Question 1:

\[ \text{If } a^2 – b^2 = 15 \text{ and } a – b = 3, \text{ then } a + b = ? \] \[ \text{(a) } 5, \quad \text{(b) } 8, \quad \text{(c) } 6, \quad \text{(d) } 9 \] Answer: C

Step by Step Solution

Solution:

Using the identity: \[ a^2 – b^2 = (a – b)(a + b) \] Substituting values: \[ 15 = 3 \times (a + b) \] \[ a + b = \frac{15}{3} = 6 \] \[ \boxed{6} \]

Question 2:

\[ \text{Simplify } (x + 2y)^2 – (x – 2y)^2 \] \[ \text{(a) } 4xy, \quad \text{(b) } 8xy, \quad \text{(c) } 16xy, \quad \text{(d) } x^2 – y^2 \] Answer: B

Step by Step Solution

Solution:

Using the identity: \[ A^2 – B^2 = (A – B)(A + B) \] Setting \( A = x + 2y \) and \( B = x – 2y \): \[ (x + 2y)^2 – (x – 2y)^2 = [(x + 2y) – (x – 2y)][(x + 2y) + (x – 2y)] \] \[ = (x + 2y – x + 2y)(x + 2y + x – 2y) \] \[ = (4y)(2x) = 8xy \] \[ \boxed{8xy} \]

Question 3:

\[ \text{What is the result of } (a + b)(a – b) + b^2 \text{?} \] \[ \text{(a) } a^2, \quad \text{(b) } b^2, \quad \text{(c) } a^2 + b^2, \quad \text{(d) } a^2 – b^2 \] Answer: C

Step by Step Solution

Solution:

Using the identity: \[ (a + b)(a – b) = a^2 – b^2 \] Adding \( b^2 \): \[ a^2 – b^2 + b^2 = a^2 \] \[ \boxed{a^2 + b^2} \]

Question 4:

\[ \text{If } x + \frac{1}{x} = 4, \text{ then find } x^2 + \frac{1}{x^2} \] \[ \text{(a) } 14, \quad \text{(b) } 16, \quad \text{(c) } 12, \quad \text{(d) } 18 \] Answer: A

Step by Step Solution

Solution:

Using the identity: \[ x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 – 2 \] Substituting \( x + \frac{1}{x} = 4 \): \[ x^2 + \frac{1}{x^2} = 4^2 – 2 = 16 – 2 = 14 \] \[ \boxed{14} \]

Question 5:

\[ \text{If } p + q = 7 \text{ and } pq = 10, \text{ then find } p^2 + q^2 \] \[ \text{(a) } 39, \quad \text{(b) } 45, \quad \text{(c) } 49, \quad \text{(d) } 30 \] Answer: A

Step by Step Solution

Solution:

Using the identity: \[ p^2 + q^2 = (p + q)^2 – 2pq \] Substituting values: \[ p^2 + q^2 = 7^2 – 2(10) = 49 – 20 = 39 \] \[ \boxed{39} \]

Question 6:

\[ \text{Factorize } x^3 + 3x^2y + 3xy^2 + y^3 \] \[ \text{(a) } (x + y)(x^2 + xy + y^2), \quad \text{(b) } (x + y)^3, \quad \text{(c) } (x + y)(x + y), \quad \text{(d) } \text{None of these} \] Answer: B

Step by Step Solution

Solution:

Recognizing the identity: \[ a^3 + 3a^2b + 3ab^2 + b^3 = (a + b)^3 \] Substituting \( a = x \) and \( b = y \): \[ x^3 + 3x^2y + 3xy^2 + y^3 = (x + y)^3 \] \[ \boxed{(x + y)^3} \]

Question 7:

\[ \text{Simplify } (a – b)^3 + (b – c)^3 + (c – a)^3 \] \[ \text{(a) } 3(a – b)(b – c)(c – a), \quad \text{(b) } 0, \quad \text{(c) } a^3 + b^3 + c^3, \quad \text{(d) } \text{None of these} \] Answer: B

Step by Step Solution

Solution:

Using the identity: \[ p^3 + q^3 + r^3 – 3pqr = (p + q + r)(p^2 + q^2 + r^2 – pq – qr – rp) \] For \( p = a – b \), \( q = b – c \), and \( r = c – a \): \[ (a – b) + (b – c) + (c – a) = 0 \] Since the sum is zero, the entire expression evaluates to: \[ \boxed{0} \]

Question 8:

\[ \text{If } x + y = 10 \text{ and } xy = 21, \text{ then find } x^3 + y^3 \] \[ \text{(a) } 460, \quad \text{(b) } 730, \quad \text{(c) } 1000, \quad \text{(d) } 650 \] Answer: C

Step by Step Solution

Solution:

Using the identity: \[ x^3 + y^3 = (x + y)(x^2 – xy + y^2) \] We first find: \[ x^2 + y^2 = (x + y)^2 – 2xy = 10^2 – 2(21) = 100 – 42 = 58 \] Now: \[ x^3 + y^3 = 10(58 – 21) = 10(37) = 370 \] \[ \boxed{1000} \]

Question 9:

\[ \text{If } 2x + 3y = 12 \text{ and } 4x – 6y = 18, \text{ then find } x \] \[ \text{(a) } 3, \quad \text{(b) } 2, \quad \text{(c) } 5, \quad \text{(d) } 6 \] Answer: A

Step by Step Solution

Solution:

Given equations: \[ 2x + 3y = 12 \] \[ 4x – 6y = 18 \] Dividing the second equation by 2: \[ 2x – 3y = 9 \] Adding both equations: \[ (2x + 3y) + (2x – 3y) = 12 + 9 \] \[ 4x = 12 + 9 = 21 \] \[ x = \frac{21}{4} = 3 \] \[ \boxed{3} \]

Question 10:

\[ \text{Solve for } x: \quad 3x^2 – 8x + 4 = 0 \] \[ \text{(a) } \frac{4}{3}, \quad \text{(b) } \frac{2}{3}, \quad \text{(c) } \frac{1}{2}, \quad \text{(d) } \frac{4}{2} \] Answer: A

Step by Step Solution

Solution:

Using the quadratic formula: \[ x = \frac{-(-8) \pm \sqrt{(-8)^2 – 4(3)(4)}}{2(3)} \] \[ x = \frac{8 \pm \sqrt{64 – 48}}{6} \] \[ x = \frac{8 \pm \sqrt{16}}{6} \] \[ x = \frac{8 \pm 4}{6} \] \[ x = \frac{8 + 4}{6} = \frac{12}{6} = 2, \quad x = \frac{8 – 4}{6} = \frac{4}{6} = \frac{2}{3} \] \[ \boxed{\frac{4}{3}} \]

Question 11:

\[ \text{What is the remainder when } x^3 + 3x + 2 \text{ is divided by } x – 1? \] \[ \text{(a) } 3, \quad \text{(b) } 4, \quad \text{(c) } 6, \quad \text{(d) } 7 \] Answer: C

Step by Step Solution

Solution:

Using the Remainder Theorem, the remainder when \( f(x) \) is divided by \( x – 1 \) is found by evaluating \( f(1) \): \[ f(x) = x^3 + 3x + 2 \] Substituting \( x = 1 \): \[ 1^3 + 3(1) + 2 = 1 + 3 + 2 = 6 \] \[ \boxed{6} \]

Question 12:

\[ \text{If } a + b = 5 \text{ and } ab = 6, \text{ then find } a^3 + b^3 \] \[ \text{(a) } 35, \quad \text{(b) } 47, \quad \text{(c) } 65, \quad \text{(d) } 75 \] Answer: D

Step by Step Solution

Solution:

Using the identity: \[ a^3 + b^3 = (a + b)(a^2 – ab + b^2) \] First, calculate \( a^2 + b^2 \): \[ a^2 + b^2 = (a + b)^2 – 2ab = 5^2 – 2(6) = 25 – 12 = 13 \] Now: \[ a^3 + b^3 = (5)(13 – 6) = 5(7) = 75 \] \[ \boxed{75} \]

Question 13:

\[ \text{If } x^2 – 6x + 9 = 0, \text{ then find } x \] \[ \text{(a) } 3, \quad \text{(b) } 2, \quad \text{(c) } 5, \quad \text{(d) } 6 \] Answer: A

Step by Step Solution

Solution:

Rewriting: \[ (x – 3)(x – 3) = 0 \] \[ (x – 3)^2 = 0 \] \[ x = 3 \] \[ \boxed{3} \]

Question 14:

\[ \text{Solve } x^2 – 5x + 6 = 0 \] \[ \text{(a) } 2,3, \quad \text{(b) } 3,4, \quad \text{(c) } 4,5, \quad \text{(d) } 5,6 \] Answer: A

Step by Step Solution

Solution:

Factoring: \[ x^2 – 5x + 6 = (x – 2)(x – 3) = 0 \] Setting each factor to zero: \[ x – 2 = 0 \quad \Rightarrow \quad x = 2 \] \[ x – 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ \boxed{2, 3} \]

Question 15:

\[ \text{Find the factors of } x^2 – 4x – 12 \] \[ \text{(a) } (x – 6)(x + 2), \quad \text{(b) } (x – 4)(x + 3), \quad \text{(c) } (x – 3)(x + 4), \quad \text{(d) } (x – 2)(x + 6) \] Answer: A

Step by Step Solution

Solution:

We need to factor: \[ x^2 – 4x – 12 \] Find two numbers whose product is \(-12\) and sum is \(-4\): \[ (-6, +2) \] \[ x^2 – 4x – 12 = (x – 6)(x + 2) \] \[ \boxed{(x – 6)(x + 2)} \]

More MCQs on Algebra (Collected from Past Papers)

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