Algebra Practice Questions — JAMB UTME — Page 4

Question 61Question →

Given the universal set $\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ , set $A = \{1, 2, 4, 5, 8\}$ , and set $B = \{2, 3, 5, 7, 9\}$ , what is $(A \cup B)'$ ?

\{6, 10\}

\{1, 4, 8\}

\{3, 6, 7, 9, 10\}

\{2, 5\}

Show answer & explanation

Answer: $\{6, 10\}$

Answer

The set

\{6, 10\}

Combining set

A

and set

B

gives

A \cup B = \{1, 2, 3, 4, 5, 7, 8, 9\}

. The complement

(A \cup B)'

contains the elements of the universal set

\mathcal{E}

that are not in this union, which are

6

and

10

, giving

\{6, 10\}

.

Step-by-Step Solution

1

Find the union of set

A

and set

B

(

A \cup B

)

A \cup B = \{1, 2, 3, 4, 5, 7, 8, 9\}

The union combines all distinct elements present in set

A

, set

B

, or both.

2

Determine the complement of

(A \cup B)

with respect to the universal set

\mathcal{E}

(A \cup B)' = \mathcal{E} \setminus (A \cup B) = \{6, 10\}

The complement consists of all elements in the universal set

\mathcal{E}

that are not present in

A \cup B

.

Key Concept

Set Union and Set Complement

Question 62Question →

Given that the determinant of the $3 \times 3$ matrix $M = \begin{pmatrix} 3 & 1 & 0 \\ 2 & x & -1 \\ 0 & 4 & 2 \end{pmatrix}$ is equal to $20$ , calculate the value of $x$ .

Show answer & explanation

Answer: 2

Answer

The value of

x

is

2

.

Expanding the matrix

M

along its top row yields

\det(M) = 3(2x + 4) - 1(4) = 6x + 8

. Setting this expression equal to

20

gives

6x + 8 = 20

, which simplifies to

6x = 12

, yielding

x = 2

.

Step-by-Step Solution

1

Expand the 3x3 matrix along the first row

\det(M) = 3(2x - (-4)) - 1(4 - 0) + 0

Cofactor expansion along a row containing a zero simplifies the computation of a 3x3 determinant.

2

Simplify the algebraic expression for the determinant

\det(M) = 6x + 8

Distribute the coefficients and combine like constant terms.

3

Solve the linear equation for x

x = 2

Subtract 8 from 20 to get 12, then divide by 6.

Key Concept

Determinant of a 3x3 Matrix via Cofactor Expansion

Question 63Question →

If set $A = \{a, b, c, d, e\}$ and set $B = \{c, d, e, f, g\}$ , what is the number of elements in the set $(A \setminus B) \cup (B \setminus A)$ ?

Show answer & explanation

Answer: 4; four

Answer

The number of elements in

(A \setminus B) \cup (B \setminus A)

is 4.

The set difference

A \setminus B

consists of elements in

A

that are not in

B

, which gives

\{a, b\}

. Similarly,

B \setminus A

consists of elements in

B

that are not in

A

, giving

\{f, g\}

. The union

(A \setminus B) \cup (B \setminus A)

is

\{a, b, f, g\}

, which has a cardinality of 4.

Step-by-Step Solution

1

Find the relative difference

A \setminus B

A \setminus B = \{a, b\}

Remove elements of

B

present in

A

.

2

Find the relative difference

B \setminus A

B \setminus A = \{f, g\}

Remove elements of

A

present in

B

.

3

Take the union of the two set differences

(A \setminus B) \cup (B \setminus A) = \{a, b, f, g\}

Combine elements from both set differences.

4

Count the number of elements in the resulting set

4 elements

The set

\{a, b, f, g\}

contains 4 distinct elements.

Key Concept

Symmetric Difference of Two Sets

Estimated Time:45s

Question 64Question →

What is the remainder when the polynomial $P(x) = x^3 + 3x^2 - 2x + 4$ is divided by $x - 1$ ?

6

8

10

4

Show answer & explanation

Answer: $6$

Answer

The remainder when

P(x)

is divided by

x - 1

is

6

.

According to the Remainder Theorem, dividing a polynomial

P(x)

by a linear divisor

x - a

leaves a remainder equal to

P(a)

. For the divisor

x - 1

, setting

x - 1 = 0

yields

x = 1

. Substituting

x = 1

into

P(x) = x^3 + 3x^2 - 2x + 4

gives

1 + 3 - 2 + 4 = 6

. Therefore, the value

6

is the correct remainder.

Step-by-Step Solution

1

Apply the Remainder Theorem

To find the remainder when

P(x)

is divided by

x - a

, set

x - 1 = 0

, giving

x = 1

. The remainder is equal to

P(1)

.

By the Remainder Theorem, dividing a polynomial

P(x)

by

(x - a)

yields a remainder of

P(a)

.

2

Substitute

x = 1

into

P(x) = x^3 + 3x^2 - 2x + 4

P(1) = (1)^3 + 3(1)^2 - 2(1) + 4 = 1 + 3 - 2 + 4 = 6

.

Direct evaluation of the expression at

x = 1

yields the numerical value of the remainder.

Key Concept

The Remainder Theorem states that when a polynomial

P(x)

is divided by a linear factor

(x - a)

, the remainder is

P(a)

.

Question 65Question →

If $\frac{4 + 3\sqrt{2}}{3 - 2\sqrt{2}}$ is expressed in the simplified form $a + b\sqrt{2}$ , where $a$ and $b$ are integers, what is the value of $a + b$ ?

Show answer & explanation

Answer: 41

Answer

The value of

a + b

is 41.

To rationalise

\frac{4 + 3\sqrt{2}}{3 - 2\sqrt{2}}

, multiply both numerator and denominator by the conjugate

3 + 2\sqrt{2}

. The denominator becomes

3^2 - (2\sqrt{2})^2 = 9 - 8 = 1

. Expanding the numerator gives

(4)(3) + 4(2\sqrt{2}) + 3\sqrt{2}(3) + 3\sqrt{2}(2\sqrt{2}) = 12 + 8\sqrt{2} + 9\sqrt{2} + 12 = 24 + 17\sqrt{2}

. Comparing with

a + b\sqrt{2}

gives

a = 24

and

b = 17

, so

a + b = 24 + 17 = 41

.

Step-by-Step Solution

1

Multiply the numerator and denominator by the conjugate of the denominator.

The expression becomes

\frac{(4 + 3\sqrt{2})(3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})}

.

Multiplying by the conjugate eliminates surds from the denominator using the difference of two squares identity

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

.

2

Simplify the denominator.

(3)^2 - (2\sqrt{2})^2 = 9 - (4 \times 2) = 9 - 8 = 1

.

Squaring

2\sqrt{2}

yields

2^2 \times (\sqrt{2})^2 = 4 \times 2 = 8

.

3

Expand the numerator.

(4 \times 3) + (4 \times 2\sqrt{2}) + (3\sqrt{2} \times 3) + (3\sqrt{2} \times 2\sqrt{2}) = 12 + 8\sqrt{2} + 9\sqrt{2} + 12 = 24 + 17\sqrt{2}

.

Applying the distributive law and grouping rational terms together and like surd terms together.

4

Identify the values of

a

and

b

and calculate

a + b

.

a = 24

,

b = 17

, so

a + b = 24 + 17 = 41

.

Matching coefficients of the simplified surd form

a + b\sqrt{2}

.

Key Concept

Rationalisation of binomial surd denominators using conjugates

Estimated Time:2m 0s

Question 66Question →

If $(x, y)$ satisfies the simultaneous equations $x + 2y = 5$ and $x^2 + y^2 = 10$ , what is the positive value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The positive value of

x

is 3.

Isolating

x

in the linear equation gives

x = 5 - 2y

. Substituting this expression into

x^2 + y^2 = 10

yields

(5 - 2y)^2 + y^2 = 10

. Expanding gives

25 - 20y + 4y^2 + y^2 = 10 \implies 5y^2 - 20y + 15 = 0

. Dividing all terms by

5

produces

y^2 - 4y + 3 = 0

, which factors as

(y - 1)(y - 3) = 0

, so

y = 1

or

y = 3

. Substituting these into

x = 5 - 2y

gives

x = 3

when

y = 1

and

x = -1

when

y = 3

. The positive value of

x

is 3.

Step-by-Step Solution

1

Express

x

from the linear equation

x = 5 - 2y

Isolating

x

allows substitution into the quadratic equation.

2

Substitute into the quadratic equation

(5 - 2y)^2 + y^2 = 10

Eliminates variable

x

to create a single-variable equation in

y

.

3

Expand and simplify

5y^2 - 20y + 15 = 0 \implies y^2 - 4y + 3 = 0

Transforms the equation into standard quadratic form for easy factorization.

4

Solve for

y

y = 1 \text{ or } y = 3

Factoring

(y - 1)(y - 3) = 0

yields the two possible values for

y

.

5

Determine corresponding

x

values and select the positive one

x = 3

(from

y = 1

)

Evaluating

x = 5 - 2y

gives

x = 3

and

x = -1

; the positive result requested is 3.

Key Concept

Solving simultaneous linear and quadratic equations by substitution

Estimated Time:1m 30s

Question 67Question →

Given the universal set

\mathcal{E} = \{x \in \mathbb{Z} : 1 \le x \le 20\}

, and subsets:

A = \{x \in \mathcal{E} : x \text{ is a multiple of } 3\}

B = \{x \in \mathcal{E} : x \text{ is a multiple of } 4\}

C = \{x \in \mathcal{E} : x \text{ is a prime number}\}

What is the number of elements in the set $(A \cup B)' \cap C$ ?

7

8

5

10

Show answer & explanation

Answer: 7

Answer

7

The universal set contains integers from 1 to 20. Subset

A

has multiples of 3, subset

B

has multiples of 4, and subset

C

contains the prime numbers

\{2, 3, 5, 7, 11, 13, 17, 19\}

. The set

A \cup B

contains all multiples of 3 or 4 within the range, namely

\{3, 4, 6, 8, 9, 12, 15, 16, 18, 20\}

. The complement

(A \cup B)'

consists of elements in the universal set not in

A \cup B

:

\{1, 2, 5, 7, 10, 11, 13, 14, 17, 19\}

. Taking the intersection of

(A \cup B)'

with

C

filters this list to only the prime numbers:

\{2, 5, 7, 11, 13, 17, 19\}

, giving a cardinality of 7.

Step-by-Step Solution

1

List the elements of the universal set and the given subsets.

\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}

A = \{3, 6, 9, 12, 15, 18\}

B = \{4, 8, 12, 16, 20\}

C = \{2, 3, 5, 7, 11, 13, 17, 19\}

Explicit listing allows accurate evaluation of set operations.

2

Find the union

A \cup B

.

A \cup B = \{3, 4, 6, 8, 9, 12, 15, 16, 18, 20\}

The union combines all elements present in either

A

or

B

.

3

Determine the complement

(A \cup B)'

relative to

\mathcal{E}

.

(A \cup B)' = \{1, 2, 5, 7, 10, 11, 13, 14, 17, 19\}

The complement contains all elements of the universal set

\mathcal{E}

that are not in

A \cup B

.

4

Find the intersection

(A \cup B)' \cap C

and count its cardinality.

(A \cup B)' \cap C = \{2, 5, 7, 11, 13, 17, 19\}

, which contains 7 elements.

The intersection yields elements common to both

(A \cup B)'

and prime set

C

.

Key Concept

Set Complement and Intersections

Question 68Question →

A binary operation $\Delta$ defined on the set of rational numbers $\mathbb{Q}$ is given by $a \Delta b = \frac{ab}{4}$ . What is the inverse element of $6$ under this operation?

\frac{2}{3}

\frac{8}{3}

\frac{1}{6}

\frac{3}{8}

Show answer & explanation

Answer: $\frac{8}{3}$

Answer

The inverse element of

6

under the given binary operation is

\frac{8}{3}

.

To find the inverse of an element under a binary operation, the identity element

e

must first be found using

a \Delta e = a

, which yields

\frac{ae}{4} = a \implies e = 4

. Then, setting

6 \Delta x = 4

gives

\frac{6x}{4} = 4

, which simplifies to

3x = 8

, giving the inverse

x = \frac{8}{3}

.

Step-by-Step Solution

1

Find the identity element

e

of the binary operation.

e = 4

By definition of identity element,

a \Delta e = a

. Substituting into the definition gives

\frac{ae}{4} = a \implies ae = 4a \implies e = 4

.

2

Set up the inverse equation for the element

6

.

6 \Delta x = 4

, where

x

is the inverse of

6

.

By definition of inverse element,

a \Delta a^{-1} = e

.

3

Solve for the inverse

x

.

x = \frac{8}{3}

Applying the operation rule:

\frac{6x}{4} = 4 \implies \frac{3x}{2} = 4 \implies 3x = 8 \implies x = \frac{8}{3}

.

Key Concept

Identity and Inverse Elements of a Binary Operation

Question 69Question →

In a group of 100 candidates preparing for an entrance examination, 48 registered for Mathematics, 45 for Physics, and 40 for Chemistry. If 18 candidates registered for both Mathematics and Physics, 15 for both Physics and Chemistry, 20 for both Mathematics and Chemistry, and 8 registered for none of these three subjects, how many candidates registered for all three subjects?

Show answer & explanation

Answer: 12; 12 candidates; 12 students

Answer

12 candidates registered for all three subjects.

By the Principle of Inclusion-Exclusion, the total number of candidates taking at least one subject is

100 - 8 = 92

. Summing the individual totals gives

48 + 45 + 40 = 133

. Subtracting the pairwise intersections gives

133 - (18 + 15 + 20) = 80

. Adding the intersection of all three sets must equal 92, yielding

92 - 80 = 12

.

Step-by-Step Solution

1

Determine the total number of candidates who registered for at least one of the three subjects.

n(M \cup P \cup C) = 100 - 8 = 92

Subtracting the number of candidates who registered for none of the subjects from the universal set gives the cardinality of the union.

2

Apply the Principle of Inclusion-Exclusion for three sets.

n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C)

This formula relates the individual set sizes, pair intersections, and triple intersection to the union.

3

Substitute the known values into the inclusion-exclusion equation.

92 = 48 + 45 + 40 - 18 - 15 - 20 + n(M \cap P \cap C)

Insert the given cardinalities into the formula.

4

Simplify and solve for

n(M \cap P \cap C)

.

92 = 80 + n(M \cap P \cap C) \implies n(M \cap P \cap C) = 92 - 80 = 12

Isolate the unknown triple intersection term.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Question 70Question →

A binary operation $\circ$ on the set of real numbers $\mathbb{R}$ is defined by $a \circ b = a + b + 2ab$ . If the identity element of the operation is $e$ , what is the value of $x$ such that the inverse of $x$ under $\circ$ is equal to $2$ ?

Show answer & explanation

Answer: -0.4

Answer

The value of

x

is

-0.4

.

First find the identity element

e

by solving

a \circ e = a

, which yields

a + e + 2ae = a \implies e(1 + 2a) = 0 \implies e = 0

. Next, by definition of an inverse,

x \circ x^{-1} = e

. Substituting

x^{-1} = 2

and

e = 0

gives

x \circ 2 = 0

. Expanding this using the binary operation rule yields

x + 2 + 2(x)(2) = 0 \implies 5x + 2 = 0 \implies x = -0.4

.

Step-by-Step Solution

1

Find the identity element

e

of the operation

\circ

e = 0

By definition of identity element,

a \circ e = a \implies a + e + 2ae = a

, which simplifies to

e(1 + 2a) = 0

, giving

e = 0

.

2

Set up the inverse equation using

x^{-1} = 2

x \circ 2 = 0

The inverse of

x

satisfies

x \circ x^{-1} = e

. Since

x^{-1} = 2

and

e = 0

,

x \circ 2 = 0

.

3

Solve for

x

x = -0.4

Expanding

x \circ 2 = 0

gives

x + 2 + 4x = 0 \implies 5x = -2 \implies x = -0.4

.

Key Concept

Identity and Inverse Elements in Binary Operations

Question 71Question →

If $2x + y = 7$ and $x^2 + xy = 6$ , what are the possible values of $x$ ?

1

or

6

-1

or

-6

2

or

3

-2

or

-3

Show answer & explanation

Answer: $1$ or $6$

Answer

The possible values of

x

are

1

or

6

.

From the linear equation

2x + y = 7

, we get

y = 7 - 2x

. Substituting this into

x^2 + xy = 6

gives

x^2 + x(7 - 2x) = 6

, which simplifies to

-x^2 + 7x = 6

, or

x^2 - 7x + 6 = 0

. Factoring gives

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

, leading to

x = 1

or

x = 6

.

Step-by-Step Solution

1

Express

y

in terms of

x

using the linear equation.

y = 7 - 2x

Isolation of one variable allows substitution into the non-linear equation.

2

Substitute

y = 7 - 2x

into the second equation

x^2 + xy = 6

.

x^2 + x(7 - 2x) = 6 \implies x^2 + 7x - 2x^2 = 6 \implies -x^2 + 7x - 6 = 0

This reduces the system to a single quadratic equation in

x

.

3

Rearrange into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - 7x + 6 = 0

Standard quadratic form enables easy factorization.

4

Factor the quadratic equation and solve for

x

.

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

Setting each factor to zero yields the values of

x

.

Key Concept

Solving simultaneous linear and quadratic equations by substitution

Question 72Question →

The polynomial $P(x) = 2x^3 + px^2 + qx - 6$ has $(x - 2)$ as a factor. When $P(x)$ is divided by $(x + 1)$ , the remainder is $-12$ . What is the value of $p + q$ ?

-2

-8

4

-4

Show answer & explanation

Answer: $-2$

Answer

The value of

p + q

is

-2

.

Using the Factor Theorem,

P(2) = 0

yields

2p + q = -5

. Using the Remainder Theorem,

P(-1) = -12

yields

p - q = -4

. Solving these two equations simultaneously gives

p = -3

and

q = 1

. Therefore,

p + q = -3 + 1 = -2

.

Step-by-Step Solution

1

Apply the Factor Theorem for divisor

(x - 2)

2p + q = -5

Since

(x - 2)

is a factor of

P(x)

,

P(2) = 0

. Substituting

x = 2

gives

2(2)^3 + p(2)^2 + q(2) - 6 = 0 \implies 16 + 4p + 2q - 6 = 0 \implies 4p + 2q = -10 \implies 2p + q = -5

.

2

Apply the Remainder Theorem for divisor

(x + 1)

p - q = -4

Dividing

P(x)

by

(x + 1)

leaves a remainder of

-12

, so

P(-1) = -12

. Substituting

x = -1

gives

2(-1)^3 + p(-1)^2 + q(-1) - 6 = -12 \implies -2 + p - q - 6 = -12 \implies p - q = -4

.

3

Solve the simultaneous linear equations for

p

and

q

p = -3

and

q = 1

Adding the two equations

(2p + q) + (p - q) = -5 + (-4)

yields

3p = -9 \implies p = -3

. Substituting

p = -3

into

p - q = -4

gives

-3 - q = -4 \implies q = 1

.

4

Calculate the required expression

p + q

p + q = -2

Summing the calculated constants:

p + q = -3 + 1 = -2

.

Key Concept

Factor and Remainder Theorems for Polynomials

Estimated Time:2m 0s

Question 73Question →

A binary operation $\ast$ is defined on the set of real numbers $\mathbb{R}$ by $a \ast b = a + b - k$ , where $k$ is a constant. If the identity element of the operation is $4$ , what is the inverse of $7$ under this operation?

1

-1

-3

7

Show answer & explanation

Answer: 1

Answer

The inverse of 7 under the binary operation is 1.

First, use the identity property a * e = a with e = 4 to determine that k = 4. Then, set 7 * x = 4 using the operational definition 7 + x - 4 = 4. Solving the linear equation x + 3 = 4 gives the correct inverse value of 1.

Step-by-Step Solution

1

Determine the value of the constant k using the identity element property.

k = 4

By definition of identity element e, a * e = a. Given e = 4, substituting into the definition yields a + 4 - k = a, which simplifies to k = 4.

2

Write the full operational formula.

a * b = a + b - 4

Substitute k = 4 into the original rule a * b = a + b - k.

3

Solve for the inverse element of 7.

x = 1

Let x be the inverse of 7. By definition of inverse, 7 * x = e, so 7 + x - 4 = 4. Simplifying gives x + 3 = 4, hence x = 1.

Key Concept

Identity and inverse elements of a binary operation

Question 74Question →

If $(x, y)$ satisfies the simultaneous equations $x - y = 1$ and $x^2 + y^2 = 25$ , what is the value of the product $xy$ ?

Show answer & explanation

Answer: 12

Answer

12

Expanding

(x - y)^2

yields

x^2 - 2xy + y^2

. Substituting the given values

x - y = 1

and

x^2 + y^2 = 25

into this identity gives

1 = 25 - 2xy

. Rearranging yields

2xy = 24

, which solves to

xy = 12

. Alternatively, solving by substitution gives solution pairs

(4, 3)

and

(-3, -4)

, both yielding a product of

12

.

Step-by-Step Solution

1

Apply the algebraic expansion identity

(x - y)^2 = x^2 + y^2 - 2xy

Connects the difference of terms, the sum of their squares, and their product.

2

Substitute the values given in the system of equations

1^2 = 25 - 2xy

Replaces

x - y

with 1 and

x^2 + y^2

with 25.

3

Isolate and calculate the product

xy

2xy = 24 \implies xy = 12

Simplifies

1 = 25 - 2xy

to find the exact numerical value of

xy

.

Key Concept

Simultaneous Linear and Quadratic Equations

Question 75Question →

Given the universal set $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 30\}$ , let $A = \{x \in \mathcal{U} : x \text{ is a multiple of } 4\}$ and $B = \{x \in \mathcal{U} : x \text{ is a perfect square}\}$ . What is the number of elements in $(A \cup B)'$ ?

20

18

10

22

Show answer & explanation

Answer: 20

Answer

The cardinality of

(A \cup B)'

is 20.

The universal set has 30 elements. Set

A

has 7 elements (multiples of 4 up to 30) and set

B

has 5 elements (perfect squares up to 30). The numbers 4 and 16 belong to both sets, so the intersection has 2 elements. By the principle of inclusion-exclusion, the union

A \cup B

contains

7 + 5 - 2 = 10

elements. The complement

(A \cup B)'

contains all elements in the universal set that are not in the union, giving

30 - 10 = 20

elements.

Step-by-Step Solution

1

Determine the elements and cardinality of the universal set

\mathcal{U}

.

\mathcal{U} = \{1, 2, 3, \dots, 30\}

, so

n(\mathcal{U}) = 30

.

The universal set bounds all possible elements under consideration.

2

List the elements of set

A

and set

B

, and find their individual cardinalities.

A = \{4, 8, 12, 16, 20, 24, 28\} \implies n(A) = 7

.

B = \{1, 4, 9, 16, 25\} \implies n(B) = 5

.

Identify multiples of 4 and perfect squares within the range 1 to 30.

3

Find the intersection

A \cap B

and compute the cardinality of the union

A \cup B

.

A \cap B = \{4, 16\} \implies n(A \cap B) = 2

.

n(A \cup B) = n(A) + n(B) - n(A \cap B) = 7 + 5 - 2 = 10

.

Apply the principle of inclusion-exclusion to avoid double-counting elements belonging to both sets.

4

Calculate the cardinality of the complement

(A \cup B)'

.

n((A \cup B)') = n(\mathcal{U}) - n(A \cup B) = 30 - 10 = 20

.

The complement set consists of all elements in the universal set that are not in

A \cup B

.

Key Concept

Complement of Set Union and Inclusion-Exclusion Principle

Question 76Question →

In a agricultural survey of 120 farmers in a community, 65 grow maize, 50 grow yam, and 42 grow cassava. Furthermore, 24 grow both maize and yam, 18 grow both maize and cassava, and 15 grow both yam and cassava. If 8 farmers grow none of these three crops, find the number of farmers who grow all three crops.

Show answer & explanation

Answer: 12

Answer

12 farmers grow all three crops.

Using the 3-set inclusion-exclusion principle, the total number of farmers growing at least one crop is

120 - 8 = 112

. Expanding

|M \cup Y \cup C| = |M| + |Y| + |C| - (|M \cap Y| + |M \cap C| + |Y \cap C|) + |M \cap Y \cap C|

gives

112 = 65 + 50 + 42 - 24 - 18 - 15 + x

. Simplifying yields

112 = 100 + x

, which gives

x = 12

.

Step-by-Step Solution

1

Determine the cardinality of the union of all three sets.

|M \cup Y \cup C| = 112

Subtract the farmers who grow none of the crops from the universal set size (

120 - 8 = 112

).

2

Set up the Inclusion-Exclusion equation for three sets.

112 = 65 + 50 + 42 - (24 + 18 + 15) + x

Inclusion-exclusion states that

|A \cup B \cup C| = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)

.

3

Solve for the unknown value

x

representing the intersection of all three sets.

x = 12

Simplifying gives

112 = 100 + x

, which leads directly to

x = 12

.

Key Concept

Principle of Inclusion-Exclusion for 3 Sets

Question 77Question →

Let the universal set $\mathcal{E} = \{x \in \mathbb{Z} : 1 \le x \le 25\}$ . Two subsets $A$ and $B$ of $\mathcal{E}$ are defined as $A = \{x \in \mathcal{E} : x \text{ is a perfect square}\}$ and $B = \{x \in \mathcal{E} : x \text{ is an odd number}\}$ . What is the cardinality of $(A \cup B)'$ ?

10

12

15

7

Show answer & explanation

Answer: 10

Answer

10

The correct answer is 10 because the universal set has 25 elements. The union

A \cup B

contains all 13 odd numbers in the range along with the 2 even perfect squares (

4

and

16

), giving

n(A \cup B) = 15

. Subtracting this from the universal set size gives

25 - 15 = 10

.

Step-by-Step Solution

1

Identify the elements of the universal set and subsets A and B

\mathcal{E} = \{1, 2, 3, \dots, 25\}

with

n(\mathcal{E}) = 25

.

A = \{1, 4, 9, 16, 25\}

B = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25\}

Explicitly listing elements clarifies overlapping regions.

2

Find the union set

A \cup B

and its cardinality

A \cup B = \{1, 3, 4, 5, 7, 9, 11, 13, 15, 16, 17, 19, 21, 23, 25\}

, so

n(A \cup B) = 15

.

The union includes all odd numbers from 1 to 25 plus the even perfect squares (4 and 16).

3

Calculate the cardinality of the complement

(A \cup B)'

n((A \cup B)') = n(\mathcal{E}) - n(A \cup B) = 25 - 15 = 10

.

The complement set contains all elements in

\mathcal{E}

that are neither odd nor perfect squares, which are the 10 non-square even numbers:

\{2, 6, 8, 10, 12, 14, 18, 20, 22, 24\}

.

Key Concept

Set Complements and De Morgan's Laws / Set Operations

Estimated Time:1m 30s

Question 78Question →

Consider the system of simultaneous equations $x - 3y = 2$ and $x^2 - 2xy - 4y^2 = 19$ . What is the sum of the $x$ -values of the real solution pairs $(x, y)$ ?

28

8

20

-20

Show answer & explanation

Answer: $28$

Answer

The sum of the

x

-values of the real solution pairs is

28

.

Rearranging the linear equation gives

x = 3y + 2

. Substituting this into

x^2 - 2xy - 4y^2 = 19

results in

(3y + 2)^2 - 2(3y + 2)y - 4y^2 = 19

, which simplifies to

y^2 - 8y + 15 = 0

. The roots are

y = 5

and

y = 3

. Substituting these back into

x = 3y + 2

yields

x = 17

and

x = 11

. Their sum is

17 + 11 = 28

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the linear equation.

x = 3y + 2

Isolation of

x

allows direct substitution into the quadratic equation.

2

Substitute

x = 3y + 2

into the quadratic equation

x^2 - 2xy - 4y^2 = 19

.

(3y + 2)^2 - 2(3y + 2)y - 4y^2 = 19

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

y

.

3

Expand and simplify the quadratic equation.

(9y^2 + 12y + 4) - (6y^2 + 4y) - 4y^2 = 19 \implies -y^2 + 8y + 4 = 19 \implies y^2 - 8y + 15 = 0

Putting the quadratic expression into standard form

ay^2 + by + c = 0

facilitates finding its roots.

4

Solve the quadratic equation

y^2 - 8y + 15 = 0

for

y

.

(y - 5)(y - 3) = 0 \implies y_1 = 5, y_2 = 3

Factoring determines the ordinate values for the solution pairs.

5

Calculate the corresponding

x

-values and find their sum.

For

y_1 = 5

:

x_1 = 3(5) + 2 = 17

.
For

y_2 = 3

:

x_2 = 3(3) + 2 = 11

.
Sum =

17 + 11 = 28

.

Plugging the

y

-values back into

x = 3y + 2

gives the abscissas, which are added to answer the question.

Key Concept

Solving simultaneous linear and quadratic equations by algebraic substitution

Estimated Time:2m 0s

Question 79Question →

If $x - 1$ , $x + 2$ , and $3x$ are three consecutive terms of a geometric progression (G.P.) with positive terms, what is the common ratio of the progression?

2

4

3

\frac{1}{2}

Show answer & explanation

Answer: $2$

Answer

The common ratio of the progression is

2

.

For any three consecutive terms in a G.P., the square of the middle term equals the product of the first and third terms. Solving

(x+2)^2 = (x-1)(3x)

gives

2x^2 - 7x - 4 = 0

, which yields

x = 4

for positive terms. Substituting

x = 4

gives the terms

3, 6, 12

, which have a common ratio of

6 \div 3 = 2

.

Step-by-Step Solution

1

Set up the condition for consecutive terms in a Geometric Progression.

(x + 2)^2 = (x - 1)(3x)

For three consecutive terms

a, b, c

in G.P., the middle term squared equals the product of the outer terms (

b^2 = ac

).

2

Expand both sides and rearrange into a quadratic equation.

x^2 + 4x + 4 = 3x^2 - 3x \implies 2x^2 - 7x - 4 = 0

Expanding allows gathering all terms on one side to solve for

x

.

3

Factorize the quadratic equation to find

x

.

(2x + 1)(x - 4) = 0 \implies x = 4

(since terms are positive,

x > 1

).

The solution

x = -\frac{1}{2}

gives negative terms, so

x = 4

is chosen.

4

Find the consecutive terms and calculate the common ratio

r

.

Terms are

4 - 1 = 3

,

4 + 2 = 6

, and

3(4) = 12

. Common ratio

r = \frac{6}{3} = 2

.

Dividing the second term by the first term gives the common ratio

r

.

Key Concept

Geometric Progression Consecutive Terms Property (

b^2 = ac

)

Estimated Time:1m 30s

Question 80Question →

When the polynomial $P(x) = 3x^3 + ax^2 + bx - 10$ is divided by $(x - 2)$ , the remainder is $14$ , and when it is divided by $(x + 1)$ , the remainder is $-16$ . What is the value of $a + b$ ?

Show answer & explanation

Answer: 1

Answer

The value of

a + b

is

1

.

According to the Remainder Theorem, dividing

P(x)

by

(x - 2)

and

(x + 1)

gives

P(2) = 14

and

P(-1) = -16

respectively. Expanding these expressions forms two linear equations:

2a + b = 0

and

a - b = -3

. Solving these simultaneously gives

a = -1

and

b = 2

, so

a + b = 1

.

Step-by-Step Solution

1

Apply the Remainder Theorem for the first divisor

(x - 2)

2a + b = 0

By the Remainder Theorem,

P(2) = 14

. Substituting

x = 2

into

P(x)

yields

3(8) + 4a + 2b - 10 = 14

, which simplifies to

2a + b = 0

.

2

Apply the Remainder Theorem for the second divisor

(x + 1)

a - b = -3

By the Remainder Theorem,

P(-1) = -16

. Substituting

x = -1

into

P(x)

yields

3(-1) + a - b - 10 = -16

, which simplifies to

a - b = -3

.

3

Solve the simultaneous equations for

a

and

b

a = -1

and

b = 2

Adding

2a + b = 0

and

a - b = -3

yields

3a = -3

, giving

a = -1

. Substituting

a = -1

into

2a + b = 0

gives

b = 2

.

4

Calculate the value of

a + b

1

Adding the computed values yields

a + b = -1 + 2 = 1

.

Key Concept

Polynomial Remainder Theorem and Systems of Linear Equations

Estimated Time:2m 0s

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