CA Foundation P3 · Chapter 2

Equations MCQs with Answers

9 multiple-choice questions on Equations for CA Foundation P3 Quantitative Aptitude. Try each one before revealing the answer and explanation.

1. Question 1

   A cashier has ₹1,200 made up of 90 notes, some of ₹10 and the rest of ₹20. The number of ₹20 notes is:

   • A) 60
   • B) 45
   • C) 40
   • D) 30
   Show answer & explanation

   Answer: D) 30

   Let the number of ₹20 notes be b, so ₹10 notes = 90 − b. Then 10(90 − b) + 20b = 1,200, so 900 + 10b = 1,200 and b = 30. There are 60 notes of ₹10; check: 600 + 600 = 1,200.

2. Question 2

   The solution of the equation 5x − 7 = 3x + 11 is:

   • A) x = 9
   • B) x = 2
   • C) x = 0.5
   • D) x = 18
   Show answer & explanation

   Answer: A) x = 9

   Transposing terms: 5x − 3x = 11 + 7, so 2x = 18 and x = 9. Check: 5(9) − 7 = 38 and 3(9) + 11 = 38. Writing 2x = 4 (wrong sign on 7) gives x = 2.

3. Question 3

   The roots of the equation x² − 11x + 28 = 0 are:

   • A) −4 and −7
   • B) 2 and 14
   • C) −4 and 7
   • D) 4 and 7
   Show answer & explanation

   Answer: D) 4 and 7

   We need two numbers with sum 11 and product 28: 4 and 7. So x² − 11x + 28 = (x − 4)(x − 7) = 0, giving x = 4 or x = 7. The pair 2 and 14 has product 28 but sum 16.

4. Question 4

   If α and β are the roots of 3x² − 12x + 7 = 0, the value of α² + β² is:

   • A) 16
   • B) 62/3
   • C) 34/3
   • D) 41/3
   Show answer & explanation

   Answer: C) 34/3

   α + β = 12/3 = 4 and αβ = 7/3. Then α² + β² = (α + β)² − 2αβ = 16 − 14/3 = 34/3. Adding 2αβ instead of subtracting gives 62/3, and forgetting the factor 2 gives 41/3.

5. Question 5

   The roots of the equation 2x² + 5x + 4 = 0 are:

   • A) Imaginary (complex conjugates)
   • B) Real and distinct
   • C) Real and equal
   • D) Rational and unequal
   Show answer & explanation

   Answer: A) Imaginary (complex conjugates)

   The discriminant is b² − 4ac = 25 − 4(2)(4) = 25 − 32 = −7. Since the discriminant is negative, the roots are imaginary and occur as a conjugate pair.

6. Question 6

   The quadratic equation whose roots are 3 + √2 and 3 − √2 is:

   • A) x² + 6x + 7 = 0
   • B) x² − 6x + 7 = 0
   • C) x² − 6x + 11 = 0
   • D) x² − 6x − 7 = 0
   Show answer & explanation

   Answer: B) x² − 6x + 7 = 0

   Sum of roots = 6 and product = (3 + √2)(3 − √2) = 9 − 2 = 7. The equation is x² − (sum)x + product = 0, i.e. x² − 6x + 7 = 0. Using 9 + 2 for the product gives x² − 6x + 11 = 0.

7. Question 7

   The solution of the simultaneous equations 3x + 4y = 26 and 5x − 2y = 13 is:

   • A) x = 3.5, y = 4
   • B) x = 4, y = 3
   • C) x = 4, y = 3.5
   • D) x = 2, y = 5
   Show answer & explanation

   Answer: C) x = 4, y = 3.5

   Multiply the second equation by 2: 10x − 4y = 26. Adding to the first: 13x = 52, so x = 4. Substituting, 3(4) + 4y = 26 gives 4y = 14 and y = 3.5. Check: 5(4) − 2(3.5) = 20 − 7 = 13.

8. Question 8

   One root of x³ − 6x² + 11x − 6 = 0 is 1. The other two roots are:

   • A) −2 and −3
   • B) 2 and −3
   • C) 2 and 3
   • D) 1 and 6
   Show answer & explanation

   Answer: C) 2 and 3

   Dividing by (x − 1) gives x² − 5x + 6 = 0, which factorises as (x − 2)(x − 3) = 0. Hence the other roots are 2 and 3. Check: the sum of all roots 1 + 2 + 3 = 6 equals the negative of the x² coefficient.

9. Question 9

   If α and β are the roots of x² − 5x + 3 = 0, the equation whose roots are α² and β² is:

   • A) x² − 25x + 9 = 0
   • B) x² − 19x + 9 = 0
   • C) x² − 19x + 3 = 0
   • D) x² + 19x + 9 = 0
   Show answer & explanation

   Answer: B) x² − 19x + 9 = 0

   α + β = 5 and αβ = 3. Sum of new roots: α² + β² = 25 − 2(3) = 19. Product of new roots: α²β² = 3² = 9. The equation is x² − 19x + 9 = 0. Taking (α + β)² as the sum gives the wrong option with 25.