Edexcel GCSE Mathematics – Nov 2017 Paper 3 (Foundation)

What are we being asked to find?

We need to round the number \(3758\) to the nearest thousand. This means deciding whether it is closer to \(3000\) or \(4000\).

Why we do this:

To round to the nearest thousand, we identify the digit in the thousands place and look at the digit directly to its right (the hundreds place). If that digit is \(5\) or more, we round up. If it is \(4\) or less, we stay the same.

What this tells us:

\(3758\) is closer to \(4000\) than it is to \(3000\). [cite: 85]

What are we being asked to find?

We are asked to simplify an algebraic expression by collecting ‘like terms’. All the terms have the same letter (\(y\)), so they can be combined.

Why we do this:

Remember that a letter on its own, like \(y\), is the same as \(1y\). We will work through the addition and subtraction from left to right, just like regular numbers.

What this tells us:

The entire expression simplifies down to a single term: \(2y\). [cite: 85]

What are we being asked to find?

A “factor” is a whole number that divides exactly into another number without leaving a remainder. We need to find every number that divides perfectly into \(18\).

Why we do this:

Finding factors in pairs is the safest way to make sure we don’t miss any. We start at \(1\) and work our way up.

What this tells us:

Our complete list of factors, writing them out in order, is \(1, 2, 3, 6, 9\), and \(18\).

Why we do this:

To find out how much cheaper the family ticket is, we first need to know exactly how much it would cost to buy \(1\) adult ticket and \(3\) child tickets individually.

Why we do this:

We subtract the cost of the family ticket from the cost of the separate tickets to see the savings.

What this tells us:

The family saves £\(0.90\) (or \(90\)p). Always ensure you write your final answer in proper currency notation.

Why we do this:

Because the film lasts \(102\) minutes, it’s easier to add it to the starting time if we break it down into hours (\(60\) minutes) and leftover minutes.

Why we do this:

We add \(1\) hour and \(42\) minutes to \(6.45\) pm. Let’s add the hour first, then the minutes.

What this tells us:

The film ends at \(8.27\) pm.

What are we being asked to find?

We are dividing a large total amount (\(2\) litres) into smaller portions (\(150\) ml). Before we can divide, both amounts must be in the same unit. Let’s convert litres to millilitres (ml).

Why we do this:

We want to find out how many times \(150\) ml fits entirely into \(2000\) ml. Since this is a calculator paper, we can divide \(2000\) by \(150\).

What this tells us:

We can fill \(13\) whole bottles. The decimal \(0.333…\) means there is some shampoo left over, but not enough to “completely fill” a \(14\)th bottle. Therefore, the answer is exactly \(13\).

What are we being asked to find?

Before we can draw the symbols for Friday and Saturday, we need to know exactly how many cycles one full circle represents. We can use the total for Tuesday, Wednesday, and Thursday to find this out.

What this tells us:

Our key is: \(1\) full circle represents \(4\) cycles.

This means a half circle represents \(2\) cycles, and a quarter circle represents \(1\) cycle.

Why we do this:

Now that we know the scale (\(1\) circle = \(4\) cycles), we can divide the daily sales by \(4\) to find out how many circles to draw for each missing day.

What this tells us:

To complete the pictogram, we must draw:

• Friday: \(2\) complete circles.
• Saturday: \(3\) complete circles and one \(\frac{3}{4}\) circle (a circle missing one quarter).
• Key: Write “4” in the key space.

What are we being asked to find?

We need to prove that triangle \(ABC\) is isosceles. An isosceles triangle has two sides of equal length, which means it must also have two angles of equal size. Therefore, our strategy is to calculate all the missing interior angles of the triangle to see if two of them are the same. We must provide geometric reasons for every calculation.

Why we do this:

We know the exterior angle at \(C\) is \(117^\circ\), and that \(BCD\) is a straight line. Angles on a straight line add up to \(180^\circ\).

Why we do this:

Now we know two angles inside the triangle: Angle \(B = 54^\circ\) (given) and Angle \(C = 63^\circ\) (just calculated). The angles inside any triangle always add up to \(180^\circ\).

What this tells us:

Looking at the three angles in triangle \(ABC\), we have \(54^\circ\), \(63^\circ\), and \(63^\circ\). Two of the angles are exactly the same size.

What are we being asked to find?

We need to estimate the real-life height of the building based on the visual comparison between the building and the bus. Since we know the bus is \(12\) m long in real life, we can use the length of the bus as a “ruler” to measure the building’s height.

Why we do this:

By visually stacking the length of the bus vertically against the building, we can estimate how many “bus lengths” fit into the building’s height.

What this tells us:

To find the estimated height, we multiply the real length of the bus by our estimated scale factor.

What this tells us:

A reasonable estimate for the building is \(30\) m. (Note: The mark scheme allows any answer between \(27\) m and \(33\) m, recognizing that estimations may vary slightly).

What are we being asked to find?

We need to find two numbers that meet three specific conditions:

1. They must both be prime numbers.
2. They must be different from each other.
3. When added together, they make a square number that is smaller than \(30\).

Why we do this:

Listing out our options makes it much easier to test combinations. A prime number has only two factors: \(1\) and itself. A square number is a number multiplied by itself.

Why we do this:

We systematically pick a square number from our list and see if we can find two primes that add up to it.

What this tells us:

There are multiple valid answers to this question. Any correct pair of primes will get full marks.

What are we being asked to find?

Before we can find \(\frac{5}{6}\) of Jim’s number, we need to know what his whole number is. We are told that \(2\) parts out of \(3\) equal \(48\).

Why we do this:

Now that we know Jim’s number is \(72\), the question asks us to find \(\frac{5}{6}\) of it. We divide by the denominator (\(6\)) to find one-sixth, then multiply by the numerator (\(5\)).

What this tells us:

\(\frac{5}{6}\) of Jim’s original number is \(60\).

What are we being asked to find?

We need to find the total price for \(12\) lessons under each offer and then compare them. Let’s start with Offer 1.

Why we do this:

Offer 1 gives the first lesson free. This means Douglas only pays the normal price for the remaining lessons. If he wants \(12\) lessons in total, he pays for \(11\) of them.

Why we do this:

Offer 2 gives \(5\%\) off the normal price of every lesson. We can find \(5\%\) of £\(24\) and subtract it, or we can multiply by \(0.95\) (since taking \(5\%\) off leaves \(95\%\) of the price). Let’s find the new price of one lesson first, then multiply by \(12\).

What this tells us:

Offer 1 costs £\(264.00\) and Offer 2 costs £\(273.60\). Offer 1 is the cheaper option.

What are we being asked to find?

This is a direct proportion question. To find the cost of \(3.5\) kg, it’s often easiest to find the “unit price” first—the cost of exactly \(1\) kg of apples.

Why we do this:

If \(2.5\) kg costs £\(3.60\), dividing the total cost by the number of kilograms will give us the cost per single kilogram.

Why we do this:

Now that we know \(1\) kg costs £\(1.44\), we can multiply this unit price by \(3.5\) to find the cost of \(3.5\) kg.

What this tells us:

\(3.5\) kg of apples will cost exactly £\(5.04\).

What are we being asked to find?

We are given an equation: \(y = \frac{1}{2}x – 1\). This means to find \(y\), we take our \(x\) value, halve it (multiply by \(0.5\)), and then subtract \(1\).

What this tells us:

The completed table gives us a set of coordinates \((x, y)\) to plot on our graph: \((-2, -2), (-1, -1.5), (0, -1), (1, -0.5), (2, 0), (3, 0.5)\).

Why we do this:

By plotting the coordinates from our table onto the grid, we will see they form a perfectly straight line because this is a linear equation.

Why we do this:

We are told \(y = 0.3\). We can go to \(0.3\) on the y-axis, draw a horizontal line to our graph, and read straight down to the x-axis. Alternatively, we can substitute \(y=0.3\) into our equation to find \(x\) perfectly.

What this tells us:

Both our graph reading and our algebra give us the same answer: when \(y = 0.3\), \(x = 2.6\).

What are we being asked to find?

We need to name the transformation that turns Shape A into Shape B. There are four types of transformations: translation (slide), rotation (turn), reflection (flip), and enlargement (resize).

Why we do this:

Looking at Shape B, it’s upside-down compared to Shape A, but hasn’t been rotated (the straight vertical edge is still on the left). It’s a mirror image, flipped perfectly across a line.

Why we do this:

To fully describe a reflection, we must state exactly where the mirror line is. The mirror line is always exactly halfway between the original shape and the reflected shape.

What this tells us:

The full description requires both the type of transformation and its specific feature. (Note: Never give two types of transformations, like “a reflection and a translation,” or you will get zero marks for the question!).

What are we being asked to find?

We need to fill in all the blanks in the table. The table tells us that \(2\)m of cotton costs £\(6\).

Why we do this:

If we find the cost of just \(1\)m of cotton, we can easily multiply that unit price to find the cost of \(6\)m, \(8\)m, and \(9\)m.

Why we do this:

We are given the ratio: Cotton : Silk = \(2:5\). This means whatever cotton costs per metre, silk costs \(\frac{5}{2}\) times as much.

What this tells us:

We have calculated all the missing values to complete the table.

What are we being asked to find?

We need to figure out how long it takes to pack the van. To do this, we first need to know exactly how many boxes fit inside. Since the boxes are in centimetres (cm) and the van is in metres (m), we must convert the van’s measurements into cm so they match.

Why we do this:

We cannot just divide the total volumes, because boxes are rigid and cannot be “melted” to fill gaps. We must see how many boxes fit neatly along the length, width, and height.

Why we do this:

By multiplying the number of boxes along the length, width, and height, we get the total number of boxes in the van. Then we use her packing speed (\(3\) boxes per minute) to find the time.

Why we do this:

The question asks what happens if the van’s cargo area isn’t a perfect cuboid (for example, if wheel arches get in the way).

What are we being asked to find?

To “factorise” means to put into brackets. We need to find the highest common factor (HCF) of both terms (\(4m\) and \(12\)) and pull it outside the brackets.

What this tells us:

We can check our answer by expanding the bracket: \(4 \times m = 4m\), and \(4 \times 3 = 12\). It matches the original expression.

Why we do this:

We need to correctly identify the algebraic names for the parts of \(3x + 5y\).

What are we being asked to find?

We need an algebraic formula (the \(n\)th term) that calculates how many counters are in any pattern number \(n\).

Why we do this:

The \(3\) times table goes: \(3, 6, 9…\). Our sequence goes: \(4, 7, 10…\). We need to see how to adjust the \(3\) times table to match our sequence.

What are we being asked to find?

We need to find out if there is a whole number \(n\) (a pattern number) where the number of counters equals exactly \(90\). We can do this by setting our formula equal to \(90\) and solving for \(n\).

What this tells us:

Since \(n\) must be a whole pattern number (you can’t have “pattern number \(29.66\)”), \(90\) counters cannot make a perfect pattern in this sequence. Pattern \(29\) would use \(88\) counters, and Pattern \(30\) would use \(91\) counters.

What are we being asked to find?

The median is the middle value when all the data is lined up in order. Since there are \(80\) children, we want to find out which class interval holds the middle child.

What this tells us:

The running total reaches \(39\) at the end of the \(150-160\) group. The very next child, the \(40\)th child, falls into the \(160-170\) group.

Why we do this:

To draw a frequency polygon, we plot points. The x-coordinate must be the exact midpoint of each class interval, and the y-coordinate is the frequency.

What are we being asked to find?

We are comparing the cost of petrol in two different cities using two different currencies (Pounds vs Dollars) and two different volumes (Litres vs Gallons). We need to convert them so they are both in the same currency and same volume unit. Let’s convert everything into Pounds (£) per litre.

Why we do this:

The NY price is in dollars ($\(2.83\)). Since £\(1\) equals $\(1.46\), we divide the dollar amount by \(1.46\) to find out how many pounds it is.

Why we do this:

Now we know NY petrol costs £\(1.938…\) for a whole gallon. Since a gallon is \(3.785\) litres, we divide by \(3.785\) to find the cost of just one single litre.

What this tells us:

Now we can easily compare:

• London: £\(1.089\) per litre
• New York: £\(0.512…\) per litre

New York is significantly cheaper per litre.

What are we being asked to find?

We need to calculate the volume. The formula linking mass, density, and volume is: \(\text{Density} = \text{Mass} \div \text{Volume}\). We can rearrange this to find Volume: \(\text{Volume} = \text{Mass} \div \text{Density}\).

Why we do this:

The density is given in grams per cubic centimetre (\(\text{g/cm}^3\)), but the mass is given in kilograms (kg). We must convert the mass into grams before we can do any calculations.

Why we do this:

Now that our units match, we divide the mass by the density to find the volume.

Why we do this:

The question specifically asks for \(3\) significant figures. We count the first three important digits from the left.

What this tells us:

The volume of the gold bar is \(648 \text{ cm}^3\).

What are we being asked to find?

We are given two separate ratios that share a common colour (green). To compare all three colours, we need to create a single, combined ratio for Blue : Green : Red.

Why we do this:

In the first ratio, green is \(5\) parts. In the second ratio, green is \(4\) parts. We need to find a common multiple for \(5\) and \(4\) (which is \(20\)) so we can link them together.

Why we do this:

The combined ratio tells us the simplest proportion of pens. The actual total number of pens must be a multiple of the total number of parts in this ratio.

What this tells us:

The total number of pens in the box must be a multiple of \(33\) (e.g., \(33\), \(66\), \(99\), \(132\)…).

Why we do this:

The question states there are “less than 100 pens”. We want to find the greatest possible number of red pens, which means we want the total number of pens to be as close to \(100\) as possible without going over.

What this tells us:

If the box has exactly \(99\) pens, \(15\) of them will be red.

What are we being asked to find?

The “reciprocal” of a number is \(1\) divided by that number. For a fraction, it means flipping the fraction upside down.

What this tells us:

The decimal value of the reciprocal is exactly \(0.625\).

Why we do this:

When a number is rounded to one decimal place to become \(9.8\), it could have been any number that was closer to \(9.8\) than to \(9.7\) or \(9.9\).

We find the boundaries by going exactly halfway to the next possible rounded numbers above and below.

What this tells us:

The error interval is written using inequalities to show that \(x\) can be equal to the lower bound, but must be just under the upper bound: \(9.75 \le x < 9.85\).

What are we being asked to find?

We are given the total perimeter of the new shape (\(70\) cm). To find the area, we first need to figure out the value of \(x\) (the width of one rectangle). We can do this by creating an algebra equation for the perimeter.

Why we do this:

By tracing the outer edge of the 8-sided shape, we can add up all the sections in terms of \(x\) and \(x+7\). The arrangement creates an outer boundary made up of the full lengths and full widths of the original rectangles.

Why we do this:

We can simplify our long algebra expression and set it equal to \(70\) cm to solve for \(x\).

What this tells us:

The width of one rectangle is \(3.5\) cm. The length is \(3.5 + 7 = 10.5\) cm.

Why we do this:

The question asks for the area of the entire 8-sided shape. Because the shape is made of exactly \(4\) identical rectangles, we can find the area of one rectangle and multiply by \(4\).

What are we being asked to find?

We are multiplying two numbers written in standard form, and we need to output a normal, non-standard form decimal number.

Why we do this:

In standard form multiplication, you can rearrange the calculation to multiply the regular numbers together, and multiply the powers of \(10\) together using index laws.

Why we do this:

The question specifically asks for an “ordinary number,” meaning we cannot leave our answer with a \(\times 10\) attached to it.

What this tells us:

The final ordinary number is \(0.0007452\). (You can also get this directly by typing the original expression into a scientific calculator and pressing equals, depending on your calculator’s settings!).

What are we being asked to find?

We need to decide which person’s experiment gives the most reliable probability estimate. In experimental probability (relative frequency), the more times you do an experiment, the more reliable your results become.

What this tells us:

Mel dropped the pin the most times (\(80\) times), so her results are the most reliable estimate.

Why we do this:

The question asks us to use “all the results in the table.” To get the absolute best estimate possible, we pool everyone’s data together to create one giant experiment.

What this tells us:

We can simplify these fractions to make the next step easier:

\(P(\text{point up}) = \frac{1}{3}\)

\(P(\text{point down}) = \frac{2}{3}\)

Why we do this:

Stuart drops the pin twice. We want the probability of “Point Up AND THEN Point Down”. When we have independent events happening one after the other (AND), we multiply their probabilities together.

What are we being asked to find?

We need to find the specific values for \(x\) and \(y\) that make both equations true at the same time. We will use the elimination method.

Why we do this:

To eliminate a letter, the number in front of that letter (the coefficient) must be the same in both equations. Right now, we have \(3y\) in the top equation and \(-1y\) in the bottom equation. If we multiply the entire bottom equation by \(3\), we will have \(3y\) and \(-3y\), allowing us to eliminate \(y\).

Why we do this:

Because we have a \(+3y\) in Eq 1 and a \(-3y\) in Eq 3, adding the two equations together will make the \(y\) terms cancel out (\(3y + (-3y) = 0\)).

What this tells us:

We have successfully found half the answer: \(x = 1.5\).

Why we do this:

Now that we know \(x = 1.5\), we can substitute this value back into either of our original equations to find the value of \(y\). Equation 1 looks easier because everything is positive.

Why we do this:

With simultaneous equations, you can always know if you’re right by plugging both answers into the other original equation (Equation 2).