Mathematics (Syllabus D) 4024/21 — May/June 2018

Cambridge O-Level · Calculator · worked solutions for every part, with the mark scheme

questions

100

marks

150

minutes

Topics Number · Algebra and Graphs · Geometry · Mensuration · Trigonometry · Probability · +3 more

Q1NumberFree sample

(a)

Use set notation to describe the shaded region in the Venn diagram.

Fig. 1

______

DifficultyEasy

Worked solution

Approach

Identify the unshaded and shaded areas. The two circles together represent the union $P \cup Q$ . The shaded region consists of all elements outside both circles, which is the complement of their union.

Working

The set containing all elements in either $P$ or $Q$ or both is $P \cup Q$ .

The region shaded is everything outside $P \cup Q$ , which is represented as:

(P \cup Q)'

Alternatively, by De Morgan's laws, this is the region outside $P$ and simultaneously outside $Q$ :

P' \cap Q'

Answer

(P \cup Q)'

Final answer

(P ∪ Q)'

Detailed explanation

Walkthrough

Look at the two circles labelled $P$ and $Q$ . The combined area inside either circle represents the union of the two sets, written as $P \cup Q$ .
The shaded region covers the area of the universal set $\mathscr{E}$ that lies completely outside both circles.
The complement of a set refers to all elements not in that set, indicated with an apostrophe/prime symbol ( $'$ ).
Therefore, everything outside $P \cup Q$ is written as $(P \cup Q)'$ . Equivalently, it can be written as the intersection of elements not in $P$ and not in $Q$ , which is $P' \cap Q'$ .

Key Takeaways

$\cup$ denotes the union ("or").
$\cap$ denotes the intersection ("and").
$'$ denotes the complement ("not").
The region outside the union of two sets is $(A \cup B)' = A' \cap B'$ .

Common Mistakes

Confusing union $(\cup)$ with intersection $(\cap)$ , e.g. incorrectly writing $(P \cap Q)'$ , which would represent everything outside just the overlap.
Forgetting parentheses when taking the complement of a union: writing $P \cup Q'$ means "elements in $P$ or outside $Q$ ", not "outside both $P$ and $Q$ ".

Things to Be Careful About

Either $(P \cup Q)'$ or $P' \cap Q'$ is acceptable, but ensure brackets and prime symbols are placed accurately.

Techniques used

identify shaded regions in Venn diagramsapply set complement and union notation

(b)

$\mathscr{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$
$A = \{x : x \text{ is a factor of } 12\}$
$B = \{x : x \text{ is a multiple of } 2\}$
$C = \{x : x \text{ is a square number\}}$

(i)

Show this information on the Venn diagram below.

Fig. 2

DifficultyMedium-Easy

Worked solution

Approach

First, list the elements belonging to the universal set $\mathscr{E}$ and each of the sets $A$ , $B$ , and $C$ . Then determine the exact region in the Venn diagram for every integer from $1$ to $12$ .

Working

Given:

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

List the elements of each set:

Factors of $12$ within $\mathscr{E}$ :

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

Multiples of $2$ within $\mathscr{E}$ :

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

Square numbers within $\mathscr{E}$ :

C = \{1, 4, 9\}

Now place each number into its unique region:

In all three sets ( $A \cap B \cap C$ ): $4$
In $A$ and $B$ only ( $A \cap B \cap C'$ ): $2, 6, 12$
In $A$ and $C$ only ( $A \cap C \cap B'$ ): $1$
In $B$ and $C$ only ( $B \cap C \cap A'$ ): none
In $A$ only ( $A \cap B' \cap C'$ ): $3$
In $B$ only ( $B \cap A' \cap C'$ ): $8, 10$
In $C$ only ( $C \cap A' \cap B'$ ): $9$
Outside all three sets ( $(A \cup B \cup C)'$ ): $5, 7, 11$

Answer

Final answer

Venn diagram completed with elements: A only: {3}; B only: {8, 10}; C only: {9}; A and B only: {2, 6, 12}; A and C only: {1}; B and C only: empty; A, B and C: {4}; Outside: {5, 7, 11}

Detailed explanation

Walkthrough

Define the sets explicitly from the universal set $\mathscr{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$ $E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}$ :
- Set $A$ (factors of $12$ ): $\{1, 2, 3, 4, 6, 12\}$
- Set $B$ (multiples of $2$ / even numbers): $\{2, 4, 6, 8, 10, 12\}$
- Set $C$ (square numbers): $\{1, 4, 9\}$
Identify intersections starting from the central triple intersection:
- Numbers appearing in $A$ , $B$ , and $C$ : $4$ is a factor of $12$ , a multiple of $2$ , and a square number. Place $4$ in the centre region ( $A \cap B \cap C$ ).
- Numbers in $A$ and $B$ but not $C$ : $2, 6, 12$ . Place these in the region common to $A$ and $B$ only.
- Numbers in $A$ and $C$ but not $B$ : $1$ . Place $1$ in the region common to $A$ and $C$ only.
- Numbers in $B$ and $C$ but not $A$ : none.
Identify elements unique to each set:
- Set $A$ only: $3$
- Set $B$ only: $8, 10$
- Set $C$ only: $9$
Identify elements in neither $A$ $A$ , $B$ $B$ , nor $C$ $C$ :
- Check remaining numbers: $5, 7, 11$ are not in any of the three sets. Place them in the region outside the three circles, inside the rectangle.

Key Takeaways

Always start filling a 3-set Venn diagram from the innermost intersection ( $A \cap B \cap C$ ) and work outwards.
Verify that every element of the universal set appears exactly once in the diagram.

Common Mistakes

Repeating elements across multiple sections rather than placing them exclusively in intersection regions.
Forgetting elements that do not belong to any subset (e.g. $5, 7, 11$ ) and omitting them from the rectangle.
Missing factors of 12 (often missing $1$ or $12$ ).

Things to Be Careful About

Ensure all 12 elements are accounted for with no duplicates.

Techniques used

list elements of sets from definitionsplace elements in correct Venn diagram regions

(ii)

Find $n(A \cap B)$ .

______

DifficultyEasy

Worked solution

Approach

Find the set of elements that belong to both $A$ and $B$ , which corresponds to the entire overlap of circles $A$ and $B$ , and count how many elements it contains.

Working

The elements in $A \cap B$ are the elements present in both $A$ and $B$ :

A \cap B = \{2, 4, 6, 12\}

Counting the number of elements:

n(A \cap B) = 4

Answer

4

Final answer

Detailed explanation

Walkthrough

The notation $n(A \cap B)$ means "the number of elements in the intersection of set $A$ and set $B$ ".
The intersection $A \cap B$ includes all elements that lie inside circle $A$ and inside circle $B$ simultaneously (this includes both the region for $A$ and $B$ only, and the central region for $A$ , $B$ , and $C$ ).
The elements in this intersection are $2, 6, 12,$ and $4$ .
There are $4$ elements in total, so $n(A \cap B) = 4$ .

Key Takeaways

$n(X)$ asks for the count (cardinality) of set $X$ , not the list of elements.
$A \cap B$ comprises all elements in the overlap between circle $A$ and circle $B$ , regardless of whether they also belong to $C$ .

Common Mistakes

Listing the elements $\{2, 4, 6, 12\}$ instead of giving the count $4$ .
Excluding the central element $4$ (confusing $A \cap B$ with $A \cap B \cap C'$ ).

Things to Be Careful About

Give a single integer as the answer representing the count.

Techniques used

count elements in an intersection

(iii)

Find $n(A \cap (B \cup C)')$ .

______

DifficultyMedium-Easy

Worked solution

Approach

Identify the region described by $A \cap (B \cup C)'$ . This means elements that are inside set $A$ and NOT inside set $B$ or set $C$ (i.e. elements in $A$ only).

Working

First, identify $B \cup C$ :

B \cup C = \{1, 2, 4, 6, 8, 9, 10, 12\}

Its complement $(B \cup C)'$ contains all elements outside $B$ and $C$ :

(B \cup C)' = \{3, 5, 7, 11\}

Intersecting with $A = \{1, 2, 3, 4, 6, 12\}$ :

A \cap (B \cup C)' = \{3\}

Counting the elements:

n(A \cap (B \cup C)') = 1

Answer

1

Final answer

Detailed explanation

Walkthrough

The expression $(B \cup C)'$ represents everything that is outside both circle $B$ and circle $C$ .
Intersecting this with $A$ , written as $A \cap (B \cup C)'$ , isolates the part of circle $A$ that does not touch $B$ or $C$ . This is the " $A$ only" region.
Looking at the Venn diagram, the only element in the " $A$ only" region is $3$ .
The notation $n(\dots)$ asks for the number of elements in this set, which is $1$ .

Key Takeaways

$A \cap (B \cup C)'$ is the formal set notation for the elements that belong exclusively to set $A$ .

Common Mistakes

Writing the element itself ( $3$ ) rather than the number of elements ( $1$ ).
Misinterpreting the complement bracket and including elements outside the entire diagram.

Things to Be Careful About

Ensure the final answer is the integer count of elements.

Techniques used

interpret complex set expressionscount elements in a specified region

(iv)

One subset in the Venn diagram in part (b)(i) has no elements.
Use set notation to describe this subset.

______

DifficultyMedium-Easy

Worked solution

Approach

Inspect the Venn diagram from part (b)(i) to locate the region that contains no elements, then write the standard set notation for that specific region.

Working

Looking at the 8 distinct regions in the Venn diagram:

$A \cap B \cap C'$ has $\{2, 6, 12\}$
$A \cap C \cap B'$ has $\{1\}$
$A \cap B \cap C$ has $\{4\}$
$A \cap B' \cap C'$ has $\{3\}$
$B \cap A' \cap C'$ has $\{8, 10\}$
$C \cap A' \cap B'$ has $\{9\}$
$(A \cup B \cup C)'$ has $\{5, 7, 11\}$
$B \cap C \cap A'$ (the region representing elements in $B$ and $C$ but not $A$ ) contains no elements.

In set notation, this subset is:

A' \cap B \cap C

Answer

A' \cap B \cap C

Final answer

A' ∩ B ∩ C

Detailed explanation

Walkthrough

From the completed Venn diagram in part (b)(i), check each of the 8 disjoint regions.
The region corresponding to elements in $B$ and $C$ but not $A$ (the overlap between circles $B$ and $C$ excluding circle $A$ ) has no numbers written in it.
To describe this region in set notation:
- It is inside $B$ , so we include $B$ .
- It is inside $C$ , so we include $C$ .
- It is outside $A$ , so we include $A'$ .
Combining these with the intersection symbol gives $A' \cap B \cap C$ (or $B \cap C \cap A'$ ).

Key Takeaways

A specific single region in a 3-set Venn diagram is described as the intersection of three sets, where sets containing the region are unprimed and sets not containing the region are primed ( $'$ ).

Common Mistakes

Writing $B \cap C$ , which represents the entire overlap of $B$ and $C$ (including $4$ , which is in $A$ ). The subset with no elements is specifically the part outside $A$ , so $A'$ must be included.
Writing $\varnothing$ or the empty set symbol: the question asks to describe the subset using set notation involving $A, B,$ and $C$ , not just state that it is empty.

Things to Be Careful About

Ensure $A'$ is explicitly included to exclude the central region where $4$ is located.

Techniques used

identify empty regions in a Venn diagramexpress subsets using set notation

(c)

(i)

Write 540 as the product of its prime factors.

______

DifficultyMedium-Easy

Worked solution

Approach

Use successive division by prime numbers or a factor tree to break $540$ down into its prime factors, then write the result in index form.

Working

Divide $540$ successively by prime numbers:

\begin{aligned} 540 \div 2 &= 270 \\ 270 \div 2 &= 135 \\ 135 \div 3 &= 45 \\ 45 \div 3 &= 15 \\ 15 \div 3 &= 5 \\ 5 \div 5 &= 1 \end{aligned}

Collect the prime factors:

540 = 2 \times 2 \times 3 \times 3 \times 3 \times 5 = 2^2 \times 3^3 \times 5

Answer

2^2 \times 3^3 \times 5

Final answer

2^2 × 3^3 × 5

Detailed explanation

Walkthrough

Start dividing $540$ $540$ by the smallest prime number, $2$ $2$ :
- $540 \div 2 = 270$
- $270 \div 2 = 135$ (no longer divisible by 2)
Move to the next prime number, $3$ $3$ :
- $135 \div 3 = 45$
- $45 \div 3 = 15$
- $15 \div 3 = 5$ (no longer divisible by 3)
Move to the next prime number, $5$ $5$ :
- $5 \div 5 = 1$
Count the number of times each prime factor appears:
- $2$ appears twice ( $2^2$ )
- $3$ appears three times ( $3^3$ )
- $5$ appears once ( $5^1$ or $5$ )
Write the final answer as the product in index notation: $2^2 \times 3^3 \times 5$ .

Key Takeaways

Prime factorisation expresses a composite number as a product of prime numbers.
Index notation is the standard form for presenting prime factorisations.

Common Mistakes

Leaving non-prime factors in the final product (e.g. $4 \times 27 \times 5$ ).
Arithmetic errors when dividing by 3 (e.g. $135 \div 3 = 35$ ).
Forgetting to write the answer in index form.

Things to Be Careful About

Ensure all bases are prime numbers.

Techniques used

perform prime factorisation using successive division or factor treeexpress a number in index form

(ii)

$p$ is the smallest possible integer such that $540p$ is a square number.
Find $\sqrt{540p}$ , giving your answer as the product of its prime factors.

______

DifficultyMedium-Easy

Worked solution

Approach

For $540p$ to be a square number, all powers of its prime factors must be even integers. Find the minimal value of $p$ , determine the prime factorisation of $540p$ , and take its square root by halving each index.

Working

From part (c)(i):

540 = 2^2 \times 3^3 \times 5

To make all powers even, the power of $3$ needs one more factor of $3$ (to become $3^4$ ), and the power of $5$ needs one more factor of $5$ (to become $5^2$ ).

Thus, the smallest integer $p$ is:

p = 3 \times 5 = 15

Then the square number is:

540p = 2^2 \times 3^4 \times 5^2

Now find the square root by halving all powers:

\sqrt{540p} = \sqrt{2^2 \times 3^4 \times 5^2} = 2^1 \times 3^2 \times 5^1 = 2 \times 3^2 \times 5

Answer

2 \times 3^2 \times 5

Final answer

2 × 3^2 × 5

Detailed explanation

Walkthrough

A number is a perfect square if and only if all the exponents in its prime factorisation are even.
The prime factorisation of $540$ $540$ is $2^2 \times 3^3 \times 5^1$ $2^{2} \times 3^{3} \times 5^{1}$ .
- The power of $2$ is $2$ (already even).
- The power of $3$ is $3$ (odd), so we need one more $3$ to make it $3^4$ .
- The power of $5$ is $1$ (odd), so we need one more $5$ to make it $5^2$ .
Therefore, the smallest integer multiplier is $p = 3 \times 5 = 15$ .
The product $540p$ is $2^2 \times 3^4 \times 5^2$ .
To find $\sqrt{540p}$ $540 p$ , divide each exponent by $2$ $2$ :
- $\sqrt{2^2} = 2^1 = 2$
- $\sqrt{3^4} = 3^2$
- $\sqrt{5^2} = 5^1 = 5$
The question specifies giving the answer as the product of its prime factors, so the result is $2 \times 3^2 \times 5$ (which evaluates to $90$ ).

Key Takeaways

To make a number a square number, multiply by the missing prime factors needed to make every exponent even.
Taking the square root of a prime-factorised number corresponds to halving every exponent.

Common Mistakes

Evaluating the numerical answer as $90$ and forgetting to leave it as a product of prime factors as demanded by the question.
Finding $p$ (which is $15$ ) instead of finding $\sqrt{540p}$ .
Giving $540p = 2^2 \times 3^4 \times 5^2$ instead of taking its square root.

Things to Be Careful About

Check the question demand: "giving your answer as the product of its prime factors". Do not write just $90$ .

Techniques used

use prime factorisation to find the multiplier for a square numberevaluate the square root in prime factor form

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