Section A [52 marks] Answer all questions in this section. (a) Simplify. $$ \frac{4x - 1}{3} + \frac{3x + 5}{2} $$ \_\_\_\_\_\_ (b) (i) Find the gradient of line $J$. \_\_\_\_\_\_ (ii) Write down the equation of line $K$. \_\_\_\_\_\_ (iii) Draw a line, $L$, through $(6, 1)$ such that the area enclosed between $J$, $K$ and $L$ is $6\,\text{cm}^2$. (iv) Find the equation of line $L$. \_\_\_\_\_\_ (v) The line $N$ is perpendicular to line $J$ at $(2, 2)$. Find the coordinates…

(a) (17x + 13) / 6 · (b)(i) 1/2 · (b)(ii) y = 1 · (b)(iii) Line from (6, 1) to (4, 3) · (b)(iv) y = -x + 7 · (b)(v) (0, 6)

Mathematics (Syllabus D) 4024/22 — May/June 2015

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

questions

100

marks

150

minutes

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

Q1NumberAlgebra and GraphsCoordinate GeometryMensurationFree sample

Section A [52 marks]

Answer all questions in this section.

(a)

Simplify.

\frac{4x - 1}{3} + \frac{3x + 5}{2}

______

DifficultyMedium-Easy

Worked solution

Approach

Find a common denominator for the two fractions, which is $6$ . Expand the numerators by multiplying each by the necessary factor, then combine and simplify.

Working

\frac{4x - 1}{3} + \frac{3x + 5}{2}

Multiply the first fraction by $\frac{2}{2}$ and the second by $\frac{3}{3}$ :

= \frac{2(4x - 1)}{6} + \frac{3(3x + 5)}{6}

Expand the numerators:

= \frac{8x - 2 + 9x + 15}{6}

Collect like terms in the numerator:

= \frac{17x + 13}{6}

Answer

\frac{17x + 13}{6}

Final answer

(17x + 13) / 6

Detailed explanation

Walkthrough

To add two algebraic fractions, we first need a common denominator. The denominators are $3$ and $2$ , so the least common denominator is $6$ . We multiply the numerator and denominator of the first fraction by $2$ and the second by $3$ to achieve this common denominator. This gives us $\frac{2(4x - 1)}{6} + \frac{3(3x + 5)}{6}$ . Next, we expand the brackets in the numerators: $2 \times 4x = 8x$ , $2 \times -1 = -2$ , $3 \times 3x = 9x$ , and $3 \times 5 = 15$ . Combining these over the single denominator gives $\frac{8x - 2 + 9x + 15}{6}$ . Finally, we collect the $x$ terms ( $8x + 9x = 17x$ ) and the constant terms ( $-2 + 15 = 13$ ) to get the simplified result $\frac{17x + 13}{6}$ .

Key Takeaways

When adding algebraic fractions, always find the lowest common denominator first. Be careful with the signs when expanding the numerators, especially when multiplying by negative constants.

Common Mistakes

Forgetting to multiply the entire numerator by the scaling factor (e.g., writing $8x - 1$ instead of $8x - 2$ ).
Sign errors when combining the constants (e.g., $-2 + 15 = 13$ , not $17$ ).
Not simplifying the final fraction if possible (here it is already in simplest form).

Things to Be Careful About

The mark scheme specifies "cao" (correct answer only), meaning no method marks are awarded if the final answer is wrong. Ensure all expansion steps are shown clearly to avoid arithmetic mistakes. The answer must be presented as a single fraction $\frac{17x + 13}{6}$ , not as separate terms like $\frac{17x}{6} + \frac{13}{6}$ .

Techniques used

find a common denominator for algebraic fractionsexpand numerators using distributive propertycollect like terms in the numerator

(b)

Fig. 1

(i)

Find the gradient of line $J$ .

______

DifficultyEasy

Worked solution

Approach

Identify two points on line $J$ from the graph and use the gradient formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ .

Working

From the graph, line $J$ passes through $(0, 1)$ and $(10, 6)$ .

m = \frac{6 - 1}{10 - 0} = \frac{5}{10} = \frac{1}{2}

Answer

\frac{1}{2}

Final answer

1/2

Detailed explanation

Walkthrough

The gradient (or slope) of a straight line is the change in $y$ divided by the change in $x$ between any two points on the line. From Fig. 1, we can clearly read that line $J$ passes through the $y$ -intercept $(0, 1)$ and the point $(10, 6)$ . Using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ , we calculate $m = \frac{6 - 1}{10 - 0} = \frac{5}{10} = \frac{1}{2}$ .

Key Takeaways

The gradient formula $m = \frac{\text{rise}}{\text{run}}$ is fundamental for finding the slope of a line from a graph. Always pick points where the line crosses grid intersections to ensure accuracy.

Common Mistakes

Reading the coordinates incorrectly (e.g., swapping $x$ and $y$ values).
Calculating $\frac{x_2 - x_1}{y_2 - y_1}$ instead of $\frac{y_2 - y_1}{x_2 - x_1}$ .
Not simplifying the fraction to its lowest terms.

Things to Be Careful About

Ensure you are reading the coordinates from the correct axes. The mark scheme accepts $\frac{1}{2}$ or $0.5$ . Both are correct, but fractions are generally preferred in exact form.

Techniques used

identify coordinates of two points on the lineapply the gradient formula m = (y2 - y1) / (x2 - x1)

(ii)

Write down the equation of line $K$ .

______

DifficultyEasy

Worked solution

Approach

Line $K$ is a horizontal line. The equation of any horizontal line is $y = c$ , where $c$ is the $y$ -intercept.

Working

From the graph, line $K$ crosses the $y$ -axis at $y = 1$ and is parallel to the $x$ -axis.

y = 1

Answer

y = 1

Final answer

y = 1

Detailed explanation

Walkthrough

Line $K$ is drawn horizontally across the graph, passing through $y = 1$ for all values of $x$ . The equation of a horizontal line is always $y = c$ , where $c$ is the constant $y$ -value. Since the line crosses the $y$ -axis at $1$ , the equation is simply $y = 1$ .

Key Takeaways

Horizontal lines have a gradient of $0$ and their equation is $y = c$ . Vertical lines have an undefined gradient and their equation is $x = c$ .

Common Mistakes

Writing the equation as $x = 1$ (confusing horizontal and vertical lines).
Adding unnecessary terms like $y = 0x + 1$ when $y = 1$ is the expected simplified form.

Things to Be Careful About

The mark scheme specifies "final answer" only, meaning no working is required. Simply write $y = 1$ .

Techniques used

recognize the equation form of a horizontal line

(iii)

Draw a line, $L$ , through $(6, 1)$ such that the area enclosed between $J$ , $K$ and $L$ is $6\,\text{cm}^2$ .

DifficultyMedium

Worked solution

Approach

The region enclosed by lines $J$ , $K$ , and $L$ forms a triangle. Line $K$ is $y = 1$ and line $J$ is $y = \frac{1}{2}x + 1$ , which intersect at $(0, 1)$ . Line $L$ passes through $(6, 1)$ on line $K$ . Calculate the required height of the triangle to achieve an area of $6\,\text{cm}^2$ , then find the corresponding point on line $J$ to draw line $L$ .

Working

The base of the triangle lies along line $K$ from $x = 0$ to $x = 6$ , so the base length is $6$ .

Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height}$

6 = \frac{1}{2} \times 6 \times h

6 = 3h \implies h = 2

The height is the vertical distance from line $K$ ( $y = 1$ ) to the third vertex on line $J$ . Thus, the $y$ -coordinate of the third vertex is $1 + 2 = 3$ .

Substitute $y = 3$ into the equation of line $J$ ( $y = \frac{1}{2}x + 1$ ):

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

2 = \frac{1}{2}x \implies x = 4

The third vertex is $(4, 3)$ . Draw line $L$ from $(6, 1)$ to $(4, 3)$ .

Answer

Line drawn from $(6, 1)$ to $(4, 3)$ .

Final answer

Line from (6, 1) to (4, 3)

Detailed explanation

Walkthrough

First, identify the vertices of the triangle. Lines $J$ and $K$ intersect at $(0, 1)$ . Line $L$ must pass through $(6, 1)$ , which lies on line $K$ . This means two vertices of the triangle are $(0, 1)$ and $(6, 1)$ , giving a horizontal base of length $6$ along line $K$ . The area of the triangle is given as $6\,\text{cm}^2$ . Using the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ , we have $6 = \frac{1}{2} \times 6 \times h$ , which gives $h = 2$ . The height is the perpendicular distance from the base (line $K$ , $y = 1$ ) to the third vertex, which must lie on line $J$ . Therefore, the $y$ -coordinate of the third vertex is $1 + 2 = 3$ . To find the $x$ -coordinate, substitute $y = 3$ into the equation of line $J$ . From part (i), the gradient is $\frac{1}{2}$ and the $y$ -intercept is $1$ , so the equation is $y = \frac{1}{2}x + 1$ . Solving $3 = \frac{1}{2}x + 1$ gives $x = 4$ . The third vertex is $(4, 3)$ . Draw a straight line from $(6, 1)$ to $(4, 3)$ .

Key Takeaways

When a problem involves finding a line to enclose a specific area, identify the fixed vertices first, use the area formula to find missing dimensions, and then use the equations of the given lines to locate the exact coordinates.

Common Mistakes

Assuming the triangle is right-angled and misidentifying the base and height.
Calculating the height incorrectly (e.g., using $y = 6$ instead of $y = 3$ ).
Forgetting that the third vertex must lie on line $J$ , not just anywhere at $y = 3$ .

Things to Be Careful About

The mark scheme awards 1 mark for the correct line from $(6, 1)$ to $(4, 3)$ . Ensure the line is drawn accurately on the grid. The area must be exactly $6\,\text{cm}^2$ , which corresponds to a height of $2$ units on this grid.

Techniques used

calculate the required height for a given triangle areafind the intersection point on line J using the area constraintdraw the line segment connecting the two vertices

(iv)

Find the equation of line $L$ .

______

DifficultyMedium-Easy

Worked solution

Approach

Use the two points on line $L$ , $(6, 1)$ and $(4, 3)$ , to calculate its gradient, then use the point-slope form to find the equation.

Working

Points on $L$ : $(6, 1)$ and $(4, 3)$ .

\text{Gradient } m = \frac{3 - 1}{4 - 6} = \frac{2}{-2} = -1

Using point $(6, 1)$ and $m = -1$ in $y - y_1 = m(x - x_1)$ :

y - 1 = -1(x - 6)

y - 1 = -x + 6

y = -x + 7

Answer

y = -x + 7

Final answer

y = -x + 7

Detailed explanation

Walkthrough

Line $L$ passes through $(6, 1)$ and $(4, 3)$ . First, calculate the gradient $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 1}{4 - 6} = \frac{2}{-2} = -1$ . Next, use the point-slope formula $y - y_1 = m(x - x_1)$ with the point $(6, 1)$ and $m = -1$ . This gives $y - 1 = -1(x - 6)$ , which simplifies to $y - 1 = -x + 6$ , and finally $y = -x + 7$ .

Key Takeaways

To find the equation of a line when two points are known, calculate the gradient first, then use one of the points to find the $y$ -intercept or use the point-slope form directly.

Common Mistakes

Sign errors when calculating the gradient (e.g., $\frac{2}{2}$ instead of $\frac{2}{-2}$ ).
Algebra errors when expanding and rearranging the equation (e.g., $y = -x + 5$ instead of $y = -x + 7$ ).
Not writing the equation in the form $y = mx + c$ .

Things to Be Careful About

The mark scheme awards B1 for any equation with gradient $-1$ and/or intercept $7$ . Ensure the final answer is clearly stated as $y = -x + 7$ . Working must be shown (www) to earn the method marks.

Techniques used

calculate gradient from two known pointsuse point-slope form to find the equation

(v)

The line $N$ is perpendicular to line $J$ at $(2, 2)$ .

Find the coordinates of the point where line $N$ crosses the $y$ -axis.

______

DifficultyMedium

Worked solution

Approach

Line $N$ is perpendicular to line $J$ . The gradient of $J$ is $\frac{1}{2}$ , so the gradient of $N$ is the negative reciprocal, $-2$ . Use the point $(2, 2)$ and $m = -2$ to find the equation of $N$ , then find where it crosses the $y$ -axis (set $x = 0$ ).

Working

Gradient of $J$ : $m_J = \frac{1}{2}$ .

Gradient of $N$ (perpendicular to $J$ ): $m_N = -\frac{1}{m_J} = -2$ .

Line $N$ passes through $(2, 2)$ with $m = -2$ :

y - 2 = -2(x - 2)

y - 2 = -2x + 4

y = -2x + 6

To find the $y$ -intercept, set $x = 0$ :

y = -2(0) + 6 = 6

The line crosses the $y$ -axis at $(0, 6)$ .

Answer

(0, 6)

Final answer

(0, 6)

Detailed explanation

Walkthrough

First, recall that the product of the gradients of two perpendicular lines is $-1$ . Since the gradient of line $J$ is $\frac{1}{2}$ , the gradient of line $N$ is $-2$ . Line $N$ passes through $(2, 2)$ , so use the point-slope form $y - y_1 = m(x - x_1)$ with $m = -2$ and $(x_1, y_1) = (2, 2)$ . This gives $y - 2 = -2(x - 2)$ , which simplifies to $y = -2x + 6$ . The $y$ -intercept occurs where $x = 0$ , so $y = 6$ . The coordinates of the point where line $N$ crosses the $y$ -axis are $(0, 6)$ .

Key Takeaways

Perpendicular lines have gradients that are negative reciprocals of each other ( $m_1 \times m_2 = -1$ ). To find where a line crosses the $y$ -axis, set $x = 0$ in the equation and solve for $y$ .

Common Mistakes

Forgetting the negative sign when finding the perpendicular gradient (e.g., using $2$ instead of $-2$ ).
Algebra errors when expanding $-2(x - 2)$ (e.g., $-2x - 4$ instead of $-2x + 4$ ).
Reading the $y$ -intercept from the graph incorrectly instead of calculating it.

Things to Be Careful About

The mark scheme awards B1 for a line from $(2, 2)$ with $y$ -intercept between $5$ and $7$ (supported by an incorrect equation), or for the correct unsimplified equation $y = -2x + 6$ . The final answer must be the coordinates $(0, 6)$ . Ensure working is shown to earn the method marks.

Techniques used

find the negative reciprocal of the gradient for a perpendicular lineuse point-slope form to find the equation of the perpendicular linecalculate the y-intercept

The rest of this paper

10 more questions

Q2Algebra and Graphs · Number8M
Q3Statistics7M
Q4Geometry · Mensuration · Algebra and Graphs9M
Q5Trigonometry11M
Q6Number8M
Q7Algebra and Graphs12M
Q8Geometry · Trigonometry12M
Q9Algebra and Graphs · Trigonometry · Mensuration12M
Q10Transformations and Vectors · Geometry · Mensuration12M
Q11Probability · Statistics · Number12M

Loading the full paper…