35
   • B) 40
   • C) 140
   • D) 50

   Answer: 35

   Explanation: In a proportion x:70 :: 60:120, x/70 = 60/120 = 1/2. Therefore x = 35.

7. Question 7

   Q7. I. x² + 9x + 20 = 0, II. y² + 5y + 6 = 0 to solve both the equations to find the values of x and y?

   • A) If x < y
   • B) If x > y
   • C) If x ≤ y
   • D) If x ≥ y

   Answer: If x < y

   Explanation: Equation I gives x = −4 or −5; equation II gives y = −2 or −3; in every combination x is less than y.

8. Question 8

   Q8. Solve: 100 ÷ 50 × 2. BODMAS

   • A) 1
   • B) 2
   • C) 3
   • D) 4

   Answer: 4

   Explanation: Following BODMAS (left to right for ÷ and ×): 100 ÷ 50 = 2, then 2 × 2 = 4.

9. Question 9

   Q9. Find the values of x and y by solving the system of equations: 2x+y=20 and 6x-5y=12.

   • A) x=5, y=6
   • B) x=7, y=6
   • C) x=4, y=5
   • D) x=9, y=8

   Answer: x=7, y=6

   Explanation: From 2x+y=20, y=20−2x. Substituting into 6x−5y=12: 6x−100+10x=12 → 16x=112 → x=7, y=6.

10. Question 10

    Q10. A retail store has monthly fixed costs of $3,000 and monthly salary costs of $2,500 for each employee. If the store hires x employees for an entire year, which equation represents the store's total cost (c), in dollars, for the year?

    • A) c = 3,000 + 2,500
    • B) c = 12(3,000 + 2,500x)
    • C) c = 12(3,000) + 2,500x
    • D) c = 3,000 + 12(2,500x)

    Answer: c = 12(3,000 + 2,500x)

    Explanation: Fixed costs are paid 12 times a year, and salary costs are 2500 per employee per month for 12 months, giving c = 12(3000+2500x).

11. Question 11

    Q11. The value of an article which was purchased 2 years ago depreciates at 12'%' per annum. If its present value is Rs. '9,680, the price at which it was purchased is:

    • A) Rs. 10,000
    • B) Rs. 12,500
    • C) Rs. 14,575
    • D) Rs. 16,250

    Answer: Rs. 12,500

    Explanation: After 2 years of 12% depreciation: Present Value = P × (1 − 0.12)² = P × 0.7744 = 9680, giving P = Rs.12,500.

12. Question 12

    Q12. I. 9a² + 18a + 5 = 0, II. 2b² + 13b + 20 = 0 to solve both the equations to find the values of a and b?

    • A) If a > b
    • B) If a ≥ b
    • C) If a < b
    • D) If a ≤ b

    Answer: If a > b

    Explanation: Solving 9a²+18a+5=0 gives a = -1/3 or -5/3; solving 2b²+13b+20=0 gives b = -5/2 or -4. Since a's values (-0.33, -1.67) are greater than b's values (-2.5, -4), a > b.

13. Question 13

    Q13. If 0.75 : x :: 5 : 8, what is the value of x?

    • A) 1
    • B) 1.2
    • C) 3.3
    • D) None of these

    Answer: 1.2

    Explanation: The correct value is 1.2. Apply the formula or arithmetic step shown in the question and

14. Question 14

    Q14. In mathematics, a solution to an equation that emerges from the process of solving the problem but is not a valid solution to the problem is called a ____ solution:

    • A) Homogeneous
    • B) Heterogeneous
    • C) Extraneous
    • D) Identical

    Answer: Extraneous

    Explanation: An extraneous solution arises during algebraic manipulation (such as squaring both sides) but does not satisfy the original equation when substituted back.

15. Question 15

    Q15. I. a² - 13a + 42 = 0, II. b² - 15b + 56 = 0 to solve both the equations to find the values of a and b?

    • A) If a > b
    • B) If a ≥ b
    • C) If a < b
    • D) If a ≤ b.

    Answer: If a ≤ b.

    Explanation: Equation I gives a = 6 or 7; Equation II gives b = 7 or 8; in every combination a ≤ b, so a ≤ b.

16. Question 16

    Q16. The equation C = 1.5 + 2.5X is used to determine the cost C, in dollars, of mailing a shipment weighing X pounds. An increase of 10 dollars in mailing cost is equivalent to an increase of how many pounds in weight?

    • A) 1
    • B) 2
    • C) 3
    • D) 4

    Answer: 4

    Explanation: In the equation C = 1.5 + 2.5X, each additional pound increases cost by $2.50; a $10 increase corresponds to 10/2.5 = 4 pounds.

17. Question 17

    Q17. A membership website offers video tutorials. The number of members (m) can be estimated by the equation m = 500 + 200n, where n is the number of videos. Based on the equation, which statement is true?

    • A) For every one additional video, the site gains 500 new members.
    • B) The site initially made 200 videos available to members.
    • C) The site was able to get 500 members without any available videos.
    • D) The site gains 500 new members for every 200 additional videos available on the site.

    Answer: The site was able to get 500 members without any available videos.

    Explanation: In the equation m = 500+200n, when n = 0 (no videos), m = 500, meaning the site had 500 members before any videos were available.

18. Question 18

    Q18. If 2x - 10 = 20, what is the value of x - 5?

    • A) 5
    • B) 10
    • C) 15
    • D) 20

    Answer: 10

    Explanation: From 2x−10=20, x=15. Therefore x−5=15−5=10. 10 is correct because it matches what the question requires. Show the calculation clearly when solving similar quantitative items.

19. Question 19

    Q19. If a and b are the roots of the equation x² - 9x + 20 = 0, find the value of a² + b² + ab?

    • A) -21
    • B) 1
    • C) 61
    • D) 21

    Answer: 61

    Explanation: For roots a and b: a+b = 9 and ab = 20; a²+b²+ab = (a+b)² − ab = 81 − 20 = 61.

20. Question 20

    Q20. I. x² + 11x + 30 = 0, II. y² + 15y + 56 = 0 to solve both the equations to find the values of x and y?

    • A) If x < y
    • B) If x > y
    • C) If x ≤ y
    • D) If x ≥ y

    Answer: If x > y

    Explanation: Roots of first equation: x = −5 or −6; roots of second: y = −7 or −8; since both values of x are greater than both values of y, x > y.