Algebra Practice Questions — JAMB UTME — Page 11

Question 201Question →

Given the simultaneous equations $2x + y = 5$ and $x^2 + xy - y^2 = -5$ , let $(x_1, y_1)$ and $(x_2, y_2)$ be the real solution pairs. What is the value of $x_1 x_2 + y_1 y_2$ ?

-5

13

5

95

Show answer & explanation

Answer: $-5$

Answer

-5

Substituting

y = 5 - 2x

into

x^2 + xy - y^2 = -5

yields

-5x^2 + 25x - 20 = 0

, which simplifies to

x^2 - 5x + 4 = 0

. The roots are

x_1 = 1

and

x_2 = 4

, giving corresponding

y

-values

y_1 = 3

and

y_2 = -3

. Evaluating

x_1 x_2 + y_1 y_2

gives

(1)(4) + (3)(-3) = 4 - 9 = -5

.

Step-by-Step Solution

1

Express

y

in terms of

x

using the linear equation.

y = 5 - 2x

Substitution method requires isolating one variable from the linear equation.

2

Substitute

y = 5 - 2x

into the quadratic equation

x^2 + xy - y^2 = -5

.

x^2 + x(5 - 2x) - (5 - 2x)^2 = -5

Form a single quadratic equation in terms of

x

.

3

Expand and simplify the quadratic equation.

x^2 + 5x - 2x^2 - (25 - 20x + 4x^2) = -5 \implies -5x^2 + 25x - 20 = 0 \implies x^2 - 5x + 4 = 0

Convert the equation to standard quadratic form

ax^2 + bx + c = 0

.

4

Factorize the quadratic equation to find the values of

x

.

(x - 1)(x - 4) = 0 \implies x_1 = 1, x_2 = 4

Obtain the two roots for

x

.

5

Find the corresponding

y

-values using

y = 5 - 2x

.

For

x_1 = 1

:

y_1 = 5 - 2(1) = 3

. For

x_2 = 4

:

y_2 = 5 - 2(4) = -3

. Solution pairs are

(1, 3)

and

(4, -3)

.

Calculate corresponding coordinate values for each root.

6

Compute the required expression

x_1 x_2 + y_1 y_2

.

x_1 x_2 + y_1 y_2 = (1)(4) + (3)(-3) = 4 - 9 = -5

Perform the final calculation requested in the stem.

Key Concept

Simultaneous Linear and Quadratic Equations

Question 202Question →

Given that $(x - 2)$ is a factor of the polynomial $P(x) = x^3 + kx^2 - 5x + 6$ , find the remainder when $P(x)$ is divided by $(x + 3)$ .

Show answer & explanation

Answer: -15

Answer

The remainder when

P(x)

is divided by

(x + 3)

is

-15

.

According to the Factor Theorem, since

(x - 2)

is a factor of

P(x) = x^3 + kx^2 - 5x + 6

, setting

x = 2

yields

P(2) = 0

. This gives

8 + 4k - 10 + 6 = 0

, which simplifies to

4k + 4 = 0

, giving

k = -1

. The polynomial is therefore

P(x) = x^3 - x^2 - 5x + 6

. By the Remainder Theorem, dividing

P(x)

by

(x + 3)

produces a remainder of

P(-3)

. Evaluating

P(-3) = (-3)^3 - (-3)^2 - 5(-3) + 6 = -27 - 9 + 15 + 6 = -15

.

Step-by-Step Solution

1

Apply the Factor Theorem to determine the unknown constant

k

.

k = -1

If

(x - 2)

is a factor of

P(x)

, then

P(2) = 0

. Substituting

x = 2

gives

2^3 + k(2)^2 - 5(2) + 6 = 0 \implies 4k + 4 = 0 \implies k = -1

.

2

Substitute

k = -1

into the original polynomial to get the full expression.

P(x) = x^3 - x^2 - 5x + 6

Replacing

k

with

-1

defines

P(x)

completely.

3

Apply the Remainder Theorem to find the remainder when

P(x)

is divided by

(x + 3)

.

Remainder is

-15

By the Remainder Theorem, dividing

P(x)

by

(x + 3)

leaves a remainder equal to

P(-3)

. Calculating

P(-3) = (-3)^3 - (-3)^2 - 5(-3) + 6 = -27 - 9 + 15 + 6 = -15

.

Key Concept

Factor and Remainder Theorems for Polynomials

Estimated Time:1m 30s

Question 203Question →

What is the simplified value of $\frac{6}{\sqrt{7} - 1} - \frac{6}{\sqrt{7} + 1}$ ?

2

2\sqrt{7}

12

12\sqrt{7}

Show answer & explanation

Answer: 2

Answer

2

Combining the fractions over the common denominator

(\sqrt{7} - 1)(\sqrt{7} + 1) = 6

yields a numerator of

6(\sqrt{7} + 1) - 6(\sqrt{7} - 1) = 12

. Dividing 12 by 6 gives 2.

Step-by-Step Solution

1

Find a common denominator for the two fractions

The common denominator is

(\sqrt{7} - 1)(\sqrt{7} + 1) = (\sqrt{7})^2 - 1^2 = 7 - 1 = 6

.

Multiplying conjugate surds eliminates the radical in the denominator.

2

Combine the numerators over the common denominator

\frac{6(\sqrt{7} + 1) - 6(\sqrt{7} - 1)}{6}

Adjust each numerator by multiplying by the conjugate of its denominator.

3

Expand and simplify the numerator

6\sqrt{7} + 6 - 6\sqrt{7} + 6 = 12

The

6\sqrt{7}

terms cancel out:

6\sqrt{7} - 6\sqrt{7} = 0

, leaving

6 - (-6) = 12

.

4

Divide the simplified numerator by the denominator

\frac{12}{6} = 2

Simplify the final fraction to obtain an integer value.

Key Concept

Rationalisation and Subtraction of Surd Expressions

Estimated Time:1m 0s

Question 204Question →

Given the matrix $A = \begin{pmatrix} 1 & 2 & 0 \\ 3 & 4 & 1 \\ 2 & 0 & k \end{pmatrix}$ , if $\det(A) = 10$ , what is the value of $k$ ?

Show answer & explanation

Answer: -3

Answer

The value of

k

is

-3

.

Expanding the determinant along the first row gives

1(4k) - 2(3k - 2) + 0 = -2k + 4

. Setting

-2k + 4 = 10

leads directly to

-2k = 6

, giving

k = -3

.

Step-by-Step Solution

1

Expand the determinant of matrix

A

along the first row.

\det(A) = 1(4k - 0) - 2(3k - 2) + 0 = -2k + 4

Using cofactor expansion along the top row to find the expression for the determinant.

2

Set the calculated determinant equal to the given value and solve for

k

.

-2k + 4 = 10 \implies -2k = 6 \implies k = -3

Equating the determinant algebraic expression to 10.

Key Concept

Determinant of a 3x3 Matrix

Question 205Question →

When the polynomial $P(x) = x^3 + ax^2 + bx - 6$ is divided by $(x - 2)$ , the remainder is $0$ . When $P(x)$ is divided by $(x + 1)$ , the remainder is $12$ . What is the value of $a + b$ ?

-7

17

7

19

Show answer & explanation

Answer: $-7$

Answer

The value of

a + b

is

-7

.

According to the Remainder Theorem, dividing

P(x)

by

(x - 2)

with a remainder of

0

means

P(2) = 0

, giving the equation

2a + b = -1

. Dividing

P(x)

by

(x + 1)

with a remainder of

12

means

P(-1) = 12

, giving

a - b = 19

. Solving these two linear equations simultaneously yields

a = 6

and

b = -13

. Adding these values together gives

a + b = -7

.

Step-by-Step Solution

1

Apply the Factor/Remainder Theorem for divisor

(x - 2)

P(2) = 2^3 + a(2)^2 + b(2) - 6 = 0 \implies 4a + 2b + 2 = 0 \implies 2a + b = -1

Since dividing

P(x)

by

(x - 2)

leaves a remainder of

0

,

P(2) = 0

.

2

Apply the Remainder Theorem for divisor

(x + 1)

P(-1) = (-1)^3 + a(-1)^2 + b(-1) - 6 = 12 \implies a - b - 7 = 12 \implies a - b = 19

Setting the linear divisor

x + 1 = 0

gives

x = -1

, so

P(-1) = 12

.

3

Solve the system of simultaneous linear equations for

a

and

b

Adding

(2a + b = -1)

and

(a - b = 19)

yields

3a = 18 \implies a = 6

. Substituting

a = 6

into

a - b = 19

gives

6 - b = 19 \implies b = -13

.

Eliminating

b

allows finding the values of constants

a

and

b

.

4

Calculate

a + b

a + b = 6 + (-13) = -7

Summing the determined constants

a

and

b

gives the target expression.

Key Concept

Remainder and Factor Theorems

Estimated Time:1m 30s

Question 206Question →

Find the set of real values of $x$ that satisfies the inequality $\frac{5 - 2x}{3} \ge x - 5$ .

x \ge 4

x \le 4

x \le -4

x \ge -4

Show answer & explanation

Answer: $x \le 4$

Answer

The set of real values satisfying the inequality is

x \le 4

.

Multiplying through by

3

yields

5 - 2x \ge 3x - 15

. Rearranging terms gives

-5x \ge -20

. Dividing both sides by

-5

requires flipping the inequality sign from

\ge

to

\le

, giving

x \le 4

.

Step-by-Step Solution

1

Multiply both sides of the inequality by 3 to clear the fraction.

5 - 2x \ge 3(x - 5)

Eliminating the denominator simplifies the algebraic expression.

2

Expand the right-hand side and collect terms containing

x

on one side and constants on the other.

5 - 2x \ge 3x - 15 \implies -2x - 3x \ge -15 - 5 \implies -5x \ge -20

Group like terms to isolate the variable

x

.

3

Divide both sides by

-5

and reverse the inequality sign.

x \le \frac{-20}{-5} \implies x \le 4

Dividing or multiplying an inequality by a negative number reverses the direction of the inequality sign.

Key Concept

Solving linear inequalities involving negative coefficient division

Question 207Question →

Given the matrix $A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$ , if $A^2 - 7A + kI = \mathbf{0}$ , where $I$ is the $2 \times 2$ identity matrix and $\mathbf{0}$ is the $2 \times 2$ zero matrix, what is the value of $k$ ?

Show answer & explanation

Answer: 10

Answer

The value of

k

is 10.

By matrix multiplication and algebraic evaluation,

A^2 - 7A = -10I

. Substituting into

A^2 - 7A + kI = \mathbf{0}

yields

-10I + kI = \mathbf{0}

, which gives

k = 10

. Alternatively, by the Cayley-Hamilton Theorem, any

2 \times 2

matrix

A

satisfies

A^2 - \text{tr}(A)A + \det(A)I = \mathbf{0}

. Here

\text{tr}(A) = 3 + 4 = 7

and

\det(A) = (3)(4) - (2)(1) = 10

, directly giving

k = \det(A) = 10

.

Step-by-Step Solution

1

Calculate the matrix product

A^2

A^2 = \begin{pmatrix} (3\cdot 3 + 2\cdot 1) & (3\cdot 2 + 2\cdot 4) \\ (1\cdot 3 + 4\cdot 1) & (1\cdot 2 + 4\cdot 4) \end{pmatrix} = \begin{pmatrix} 11 & 14 \\ 7 & 18 \end{pmatrix}

Squaring matrix

A

involves multiplying rows of

A

by columns of

A

.

2

Perform scalar multiplication for

7A

7A = \begin{pmatrix} 21 & 14 \\ 7 & 28 \end{pmatrix}

Each entry of matrix

A

is multiplied by the scalar 7.

3

Subtract

7A

from

A^2

A^2 - 7A = \begin{pmatrix} 11 - 21 & 14 - 14 \\ 7 - 7 & 18 - 28 \end{pmatrix} = \begin{pmatrix} -10 & 0 \\ 0 & -10 \end{pmatrix} = -10I

Subtracting corresponding entries yields a scalar multiple of the identity matrix.

4

Solve for the unknown scalar

k

-10I + kI = \mathbf{0} \implies k = 10

Setting

(k - 10)I = \mathbf{0}

implies

k - 10 = 0

, so

k = 10

.

Key Concept

Matrix Polynomial Equations and Cayley-Hamilton Theorem

Question 208Question →

In a secondary school class of 45 students, 28 study Chemistry, 25 study Physics, and 6 study neither of the two subjects. How many students study both Chemistry and Physics?

Show answer & explanation

Answer: 14; 14 students

Answer

14 students study both Chemistry and Physics.

First, find the number of students taking at least one subject by subtracting those taking neither from the total: 45 - 6 = 39. Then, apply the formula n(C ∪ P) = n(C) + n(P) - n(C ∩ P). Substituting the values gives 39 = 28 + 25 - n(C ∩ P), which simplifies to n(C ∩ P) = 53 - 39 = 14.

Step-by-Step Solution

1

Calculate the number of students taking at least one of the two subjects.

n(Chemistry ∪ Physics) = 45 - 6 = 39

Students who study neither subject must be subtracted from the total class population to find the cardinality of the union.

2

Set up the two-set inclusion-exclusion formula.

n(Chemistry ∪ Physics) = n(Chemistry) + n(Physics) - n(Chemistry ∩ Physics)

The sum of individual set cardinalities overcounts elements present in both sets.

3

Substitute the known values into the equation and solve for the intersection.

39 = 28 + 25 - n(Chemistry ∩ Physics) ⇒ n(Chemistry ∩ Physics) = 53 - 39 = 14

Subtracting 39 from 53 gives the exact number of students taking both subjects.

Key Concept

Two-set inclusion-exclusion principle and complement of a set

Estimated Time:1m 30s

Question 209Question →

The pressure $P$ of a given mass of gas varies directly as its absolute temperature $T$ and inversely as its volume $V$ . Given that $P = 50\text{ kPa}$ when $T = 300\text{ K}$ and $V = 10\text{ m}^3$ , what is the value of $P$ when $T = 360\text{ K}$ and $V = 8\text{ m}^3$ ?

48\text{ kPa}

60\text{ kPa}

75\text{ kPa}

90\text{ kPa}

Show answer & explanation

Answer: $75\text{ kPa}$

Answer

75\text{ kPa}

The relationship is governed by

P = \frac{kT}{V}

. Substituting

P = 50

,

T = 300

, and

V = 10

yields

k = \frac{5}{3}

. Using

k = \frac{5}{3}

with

T = 360

and

V = 8

gives

P = \frac{(5/3) \times 360}{8} = 75\text{ kPa}

.

Step-by-Step Solution

1

Set up the general formula for joint and inverse variation.

P = \frac{kT}{V}

, where

k

is the constant of variation.

Pressure varies directly as temperature

T

and inversely as volume

V

.

2

Substitute initial conditions to determine

k

.

50 = \frac{k \times 300}{10} \implies 50 = 30k \implies k = \frac{5}{3}

.

The initial values

P = 50\text{ kPa}

,

T = 300\text{ K}

, and

V = 10\text{ m}^3

allow solving for

k

.

3

Calculate the new pressure with updated temperature and volume values.

P = \frac{\frac{5}{3} \times 360}{8} = \frac{600}{8} = 75\text{ kPa}

.

Substitute

k = \frac{5}{3}

,

T = 360\text{ K}

, and

V = 8\text{ m}^3

into the variation formula.

Key Concept

Joint and inverse variation in algebraic relationships

Estimated Time:1m 30s

Question 210Question →

Let $A$ be a $3 \times 3$ square matrix such that $\det(A) > 0$ . If the matrix satisfies the property $\det(3A^{-1}) = \det(A^T A)$ , what is the value of $\det(A)$ ?

Show answer & explanation

Answer: 3

Answer

The value of

\det(A)

is 3.

Using fundamental determinant identities for a

3 \times 3

matrix (

n = 3

):

\det(3A^{-1}) = 3^3 \det(A^{-1}) = \frac{27}{\det(A)}

and

\det(A^T A) = \det(A^T)\det(A) = (\det(A))^2

. Equating them yields

\frac{27}{\det(A)} = (\det(A))^2 \implies (\det(A))^3 = 27

, which yields

\det(A) = 3

.

Step-by-Step Solution

1

Express

\det(3A^{-1})

in terms of

\det(A)

using determinant scaling and inverse properties.

\det(3A^{-1}) = 3^3 \det(A^{-1}) = \frac{27}{\det(A)}.

For an

n \times n

matrix

M

, scaling by constant

k

gives

\det(kM) = k^n \det(M)

, and

\det(M^{-1}) = \frac{1}{\det(M)}

.

2

Express

\det(A^T A)

in terms of

\det(A)

using transpose and multiplication properties.

\det(A^T A) = \det(A^T)\det(A) = (\det(A))^2.

The determinant of a product is the product of determinants, and

\det(A^T) = \det(A)

.

3

Set the two simplified expressions equal to each other and solve for

\det(A)

.

\frac{27}{\det(A)} = (\det(A))^2 \implies (\det(A))^3 = 27 \implies \det(A) = 3.

Taking the cube root of both sides gives the unique real value since

\det(A) > 0

.

Key Concept

Properties of Determinants (Scalar Multiplication, Transpose, Inverse, and Matrix Products)

Estimated Time:1m 30s

Question 211Question →

Given that $(x - 3)$ is a factor of the polynomial $P(x) = x^3 + kx^2 + mx - 6$ and that dividing $P(x)$ by $(x + 1)$ leaves a remainder of $-12$ , what is the value of $k + m$ ?

-1

-7

-5

5

Show answer & explanation

Answer: $-1$

Answer

The value of

k + m

is

-1

.

Applying the Factor Theorem

P(3) = 0

gives

3k + m = -7

, and applying the Remainder Theorem

P(-1) = -12

gives

k - m = -5

. Solving these simultaneous linear equations gives

k = -3

and

m = 2

, which sums to

k + m = -1

.

Step-by-Step Solution

1

Apply the Factor Theorem for the linear factor

(x - 3)

3k + m = -7

By the Factor Theorem, if

(x - 3)

is a factor, then

P(3) = 0

. Substituting

x = 3

gives

3^3 + k(3)^2 + m(3) - 6 = 0 \implies 27 + 9k + 3m - 6 = 0 \implies 9k + 3m = -21 \implies 3k + m = -7

.

2

Apply the Remainder Theorem for the divisor

(x + 1)

k - m = -5

By the Remainder Theorem, dividing

P(x)

by

(x + 1)

gives remainder

P(-1) = -12

. Substituting

x = -1

gives

(-1)^3 + k(-1)^2 + m(-1) - 6 = -12 \implies -1 + k - m - 6 = -12 \implies k - m - 7 = -12 \implies k - m = -5

.

3

Solve the system of linear equations for

k

and

m

k = -3

and

m = 2

Adding the two equations

(3k + m) + (k - m) = -7 + (-5)

gives

4k = -12 \implies k = -3

. Substituting

k = -3

into

k - m = -5

gives

-3 - m = -5 \implies m = 2

.

4

Calculate the target value

k + m

k + m = -1

Summing

k = -3

and

m = 2

yields

k + m = -3 + 2 = -1

.

Key Concept

Factor and Remainder Theorems

Estimated Time:1m 30s

Question 212Question →

Which of the following is the set of real values of $x$ that satisfies the inequality $\frac{3 - x}{4} \le \frac{2x + 5}{3}$ ?

x \ge -1

x \le -1

x \ge 1

x \le 1

Show answer & explanation

Answer: $x \ge -1$

Answer

The set of real values of

x

that satisfies the inequality is

x \ge -1

.

Multiplying through by 12 gives

9 - 3x \le 8x + 20

. Grouping terms results in

-11x \le 11

. Dividing by

-11

requires reversing the inequality sign from

\le

to

\ge

, giving the solution

x \ge -1

.

Step-by-Step Solution

1

Clear the denominators by multiplying both sides of the inequality by the lowest common multiple, 12.

3(3 - x) \le 4(2x + 5)

Eliminating fractions simplifies the algebraic expression.

2

Expand both sides by distributing the multipliers.

9 - 3x \le 8x + 20

Prepares terms for grouping variables on one side and constants on the other.

3

Collect all terms containing

x

on the left side and constant terms on the right side.

-3x - 8x \le 20 - 9 \implies -11x \le 11

Isolates the linear variable term.

4

Divide both sides by

-11

and flip the inequality sign.

x \ge -1

Dividing or multiplying an inequality by a negative number reverses the direction of the inequality sign.

Key Concept

Linear Inequalities and Reversing Inequality Sign on Division by Negative Numbers

Question 213Question →

Given the matrices $A = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -2 \\ 0 & 3 \end{pmatrix}$ , if $C = AB - 2I$ , where $I$ is the $2 \times 2$ identity matrix, what is the determinant of matrix $C$ ?

3

-3

0

15

Show answer & explanation

Answer: $3$

Answer

The determinant of matrix C is 3.

Multiplying matrices A and B using standard row-by-column multiplication yields \begin{pmatrix} 2 & -1 \\ 3 & 6 \end{pmatrix}. Subtracting the scaled identity matrix \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} produces matrix C = \begin{pmatrix} 0 & -1 \\ 3 & 4 \end{pmatrix}. Taking the determinant ad - bc gives (0)(4) - (-1)(3) = 3.

Step-by-Step Solution

1

Compute the matrix product AB.

AB = \begin{pmatrix} (2)(1) + (1)(0) & (2)(-2) + (1)(3) \\ (3)(1) + (4)(0) & (3)(-2) + (4)(3) \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 3 & 6 \end{pmatrix}

Matrix multiplication uses dot products of rows of the first matrix and columns of the second matrix.

2

Subtract 2I from AB to find matrix C.

C = \begin{pmatrix} 2 & -1 \\ 3 & 6 \end{pmatrix} - \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 3 & 4 \end{pmatrix}

The identity matrix I multiplied by scalar 2 has 2 on its main diagonal and 0 elsewhere.

3

Calculate the determinant of matrix C.

\det(C) = (0)(4) - (-1)(3) = 0 + 3 = 3

For a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is ad - bc.

Key Concept

Matrix multiplication, matrix arithmetic operations, and evaluation of 2x2 determinants.

Estimated Time:1m 30s

Question 214Question →

Find the real value of $x$ that satisfies the logarithmic equation $\log_2(x^2 + 3x - 22) - \log_2(x - 2) = 3$ .

Show answer & explanation

Answer: 6

Answer

The real value of

x

that satisfies the equation is

6

.

Applying the logarithmic quotient rule transforms

\log_2(x^2 + 3x - 22) - \log_2(x - 2) = 3

into

\log_2\left(\frac{x^2 + 3x - 22}{x - 2}\right) = 3

. Expressing this in exponential form yields

\frac{x^2 + 3x - 22}{x - 2} = 8

, which simplifies to

x^2 - 5x - 6 = 0

. Factoring gives

(x - 6)(x + 1) = 0

. Since the argument of a logarithm must be strictly positive (

x - 2 > 0 \implies x > 2

),

x = -1

is extraneous and

x = 6

is the only valid solution.

Step-by-Step Solution

1

Combine the logarithmic terms using the quotient law of logarithms.

\log_2\left(\frac{x^2 + 3x - 22}{x - 2}\right) = 3

The difference of two logarithms with the same base equals the logarithm of their quotient:

\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)

.

2

Convert the logarithmic equation into its equivalent exponential form.

\frac{x^2 + 3x - 22}{x - 2} = 2^3 = 8

If

\log_b Y = c

, then

Y = b^c

.

3

Clear the denominator and simplify to form a quadratic equation.

x^2 - 5x - 6 = 0

Multiplying both sides by

(x - 2)

yields

x^2 + 3x - 22 = 8x - 16

, which rearranges to

x^2 - 5x - 6 = 0

.

4

Solve the quadratic equation by factoring.

(x - 6)(x + 1) = 0 \implies x = 6 \text{ or } x = -1

The factors of

-6

that sum to

-5

are

-6

and

1

.

5

Test roots against domain restrictions to eliminate extraneous solutions.

x = 6

Logarithmic arguments must be strictly positive. For

\log_2(x - 2)

to be defined,

x > 2

. Thus,

x = -1

is extraneous, leaving

x = 6

as the unique valid solution.

Key Concept

Logarithmic Equations and Domain Restrictions

Question 215Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z}^+ : 1 \le x \le 15\}$ . Consider two subsets of $\mathcal{U}$ defined as $A = \{x \in \mathcal{U} : x \text{ is prime}\}$ and $B = \{x \in \mathcal{U} : x \text{ is an odd integer greater than } 1\}$ . What is the number of elements in $(A \cap B)' \setminus (A \cup B)'$ ?

3

5

7

10

Show answer & explanation

Answer: $3$

Answer

3

Evaluating

(A \cap B)'

gives

\{1, 2, 4, 6, 8, 9, 10, 12, 14, 15\}

and evaluating

(A \cup B)'

gives

\{1, 4, 6, 8, 10, 12, 14\}

. Subtracting

(A \cup B)'

from

(A \cap B)'

leaves

\{2, 9, 15\}

, which has 3 elements.

Step-by-Step Solution

1

List the elements of the universal set

\mathcal{U}

and subsets

A

and

B

.

\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}

,

A = \{2, 3, 5, 7, 11, 13\}

, and

B = \{3, 5, 7, 9, 11, 13, 15\}

.

Explicitly writing the elements helps ensure set operations are performed accurately.

2

Find the intersection

A \cap B

and union

A \cup B

.

A \cap B = \{3, 5, 7, 11, 13\}

and

A \cup B = \{2, 3, 5, 7, 9, 11, 13, 15\}

.

These intermediate set operations are required to evaluate their complements.

3

Determine the complements

(A \cap B)'

and

(A \cup B)'

relative to

\mathcal{U}

.

(A \cap B)' = \{1, 2, 4, 6, 8, 9, 10, 12, 14, 15\}

and

(A \cup B)' = \{1, 4, 6, 8, 10, 12, 14\}

.

The complement of a set consists of all elements in

\mathcal{U}

not present in that set.

4

Calculate the set difference

(A \cap B)' \setminus (A \cup B)'

and count its elements.

(A \cap B)' \setminus (A \cup B)' = \{2, 9, 15\}

, which contains 3 elements.

Set difference removes all elements of

(A \cup B)'

from

(A \cap B)'

, leaving elements present in

A \cup B

but not in

A \cap B

.

Key Concept

Set Complements and Relative Difference

Estimated Time:1m 30s

Question 216Question →

If $P$ varies directly as the square root of $q$ and inversely as the square of $r$ , what is the percentage change in $P$ when $q$ is increased by $44\%$ and $r$ is decreased by $20\%$ ?

50\%

increase

125\%

increase

87.5\%

increase

23.2\%

decrease

Show answer & explanation

Answer: $87.5\%$ increase

Answer

An increase of

87.5\%

The variation relation is

P = k \frac{\sqrt{q}}{r^2}

. Increasing

q

by

44\%

scales

\sqrt{q}

by

\sqrt{1.44} = 1.2

. Decreasing

r

by

20\%

scales

r^2

by

(0.8)^2 = 0.64

. Dividing

1.2

by

0.64

gives

1.875

, meaning the new value

P'

is

187.5\%

of the original

P

, which corresponds to an increase of

87.5\%

.

Step-by-Step Solution

1

Set up the joint variation equation

P = k \frac{\sqrt{q}}{r^2}

, where

k

is a constant of variation.

Direct variation puts

\sqrt{q}

in the numerator, and inverse variation puts

r^2

in the denominator.

2

Express new values

q'

and

r'

in terms of original variables

q' = 1.44q

and

r' = 0.80r

.

An increase of

44\%

gives

1 + 0.44 = 1.44

, and a decrease of

20\%

gives

1 - 0.20 = 0.80

.

3

Substitute new variables into the variation formula to find the new value

P'

P' = k \frac{\sqrt{1.44q}}{(0.80r)^2} = k \frac{1.2\sqrt{q}}{0.64r^2} = \frac{1.2}{0.64} \left(k \frac{\sqrt{q}}{r^2}\right) = 1.875P

.

Evaluating

\sqrt{1.44} = 1.2

and

(0.80)^2 = 0.64

gives the multiplier for

P

.

4

Calculate the percentage change in

P

\text{Percentage Change} = (1.875 - 1) \times 100\% = 87.5\%

increase.

Subtracting

1

converts the multiplier to a relative increase.

Key Concept

Joint Variation with Percentage Changes

Estimated Time:1m 30s

Question 217Question →

Given the matrices $A = \begin{pmatrix} 3 & y \\ 2 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ x & 2 \end{pmatrix}$ , if $AB = \begin{pmatrix} 9 & 16 \\ 5 & 10 \end{pmatrix}$ , what is the value of $x + y$ ?

5

6

7

9

Show answer & explanation

Answer: $5$

Answer

The value of

x + y

is

5

.

Multiplying matrix

A

by matrix

B

using standard row-by-column multiplication yields

AB = \begin{pmatrix} 3 + xy & 12 + 2y \\ 2 + x & 10 \end{pmatrix}

. Setting this equal to

\begin{pmatrix} 9 & 16 \\ 5 & 10 \end{pmatrix}

gives

2 + x = 5 \implies x = 3

and

12 + 2y = 16 \implies y = 2

. Thus,

x + y = 3 + 2 = 5

.

Step-by-Step Solution

1

Compute the product matrix

AB

using matrix multiplication rules.

AB = \begin{pmatrix} 3(1) + y(x) & 3(4) + y(2) \\ 2(1) + 1(x) & 2(4) + 1(2) \end{pmatrix} = \begin{pmatrix} 3 + xy & 12 + 2y \\ 2 + x & 10 \end{pmatrix}

Matrix multiplication requires taking the dot product of rows from the first matrix and columns from the second matrix.

2

Equate the elements of

AB

with the given matrix

\begin{pmatrix} 9 & 16 \\ 5 & 10 \end{pmatrix}

to solve for

x

and

y

.

From row 2, column 1:

2 + x = 5 \implies x = 3

. From row 1, column 2:

12 + 2y = 16 \implies 2y = 4 \implies y = 2

.

Two matrices are equal if and only if their corresponding elements are equal.

3

Calculate the sum

x + y

.

x + y = 3 + 2 = 5

Substituting the values found for

x

and

y

into the requested expression.

Key Concept

Matrix Multiplication and Equality of Matrices

Alternative Method

Verification can be done by checking row 1, column 1:

3 + xy = 3 + (3)(2) = 9

, which matches the given matrix entry.

Estimated Time:1m 15s

Question 218Question →

If $x$ and $y$ are real numbers satisfying the simultaneous equations $x - y = 4$ and $x^2 + y^2 = 26$ , what is the value of the product $xy$ ?

Show answer & explanation

Answer: 5

Answer

The value of the product

xy

is 5.

Expressing

x

from the linear equation yields

x = y + 4

. Substituting this into

x^2 + y^2 = 26

gives

(y + 4)^2 + y^2 = 26

, which expands and simplifies to

y^2 + 4y - 5 = 0

. Factoring gives solutions

y = 1

(with

x = 5

) and

y = -5

(with

x = -1

). Both solution pairs

(5, 1)

and

(-1, -5)

result in the product

xy = 5

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the linear equation

x = y + 4

Isolate

x

to substitute into the second equation.

2

Substitute

x = y + 4

into the quadratic equation

x^2 + y^2 = 26

(y + 4)^2 + y^2 = 26

Reduce the system to a single quadratic equation in

y

.

3

Expand and simplify into standard quadratic form

y^2 + 4y - 5 = 0

Expanding gives

2y^2 + 8y + 16 = 26

, which simplifies by subtracting 26 and dividing by 2.

4

Solve for

y

by factoring

y = 1

or

y = -5

The factors of

y^2 + 4y - 5

are

(y + 5)(y - 1) = 0

.

5

Compute corresponding

x

values and the product

xy

For

y = 1

,

x = 5 \implies xy = 5

; for

y = -5

,

x = -1 \implies xy = 5

Substitute each

y

back into

x = y + 4

and evaluate

xy

.

Key Concept

Simultaneous Linear and Quadratic Equations

Question 219Question →

Given the simultaneous equations $x - y = 1$ and $x^2 + y^2 = 25$ , where $x$ and $y$ are both positive real numbers, what is the value of $x + y$ ?

7

9

1

3

Show answer & explanation

Answer: $7$

Answer

The value of

x + y

is

7

.

Expressing

x

as

y + 1

and substituting into

x^2 + y^2 = 25

yields

2y^2 + 2y - 24 = 0

. Factoring gives

y = 3

(rejecting

y = -4

as

y > 0

). Substituting

y = 3

back gives

x = 4

. Adding these values together yields

x + y = 7

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the linear equation.

x = y + 1

Rearranging

x - y = 1

allows substitution into the quadratic equation.

2

Substitute

x = y + 1

into the quadratic equation

x^2 + y^2 = 25

.

(y + 1)^2 + y^2 = 25 \implies y^2 + 2y + 1 + y^2 = 25 \implies 2y^2 + 2y - 24 = 0

This reduces the system to a single quadratic equation in terms of

y

.

3

Solve the quadratic equation for

y

.

y^2 + y - 12 = 0 \implies (y + 4)(y - 3) = 0 \implies y = 3

or

y = -4

Dividing by

2

simplifies the equation, and factoring gives the potential roots for

y

.

4

Select the positive value of

y

and calculate

x

and

x + y

.

Since

y > 0

,

y = 3

. Then

x = 3 + 1 = 4

, so

x + y = 4 + 3 = 7

.

The question specifies that both

x

and

y

are positive real numbers.

Key Concept

Solving Simultaneous Linear and Quadratic Equations by Substitution

Question 220Question →

Find the real value of $x$ that satisfies the logarithmic equation $\log_5(x^2 - 4) - \log_5(x - 2) = 2$ .

Show answer & explanation

Answer: 23

Answer

The value of

x

that satisfies the equation is 23.

Applying the logarithmic quotient rule gives

\log_5 \left(\frac{x^2 - 4}{x - 2}\right) = 2

. Factoring the numerator gives

\log_5 \left(\frac{(x - 2)(x + 2)}{x - 2}\right) = \log_5(x + 2) = 2

. Converting to exponential form yields

x + 2 = 5^2 = 25

, which simplifies to

x = 23

.

Step-by-Step Solution

1

Apply the logarithmic quotient rule

\log_5\left(\frac{x^2 - 4}{x - 2}\right) = 2

The difference of two logarithms of the same base equals the logarithm of their quotient.

2

Factor the numerator and simplify the expression

\log_5(x + 2) = 2

Factoring

x^2 - 4

into

(x-2)(x+2)

allows canceling the common factor

(x-2)

in the denominator.

3

Convert the logarithmic equation into exponential form

x + 2 = 5^2 = 25

By definition,

\log_b(y) = c

is equivalent to

b^c = y

.

4

Solve the linear equation for

x

x = 23

Subtracting 2 from both sides gives

x = 23

.

Key Concept

Logarithmic Quotient Rule and Logarithm-to-Exponent Conversion

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