All practice questions

Question 7401Question →

A vessel contains a mixture of two liquids, $A$ and $B$ , in the ratio $5 : 3$ . If $16\text{ litres}$ of the mixture is drawn off and replaced with an equal volume of liquid $B$ , the ratio of liquid $A$ to liquid $B$ in the vessel becomes $1 : 1$ . What was the initial total volume of the mixture in the vessel, in litres?

Show answer & explanation

Answer: 80

Answer

The initial total volume of the mixture in the vessel was

80\text{ litres}

.

Let the initial volume of the mixture be

V

litres. Liquid

A

initially comprises

\frac{5}{8}V

litres and liquid

B

comprises

\frac{3}{8}V

litres. When

16\text{ litres}

of mixture is removed, the volume of liquid

A

removed is

\frac{5}{8} \times 16 = 10\text{ litres}

, and the volume of liquid

B

removed is

\frac{3}{8} \times 16 = 6\text{ litres}

. After adding

16\text{ litres}

of pure liquid

B

, the new volume of liquid

A

is

\frac{5}{8}V - 10

and the new volume of liquid

B

is

\frac{3}{8}V - 6 + 16 = \frac{3}{8}V + 10

. Since the new ratio is

1 : 1

, setting

\frac{5}{8}V - 10 = \frac{3}{8}V + 10

gives

\frac{2}{8}V = 20

, which simplifies to

V = 80\text{ litres}

.

Step-by-Step Solution

1

Define total initial volume as

V

and express the initial quantities of liquids

A

and

B

.

Liquid

A = \frac{5}{8}V

litres, Liquid

B = \frac{3}{8}V

litres.

The total ratio parts equal

5 + 3 = 8

.

2

Calculate the volume of each liquid removed in

16\text{ litres}

of mixture.

Volume of

A

removed

= 10

litres; Volume of

B

removed

= 6

litres.

The drawn-off mixture retains the original

5 : 3

ratio of liquids.

3

Formulate expressions for the quantities of

A

and

B

after adding

16\text{ litres}

of pure liquid

B

.

New amount of

A = \frac{5}{8}V - 10

; New amount of

B = \frac{3}{8}V + 10

.

16

litres of liquid

B

is added to the remaining quantity of

B

, which was

\frac{3}{8}V - 6

.

4

Equate the new quantities of

A

and

B

since the final ratio is

1 : 1

.

\frac{5}{8}V - 10 = \frac{3}{8}V + 10 \implies \frac{2}{8}V = 20 \implies V = 80

.

A final ratio of

1 : 1

implies equal quantities of both liquids.

Key Concept

Ratio modification through mixture removal and replacement

Estimated Time:2m 0s

Question 7402Question →

If $2^{x-1} + 2^{x+1} = 160$ , what is the value of $3^{x-2}$ ?

27

81

243

12

Show answer & explanation

Answer: 81

Answer

81

Factoring

2^{x-1}

from the left-hand side of

2^{x-1} + 2^{x+1} = 160

yields

2^{x-1}(1 + 2^2) = 160

, which reduces to

5 \cdot 2^{x-1} = 160

. Dividing by

5

gives

2^{x-1} = 32 = 2^5

. Equating exponents gives

x - 1 = 5

, so

x = 6

. Substituting

x = 6

into

3^{x-2}

gives

3^{6-2} = 3^4 = 81

.

Step-by-Step Solution

1

Factor out the common term

2^{x-1}

from the expression

2^{x-1} + 2^{x+1}

.

2^{x-1}(1 + 2^2) = 160

Applying index law

2^{x+1} = 2^{x-1} \cdot 2^2

allows factoring out

2^{x-1}

.

2

Simplify the bracketed terms and solve for

2^{x-1}

.

5 \cdot 2^{x-1} = 160 \implies 2^{x-1} = 32

Dividing both sides by

5

isolates the term with the unknown exponent.

3

Express

32

as a power of

2

and solve for

x

.

2^{x-1} = 2^5 \implies x - 1 = 5 \implies x = 6

Equating exponents since the bases are identical.

4

Substitute

x = 6

into the target expression

3^{x-2}

.

3^{6-2} = 3^4 = 81

Evaluating the power gives the final answer.

Key Concept

Factoring exponential expressions with common bases and applying laws of indices

Estimated Time:1m 30s

Question 7403Question →

In how many distinct ways can the letters of the word $\text{KADUNA}$ be arranged?

720

15

360

180

Show answer & explanation

Answer: $360$

Answer

The total number of distinct arrangements is

360

.

The word

\text{KADUNA}

consists of

6

total letters with the letter '\text{A}' repeated

2

times. Using the formula for permutations with identical items, the number of distinct arrangements is

\frac{6!}{2!} = \frac{720}{2} = 360

.

Step-by-Step Solution

1

Count the total number of letters in the word

\text{KADUNA}

and identify repetitions.

Total letters

n = 6

. The letter '\text{A}' appears

2

times, while '\text{K}', '\text{D}', '\text{U}', and '\text{N}' each appear

1

time.

Arrangements of

n

items with repeated elements require dividing

n!

by the factorial of the count of each repeated element.

2

Apply the permutation formula for repeated elements:

P = \frac{n!}{p!}

where

p

is the frequency of the repeated letter.

P = \frac{6!}{2!} = \frac{720}{2} = 360

Dividing by

2!

eliminates duplicate arrangements caused by swapping identical letters.

Key Concept

Permutations of items with repeated elements

Estimated Time:45s

Question 7404Question →

What is the value of $\frac{\log_4 81}{\log_4 3}$ ?

Show answer & explanation

Answer: 4

Answer

4

Applying the change of base property

\frac{\log_c b}{\log_c a} = \log_a b

, the ratio

\frac{\log_4 81}{\log_4 3}

reduces directly to

\log_3 81

. Since

3^4 = 81

, the value is 4.

Step-by-Step Solution

1

Apply the change of base formula to rewrite the ratio

\frac{\log_4 81}{\log_4 3} = \log_3 81

By the change of base identity,

\frac{\log_c b}{\log_c a} = \log_a b

.

2

Evaluate the logarithm

4

Since

3^4 = 81

,

\log_3 81 = 4

.

Key Concept

Change of Base Formula for Logarithms

Question 7405Question →

A vessel initially contains $80\text{ litres}$ of water. After standing in the sun, $12\text{ litres}$ of water evaporates. What percentage of the original volume of water evaporated?

Show answer & explanation

Answer: 15

Answer

The percentage of the original volume of water that evaporated is

15\%

.

To find the percentage of evaporated water, divide the evaporated amount (

12\text{ litres}

) by the original total volume (

80\text{ litres}

) and multiply by

100\%

. This yields

\frac{12}{80} \times 100\% = 15\%

.

Step-by-Step Solution

1

Determine the fractional portion of the evaporated water relative to the initial total volume.

The fraction is

\frac{12}{80} = \frac{3}{20}

.

Percentage loss must be evaluated relative to the original starting amount.

2

Convert the resulting fraction into a percentage.

\frac{3}{20} \times 100\% = 15\%.

Multiplying a dimensionless ratio by 100 yields its percentage equivalent.

Key Concept

Calculating percentage change or loss relative to an initial amount

Estimated Time:45s

Question 7406Question →

If $123_x = 38_{10}$ , what is the value of the base $x$ ?

Show answer & explanation

Answer: 5

Answer

The base

x

is 5.

Expanding

123_x

in terms of powers of

x

yields

1 \cdot x^2 + 2 \cdot x^1 + 3 \cdot x^0 = x^2 + 2x + 3

. Setting this equal to the decimal value 38 produces the quadratic equation

x^2 + 2x + 3 = 38

, which simplifies to

x^2 + 2x - 35 = 0

. Factoring gives

(x + 7)(x - 5) = 0

, yielding solutions

x = -7

and

x = 5

. Since a number base must be a positive integer, the correct value for

x

is 5.

Step-by-Step Solution

1

Expand the base

x

number into decimal form using place values

123_x = 1 \cdot x^2 + 2 \cdot x^1 + 3 \cdot x^0 = x^2 + 2x + 3

Each digit position in base

x

corresponds to a power of

x

, starting from

x^0

on the right.

2

Set up and rearrange the quadratic equation

x^2 + 2x + 3 = 38 \implies x^2 + 2x - 35 = 0

Subtracting 38 from both sides converts the equation into standard quadratic form

ax^2 + bx + c = 0

.

3

Solve the quadratic equation for

x

(x + 7)(x - 5) = 0 \implies x = -7 \text{ or } x = 5

Factoring gives the roots of the quadratic equation.

4

Select the valid positive base

x = 5

A base must be a positive integer greater than the largest digit appearing in the number (which is 3).

Key Concept

Place value expansion and base conversion to base 10

Question 7407Question →

The mean of five numbers $3, 5, 7, x,$ and $y$ is $6$ . Given that the variance of the numbers is $8$ and $x < y$ , calculate the value of $y$ .

Show answer & explanation

Answer: 11

Answer

11

The total sum of the five numbers is

5 \times 6 = 30

, giving

x + y = 15

. The sum of squared deviations from the mean

6

is

5 \times 8 = 40

. The known numbers

3, 5, 7

contribute

(-3)^2 + (-1)^2 + 1^2 = 11

to this sum, leaving

(x-6)^2 + (y-6)^2 = 29

. Substituting

y = 15 - x

yields

2x^2 - 30x + 88 = 0 \implies x^2 - 15x + 44 = 0

. Factoring gives roots

4

and

11

. Since

x < y

, we find

y = 11

.

Step-by-Step Solution

1

Use the definition of the arithmetic mean to write a linear relationship between

x

and

y

.

x + y = 15

, or

y = 15 - x

.

The total sum of 5 numbers with a mean of 6 is

5 \times 6 = 30

. Subtracting the known numbers

3 + 5 + 7 = 15

leaves

x + y = 15

.

2

Apply the variance formula for population data.

\sum (x_i - 6)^2 = 40

.

Variance is the mean of squared deviations from the mean:

\frac{\sum (x_i - \bar{x})^2}{5} = 8 \implies \sum (x_i - 6)^2 = 40

.

3

Compute the sum of squared deviations for the known elements and simplify the variance equation.

(x - 6)^2 + (y - 6)^2 = 29

.

The squared deviations for

3, 5, 7

are

(-3)^2 = 9

,

(-1)^2 = 1

, and

1^2 = 1

. Subtracting

9 + 1 + 1 = 11

from

40

gives

29

.

4

Substitute

y = 15 - x

into the simplified equation and solve the resulting quadratic equation.

x = 4

or

x = 11

.

Substituting

y = 15 - x

gives

(x - 6)^2 + (9 - x)^2 = 29 \implies 2x^2 - 30x + 88 = 0 \implies x^2 - 15x + 44 = 0 \implies (x - 4)(x - 11) = 0

.

5

Select the correct pair

(x, y)

using the condition

x < y

.

x = 4

and

y = 11

.

Since

x < y

,

x

must be the smaller value (

4

) and

y

must be the larger value (

11

).

Key Concept

Calculation of variance and mean for ungrouped data containing unknown elements

Question 7408Question →

The frequency distribution table below shows the marks obtained by $40$ students in a mathematics quiz:

Mark Interval	Frequency ( $f$ )
$1 - 10$	$5$
$11 - 20$	$8$
$21 - 30$	$12$
$31 - 40$	$10$
$41 - 50$	$5$

From a cumulative frequency curve (ogive) constructed for this data, what is the score corresponding to the $40^{\text{th}}$ percentile?

23.0

23.5

25.5

28.0

Show answer & explanation

Answer: $23.0$

Answer

23.0

To find the 40th percentile score, calculate

40\%

of the total

40

students, which gives the

16^{\text{th}}

cumulative student. The

16^{\text{th}}

student falls in the

21 - 30

mark interval, which has a lower class boundary of

20.5

. Interpolating gives

20.5 + \left(\frac{16 - 13}{12}\right) \times 10 = 20.5 + 2.5 = 23.0

.

Step-by-Step Solution

1

Construct the cumulative frequency distribution table.

Cumulative frequencies (

CF

):

1-10 \rightarrow 5

;

11-20 \rightarrow 13

;

21-30 \rightarrow 25

;

31-40 \rightarrow 35

;

41-50 \rightarrow 40

. Total frequency

N = 40

.

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

2

Determine the rank position for the 40th percentile (

P_{40}

).

Rank position

= 0.40 \times 40 = 16^{\text{th}}

position.

The 40th percentile represents the value below which

40\%

of the observations fall.

3

Identify the percentile class interval and its boundaries.

Since

13 < 16 \leq 25

, the

16^{\text{th}}

position lies in the interval

21 - 30

. Lower class boundary

L = 20.5

, cumulative frequency preceding the class

CF_b = 13

, class frequency

f = 12

, and class width

c = 10

.

Interpolation on an ogive relies on the lower class boundary of the target interval.

4

Apply the linear interpolation formula for percentiles.

P_{40} = L + \left( \frac{\text{Rank} - CF_b}{f} \right) \times c = 20.5 + \left( \frac{16 - 13}{12} \right) \times 10 = 20.5 + 2.5 = 23.0

.

This yields the exact mark corresponding to the 40th percentile reading from the ogive.

Key Concept

Calculating percentiles from a cumulative frequency distribution or ogive

Estimated Time:1m 30s

Question 7409Question →

If $\log_{10} 2 = p$ and $\log_{10} 3 = q$ , which of the following expressions represents $\log_{10} 18$ in terms of $p$ and $q$ ?

p + 2q

2pq

2p + q

p + q^2

Show answer & explanation

Answer: $p + 2q$

Answer

The expression for

\log_{10} 18

in terms of

p

and

q

is

p + 2q

.

Expanding

\log_{10} 18

as

\log_{10}(2 \times 3^2)

gives

\log_{10} 2 + 2\log_{10} 3

, which simplifies directly to

p + 2q

.

Step-by-Step Solution

1

Factorize 18 into prime factors.

18 = 2 \times 3^2

Decomposing 18 into prime factors allows the application of the given logarithm values for 2 and 3.

2

Apply the product rule of logarithms:

\log_{10}(a \times b) = \log_{10} a + \log_{10} b

.

\log_{10} 18 = \log_{10}(2 \times 3^2) = \log_{10} 2 + \log_{10}(3^2)

The logarithm of a product equals the sum of the logarithms of individual factors.

3

Apply the power rule of logarithms:

\log_{10}(b^n) = n \log_{10} b

.

\log_{10}(3^2) = 2 \log_{10} 3

The logarithm of a power expression allows bringing the exponent to the front as a multiplier.

4

Substitute the defined values

p = \log_{10} 2

and

q = \log_{10} 3

.

\log_{10} 18 = p + 2q

Replaces logarithmic terms with their algebraic representations.

Key Concept

Logarithm Expansion Laws (Product and Power Rules)

Question 7410Question →

Three consecutive terms of an arithmetic progression are $x + 2$ , $3x - 1$ , and $4x + 1$ . What is the $10^{\text{th}}$ term of the progression?

63

77

70

35

Show answer & explanation

Answer: 70

Answer

The

10^{\text{th}}

term of the progression is 70.

Equating the differences between consecutive terms gives

(3x - 1) - (x + 2) = (4x + 1) - (3x - 1)

, which simplifies to

2x - 3 = x + 2

, giving

x = 5

. The first term is

a = 5 + 2 = 7

and the common difference is

d = 14 - 7 = 7

. Substituting these into

T_n = a + (n - 1)d

for

n = 10

yields

T_{10} = 7 + 9(7) = 70

.

Step-by-Step Solution

1

Set up an equation using the common difference property of an arithmetic progression.

(3x - 1) - (x + 2) = (4x + 1) - (3x - 1)

In any arithmetic progression, the difference between consecutive terms is constant (

T_2 - T_1 = T_3 - T_2

).

2

Simplify and solve for

x

.

2x - 3 = x + 2 \implies x = 5

Subtract

x

from both sides and add 3 to both sides.

3

Find the first term

a

and the common difference

d

.

First term

a = 5 + 2 = 7

; second term

T_2 = 3(5) - 1 = 14

; common difference

d = 14 - 7 = 7

.

Substitute

x = 5

into the expressions for the terms.

4

Calculate the

10^{\text{th}}

term using

T_n = a + (n - 1)d

.

T_{10} = 7 + (10 - 1)(7) = 7 + 9(7) = 7 + 63 = 70

Apply

n = 10

,

a = 7

, and

d = 7

to the

n^{\text{th}}

term formula.

Key Concept

Arithmetic Progression - Consecutive terms and nth term evaluation

Question 7411Question →

The operating lifespans (in hours) of a sample of $60$ newly manufactured micro-components tested under laboratory conditions are presented in the frequency table below:

Lifespan (hours)	Frequency ( $f$ )
$10 - 19$	$5$
$20 - 29$	$8$
$30 - 39$	$12$
$40 - 49$	$20$
$50 - 59$	$15$

Using linear interpolation on class boundaries, calculate the median lifespan of the components in hours.

Show answer & explanation

Answer: 42

Answer

The median lifespan of the micro-components is 42 hours.

To calculate the median of grouped data, first determine cumulative frequencies: 5, 13, 25, 45, 60. Total frequency is

N = 60

, placing the median at

N/2 = 30

. The median class is

40 - 49

with lower class boundary

L = 39.5

, class frequency

f = 20

, cumulative frequency prior to median class

c.f. = 25

, and class width

c = 10

. Substituting into

\text{Median} = L + \left(\frac{N/2 - c.f.}{f}\right) \times c

gives

39.5 + \left(\frac{30 - 25}{20}\right) \times 10 = 39.5 + 2.5 = 42

.

Step-by-Step Solution

1

Calculate the total frequency

N

and the median rank

N/2

.

N = 5 + 8 + 12 + 20 + 15 = 60

, so

N/2 = 30

.

The median corresponds to the value at position 30 in the ordered cumulative frequency distribution.

2

Determine cumulative frequencies to locate the median class.

Cumulative frequencies are 5, 13, 25, 45, 60. The 30th value falls in the interval

40 - 49

.

The cumulative frequency first reaches or exceeds 30 at the

40 - 49

class.

3

Identify boundary parameters for the median class.

Lower boundary

L = 39.5

, class width

c = 10

, class frequency

f = 20

, and cumulative frequency of preceding class

c.f. = 25

.

Continuous data analysis requires using lower class boundaries rather than class limits.

4

Apply the grouped median linear interpolation formula.

\text{Median} = 39.5 + \left(\frac{30 - 25}{20}\right) \times 10 = 39.5 + 2.5 = 42

.

Linear interpolation within the median class yields the precise median value.

Key Concept

Grouped Data Median via Class Boundary Interpolation

Estimated Time:2m 0s

Question 7412Question →

If $y = e^{\sin^2(3x)}$ , what is $\frac{dy}{dx}$ ?

3\sin(6x)e^{\sin^2(3x)}

-3\sin(6x)e^{\sin^2(3x)}

\sin(6x)e^{\sin^2(3x)}

6\cos(3x)e^{\sin^2(3x)}

Show answer & explanation

Answer: $3\sin(6x)e^{\sin^2(3x)}$

Answer

The derivative

\frac{dy}{dx}

is

3\sin(6x)e^{\sin^2(3x)}

.

Differentiating

y = e^{\sin^2(3x)}

requires applying the chain rule step-by-step: first differentiating the exponential function to get

e^{\sin^2(3x)}

, then differentiating

\sin^2(3x)

to obtain

2\sin(3x) \cdot 3\cos(3x) = 6\sin(3x)\cos(3x)

. Multiplying these together and applying the double-angle identity

2\sin(3x)\cos(3x) = \sin(6x)

yields

3\sin(6x)e^{\sin^2(3x)}

.

Step-by-Step Solution

1

Identify the inner function

u

and outer function

y

for applying the chain rule.

Let

u = \sin^2(3x) = (\sin(3x))^2

, so

y = e^u

.

The function is an exponential function whose exponent is a composite trigonometric function.

2

Differentiate

u = (\sin(3x))^2

with respect to

x

using the chain rule.

\frac{du}{dx} = 2\sin(3x) \cdot \frac{d}{dx}(\sin(3x)) = 2\sin(3x) \cdot 3\cos(3x) = 6\sin(3x)\cos(3x).

The derivative of

[g(x)]^2

is

2g(x)g'(x)

, and

\frac{d}{dx}(\sin(3x)) = 3\cos(3x)

.

3

Differentiate

y = e^u

with respect to

x

using

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

.

\frac{dy}{dx} = e^{\sin^2(3x)} \cdot 6\sin(3x)\cos(3x).

The derivative of

e^u

with respect to

u

is

e^u

.

4

Simplify the expression using the trigonometric double-angle identity

2\sin\theta\cos\theta = \sin(2\theta)

, where

\theta = 3x

.

\frac{dy}{dx} = 3 \cdot (2\sin(3x)\cos(3x)) e^{\sin^2(3x)} = 3\sin(6x)e^{\sin^2(3x)}.

Rewriting

6\sin(3x)\cos(3x)

as

3\sin(6x)

simplifies the expression to standard examination form.

Key Concept

Chain Rule for Composite Exponential and Trigonometric Functions

Estimated Time:2m 0s

Question 7413Question →

A rectangular lawn was measured to have a length of $15.0\text{ m}$ and a width of $8.0\text{ m}$ . If the actual length of the lawn is $16.0\text{ m}$ and the measured width is exact, what is the percentage error in the calculated area of the lawn?

Show answer & explanation

Answer: 6.25

Answer

The percentage error in the calculated area is

6.25\%

.

The actual area of the lawn is

16.0\text{ m} \times 8.0\text{ m} = 128\text{ m}^2

, while the measured area is

15.0\text{ m} \times 8.0\text{ m} = 120\text{ m}^2

. The error is

128 - 120 = 8\text{ m}^2

. Expressed as a percentage of the actual area,

\frac{8}{128} \times 100\% = 6.25\%

.

Step-by-Step Solution

1

Calculate the measured area of the lawn using the measured length and width

\text{Measured Area} = 15.0\text{ m} \times 8.0\text{ m} = 120\text{ m}^2

Area of a rectangle is length multiplied by width.

2

Calculate the true (actual) area of the lawn using the actual length and exact width

\text{Actual Area} = 16.0\text{ m} \times 8.0\text{ m} = 128\text{ m}^2

The actual dimensions determine the true surface area.

3

Determine the magnitude of the error in area

\text{Error} = |128\text{ m}^2 - 120\text{ m}^2| = 8\text{ m}^2

Error is defined as the absolute difference between actual and measured values.

4

Compute the percentage error relative to the actual area

\text{Percentage Error} = \frac{8}{128} \times 100\% = 6.25\%

Percentage error is always calculated as

\frac{\text{Error}}{\text{Actual Value}} \times 100\%

.

Key Concept

Percentage Error Calculation in Compound Quantities

Estimated Time:1m 30s

Question 7414Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 12\}$ . Consider the subsets $P = \{x : x \text{ is a factor of } 12\}$ and $Q = \{x : x \text{ is an even number}, 1 \le x \le 12\}$ . Which of the following sets represents $(P \cup Q)'$ ?

\{5, 7, 9, 11\}

\{2, 4, 6, 12\}

\{5, 7, 11\}

\{1, 3, 5, 7, 9, 11\}

Show answer & explanation

Answer: $\{5, 7, 9, 11\}$

Answer

\{5, 7, 9, 11\}

The set of factors of 12 within the given range is {1, 2, 3, 4, 6, 12} and the set of even numbers up to 12 is {2, 4, 6, 8, 10, 12}. Their union contains all numbers that are either even or factors of 12, which is {1, 2, 3, 4, 6, 8, 10, 12}. Subtracting this union from the universal set {1, 2, ..., 12} yields {5, 7, 9, 11}.

Step-by-Step Solution

1

List all elements of the universal set and the subsets

P

and

Q

.

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

,

P = \{1, 2, 3, 4, 6, 12\}

, and

Q = \{2, 4, 6, 8, 10, 12\}

.

Identify the element listings explicitly from the given definitions.

2

Find the union of sets

P

and

Q

, denoted by

P \cup Q

.

P \cup Q = \{1, 2, 3, 4, 6, 8, 10, 12\}

.

Combine all unique elements belonging to either

P

,

Q

, or both.

3

Find the complement

(P \cup Q)'

relative to the universal set

\mathcal{U}

.

(P \cup Q)' = \mathcal{U} \setminus (P \cup Q) = \{5, 7, 9, 11\}

.

Select all elements in

\mathcal{U}

that do not appear in

P \cup Q

.

Key Concept

Complement of a Set Union

Estimated Time:50s

Question 7415Question →

Five numbers $3, 7, 8, x,$ and $12$ are arranged in ascending order, where $x > 8$ . If the mean of the data set is equal to its median, what is the value of $x$ ?

10

5

14

18

Show answer & explanation

Answer: $10$

Answer

The value of

x

is

10

.

For five numbers arranged in ascending order with

x > 8

, the middle position (3rd value) is

8

, so the median is

8

. Summing all five values gives

3 + 7 + 8 + 12 + x = 30 + x

. Equating the mean

\frac{30 + x}{5}

to the median

8

yields

30 + x = 40

, which gives

x = 10

.

Step-by-Step Solution

1

Determine the median of the ordered data set.

Since there are

5

numbers arranged in ascending order (

3, 7, 8, x, 12

) with

x > 8

, the median is the middle (3rd) number, which is

8

.

For an odd count of data points

n = 5

, the median position is

\frac{5+1}{2} = 3

.

2

Formulate the expression for the arithmetic mean.

\text{Mean} = \frac{3 + 7 + 8 + 12 + x}{5} = \frac{30 + x}{5}

.

The mean of an ungrouped data set is the sum of all values divided by the total number of items.

3

Equate the mean to the median and solve for

x

.

\begin{align*} \frac{30 + x}{5} &= 8 \\ 30 + x &= 40 \\ x &= 10 \end{align*}

The question specifies that the mean of the data set is equal to its median.

Key Concept

Measures of Central Tendency for Ungrouped Data

Estimated Time:1m 30s

Question 7416Question →

The frequency distribution table below shows the recorded speeds (in $\text{km/h}$ ) of a sample of commercial buses passing through a highway toll checkpoint:

Speed ( $\text{km/h}$ )	Frequency
$40 - 49$	$6$
$50 - 59$	$10$
$60 - 69$	$f$
$70 - 79$	$12$
$80 - 89$	$8$

If the mean speed of the buses is $66.5\text{ km/h}$ , find the value of the missing frequency $f$ .

Show answer & explanation

Answer: 6

Answer

The value of the missing frequency

f

is

6

.

Each class interval's midpoint is calculated by averaging its lower and upper limits. The total frequency is

\sum f = 36 + f

and the total sum of products is

\sum fx = 2382 + 64.5f

. Applying the grouped mean formula

\bar{x} = \frac{\sum fx}{\sum f} = 66.5

gives

2394 + 66.5f = 2382 + 64.5f

, which yields

2f = 12

, so

f = 6

.

Step-by-Step Solution

1

Find the class midpoints (

x

) for each grouped interval.

Midpoints are

44.5, 54.5, 64.5, 74.5,

and

84.5

.

Midpoints represent the central values of each interval for grouped mean calculations.

2

Calculate expressions for

\sum f

and

\sum fx

.

\sum f = 36 + f

and

\sum fx = 2382 + 64.5f$.

Summing the frequencies and the products of midpoints and frequencies gives the components required for the mean equation.

3

Substitute known values into the mean formula and solve for

f

.

66.5(36 + f) = 2382 + 64.5f \implies 2f = 12 \implies f = 6$.

Equating the formula expression to the given mean of

66.5\text{ km/h}

allows solving for the unknown frequency.

Key Concept

Grouped Mean with Missing Frequency

Question 7417Question →

The $1^{\text{st}}$ , $2^{\text{nd}}$ , and $5^{\text{th}}$ terms of an arithmetic progression (A.P.) with a non-zero common difference form three consecutive terms of a geometric progression (G.P.). If the sum of the first $4$ terms of the A.P. is $40$ , what is the $5^{\text{th}}$ term of the G.P.?

\frac{405}{2}

\frac{1215}{2}

162

\frac{45}{2}

Show answer & explanation

Answer: $\frac{405}{2}$

Answer

\frac{405}{2}

By setting up the geometric mean property

(a+d)^2 = a(a+4d)

, we find

d = 2a

, which establishes that the G.P. has a common ratio

r = 3

. Substituting

d = 2a

into the A.P. sum formula

S_4 = 2[2a + 3d] = 16a = 40

gives

a = \frac{5}{2}

. Finally, evaluating the

5^{\text{th}}

term of the G.P. using

a r^4 = \frac{5}{2} \times 3^4

yields

\frac{405}{2}

.

Step-by-Step Solution

1

Express the given A.P. terms in terms of first term

a

and common difference

d

, and set up the G.P. condition.

The terms are

T_1 = a

,

T_2 = a + d

, and

T_5 = a + 4d

. Since they form a G.P.,

(a + d)^2 = a(a + 4d)

.

Three terms

x, y, z

form a G.P. if

y^2 = xz

.

2

Solve for the relationship between

d

and

a

.

a^2 + 2ad + d^2 = a^2 + 4ad \implies d^2 = 2ad \implies d = 2a

(since

d \neq 0

).

Expanding and simplifying the equation yields the ratio of

d

to

a

.

3

Determine the common ratio

r

of the G.P.

r = \frac{T_2}{T_1} = \frac{a + d}{a} = \frac{a + 2a}{a} = 3

.

The common ratio is the quotient of consecutive terms of the G.P.

4

Use the sum of the first

4

terms of the A.P. to find

a

.

S_4 = \frac{4}{2}[2a + (4-1)d] = 2[2a + 3(2a)] = 16a = 40 \implies a = \frac{40}{16} = \frac{5}{2}

.

Applying

S_n = \frac{n}{2}[2a + (n-1)d]

with

S_4 = 40

allows solving for

a

.

5

Calculate the

5^{\text{th}}

term of the G.P.

G_5 = g_1 \cdot r^{5-1} = a \cdot r^4 = \frac{5}{2} \cdot 3^4 = \frac{5}{2} \cdot 81 = \frac{405}{2}

.

The

n^{\text{th}}

term of a G.P. is

g_n = g_1 r^{n-1}

.

Key Concept

Connecting Arithmetic and Geometric Progressions using term definitions and sum formulas.

Question 7418Question →

The frequency table below shows the distribution of masses (in kg) of $20$ packages delivered to a warehouse:

Mass (kg)	Frequency ( $f$ )
10 – 14	3
15 – 19	5
20 – 24	8
25 – 29	4

What is the mean mass of the packages?

18.25\text{ kg}

19.50\text{ kg}

20.25\text{ kg}

22.25\text{ kg}

Show answer & explanation

Answer: $20.25\text{ kg}$

Answer

The mean mass of the packages is

20.25\text{ kg}

.

The mean of grouped frequency distribution is found using the formula

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

, where

x

represents the midpoint of each class interval. Calculating the midpoints yields

12

,

17

,

22

, and

27

. Multiplying each midpoint by its frequency gives products of

36

,

85

,

176

, and

108

, totaling

405

. Dividing

405

by the total frequency of

20

produces

20.25\text{ kg}

.

Step-by-Step Solution

1

Find the class midpoint (

x

) for each class interval.

Midpoints are: 10–14

\rightarrow x = 12

; 15–19

\rightarrow x = 17

; 20–24

\rightarrow x = 22

; 25–29

\rightarrow x = 27

.

The mean of grouped data requires using the midpoint of each interval to represent its data values.

2

Multiply each class midpoint (

x

) by its frequency (

f

) to find

fx

.

3 \times 12 = 36

;

5 \times 17 = 85

;

8 \times 22 = 176

;

4 \times 27 = 108

.

This calculates the total estimated mass for each class interval.

3

Sum the frequencies (

\sum f

) and the products (

\sum fx

).

\sum f = 3 + 5 + 8 + 4 = 20

and

\sum fx = 36 + 85 + 176 + 108 = 405

.

These sums give the total number of packages and the total estimated mass.

4

Calculate the mean using the formula

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

.

\bar{x} = \frac{405}{20} = 20.25\text{ kg}

.

Dividing the total mass by the total number of items yields the grouped mean.

Key Concept

Mean of Grouped Data

Estimated Time:1m 0s

Question 7419Question →

If $\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} = k\sqrt{6}$ , find the value of the integer $k$ .

Show answer & explanation

Answer: 4

Answer

The value of the integer

k

is 4.

Combining the fractions over the common denominator

(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2}) = 3-2 = 1

yields a numerator of

(3+2+2\sqrt{6}) - (3+2-2\sqrt{6}) = 4\sqrt{6}

. Thus,

k\sqrt{6} = 4\sqrt{6}

, which gives

k = 4

.

Step-by-Step Solution

1

Combine the fractions using their common denominator

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

Subtracting algebraic fractions requires finding the least common denominator, which is the product of the conjugate pair.

2

Expand the terms in the numerator

(\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6} \text{ and } (\sqrt{3} - \sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6}

Use the perfect square expansion formula

(a \pm b)^2 = a^2 \pm 2ab + b^2

.

3

Subtract the expanded terms in the numerator and simplify the denominator

\text{Numerator: } (5 + 2\sqrt{6}) - (5 - 2\sqrt{6}) = 4\sqrt{6}, \text{ Denominator: } (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1

Apply the difference of two squares identity

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

to the denominator and carefully distribute the negative sign across terms in the numerator.

4

Equate the simplified expression to

k\sqrt{6}

and solve for

k

k = 4

Comparing

\frac{4\sqrt{6}}{1} = 4\sqrt{6}

with

k\sqrt{6}

yields

k = 4

.

Key Concept

Rationalisation of surd denominators using conjugate pairs and difference of squares

Question 7420Question →

A forest ranger station $Y$ is located on a bearing of $115^\circ$ from an observation tower $X$ . What is the bearing of the observation tower $X$ from the forest ranger station $Y$ ?

065^\circ

245^\circ

295^\circ

115^\circ

Show answer & explanation

Answer: $295^\circ$

Answer

295^\circ

The bearing of

Y

from

X

is

115^\circ

. To find the bearing of

X

from

Y

(the back bearing), add

180^\circ

to the forward bearing because

115^\circ < 180^\circ

. Calculating

115^\circ + 180^\circ

yields

295^\circ

.

Step-by-Step Solution

1

Identify the given forward bearing

The bearing of

Y

from

X

is

\theta = 115^\circ

.

This is the initial directional angle measured clockwise from True North at point

X

.

2

Calculate the back bearing of

X

from

Y

Back bearing

= 115^\circ + 180^\circ = 295^\circ

.

Because the forward bearing is less than

180^\circ

, the back bearing is obtained by adding

180^\circ

to find the opposite direction.

Key Concept

Back Bearing Calculation

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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