GCSE Edexcel Mathematics Higher Paper 2 – June 2017

What are we being asked to find?

We need an estimate for the number of times the dice lands on a 1 OR a 3. First, we need the probability of rolling a 1 or a 3. Since the sum of all probabilities for a dice must be exactly 1, we can find the missing probability for rolling a 1.

Why we do this:

We know the probabilities of rolling 2, 3, 4, 5, and 6. By adding these together and subtracting from 1, we get the missing probability for 1.

Why we do this:

To find the probability of rolling a 1 or a 3, we add their individual probabilities. Then, to find the estimated frequency out of 200 rolls, we multiply this combined probability by the total number of rolls.

What this tells us:

Out of 200 rolls, we can expect the dice to land on either a 1 or a 3 approximately 98 times.

What are we being asked to find?

We need to figure out the total number of people (adults + children) in the theatre and check if that total is strictly greater than 60% of the 2600 available seats. We will work backwards from the known number of children in the Circle to find the total children, then use the ratio to find the adults.

Why we do this:

We are told \(\frac{3}{4}\) of the children are in the Stalls. This means the rest of the children are in the Circle. The fraction in the Circle is \(1 – \frac{3}{4} = \frac{1}{4}\). Since we know exactly 117 children are in the Circle, we know that \(\frac{1}{4}\) of all children equals 117.

Why we do this:

The ratio of Adults to Children is 5:2. Now that we know there are 468 children, we know that “2 parts” of our ratio is equal to 468. We can use this to find the total number of people (7 parts in total, or 5 parts for adults).

Why we do this:

To answer the final question, we need to compare our total (1638) against 60% of the theatre’s maximum capacity (2600).

What this tells us:

1560 represents the 60% mark. Because 1638 people attended, the attendance was definitely higher than 60%. (In fact, it was exactly 63%).

What are we being asked to draw?

We need to draw two 2D views of this 3D prism. The scale is 2 cm to 1 m. Since standard grid squares are \(1\text{ cm} \times 1\text{ cm}\), every 1 metre on the real prism takes up 2 squares on the grid.

• Front Elevation: This is the view looking in the direction of the “front” arrow. We will see the flat trapezium face.
• Side Elevation: This is the view looking from the “side” arrow (the right side). We will see the right-hand face, plus the part of the left-hand face that sticks up higher behind it.

Why we do this:

We convert the dimensions of the front trapezium into grid squares using our scale multiplier (multiply metres by 2 to get cm/squares).

Why we do this:

Looking from the right side, the prism is \(1\text{ m}\) deep. The tallest point at the back (left side of front face) is \(2\text{ m}\) high. The shortest point in the foreground is \(0.5\text{ m}\) high. Because it’s a flat 2D projection, the slanted top appears as a flat rectangle with a horizontal line marking where the slant begins.

What this tells us:

The completed drawing should look exactly like the grid below. The exact placement on the grid doesn’t matter as long as the dimensions are correct.

What are we being asked to find?

We need the total average speed for Olly’s entire journey. To find total average speed, we must use the formula: \(\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}\).

We cannot just average the speeds of the two legs. We need to find his total distance and his total time in hours.

Why we do this:

We know the distance (\(56\text{ km}\)) and speed (\(70\text{ km/h}\)) for the Liverpool to Manchester leg, but we are missing the time. We use \(T = \frac{D}{S}\) to find the time in hours.

Why we do this:

Now we combine the distances and times of both parts of the journey. The second leg took 75 minutes, which must be converted into hours to match our km/h speed units.

Why we do this:

Divide the total distance by the total time.

What does this tell us?

Taking the mean of two speeds (e.g., adding them and dividing by 2) only gives the true average speed for the whole journey if the person spends exactly the same amount of time travelling at both speeds.

What are we being asked to find?

We are asked to find the length of the side \(AE\). Because line \(EA\) is parallel to line \(DB\), the small triangle \(CDB\) is similar to the large triangle \(CEA\). This means the large triangle is a scaled-up version of the small one.

Why we do this:

To find the missing side, we first need to know how many times bigger the large triangle is compared to the small one. We compare matching sides: the bottom side \(EC\) of the large triangle matches the bottom side \(DC\) of the small triangle.

Why we do this:

We multiply the matching small side \(DB\) by the scale factor to find the large side \(AE\).

Why we do this:

We are asked to find \(AB\), which is just a piece of the long line \(AC\). If we know the whole length \(AC\) (which is 6.15) and we find the small length \(CB\), we can subtract them to find \(AB\). To find \(CB\), we take the large side \(AC\) and divide it by our scale factor (1.5) to shrink it down to the small triangle’s size.

Why we do this:

Since the line is \(A-B-C\), the length of \(AB\) is the total length \(AC\) minus the segment \(CB\).

What this tells us:

The segment \(AB\) is exactly 2.05 cm. Because of similarity, all parallel components grow by the exact same multiplier.

Why we do this:

We need to find out how much money Anil will have in the Personal Bank after 3 years. Since it pays 2% compound interest each year, the multiplier is \(1 + 0.02 = 1.02\). Because the money is invested for 3 years, we raise this multiplier to the power of 3.

Why we do this:

Secure Bank changes its interest rate after the first year. So, we multiply by the Year 1 multiplier (4.3% increase = 1.043), and then by the multiplier for the “extra” years (0.9% increase = 1.009). Since the total investment is 3 years, there are 2 extra years.

What this tells us:

We compare the total interest (or total final amounts) to answer the specific question: “Which bank will give Anil the most interest?”

What are we being asked to find?

An error interval shows the range of all possible original numbers before rounding. Since \(n\) was rounded to 2 decimal places, the rounding “gap” or scale is \(0.01\). To find the limits, we add and subtract half of this gap (\(0.005\)) from our rounded number.

Why we do this:

The lower bound is the absolute smallest number that would round up to 4.76. The upper bound is the cut-off point where numbers start rounding up to 4.77 instead.

What this tells us:

Any number from 4.755 up to (but strictly less than) 4.765 will round to 4.76. We format this using the inequality signs \(\le\) and \(<\).

Why we do this:

We need to find how many students are 160 cm or shorter. We do this by locating 160 on the x-axis (Height), drawing a vertical line up to the curve, and then a horizontal line across to the y-axis (Cumulative frequency).

Why we do this:

The y-axis tells us the running total of students. A reading of 48 means there are 48 students who are 160 cm or shorter. To find how many are taller, we subtract this from the total number of students (which we are told is 60, and we can also see 60 is the top of the curve).

What this tells us:

There are approximately 12 students in the top section of the graph (above 160 cm).

Why we do this:

To see if Amy is right, we need to map out where the triangles actually end up. We’ll start by taking the coordinates of a vertex on Triangle A—let’s use the top right corner at \((4, 3)\)—and track its journey through Kyle’s reflections.

• Reflect in the x-axis: The y-coordinate flips sign. \((x, y) \rightarrow (x, -y)\).
• Reflect in \(y = x\): The x and y coordinates swap places. \((x, y) \rightarrow (y, x)\).

Why we do this:

Now we do the same starting point \((4, 3)\) but apply Amy’s sequence of reflections to find out where Triangle E ends up.

What this tells us:

Kyle’s point C ended up at \((-3, 4)\). Amy’s point E ended up at \((3, -4)\). Since these are completely different coordinates, the triangles are NOT in the same position. The order of transformations matters!

Why we do this:

To find the greatest mass, we look at the exponents (the powers of 10) in the “Mass” column first. The larger the power, the larger the number.

Why we do this:

We need to subtract the mass of Mercury from the mass of Venus. Since we have a calculator, we can enter these numbers directly in standard form.

Why we do this:

Nishat claims: Distance to Neptune > 100 × (Distance to Venus). We can test this by calculating 100 times the distance to Venus and seeing if Neptune’s distance is larger than that.

What this tells us:

Because Neptune’s actual distance is greater than 100 times Venus’s distance, Nishat is correct.

Why we do this:

Solving equations with fractions is tricky. We can make it much easier by multiplying every single term by the Lowest Common Multiple (LCM) of the denominators (4, 3, and 6). The lowest number they all divide into is 12.

Why we do this:

Now that we have a linear equation without fractions, we need to multiply out the brackets. Be very careful with the negative sign in front of the second bracket: \(-4 \times 5 = -20\).

Why we do this:

We want to get all the \(x\) terms on one side of the equals sign, and all the numbers on the other side.

What this tells us:

The solution is \(x = 9\frac{1}{3}\) (or \(28/3\), or \(9.333…\)).

What are we being asked to find?

We need to spot a mathematical error in the diagram. There are a few major errors to check in probability trees:

• Do the branches from a single point add up to 1?
• Did the total number of students change correctly for the second pick? (Since a student is removed, the total should drop).

Why we do this:

In probability, the word “AND” (both boys scoring) means we should multiply the independent probabilities. The word “OR” (either boy scoring) means we should add them.

What are we being asked to find?

First, we need to find the total number of members who are over 50 years of age. In a histogram, the area of the bars represents the frequency (number of people).

\[ \text{Frequency} = \text{Class Width} \times \text{Frequency Density} \]

Why we do this:

The “over 50” group splits across two bars. The first part is from 50 to 60 (which is the right half of the 40-60 bar). We read the height (frequency density) from the y-axis.

Why we do this:

The second part of the “over 50” group is the entire 60 to 90 bar. We again calculate the area of this bar.

Why we do this:

Add the two frequencies together to get the total number of people over 50. Then, apply the 20% condition to find the number of females.

What this tells us:

There are exactly 7 female members estimated to be over 50 years old in the sports club.

Why we do this:

The sine wave is a repeating wave that passes through the origin \((0,0)\) and goes up into the positive y-values first. Looking at the options, C and D are waves. Graph C passes through \((0,0)\) and goes up. (Graph D passes through \((0,0)\) but goes down, making it \(-\sin x\)).

Why we do this:

This is a cubic equation with a positive \(x^3\) coefficient. Positive cubics generally go from the bottom-left to the top-right. When \(x = 0\), \(y = 0\), so it passes through the origin. Graph F shows this classic cubic shape passing through the origin.

Why we do this:

This is an exponential growth curve. As \(x\) gets bigger, \(y\) gets much bigger, much faster. It never drops below the x-axis, and crosses the y-axis at \((0,1)\) because \(2^0 = 1\). Graph A shows standard exponential growth.

Why we do this:

This is a reciprocal graph. It has two separate branches because you cannot divide by zero (there’s an asymptote at \(x=0\)). Because it’s a positive fraction, a positive \(x\) gives a positive \(y\) (top-right quadrant), and a negative \(x\) gives a negative \(y\) (bottom-left quadrant). Graph H shows these branches in the 1st and 3rd quadrants.

What are we being asked to do?

We need to prove that the top triangle (\(ABE\)) and the bottom triangle (\(DCE\)) are similar. Two triangles are similar if all three of their corresponding angles are equal (the AAA rule). Because these triangles are formed by chords intersecting inside a circle, we must use circle theorems to justify our equal angles.

Why we do this:

Look at the points \(A\) and \(D\). The angles \(\angle ABE\) and \(\angle DCE\) both start at point \(A\), touch the circumference, and end at point \(D\). They are subtended by the same arc \(AD\).

Why we do this:

We can use the exact same circle theorem for the other side of the circle, or we can use the intersecting straight lines at point \(E\).

What this tells us:

Because we have proven that all three angles match, the triangles must be similar.

Why we do this:

To multiply these numbers exactly, we must convert both recurring decimals into fractions. For \( 0.1\dot{3}\dot{6} \) (which means \( 0.1363636… \)), we need to set up two equations with the exact same decimal part so that when we subtract them, the infinite decimal tails cancel out completely.

Why we do this:

We repeat the same algebraic process for \( 0.\dot{2} \) (which means \( 0.2222… \)).

Why we do this:

Now we substitute our two fractions back into the original multiplication problem to show it equals \( \frac{1}{33} \).

What this tells us:

By successfully turning both decimals into fractions, multiplying them, and simplifying, we have algebraically proven the statement is true.

What are we being asked to do?

We need to find the percentage of the sector’s area that is shaded. To find any area, we first need the angle of the sector. The problem states that triangle \( AOB \) is equilateral.

Why we do this:

The formula for the area of a sector is \( \frac{\theta}{360} \times \pi \times r^2 \). We know the radius \( r = 11 \) and the angle \( \theta = 60 \).

Why we do this:

The shaded area is the total sector minus the white triangle. To find the area of a triangle when we know two sides and the included angle, we use the formula: \( \text{Area} = \frac{1}{2}ab \sin(C) \).

Why we do this:

Subtract the triangle’s area from the sector’s area to find the shaded area. Then, divide the shaded area by the total sector area and multiply by 100 to find the percentage.

What this tells us:

Rounding to 1 decimal place, the shaded region takes up 66.5% of the total sector.

Why we do this:

To solve an equation with indices (powers), all the large numbers (bases) must be the same. Here, the numbers 16, 2, and 8 are all powers of 2. We can rewrite \( 16 \) as \( 2^4 \) and \( 8 \) as \( 2^3 \).

Why we do this:

We use the index law \( (x^a)^b = x^{a \times b} \) to multiply the powers together.

Why we do this:

When multiplying terms with the same base, we add their indices: \( x^a \times x^b = x^{a+b} \). Once the left side is simplified to a single base of 2, we can just set the powers equal to each other.

What this tells us:

The exact value of the missing index \( x \) is \( 1.45 \) (or \( \frac{29}{20} \)).

Why we do this:

To combine terms into a single fraction, they must all have the same denominator. The denominators are \( (x – 3) \) and \( (x + 3) \). When multiplied, these create a difference of two squares: \( (x – 3)(x + 3) = x^2 – 9 \), which perfectly matches our target format. We will rewrite the number \( 2 \) as a fraction over this denominator as well.

Why we do this:

Now we place everything over one single denominator and expand the brackets in the numerator carefully. We must be especially careful with the minus signs in front of the fractions!

Why we do this:

We distribute the negative signs and collect like terms to get our final linear expression \( ax + b \).

What this tells us:

The entire expression simplifies to \( \frac{4x – 42}{x^2 – 9} \). Since the target format is \( \frac{ax + b}{x^2 – 9} \), we can see that \( a = 4 \) and \( b = -42 \).

What are we being asked to do?

We are given the curve \( y = x^2 – 2x + 3 \) but asked to solve a completely different equation: \( x^2 – 3x – 1 = 0 \). To solve this using the graph we already have, we need to manipulate the target equation so that one side looks exactly like our curve’s equation.

Why we do this:

Because \( x^2 – 2x + 3 \) is our curve (\( y \)), we must draw the straight line \( y = x + 4 \). The points where the line crosses the curve are the solutions to our equation. Look directly down to the x-axis from those intersection points to find the answers.

What are we being asked to do?

The gradient of a curve at a specific point is the gradient of a straight tangent line drawn exactly at that point. We find \( x = 2 \) on the curve, place our ruler there, and draw a straight line that grazes the curve.

Why we do this:

To find the gradient of our tangent line, we pick two easy-to-read points on the drawn line and use the formula: \( \text{Gradient} = \frac{\text{Change in } y}{\text{Change in } x} \).

What this tells us:

Depending on how exactly you drew your tangent line by hand, you will get an answer close to 2. Any value between 1.6 and 2.5 is accepted.

What are we being asked to do?

To find the area of the rectangle, we need its total width and its total height. Looking horizontally across the top, there are exactly two circles side-by-side touching the edges. This means the width of the rectangle is equal to the width of two full circles (4 radiuses).

Why we do this:

The height of the rectangle is NOT simply the height of two circles. The circles nestle into each other. To find the exact height, we draw straight lines connecting the centres of the 3 circles. Because all the circles have a radius of 24 mm, the distance between any two centres is \( 24 + 24 = 48\text{ mm} \). This forms an equilateral triangle.

Why we do this:

The total height of the rectangle is made of three parts: the radius of the top circle, the vertical height of the triangle (between the centres), and the radius of the bottom circle.

Why we do this:

Now that we have both dimensions, we multiply them to find the area and round to 3 significant figures as requested.

What this tells us:

Rounding 8598.64 to 3 significant figures means looking at the 4th digit (8). Because 8 is 5 or larger, the previous digit (9) rounds up. 59 rounds up to 60, giving us 8600.

What are we being asked to do?

We need to find the formula for this sequence. First, we check if it goes up by the same amount every time (a linear sequence). If it doesn’t, we check the “difference of the differences” (second differences) to see if it’s a quadratic sequence (\( an^2 + bn + c \)).

Why we do this:

For any quadratic sequence, the number in front of \( n^2 \) (the coefficient \( a \)) is always exactly half of the constant second difference.

Why we do this:

We write out the sequence generated by just \( 2n^2 \) and subtract it from our actual sequence. The leftovers will form a simple linear sequence that we can easily find the formula for.

What this tells us:

By putting the two parts together, our complete formula for the sequence is \( 2n^2 + n + 1 \).

What are we being asked to do?

We need the equation of a tangent line. A tangent to a circle is always perpendicular to the radius at that point. So, first we find the gradient of the radius from the origin \( (0,0) \) to point P.

Why we do this:

Because the tangent is perpendicular to the radius, its gradient is the “negative reciprocal” of the radius’s gradient. This means we flip the fraction upside down and change its sign.

Why we do this:

We use the straight line equation formula \( y – y_1 = m(x – x_1) \), substituting our tangent gradient for \( m \), and the coordinates of point P for \( (x_1, y_1) \).

What this tells us:

This is the final equation of the straight tangent line. The exam board also accepts the version where the denominators have been rationalised: \( y = -\frac{3\sqrt{7}}{7}x + \frac{8\sqrt{7}}{7} \).