All practice questions

Question 7321Question →

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 7322Question →

Using differentiation from first principles, what is the derivative of the function $f(x) = x^2 - 5x$ with respect to $x$ ?

2x - 5

2x + 5

x - 5

2x

Show answer & explanation

Answer: $2x - 5$

Answer

The derivative of

f(x) = x^2 - 5x

with respect to

x

is

2x - 5

.

Evaluating the difference quotient

\frac{f(x+h)-f(x)}{h}

for

f(x) = x^2 - 5x

gives

\frac{x^2+2xh+h^2-5x-5h-(x^2-5x)}{h} = 2x + h - 5

. Taking the limit as

h \to 0

yields

2x - 5

.

Step-by-Step Solution

1

Write down the definition of the derivative from first principles.

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

This is the standard formula for differentiation from first principles.

2

Substitute

f(x) = x^2 - 5x

and

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

into the definition.

\frac{f(x+h) - f(x)}{h} = \frac{[(x+h)^2 - 5(x+h)] - (x^2 - 5x)}{h}

This sets up the difference quotient for the given polynomial function.

3

Expand the numerator and combine like terms.

\frac{x^2 + 2xh + h^2 - 5x - 5h - x^2 + 5x}{h} = \frac{2xh + h^2 - 5h}{h} = 2x + h - 5

Expanding allows

x^2

and

-5x

terms to cancel out, leaving terms containing

h

.

4

Evaluate the limit as

h \to 0

.

\lim_{h \to 0} (2x + h - 5) = 2x - 5

As

h

approaches 0, the term

h

vanishes, giving the final derivative.

Key Concept

Differentiation from First Principles

Question 7323Question →

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 7324Question →

Each exterior angle of a regular polygon measures $24^\circ$ . How many sides does this polygon have?

Show answer & explanation

Answer: 15

Answer

The polygon has 15 sides.

The sum of all exterior angles of any convex polygon is always

360^\circ

. Because the polygon is regular, all its exterior angles are equal. Dividing

360^\circ

by the measure of one exterior angle (

24^\circ

) yields exactly 15 sides.

Step-by-Step Solution

1

Identify the formula relating exterior angle and number of sides

The sum of exterior angles of any convex polygon is

360^\circ

. For a regular polygon with

n

sides, each exterior angle equals

\frac{360^\circ}{n}

.

All exterior angles in a regular polygon are equal.

2

Rearrange the formula to solve for the number of sides

n

n = \frac{360^\circ}{\text{Exterior Angle}} = \frac{360^\circ}{24^\circ} = 15

.

Dividing the total sum of exterior angles by the measure of one exterior angle yields the total number of sides.

Key Concept

Exterior angle property of regular polygons

Question 7325Question →

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 7326Question →

If $1101_2 + 32_4 = x_5$ , what is the value of $x$ ?

Show answer & explanation

Answer: 102

Answer

The value of

x

is

102

.

To add numbers given in different bases, convert each term to base 10:

1101_2 = 13_{10}

and

32_4 = 14_{10}

. Summing these decimal values gives

13 + 14 = 27_{10}

. Converting

27_{10}

to base 5 by repeated division by 5 gives remainders 2, 0, 1 (from least significant to most significant digit), resulting in

102_5

. Therefore,

x = 102

.

Step-by-Step Solution

1

Convert

1101_2

to decimal (base 10)

13_{10}

Expand in powers of 2:

1(2^3) + 1(2^2) + 0(2^1) + 1(2^0) = 8 + 4 + 0 + 1 = 13

.

2

Convert

32_4

to decimal (base 10)

14_{10}

Expand in powers of 4:

3(4^1) + 2(4^0) = 12 + 2 = 14

.

3

Add the two base-10 numbers together

27_{10}

Perform standard addition:

13 + 14 = 27

.

4

Convert

27_{10}

into base 5

102_5

Divide 27 repeatedly by 5 and record remainders:

27 = 5 \times 5 + 2

,

5 = 1 \times 5 + 0

,

1 = 0 \times 5 + 1

. Reading remainders upward yields

102_5

.

Key Concept

Conversion between different number bases via base 10

Question 7327Question →

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 7328Question →

If $y = e^{2x} \ln(e^x + \sin x)$ , determine the value of $\frac{d^2 y}{dx^2}$ at $x = 0$ .

Show answer & explanation

Answer: 5

Answer

5

Evaluating the second derivative of

y = e^{2x} \ln(e^x + \sin x)

at

x = 0

yields

5

. This is calculated by applying the product, chain, and quotient rules to get

\frac{d^2 y}{dx^2}

, and substituting

x = 0

, where

e^0 = 1

,

\sin 0 = 0

,

\cos 0 = 1

, and

\ln 1 = 0

.

Step-by-Step Solution

1

Differentiate

y = e^{2x} \ln(e^x + \sin x)

with respect to

x

using the product rule.

\frac{dy}{dx} = 2e^{2x} \ln(e^x + \sin x) + e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right)

The function is composed of

u(x) = e^{2x}

and

v(x) = \ln(e^x + \sin x)

. By chain rule,

v'(x) = \frac{e^x + \cos x}{e^x + \sin x}

.

2

Differentiate the first term

T_1(x) = 2e^{2x} \ln(e^x + \sin x)

to get

T_1'(x)

.

T_1'(x) = 4e^{2x} \ln(e^x + \sin x) + 2e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right)

Applying the product rule to

2e^{2x}

and

\ln(e^x + \sin x)

.

3

Differentiate the second term

T_2(x) = e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right)

using the product and quotient rules.

T_2'(x) = 2e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right) + e^{2x} \left(\frac{(e^x - \sin x)(e^x + \sin x) - (e^x + \cos x)^2}{(e^x + \sin x)^2}\right)

The derivative of

\frac{e^x + \cos x}{e^x + \sin x}

requires the quotient rule.

4

Evaluate

T_1'(0)

and

T_2'(0)

at

x = 0

.

T_1'(0) = 4(1)(0) + 2(1)(2) = 4

, and

T_2'(0) = 2(1)(2) + 1 \left(\frac{(1)(1) - 2^2}{1^2}\right) = 4 - 3 = 1

.

At

x = 0

,

e^0 = 1

,

\sin 0 = 0

,

\cos 0 = 1

, and

\ln(1) = 0

.

5

Sum the evaluated derivative components to find

\frac{d^2 y}{dx^2}\Big|_{x=0}

.

\frac{d^2 y}{dx^2}\Big|_{x=0} = T_1'(0) + T_2'(0) = 4 + 1 = 5

Combining the evaluated terms gives the final numerical value.

Key Concept

Higher-Order Derivatives of Combined Transcendental Functions

Question 7329Question →

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 7330Question →

The frequency distribution table below shows the daily rainfall (in mm) recorded across $80$ weather monitoring stations during a storm:

Daily Rainfall (mm)	Frequency ( $f$ )
$10 - 19$	$8$
$20 - 29$	$14$
$30 - 39$	$26$
$40 - 49$	$20$
$50 - 59$	$12$

Using linear interpolation from the cumulative frequency distribution (ogive), what is the 75th percentile ( $P_{75}$ ) of the daily rainfall in mm?

Show answer & explanation

Answer: 45.5

Answer

The 75th percentile of the daily rainfall distribution is

45.5\text{ mm}

.

To find the 75th percentile (

P_{75}

), determine the

60^{\text{th}}

cumulative frequency position (

0.75 \times 80 = 60

). The value lies within the

40 - 49

class interval. Applying the lower class boundary

L = 39.5

, preceding cumulative frequency

cf_b = 48

, frequency

f = 20

, and class width

c = 10

, linear interpolation yields

P_{75} = 39.5 + \frac{60 - 48}{20} \times 10 = 45.5\text{ mm}

.

Step-by-Step Solution

1

Calculate cumulative frequencies across all class intervals.

Cumulative frequencies are

8

for

10-19

,

22

for

20-29

,

48

for

30-39

,

68

for

40-49

, and

80

for

50-59

. Total frequency

N = 80

.

Cumulative frequencies are necessary to locate percentile positions on an ogive.

2

Determine the position corresponding to the 75th percentile.

Position

= 0.75 \times 80 = 60^{\text{th}}

cumulative frequency item.

The 75th percentile represents

75\%

of the total sample size.

3

Identify the target class interval parameters containing the 60th observation.

The interval

40 - 49

contains cumulative frequencies from

49

to

68

. Parameters:

L = 39.5

,

c = 10

,

f = 20

,

cf_b = 48

.

Linear interpolation requires the exact boundaries and frequencies of the container class.

4

Apply the percentile interpolation formula

P_{75} = L + \left(\frac{60 - cf_b}{f}\right) \times c

.

P_{75} = 39.5 + \left(\frac{60 - 48}{20}\right) \times 10 = 39.5 + 6 = 45.5\text{ mm}

.

Calculates the exact rainfall value corresponding to the 75th percentile.

Key Concept

Linear interpolation for percentiles using cumulative frequency distribution (ogive)

Estimated Time:1m 30s

Question 7331Question →

A library recorded the number of books borrowed over five consecutive days as $2$ , $4$ , $5$ , $6$ , and $8$ . What is the standard deviation of the number of books borrowed?

2

4

5

6

Show answer & explanation

Answer: $2$

Answer

The standard deviation of the number of books borrowed is

2

.

The mean of the five data points is

5

. The sum of the squared deviations from the mean is

9 + 1 + 0 + 1 + 9 = 20

. Dividing by

5

gives a variance of

4

. Taking the square root of

4

yields the standard deviation of

2

.

Step-by-Step Solution

1

Calculate the arithmetic mean (

\bar{x}

) of the given dataset.

\bar{x} = \frac{2 + 4 + 5 + 6 + 8}{5} = \frac{25}{5} = 5

The mean is required to determine the deviations of each data point.

2

Find the deviations from the mean

(x - \bar{x})

and square each deviation

(x - \bar{x})^2

.

(2-5)^2 = 9, \quad (4-5)^2 = 1, \quad (5-5)^2 = 0, \quad (6-5)^2 = 1, \quad (8-5)^2 = 9

Squaring ensures all deviation values are positive before summation.

3

Calculate the variance (mean of squared deviations).

\text{Variance } (\sigma^2) = \frac{9 + 1 + 0 + 1 + 9}{5} = \frac{20}{5} = 4

Variance measures the average squared spread around the mean.

4

Take the square root of the variance to find the standard deviation.

\text{Standard Deviation } (\sigma) = \sqrt{4} = 2

Standard deviation expresses dispersion in the original unit of measurement.

Key Concept

Standard Deviation of Ungrouped Data

Question 7332Question →

The table below shows the distribution of masses (in kg) of $50$ bags of cement inspected at a building construction site:

Mass (kg)	Frequency ( $f$ )
$40 - 44$	$5$
$45 - 49$	$12$
$50 - 54$	$18$
$55 - 59$	$15$

Calculate the mean mass of the bags of cement in kg.

Show answer & explanation

Answer: 51.3

Answer

The mean mass of the cement bags is 51.3 kg.

The mean mass of grouped data is computed using the formula

\bar{x} = \frac{\sum fx}{\sum f}

, where

x

represents the class midpoints and

f

represents the frequency of each class. The class midpoints are

42

,

47

,

52

, and

57

. Multiplying these midpoints by their respective frequencies yields

210

,

564

,

936

, and

855

. Summing these values gives

\sum fx = 2565

. Dividing by the total frequency

\sum f = 50

produces a mean mass of

\frac{2565}{50} = 51.3\text{ kg}

.

Step-by-Step Solution

1

Determine the midpoint (x) of each class interval

Midpoints are 42, 47, 52, and 57

For grouped data, the class midpoint represents the average value of all observations falling within that class interval.

2

Calculate the product of each midpoint and its corresponding frequency (fx)

Products are 210, 564, 936, and 855

Multiplying the midpoint by frequency gives the total estimated mass contributed by that class interval.

3

Sum all frequencies and all fx products

Total frequency sum = 50, Total product sum = 2565

These totals are required to calculate the weighted average across all intervals.

4

Divide the sum of fx by the total frequency

Mean = 51.3 kg

Applying the formula for grouped mean: Mean = (sum of fx) / (sum of f).

Key Concept

Grouped Data Mean Calculation

Question 7333Question →

What is the simplified form of the surd expression $\frac{4}{\sqrt{5} - 1}$ after rationalizing the denominator?

1 + \sqrt{5}

2

4 + 4\sqrt{5}

\sqrt{5} - 1

Show answer & explanation

Answer: $1 + \sqrt{5}$

Answer

1 + \sqrt{5}

The expression

1 + \sqrt{5}

is correct because multiplying the numerator and denominator by the conjugate

\sqrt{5} + 1

converts the denominator to

(\sqrt{5})^2 - 1^2 = 4

. Canceling the common factor of

4

in numerator and denominator simplifies the expression completely to

1 + \sqrt{5}

.

Step-by-Step Solution

1

Identify the conjugate of the denominator

The conjugate of

\sqrt{5} - 1

is

\sqrt{5} + 1

.

Multiplying a binomial surd by its conjugate eliminates the radical in the denominator using the difference of two squares identity

(a-b)(a+b) = a^2 - b^2

.

2

Multiply both the numerator and the denominator by the conjugate

\frac{4(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)} = \frac{4(\sqrt{5} + 1)}{(\sqrt{5})^2 - (1)^2}

Multiplying both numerator and denominator by the same expression preserves the value of the fraction.

3

Simplify the denominator and evaluate the fraction

\frac{4(\sqrt{5} + 1)}{5 - 1} = \frac{4(\sqrt{5} + 1)}{4} = 1 + \sqrt{5}

Dividing the numerator by

4

cancels out the factor of

4

.

Key Concept

Rationalization of Binomial Denominators

Question 7334Question →

If $x = \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}}$ and $y = \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}}$ , determine the numerical value of $x^2 + y^2$ .

Show answer & explanation

Answer: 23

Answer

The numerical value of

x^2 + y^2

is

23

.

Rationalizing

x

yields

\frac{5 + \sqrt{21}}{2}

and rationalizing

y

yields

\frac{5 - \sqrt{21}}{2}

. The sum

x + y

equals

5

and the product

xy

equals

1

. Substituting these into

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

gives

5^2 - 2(1) = 23

.

Step-by-Step Solution

1

Rationalize the denominators of

x

and

y

x = \frac{5 + \sqrt{21}}{2}

and

y = \frac{5 - \sqrt{21}}{2}

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

2

Calculate the sum

x + y

and the product

xy

x + y = 5

and

xy = 1

Summing conjugate surd expressions cancels the radical term, and multiplying them applies the difference of two squares.

3

Evaluate

x^2 + y^2

using the identity

(x + y)^2 - 2xy

x^2 + y^2 = 5^2 - 2(1) = 23

Substituting the known sum and product avoids having to square complex surd expressions directly.

Key Concept

Rationalization of binomial denominators and application of symmetric algebraic identities.

Question 7335Question →

The table below shows the frequency distribution of marks obtained by $25$ students in a mathematics quiz:

Score ( $x$ )	1	3	5	7	9	11
Frequency ( $f$ )	2	$p$	6	$q$	3	2

If the mean score of the distribution is $5.48$ , what is the median score?

5

6

3

7

Show answer & explanation

Answer: 5

Answer

The median score is 5.

By using the total student count of 25 and the mean formula, we obtain the simultaneous equations

p + q = 12

and

3p + 7q = 56

, which yield

p = 7

and

q = 5

. Computing cumulative frequencies shows that items 1 to 2 have score 1, items 3 to 9 have score 3, and items 10 to 15 have score 5. The 13th item lies in this third group, so the median score is 5.

Step-by-Step Solution

1

Set up an equation for total frequency

2 + p + 6 + q + 3 + 2 = 25 \implies p + q + 13 = 25 \implies p + q = 12

The sum of all frequencies equals the total number of students (

25

).

2

Set up an equation for the mean score

\sum fx = 1(2) + 3(p) + 5(6) + 7(q) + 9(3) + 11(2) = 3p + 7q + 81.
\text{Mean} = \frac{3p + 7q + 81}{25} = 5.48 \implies 3p + 7q + 81 = 137 \implies 3p + 7q = 56.

The mean of an ungrouped frequency distribution is calculated using \bar{x} = \frac{\sum fx}{N}.

3

Solve the system of linear equations for

p

and

q

Substitute

p = 12 - q

into

3p + 7q = 56

:
3(12 - q) + 7q = 56 \implies 36 + 4q = 56 \implies 4q = 20 \implies q = 5.
Then

p = 12 - 5 = 7

.

Finding the missing frequencies is necessary to determine cumulative frequencies.

4

Determine the position and value of the median score

Position of median = \frac{N + 1}{2} = \frac{25 + 1}{2} = 13\text{th position}.
Cumulative frequencies:
- Score 1: 2
- Score 3: 2 + 7 = 9
- Score 5: 9 + 6 = 15
Since the 13th value lies in the cumulative frequency interval up to 15, the median score is 5.

The median of

N=25

items is the score corresponding to the

\frac{N+1}{2}

th item when ordered.

Key Concept

Measures of Central Tendency for Ungrouped Data

Question 7336Question →

The table below shows the distribution of heights (in cm) of $20$ potted plants recorded during a biology experiment:

Height (cm)	Frequency ( $f$ )
10 – 14	4
15 – 19	6
20 – 24	8
25 – 29	2

What is the mean height of the potted plants?

17\text{ cm}

19\text{ cm}

19.5\text{ cm}

21\text{ cm}

Show answer & explanation

Answer: $19\text{ cm}$

Answer

The mean height of the potted plants is

19\text{ cm}

.

The mean height is calculated by multiplying each class interval's midpoint by its frequency, summing these products (

380

), and dividing by the total number of plants (

20

), giving

19\text{ cm}

.

Step-by-Step Solution

1

Calculate the midpoint (

x

) for each class interval.

Class midpoints are

12

,

17

,

22

, and

27

.

The midpoint represents the estimated mean value of data items in a grouped class interval.

2

Multiply each midpoint (

x

) by its corresponding frequency (

f

) to find

f \cdot x

.

4 \times 12 = 48

,

6 \times 17 = 102

,

8 \times 22 = 176

, and

2 \times 27 = 54

.

This determines the total estimated sum of values for each class.

3

Sum all

f \cdot x

values and calculate the total frequency

\sum f

.

\sum fx = 48 + 102 + 176 + 54 = 380

and

\sum f = 4 + 6 + 8 + 2 = 20$.

These totals are required for the grouped mean formula.

4

Compute the mean using the formula

\bar{x} = \frac{\sum fx}{\sum f}

.

\bar{x} = \frac{380}{20} = 19\text{ cm}$.

Dividing total sum of values by total frequency yields the mean.

Key Concept

Grouped Mean Calculation using Class Midpoints

Estimated Time:1m 0s

Question 7337Question →

If $1.41_x = \frac{46}{25}_{10}$ , where $x$ is a positive integer base, find the value of $x$ .

Show answer & explanation

Answer: 5

Answer

The value of the base

x

is 5.

Expanding

1.41_x

yields

1 + \frac{4}{x} + \frac{1}{x^2}

. Setting this equal to

\frac{46}{25}

gives

\frac{4x + 1}{x^2} = \frac{21}{25}

. Cross-multiplying results in the quadratic equation

21x^2 - 100x - 25 = 0

, which factors as

(21x + 5)(x - 5) = 0

. Since a number base must be a positive integer greater than 4,

x = 5

is the only valid solution.

Step-by-Step Solution

1

Expand

1.41_x

using place value powers of

x

.

1 + \frac{4}{x} + \frac{1}{x^2}

Fractional digits to the right of the radix point represent negative powers of the base (

x^{-1}, x^{-2}, \dots

).

2

Equate the expanded form to

\frac{46}{25}

and simplify.

\frac{4x + 1}{x^2} = \frac{21}{25}

Subtracting 1 from both sides isolates the fractional place values.

3

Cross-multiply and solve the quadratic equation

21x^2 - 100x - 25 = 0

.

(21x + 5)(x - 5) = 0 \implies x = 5

A base must be a positive integer strictly greater than any individual digit in the number (digits present are 1 and 4).

Key Concept

Conversion of fractional numbers in non-decimal bases to base 10 and solving polynomial equations in unknown bases.

Question 7338Question →

Four different Mathematics textbooks and three different Physics textbooks are to be arranged in a line on a shelf. In how many distinct ways can the books be arranged if all four Mathematics textbooks must be kept together?

144

576

840

5,040

Show answer & explanation

Answer: 576

Answer

576 distinct ways

Treating the four Mathematics textbooks as a single unit gives 4 items to arrange on the shelf (the Mathematics unit and the three individual Physics textbooks). These 4 items can be arranged in

4! = 24

ways. Furthermore, the four distinct Mathematics textbooks within the unit can be arranged among themselves in

4! = 24

ways. By the multiplication principle of counting, the total number of distinct arrangements is

24 \times 24 = 576

.

Step-by-Step Solution

1

Group the Mathematics textbooks into a single block

1 Mathematics block + 3 individual Physics textbooks = 4 items to arrange.

Because all four Mathematics textbooks must remain together, they act as a single composite unit.

2

Calculate the arrangements of the 4 main items

4! = 4 \times 3 \times 2 \times 1 = 24

ways.

There are 4 distinct items (the Mathematics block and 3 separate Physics books) to place in linear order.

3

Calculate internal permutations of the Mathematics textbooks inside their block

4! = 4 \times 3 \times 2 \times 1 = 24

ways.

The 4 Mathematics textbooks can be arranged in different orders among themselves.

4

Apply the fundamental counting principle

Total arrangements =

24 \times 24 = 576

.

Multiply the number of block arrangements by the internal arrangements of the Mathematics textbooks.

Key Concept

Permutations with Restricted Grouping (Block Method)

Question 7339Question →

If the surd expression $\frac{12}{\sqrt{6}}$ is simplified to the form $k\sqrt{6}$ , what is the value of $k$ ?

Show answer & explanation

Answer: 2

Answer

The value of

k

is 2.

Multiplying the numerator and denominator of

\frac{12}{\sqrt{6}}

by

\sqrt{6}

yields

\frac{12\sqrt{6}}{6}

. Dividing

12

by

6

simplifies the expression to

2\sqrt{6}

. Matching this with

k\sqrt{6}

gives

k = 2

.

Step-by-Step Solution

1

Rationalize the denominator of

\frac{12}{\sqrt{6}}

by multiplying the numerator and denominator by

\sqrt{6}

.

\frac{12\sqrt{6}}{6}

Multiplying by

\frac{\sqrt{6}}{\sqrt{6}}

removes the radical from the denominator without changing the value of the expression.

2

Divide the integer coefficient in the numerator by the denominator.

2\sqrt{6}

Simplifying

\frac{12}{6}

yields 2.

3

Compare

2\sqrt{6}

with

k\sqrt{6}

to determine the value of

k

.

k = 2

The coefficient of

\sqrt{6}

is 2.

Key Concept

Rationalization of monomial surd denominators

Question 7340Question →

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

Answer

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