Algebra Practice Questions — JAMB UTME — Page 7

Question 121Question →

The total cost $C$ per trip of operating a high-speed passenger ferry consists of a fixed administrative overhead cost and an operational cost that varies directly as the cube of its speed $v$ in knots. Given that the total cost per trip is $\$1,400$ when the ferry travels at $10\text{ knots}$ and $\$4,200$ when it travels at $20\text{ knots}$ , what is the total cost per trip when the ferry operates at a speed of $15\text{ knots}$ ?

\\

2,350$

\\

2,800$

\\

3,300$

\\

4,725$

Show answer & explanation

Answer: $\\$ 2,350$

Answer

\\

2,350$

The partial variation formula is

C = k_1 + k_2 v^3

. Substituting the given points yields

k_1 + 1000 k_2 = 1400

and

k_1 + 8000 k_2 = 4200

. Subtracting these equations gives

7000 k_2 = 2800

, so

k_2 = 0.4

and

k_1 = 1000

. Evaluating at

v = 15

gives

C = 1000 + 0.4(3375) = 1000 + 1350 = \\

2,350$.

Step-by-Step Solution

1

Set up the partial variation equation

C = k_1 + k_2 v^3

Partial variation combines a fixed constant

k_1

with a variable term

k_2 v^3

.

2

Form simultaneous linear equations using given conditions

Equation (1):

k_1 + 1000 k_2 = 1400

; Equation (2):

k_1 + 8000 k_2 = 4200

Substitute

v = 10

,

C = 1400

and

v = 20

,

C = 4200

into the variation model.

3

Solve for the constants

k_1

and

k_2

k_2 = 0.4

and

k_1 = 1000

Subtract Equation (1) from Equation (2):

7000 k_2 = 2800 \implies k_2 = 0.4

. Substitute

k_2 = 0.4

into Equation (1) to get

k_1 = 1400 - 400 = 1000

.

4

Calculate total cost

C

for speed

v = 15\text{ knots}

C = 1000 + 0.4(15)^3 = 1000 + 0.4(3375) = 1000 + 1350 = 2350

Substitute

k_1 = 1000

,

k_2 = 0.4

, and

v = 15

back into the formula.

Key Concept

Partial Variation with non-linear powers solved via simultaneous linear equations

Estimated Time:2m 0s

Question 122Question →

How many positive integer values of $x$ satisfy the quadratic inequality $x^2 - 4x - 5 < 0$ ?

Show answer & explanation

Answer: 4

Answer

There are 4 positive integer values of

x

that satisfy the inequality.

Factoring

x^2 - 4x - 5 < 0

yields

(x - 5)(x + 1) < 0

, giving the real solution interval

-1 < x < 5

. Filtering for positive integers (

x \in \mathbb{Z}^+

) yields the set

\{1, 2, 3, 4\}

, which contains exactly 4 values.

Step-by-Step Solution

1

Factor the quadratic inequality

(x - 5)(x + 1) < 0

Factoring helps find the boundary roots of the quadratic equation.

2

Determine the solution set interval

-1 < x < 5

For a quadratic inequality of the form

(x - a)(x - b) < 0

with

a < b

, the solution interval is

a < x < b

.

3

List the positive integer solutions in the range

-1 < x < 5

x \in \{1, 2, 3, 4\}

Positive integers are whole numbers strictly greater than 0.

4

Count the number of positive integers

4

Counting the elements in the set

\{1, 2, 3, 4\}

gives a total of 4.

Key Concept

Solving quadratic inequalities and finding valid integer solutions within a target domain.

Estimated Time:45s

Question 123Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 40\}$ . Subsets $P$ , $Q$ , and $R$ of $\mathcal{U}$ are defined as follows:
$P = \{x \in \mathcal{U} : x \text{ is a multiple of } 3\}$ ,
$Q = \{x \in \mathcal{U} : x \text{ is a multiple of } 4\}$ , and
$R = \{x \in \mathcal{U} : x \text{ is a perfect square\}}$ .
What is the cardinality of the set $(P \cup Q)' \cap R'$ ?

18

22

20

17

Show answer & explanation

Answer: 18

Answer

18

By De Morgan's Law, the expression

(P \cup Q)' \cap R'

is equivalent to

(P \cup Q \cup R)'

. The universal set contains 40 integers. The set

P \cup Q \cup R

consists of all integers from 1 to 40 that are multiples of 3, multiples of 4, or perfect squares, giving a total of 22 distinct elements. Subtracting this from 40 leaves 18 elements that belong to neither

P

,

Q

, nor

R

.

Step-by-Step Solution

1

Apply De Morgan's Laws to simplify the required set expression

(P \cup Q)' \cap R' = (P \cup Q \cup R)'

The intersection of the complements of two sets is equal to the complement of their union.

2

List and count the elements belonging to sets

P

,

Q

, and

R

within

\mathcal{U} = \{1, 2, \dots, 40\}

P

has 13 elements (multiples of 3);

Q

adds 7 elements not in

P

(multiples of 4 excluding 12, 24, 36);

R

adds 2 elements not in

P \cup Q

(

1

and

25

)

Counting unique elements avoids double-counting shared elements across sets.

3

Calculate the cardinality of

P \cup Q \cup R

n(P \cup Q \cup R) = 13 + 7 + 2 = 22

Summing the non-overlapping subsets gives the total size of the union.

4

Subtract the union cardinality from the universal set cardinality

n(\mathcal{U}) = 40

n((P \cup Q \cup R)') = 40 - 22 = 18

The cardinality of the complement is the size of the universal set minus the size of the set itself.

Key Concept

De Morgan's Laws and Set Complements

Question 124Question →

Determine the number of integer values of $x$ that satisfy the quadratic inequality $3x^2 - 14x - 5 < 0$ .

Show answer & explanation

Answer: 5

Answer

The total number of integer values satisfying the inequality is 5.

Factoring

3x^2 - 14x - 5 < 0

gives

(3x + 1)(x - 5) < 0

. The critical roots are

x = -\frac{1}{3}

and

x = 5

. Since the parabola opens upward, the expression is negative between the roots, yielding the interval

-\frac{1}{3} < x < 5

. The integer values contained in this interval are

0, 1, 2, 3,

and

4

, which total 5 values.

Step-by-Step Solution

1

Factor the quadratic expression

(3x + 1)(x - 5) < 0

Factoring identifies the critical values where the quadratic expression changes sign.

2

Determine the critical points

x = -\frac{1}{3}

and

x = 5

Setting each factor to zero gives the boundary roots of the equation.

3

Formulate the solution interval

-\frac{1}{3} < x < 5

Since the coefficient of

x^2

is positive, the quadratic curve is below the x-axis strictly between the two roots.

4

List and count the integer solutions

The integers are

0, 1, 2, 3, 4

, yielding 5 integer solutions.

Counting integers strictly greater than

-\frac{1}{3}

and strictly less than

5

.

Key Concept

Solving quadratic inequalities and finding integer solution counts

Estimated Time:1m 30s

Question 125Question →

The monthly operating cost $C$ (in naira) of a commercial power generator is partly constant and partly varies directly as the square of its daily operating time $h$ (in hours). When the generator operates for $3$ hours per day, the monthly cost is ₦ $5,500$ . When it operates for $5$ hours per day, the monthly cost is ₦ $13,500$ . What is the daily operating time, in hours, when the monthly operating cost is ₦ $33,000$ ?

8 hours

7 hours

9 hours

10 hours

Show answer & explanation

Answer: 8 hours

Answer

8 hours

The relationship is modeled by partial variation

C = k_1 + k_2 h^2

. Solving the simultaneous equations derived from

h = 3

(

C = 5500

) and

h = 5

(

C = 13500

) yields

k_2 = 500

and

k_1 = 1000

. Substituting

C = 33000

into

33000 = 1000 + 500 h^2

gives

500 h^2 = 32000

, so

h^2 = 64

, which yields

h = 8

hours.

Step-by-Step Solution

1

Set up the general formula for partial variation.

C = k_1 + k_2 h^2

, where

k_1

and

k_2

are constants.

The cost consists of a constant part (

k_1

) and a part that varies directly as the square of daily hours (

k_2 h^2

).

2

Form simultaneous linear equations using the given data points.

Equation (1):

5500 = k_1 + 9 k_2

Equation (2):

13500 = k_1 + 25 k_2

Substitute

h = 3, C = 5500

and

h = 5, C = 13500

into the variation equation.

3

Solve for the variation constants

k_1

and

k_2

.

Subtract Equation (1) from Equation (2):

8000 = 16 k_2 \implies k_2 = 500

.
Substitute

k_2 = 500

into Equation (1):

5500 = k_1 + 9(500) \implies k_1 = 1000

.

Eliminating

k_1

gives

k_2

, which is then used to find the constant part

k_1

.

4

Calculate

h

when

C = 33000

.

33000 = 1000 + 500 h^2 \implies 32000 = 500 h^2 \implies h^2 = 64 \implies h = 8

hours.

Substitute the values of

k_1, k_2,

and target

C

into the relation equation to solve for

h

.

Key Concept

Partial Variation with Simultaneous Equations

Estimated Time:2m 30s

Question 126Question →

Find the smallest integer $x$ that satisfies the linear inequality $5 - 3x \le -7$ .

Show answer & explanation

Answer: 4

Answer

The smallest integer value of

x

satisfying the inequality is

4

.

Subtracting 5 from both sides of

5 - 3x \le -7

gives

-3x \le -12

. Dividing both sides by

-3

requires reversing the inequality sign to obtain

x \ge 4

. Therefore, the smallest integer value in the solution set is 4.

Step-by-Step Solution

1

Subtract 5 from both sides of the inequality to isolate the variable term

-3x \le -12

Subtracting a constant from both sides maintains the inequality direction.

2

Divide both sides by

-3

and flip the inequality sign

x \ge 4

Dividing an inequality by a negative number reverses the direction of the inequality symbol.

3

Identify the smallest integer satisfying the condition

4

The solution set contains all real numbers greater than or equal to 4, making 4 the minimum integer value.

Key Concept

Solving Linear Inequalities with Negative Coefficients

Estimated Time:45s

Question 127Question →

Find the sum of all integer values of $x$ that satisfy the system of inequalities $3x + 1 > 7$ and $x^2 - 4x - 21 \le 0$ .

Show answer & explanation

Answer: 25

Answer

25

Solving

3x + 1 > 7

gives

x > 2

. Solving

x^2 - 4x - 21 \le 0

gives

-3 \le x \le 7

. The combined condition is

2 < x \le 7

. The integer values satisfying this condition are 3, 4, 5, 6, and 7. The sum of these integers is

3 + 4 + 5 + 6 + 7 = 25

.

Step-by-Step Solution

1

Solve the linear inequality

3x + 1 > 7

.

x > 2

Subtracting 1 from both sides gives

3x > 6

, and dividing by 3 yields

x > 2

.

2

Solve the quadratic inequality

x^2 - 4x - 21 \le 0

.

-3 \le x \le 7

Factoring the quadratic expression gives

(x - 7)(x + 3) \le 0

. The roots are

-3

and

7

, and the quadratic curve is non-positive between these roots.

3

Determine the intersection of the solution sets

x > 2

and

-3 \le x \le 7

.

2 < x \le 7

The lower bound is determined by the strict linear constraint

x > 2

, while the upper bound is determined by the inclusive quadratic constraint

x \le 7

.

4

List the integer values of

x

within

2 < x \le 7

and compute their sum.

Integers are

3, 4, 5, 6, 7

; their sum is

25

.

The strict inequality excludes

2

, while the inclusive inequality includes

7

.

Key Concept

System of linear and quadratic inequalities

Question 128Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z}^+ : 1 \le x \le 200\}$ . Subsets $A$ , $B$ , and $C$ of $\mathcal{U}$ are defined as follows:
- $A = \{x \in \mathcal{U} : x \text{ is a multiple of } 5\}$
- $B = \{x \in \mathcal{U} : x \text{ is a multiple of } 7\}$
- $C = \{x \in \mathcal{U} : x \text{ is a multiple of } 10\}$

Find the number of elements in the set $(A \cap B') \setminus C$ .

Show answer & explanation

Answer: 17; seventeen

Answer

The number of elements in the set

(A \cap B') \setminus C

is 17.

The set

(A \cap B') \setminus C

consists of elements in

A

that are neither in

B

nor in

C

. Taking

A \setminus C

isolates the 20 odd multiples of 5 between 1 and 200. From these 20 numbers, removing those divisible by 7 leaves only the odd multiples of 35 (35, 105, and 175) to be subtracted, giving

20 - 3 = 17

.

Step-by-Step Solution

1

Simplify the set expression

(A \cap B') \setminus C

.

(A \cap B') \setminus C = (A \setminus C) \setminus B

By set algebra, taking elements in

A

that are not in

B

and then excluding elements in

C

is equivalent to first removing elements of

C

from

A

, and then removing any remaining elements that belong to

B

.

2

Determine the cardinality of

A \setminus C

.

n(A \setminus C) = 20

Set

A

contains all multiples of 5 up to 200, of which there are

\lfloor 200/5 \rfloor = 40

. Set

C

contains all multiples of 10 up to 200, which are the even multiples of 5, amounting to

\lfloor 200/10 \rfloor = 20

. Therefore,

A \setminus C

consists of the odd multiples of 5 up to 200, which gives

40 - 20 = 20

elements.

3

Find which elements of

A \setminus C

are also in set

B

.

The common elements are 35, 105, and 175 (3 elements).

Elements in both

A

and

B

are multiples of

\text{lcm}(5, 7) = 35

. The multiples of 35 up to 200 are 35, 70, 105, 140, and 175. However, 70 and 140 are multiples of 10 (belonging to set

C

) and have already been removed. Thus, only the odd multiples of 35 (35, 105, and 175) remain in

A \setminus C

and also belong to

B

.

4

Subtract these common elements to calculate

n((A \setminus C) \setminus B)

.

20 - 3 = 17

Removing the 3 common elements from the 20 elements of

A \setminus C

leaves 17 elements.

Key Concept

Set difference, intersection with complement, and evaluation of cardinalities using divisibility properties.

Question 129Question →

What is the set of real values of $x$ that satisfies the quadratic inequality $-2x^2 + 5x + 3 \ge 0$ ?

-\frac{1}{2} \le x \le 3

x \le -\frac{1}{2}

or

x \ge 3

-3 \le x \le \frac{1}{2}

x \le -3

or

x \ge \frac{1}{2}

Show answer & explanation

Answer: $-\frac{1}{2} \le x \le 3$

Answer

The set of real values of

x

satisfying the inequality is

-\frac{1}{2} \le x \le 3

.

Multiplying

-2x^2 + 5x + 3 \ge 0

by

-1

gives

2x^2 - 5x - 3 \le 0

. Factorizing yields

(2x + 1)(x - 3) \le 0

. The product is non-positive between the roots

x = -\frac{1}{2}

and

x = 3

, giving

-\frac{1}{2} \le x \le 3

.

Step-by-Step Solution

1

Multiply or divide the inequality by

-1

to make the leading coefficient positive.

2x^2 - 5x - 3 \le 0

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

2

Factorize the quadratic expression

2x^2 - 5x - 3

.

(2x + 1)(x - 3) \le 0

Splitting the middle term

-5x

into

-6x + x

gives

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

.

3

Find the critical points by setting the expression equal to zero.

x = -\frac{1}{2}

and

x = 3

Critical points mark the boundaries where the sign of the quadratic expression changes.

4

Determine the interval satisfying the inequality

(2x + 1)(x - 3) \le 0

.

-\frac{1}{2} \le x \le 3

A quadratic expression with a positive coefficient of

x^2

is less than or equal to zero between its roots.

Key Concept

Quadratic Inequalities and Sign Reversal

Estimated Time:2m 0s

Question 130Question →

What is the solution set for the linear inequality $4 - 3x \ge 19$ ?

x \le -5

x \ge -5

x \le 5

x \ge 5

Show answer & explanation

Answer: $x \le -5$

Answer

x \le -5

Subtracting

4

from both sides simplifies the inequality to

-3x \ge 15

. Dividing both sides by

-3

requires reversing the direction of the inequality sign, yielding

x \le -5

.

Step-by-Step Solution

1

Subtract 4 from both sides of the inequality

-3x \ge 15

Isolate the variable term

-3x

on the left side of the inequality.

2

Divide both sides by

-3

and reverse the inequality direction

x \le -5

Dividing an inequality by a negative number flips the inequality symbol from

\ge

to

\le

.

Key Concept

Reversing inequality signs when dividing by a negative quantity

Estimated Time:45s

Question 131Question →

When the cubic polynomial $P(x) = 2x^3 + px^2 + qx + 6$ is divided by $x^2 - 1$ , the remainder is $3x + 2$ . What is the remainder when $P(x)$ is divided by $2x - 3$ ?

\frac{21}{4}

-\frac{45}{4}

-\frac{15}{4}

27

Show answer & explanation

Answer: $\frac{21}{4}$

Answer

The remainder when

P(x)

is divided by

2x - 3

is

\frac{21}{4}

.

Using the Remainder Theorem on the quadratic divisor

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

, we find

P(1) = 5

and

P(-1) = -1

. Substituting these values into

P(x) = 2x^3 + px^2 + qx + 6

produces the simultaneous linear equations

p + q = -3

and

p - q = -5

. Solving gives

p = -4

and

q = 1

, leading to

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

. Dividing by

(2x - 3)

requires evaluating

P\left(\frac{3}{2}\right)

, which equals

\frac{21}{4}

.

Step-by-Step Solution

1

Express the Division Algorithm for quadratic divisor

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

.

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

By the Remainder Theorem, evaluating at roots of

x^2 - 1 = 0

(

x = 1

and

x = -1

) yields the values of

P(1)

and

P(-1)

.

2

Calculate

P(1)

and

P(-1)

from the remainder expression.

P(1) = 3(1) + 2 = 5

and

P(-1) = 3(-1) + 2 = -1

The quotient term

(x^2 - 1)Q(x)

vanishes at

x = 1

and

x = -1

.

3

Substitute

x = 1

and

x = -1

into

P(x) = 2x^3 + px^2 + qx + 6

to build a system of linear equations.

P(1) = 2 + p + q + 6 = p + q + 8 = 5 \implies p + q = -3

, and

P(-1) = -2 + p - q + 6 = p - q + 4 = -1 \implies p - q = -5

This sets up two simultaneous linear equations in terms of

p

and

q

.

4

Solve the system of equations for

p

and

q

.

Adding the equations gives

2p = -8 \implies p = -4

, and substituting into

p + q = -3

gives

q = 1

. Thus,

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

.

Determining

p

and

q

gives the explicit formula for the polynomial.

5

Apply the Remainder Theorem to find the remainder when

P(x)

is divided by

2x - 3

.

Set

2x - 3 = 0 \implies x = \frac{3}{2}

. Evaluate

P\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^3 - 4\left(\frac{3}{2}\right)^2 + \frac{3}{2} + 6 = 2\left(\frac{27}{8}\right) - 4\left(\frac{9}{4}\right) + \frac{3}{2} + 6 = \frac{27}{4} - 9 + \frac{6}{4} + 6 = \frac{33}{4} - 3 = \frac{21}{4}

.

The remainder of a polynomial

P(x)

divided by

(ax - b)

is

P\left(\frac{b}{a}\right)

.

Key Concept

Polynomial Division Algorithm and Remainder Theorem for Linear and Quadratic Divisors

Question 132Question →

Let the universal set be $\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and set $P = \{2, 3, 5, 7\}$ . Which of the following is the complement of set $P$ , denoted by $P'$ ?

\{1, 4, 6, 8, 9, 10\}

\{1, 3, 5, 7, 9\}

\{4, 6, 8, 10\}

\{2, 3, 5, 7\}

Show answer & explanation

Answer: $\{1, 4, 6, 8, 9, 10\}$

Answer

\{1, 4, 6, 8, 9, 10\}

The complement of set

P

, denoted as

P'

, consists of every element in the universal set

\mathcal{U}

that is not present in set

P

. Removing

2, 3, 5, 7

from

\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

leaves

\{1, 4, 6, 8, 9, 10\}

.

Step-by-Step Solution

1

Identify elements of the universal set

\mathcal{U}

and set

P

\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

and

P = \{2, 3, 5, 7\}

The complement set requires comparing set

P

against the universal set.

2

Remove elements of set

P

from

\mathcal{U}

to find

P'

P' = \mathcal{U} \setminus P = \{1, 4, 6, 8, 9, 10\}

By definition, the complement

P'

consists of all elements in

\mathcal{U}

that do not belong to

P

.

Key Concept

Set Complement

Question 133Question →

If $\frac{\sqrt{50} + \sqrt{18}}{\sqrt{2}} = m$ , what is the value of the integer $m$ ?

Show answer & explanation

Answer: 8

Answer

The value of the integer

m

is 8.

Simplifying

\sqrt{50}

to

5\sqrt{2}

and

\sqrt{18}

to

3\sqrt{2}

gives a numerator of

8\sqrt{2}

. Dividing

8\sqrt{2}

by

\sqrt{2}

cancels out the radical part, yielding the integer 8.

Step-by-Step Solution

1

Simplify the radical expressions in the numerator.

\sqrt{50} = 5\sqrt{2}

and

\sqrt{18} = 3\sqrt{2}

.

Factor out perfect square numbers from within each radical.

2

Sum the simplified surds in the numerator.

5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}

.

Surds with identical radicands are like terms and can be added by adding their coefficients.

3

Divide the numerator by the denominator.

\frac{8\sqrt{2}}{\sqrt{2}} = 8

.

Cancel the common factor of

\sqrt{2}

present in both numerator and denominator.

Key Concept

Simplification and division of surds

Question 134Question →

The total cost $C$ of printing a school magazine is partly constant and partly varies directly as the number of copies $n$ printed. If it costs $\text{N}70,000$ to print $500$ copies and $\text{N}140,000$ to print $1,200$ copies, what is the cost of printing $2,000$ copies?

\text{N}200,000

\text{N}220,000

\text{N}240,000

\text{N}280,000

Show answer & explanation

Answer: $\text{N}220,000$

Answer

\text{N}220,000

In partial variation, the total cost

C

is represented by

C = k_1 + k_2 n

. Subtracting the two linear equations

70,000 = k_1 + 500 k_2

and

140,000 = k_1 + 1,200 k_2

gives

700 k_2 = 70,000

, so

k_2 = 100

. Substituting

k_2 = 100

yields the fixed constant

k_1 = 20,000

. Substituting

n = 2,000

into

C = 20,000 + 100n

yields

20,000 + 200,000 = \text{N}220,000

.

Step-by-Step Solution

1

Set up the partial variation equation

C = k_1 + k_2 n

, where

k_1

is the fixed cost and

k_2

is the rate per copy

Partial variation consists of a constant term and a term that varies directly with the independent variable.

2

Substitute given values to form simultaneous equations

Equation 1:

70,000 = k_1 + 500 k_2

Equation 2:

140,000 = k_1 + 1,200 k_2

Using the two given data points

(n=500, C=70,000)

and

(n=1,200, C=140,000)

creates a system of linear equations.

3

Solve for the variation constants

k_1

and

k_2

Subtracting Equation 1 from Equation 2 gives

70,000 = 700 k_2 \implies k_2 = 100

.
Substituting

k_2 = 100

into Equation 1 gives

70,000 = k_1 + 500(100) \implies k_1 = 20,000

.

Determines the specific values for the fixed overhead and rate per copy.

4

Calculate the total cost for

2,000

copies

C = 20,000 + 100(2,000) = 20,000 + 200,000 = \text{N}220,000

Evaluates the completed formula

C = 20,000 + 100n

at

n = 2,000

.

Key Concept

Partial Variation and Simultaneous Equations

Estimated Time:1m 30s

Question 135Question →

In a survey of 120 final year secondary school students in Lagos, 47 offer Physics ( $P$ ), 57 offer Chemistry ( $C$ ), and 48 offer Biology ( $B$ ). Records show that 17 students offer both Physics and Chemistry, 20 offer both Chemistry and Biology, and 15 offer both Physics and Biology. If the number of students who offer none of these three subjects is three times the number of students who offer all three subjects, find the total number of students who offer exactly one of the three subjects.

Show answer & explanation

Answer: 63

Answer

63 students offer exactly one of the three subjects.

By applying the principle of inclusion-exclusion for three sets, the total union is

|P \cup C \cup B| = 47 + 57 + 48 - (17 + 20 + 15) + x = 100 + x

, where

x

is the number of students taking all three subjects. Setting the universal set total to 120 gives

(100 + x) + 3x = 120

, solving to

x = 5

. Subtracting the relevant intersections yields 20 students for Physics only, 25 for Chemistry only, and 18 for Biology only. Summing these gives 63.

Step-by-Step Solution

1

Define unknown variables and set up the inclusion-exclusion formula for the union of three sets.

|P \cup C \cup B| = 100 + x

, where

x = |P \cap C \cap B|

.

The Principle of Inclusion-Exclusion states that

|P \cup C \cup B| = |P| + |C| + |B| - (|P \cap C| + |C \cap B| + |P \cap B|) + |P \cap C \cap B|

.

2

Formulate and solve an equation for the total number of students in the universal set.

120 = (100 + x) + 3x \implies 4x = 20 \implies x = 5

.

The total number of students equals those taking at least one subject plus those taking none (

3x

).

3

Determine the number of students taking exactly two subjects.

Physics and Chemistry only =

17 - 5 = 12

; Chemistry and Biology only =

20 - 5 = 15

; Physics and Biology only =

15 - 5 = 10

.

Subtract the triple intersection count (

x = 5

) from each pairwise intersection count.

4

Calculate the number of students offering only Physics, only Chemistry, and only Biology.

Physics only =

47 - (12 + 10 + 5) = 20

; Chemistry only =

57 - (12 + 15 + 5) = 25

; Biology only =

48 - (10 + 15 + 5) = 18

.

Subtract the sum of the two-subject-only regions and the three-subject region from each set's total cardinality.

5

Sum the counts of students offering exactly one subject.

20 + 25 + 18 = 63

.

The total offering exactly one subject is the sum of the three disjoint single-subject regions.

Key Concept

3-Set Principle of Inclusion-Exclusion and Venn Diagram Region Decomposition

Question 136Question →

The electrical resistance $R$ of a wire varies directly as its length $L$ and inversely as the square of its diameter $d$ . A wire of length $50\text{ m}$ and diameter $2\text{ mm}$ has a resistance of $5\text{ ohms}$ . Calculate the resistance (in ohms) of a wire made of the same material with a length of $80\text{ m}$ and a diameter of $4\text{ mm}$ .

Show answer & explanation

Answer: 2

Answer

The resistance of the wire is

2\text{ ohms}

.

The variation model is

R = \frac{k L}{d^2}

. Substituting

R = 5\text{ ohms}

,

L = 50\text{ m}

, and

d = 2\text{ mm}

gives

5 = \frac{50k}{4}

, so

k = 0.4

. Substituting

k = 0.4

,

L = 80\text{ m}

, and

d = 4\text{ mm}

yields

R = \frac{0.4 \times 80}{4^2} = \frac{32}{16} = 2\text{ ohms}

.

Step-by-Step Solution

1

Set up the variation formula

R = \frac{k L}{d^2}

Resistance varies directly as length

L

and inversely as the square of diameter

d

.

2

Calculate the constant of variation

k

k = 0.4

Substitute the initial values

R = 5

,

L = 50

, and

d = 2

into the variation equation.

3

Compute the new resistance

R

R = 2\text{ ohms}

Substitute

k = 0.4

,

L = 80

, and

d = 4

into the formula.

Key Concept

Direct and Inverse Joint Variation

Estimated Time:1m 30s

Question 137Question →

Find the real value of $x$ that satisfies the equation $\log_2 (2^{2x} + 12) - x = 3 - \log_2 (2^x - 1)$ .

1

2

3

0

Show answer & explanation

Answer: 1

Answer

1

By bringing all terms involving

x

together, the equation becomes

\log_2(2^{2x}+12) + \log_2(2^x-1) - \log_2(2^x) = 3

. Combining these logarithmic terms yields

\log_2\left(\frac{(2^{2x}+12)(2^x-1)}{2^x}\right) = 3

. Converting to exponential form gives

\frac{(2^{2x}+12)(2^x-1)}{2^x} = 8

. Setting

u = 2^x

converts this into the cubic equation

u^3 - u^2 + 4u - 12 = 0

. Factoring out

(u - 2)

leaves

(u - 2)(u^2 + u + 6) = 0

. Since

u^2 + u + 6 = 0

has no real solutions,

u = 2

is the only real root. Hence

2^x = 2

, which gives

x = 1

.

Step-by-Step Solution

1

Rearrange the equation to group all logarithmic terms on one side.

\log_2 (2^{2x} + 12) + \log_2 (2^x - 1) - x = 3

Grouping logarithmic terms allows the application of logarithm combination laws.

2

Express

x

as

\log_2 (2^x)

and combine the logarithms using log laws.

\log_2 \left( \frac{(2^{2x} + 12)(2^x - 1)}{2^x} \right) = 3

Product and quotient laws of logarithms allow single logarithmic expression formation:

\log_2 A + \log_2 B - \log_2 C = \log_2\left(\frac{AB}{C}\right)

.

3

Convert from logarithmic to exponential form.

\frac{(2^{2x} + 12)(2^x - 1)}{2^x} = 2^3 = 8

Definition of logarithm:

\log_b(Y) = Z \implies Y = b^Z

.

4

Substitute

u = 2^x

(where

u > 1

) to form an algebraic equation.

\frac{(u^2 + 12)(u - 1)}{u} = 8 \implies (u^2 + 12)(u - 1) = 8u

Since

2^{2x} = (2^x)^2 = u^2

, substitution simplifies the exponential expression into a polynomial.

5

Expand and simplify into a standard cubic equation.

u^3 - u^2 + 12u - 12 = 8u \implies u^3 - u^2 + 4u - 12 = 0

Standard polynomial form enables root searching via polynomial division and factoring.

6

Factor the cubic equation.

(u - 2)(u^2 + u + 6) = 0

Testing

u = 2

gives

2^3 - 2^2 + 4(2) - 12 = 8 - 4 + 8 - 12 = 0

, confirming

(u - 2)

is a factor.

7

Solve for real values of

u

and

x

.

u = 2 \implies 2^x = 2 \implies x = 1

The quadratic factor

u^2 + u + 6 = 0

has discriminant

\Delta = 1 - 24 = -23 < 0

(no real roots). Thus

u = 2

is the unique solution.

Key Concept

Logarithmic equations requiring exponential substitution and polynomial factorization

Estimated Time:2m 30s

Question 138Question →

The sum of the first $n$ terms of an arithmetic progression (A.P.) is given by $S_n = 3n^2 + 5n$ . A geometric progression (G.P.) has a first term of $1$ and a common ratio of $2$ . If the $5^{\text{th}}$ term of the A.P. is equal to the $k^{\text{th}}$ term of the G.P., what is the value of $k$ ?

5

6

4

7

Show answer & explanation

Answer: 6

Answer

The value of

k

is 6.

Evaluating the

5^{\text{th}}

term of the A.P. gives

32

. Equating this to the

k^{\text{th}}

term formula of the G.P.,

1 \cdot 2^{k-1} = 32

, leads to

2^{k-1} = 2^5

, which gives

k = 6

.

Step-by-Step Solution

1

Find the general formula for the

n^{\text{th}}

term

T_n

of the A.P. using

S_n

T_n = S_n - S_{n-1} = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)] = 6n + 2

The

n^{\text{th}}

term of a series is the difference between the sum of the first

n

terms and the sum of the first

n-1

terms.

2

Calculate the

5^{\text{th}}

term of the A.P.

T_5 = 6(5) + 2 = 32

Substitute

n = 5

into the derived expression for

T_n

.

3

Set up the equation for the

k^{\text{th}}

term of the G.P.

G_k = a \cdot r^{k-1} = 1 \cdot 2^{k-1} = 2^{k-1}

The standard formula for the

k^{\text{th}}

term of a G.P. is

G_k = a \cdot r^{k-1}

.

4

Equate

T_5

and

G_k

to solve for

k

2^{k-1} = 32 \implies 2^{k-1} = 2^5 \implies k - 1 = 5 \implies k = 6

Since the bases are equal (

2

), equate the exponents to find

k

.

Key Concept

Relating Arithmetic Progression sum formulas to term values and solving indexed Geometric Progression equations.

Estimated Time:2m 0s

Question 139Question →

Find the largest integer value of $x$ that satisfies the linear inequality $\frac{2x - 5}{3} \le \frac{x + 1}{4}$ .

Show answer & explanation

Answer: 4

Answer

The largest integer value of

x

satisfying the inequality is 4.

Multiplying the entire inequality by 12 yields 4(2x - 5) <= 3(x + 1). Expanding both sides produces 8x - 20 <= 3x + 3. Subtracting 3x and adding 20 gives 5x <= 23, which simplifies to x <= 4.6. The largest integer less than or equal to 4.6 is 4.

Step-by-Step Solution

1

Clear the denominators by multiplying both sides by 12.

4(2x - 5) \le 3(x + 1)

Multiplying by a positive number preserves the inequality direction while clearing fractions.

2

Expand both sides of the inequality using the distributive property.

8x - 20 \le 3x + 3

Multiply 4 through (2x - 5) and 3 through (x + 1).

3

Isolate the variable terms on one side and constant terms on the other.

5x \le 23 \implies x \le 4.6

Subtract 3x from both sides and add 20 to both sides, then divide by 5.

4

Determine the maximum integer value satisfying the inequality boundary.

4

Since x must be less than or equal to 4.6, the greatest whole integer satisfying this condition is 4.

Key Concept

Solving linear inequalities with fractions and finding integer bounds

Question 140Question →

The first term of an arithmetic progression (A.P.) is $7$ and its common difference is $4$ . What is the value of the $12^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 51

Answer

The

12^{\text{th}}

term of the arithmetic progression is

51

.

Applying the arithmetic progression

n^{\text{th}}

term formula

T_n = a + (n - 1)d

with first term

a = 7

, common difference

d = 4

, and term index

n = 12

gives

T_{12} = 7 + (12 - 1) \times 4 = 7 + 44 = 51

.

Step-by-Step Solution

1

Identify given variables

a = 7

,

d = 4

, and

n = 12

These are the necessary components to calculate the required term position in an arithmetic sequence.

2

Apply the

n^{\text{th}}

term formula for an AP

T_{12} = 7 + (12 - 1) \times 4

The standard formula

T_n = a + (n - 1)d

determines the value of any term in an AP.

3

Compute the numerical result

T_{12} = 7 + 44 = 51

Multiplying the common difference by 11 and adding the first term yields the correct value.

Key Concept

Calculating the nth term of an Arithmetic Progression

Algebra

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