A sequence of patterns is made using crosses and circles. (a) Draw Pattern 4. (b) Complete the table for Pattern 4 and Pattern 5. Pattern number ($n$) 1 2 3 4 5 --- --- --- --- --- --- Number of crosses 4 6 8 Number of circles 2 6 12 (c) The expression for the number of crosses in Pattern $n$ is $2n + 2$. Find the number of crosses in Pattern 35. \_\_\_\_\_\_ (d) Find an expression, in terms of $n$, for the number of circles in Pattern $n$. \_\_\_\_\_\_

(a) A 4x5 grid of circles with 10 crosses arranged in an L-shape along the right and bottom edges. · (b) Pattern 4: 10 crosses, 20 circles; Pattern 5: 12 crosses, 30 circles · (c) 72 · (d) n(n + 1)

(a) Draw the image of shape $A$ after a translation by vector $\begin{pmatrix} 3 \\ -5 \end{pmatrix}$. (b) Describe fully the single transformation that maps shape $A$ onto shape $B$. \_\_\_\_\_\_

(a) Shape drawn with vertices at (1, -1), (2, -1), (2, -2), (5, -2), (5, -3), (1, -3) · (b) Rotation 90 degrees clockwise about centre (4, 1)

(a) Simplify. $$ (16a^{20})^{\frac{3}{4}} $$ \_\_\_\_\_\_ (b) Expand and simplify. $$ (2c + 9d)(4c - 3d) $$ \_\_\_\_\_\_

(a) 8a^{15} · (b) 8c^2 + 30cd - 27d^2

$$f(x) = \frac{3x - 1}{2} \quad g(x) = 5^x$$ (a) Find $f(-7)$. \_\_\_\_\_\_ (b) Find $f^{-1}(x)$. $f^{-1}(x)$ = \_\_\_\_\_\_ (c) $$f\left(\frac{9}{25}\right) = g(x)$$ Find $x$. $x$ = \_\_\_\_\_\_

(a) -11 · (b) f^{-1}(x) = (2x + 1)/3 · (c) -2

$P$ is the point $(-1, 4)$ and $Q$ is the point $(-3, -2)$. (a) Find the coordinates of the midpoint of the line $PQ$. ( \_\_\_\_\_\_ , \_\_\_\_\_\_ ) (b) Find the equation of the line perpendicular to $PQ$ which passes through the point $P$. \_\_\_\_\_\_

(a) (-2, 1) · (b) y = -1/3x + 11/3

4024/12 Oct/Nov 2024 — Mathematics (Syllabus D) Solved Past Paper

Q11MEasyNumber

Write 43.07862 correct to 3 decimal places.

______

DifficultyEasy

Worked solution

Approach

To round $43.07862$ to 3 decimal places, identify the digit in the third decimal place (the thousandths place) and the digit immediately following it (the ten-thousandths place). If the following digit is 5 or greater, increase the third decimal digit by 1; otherwise, leave it unchanged.

Working

The number is $43.07862$ .

Identify the first three digits after the decimal point: $0$ , $7$ , and $8$ . The digit in the third decimal place is $8$ .
Look at the next digit to the right (the fourth decimal place): $6$ .
Since $6 \geq 5$ , we round up the previous digit ( $8$ ) by adding $1$ .

8 + 1 = 9

The digits before the rounded position remain unchanged. Therefore, the first three decimal digits become $.079$ .

Answer

43.079

Final answer

43.079

Detailed explanation

Walkthrough

Rounding to a specific number of decimal places involves looking at the digit immediately to the right of the desired precision.

Here, we want 3 decimal places. The digits are:

1st decimal place: $0$
2nd decimal place: $7$
3rd decimal place: $8$
4th decimal place: $6$

The rule is: if the digit to the right (the 4th decimal place) is 5 or more, we round up the last retained digit. If it is less than 5, we keep the last retained digit as it is.

In this case, the digit to the right is $6$ . Since $6 > 5$ , we add 1 to the digit in the 3rd decimal place ( $8$ ).

8 + 1 = 9

So, the number becomes $43.079$ .

Key Takeaways

Identify the target place value: Count from the decimal point to find which digit needs to be rounded.
Check the neighbor: Look at the digit immediately to the right of the target place.
Apply the rule: Round up if the neighbor is $\geq 5$ , otherwise round down (keep same).
Preserve preceding zeros: Do not drop leading zeros after the decimal point when rounding (e.g., do not write 43.79).

Common Mistakes

Incorrectly dropping zeros: Writing $43.79$ instead of $43.079$ . The zero in the hundredths place is significant for the decimal position.
Rounding too early or incorrectly: Confusing the digit to round with the digit to check. For example, seeing the 6 and thinking it rounds the 7 instead of the 8.
Off-by-one errors: Adding 1 incorrectly (e.g., $8+1=81$ ).

Things to Be Careful About

Ensure you count exactly 3 places after the decimal point.
Remember that 'cao' (correct answer only) means no working is shown, but the logic must be sound. A common trap is misidentifying the decimal positions (tenths, hundredths, thousandths).
The number $43.07862$ has more than 3 decimal places, so truncation (simply cutting off digits) is incorrect; proper rounding is required.

Techniques used

round a decimal to a specified number of decimal places

Similar questions

Q21MEasyNumber

At midnight the temperature is $-7°\text{C}$ .
At 11 am the next day the temperature is $12°\text{C}$ .

Find the increase in temperature from midnight to 11 am.

______ $°\text{C}$

Similar questions

Q32MEasyNumber

Write these numbers in order of size, starting with the smallest.

\frac{2}{3} \quad 66\% \quad 0.6 \quad \frac{16}{25} \quad 0.606

______ , ______ , ______ , ______ , ______
smallest

DifficultyEasy

Worked solution

Approach

To compare these numbers easily, we convert every number into its decimal equivalent. This allows us to compare them digit by digit directly.

Working

Convert each number:

$\frac{2}{3}$ : Performing the division $2 \div 3$ gives $0.666...$ or $0.\dot{6}$ .
$66\%$ : To convert a percentage to a decimal, divide by $100$ . So, $66\% = \frac{66}{100} = 0.66$ .
$0.6$ : This is already in decimal form. We can write it as $0.600$ for easier comparison.
$\frac{16}{25}$ : To convert this fraction, we can multiply the numerator and denominator by $4$ to get a denominator of $100$ . $\frac{16 \times 4}{25 \times 4} = \frac{64}{100} = 0.64$ .
$0.606$ : This is already in decimal form.

Now list the decimal equivalents:

\begin{aligned} \frac{2}{3} &= 0.666... \\ 66\% &= 0.66 \\ 0.6 &= 0.600 \\ \frac{16}{25} &= 0.64 \\ 0.606 &= 0.606 \end{aligned}

Compare the values:

$0.600$ is the smallest.
$0.606$ is the next smallest ( $0.606 > 0.600$ ).
$0.64$ is next ( $0.640 > 0.606$ ).
$0.66$ is next ( $0.660 > 0.640$ ).
$0.666...$ is the largest.

Ordering the original expressions from smallest to largest:

0.6, \quad 0.606, \quad \frac{16}{25}, \quad 66\%, \quad \frac{2}{3}

Answer

0.6, \ 0.606, \ \frac{16}{25}, \ 66\%, \ \frac{2}{3}

Final answer

0.6, 0.606, 16/25, 66%, 2/3

Detailed explanation

Walkthrough

The key to ordering numbers presented in mixed formats (fractions, percentages, and decimals) is to unify them into a single format. Decimals are usually the easiest because our place-value system makes comparing them intuitive.

Convert Fractions: For $\frac{2}{3}$ , we perform long division ( $2 \div 3$ ). Since $3$ goes into $20$ six times with a remainder of $2$ , the pattern $6$ repeats indefinitely, giving $0.\dot{6}$ (or $0.666...$ ). For $\frac{16}{25}$ , notice that $25 \times 4 = 100$ . Multiplying both top and bottom by $4$ gives $\frac{64}{100}$ , which is exactly $0.64$ . This avoids needing to do division.
Convert Percentages: A percentage is literally 'per hundred'. So $66\%$ means $\frac{66}{100}$ , which moves the decimal point two places to the left, resulting in $0.66$ .
Align Decimal Places: Write all numbers with the same number of decimal places to avoid errors. Using three decimal places is sufficient here:
- $0.6 \rightarrow 0.600$
- $0.606 \rightarrow 0.606$
- $\frac{16}{25} \rightarrow 0.640$
- $66\% \rightarrow 0.660$
- $\frac{2}{3} \rightarrow 0.667$ (rounded for visual check, though strictly $0.666...$ )
Compare: Look at the digits from left to right. The tenths digit is $6$ for all. The hundredths digit determines the order for most: $0$ comes first ( $0.600, 0.606$ ), then $4$ ( $0.640$ ), then $6$ ( $0.660, 0.666...$ ). Between $0.600$ and $0.606$ , the thousandths digit $0$ is less than $6$ , so $0.6$ is smaller than $0.606$ . Similarly, $0.660$ is smaller than $0.666...$ .

Key Takeaways

Always convert mixed number types to a common format before comparing.
Common fraction-to-decimal conversions (like thirds and quarters/hundredths) should be memorized or quickly derived.
When converting percentages, remember to move the decimal point two places to the left.

Common Mistakes

Rounding too early: Treating $\frac{2}{3}$ as exactly $0.67$ might make it look larger than $66\%$ ( $0.66$ ), but even $0.666$ is clearly larger than $0.66$ . However, if one rounded $\frac{2}{3}$ down incorrectly or compared $0.66$ vs $0.666$ poorly, mistakes occur. The mark scheme awards partial credit (B1) for four correct pairs, implying close calls are possible.
Ignoring the repeating nature of $\frac{2}{3}$ : Writing $0.66$ instead of $0.666...$ could lead to thinking $\frac{2}{3}$ and $66\%$ are equal, or ordering them incorrectly relative to each other.
Decimal alignment errors: Comparing $0.6$ and $0.606$ without padding zeros might lead some students to think $0.606$ is smaller because it has fewer significant digits, or confuse the position of the last digit.

Things to Be Careful About

Exactness: On the non-calculator component, you must show the method (division or multiplication to get a denominator of 100). Do not just guess the decimal value.
Order direction: The question asks for "starting with the smallest". Ensure your final list begins with the lowest value ( $0.6$ ) and ends with the highest ( $\frac{2}{3}$ ).
Formatting: The final answer requires the original forms ( $\frac{2}{3}$ , not $0.666...$ ; $66\%$ , not $0.66$ ).

Techniques used

convert fractions to decimal formconvert percentages to decimal formcompare decimal valuesorder numbers from smallest to largest

Similar questions

Q4EasyAlgebra and GraphsNumber

Simplify.

(a)

\frac{t^4 \times t^3}{t^{10}}

______

1M Similar

DifficultyEasy

Worked solution

Approach

Use the laws of indices to simplify the expression. First, combine the terms in the numerator using the product law ( $a^m \times a^n = a^{m+n}$ ), then divide by the term in the denominator using the quotient law ( $a^m \div a^n = a^{m-n}$ ). Finally, rewrite the result with a positive index if necessary.

Working

The expression is:

\frac{t^4 \times t^3}{t^{10}}

Step 1: Simplify the numerator using the product law for indices ( $x^a \times x^b = x^{a+b}$ ).

t^4 \times t^3 = t^{4+3} = t^7

So the fraction becomes:

\frac{t^7}{t^{10}}

Step 2: Simplify the fraction using the quotient law for indices ( $x^a \div x^b = x^{a-b}$ ).

t^7 \div t^{10} = t^{7-10} = t^{-3}

Step 3: Convert the negative exponent to a positive one. The rule is $x^{-n} = \frac{1}{x^n}$ .

t^{-3} = \frac{1}{t^3}

Both $t^{-3}$ and $\frac{1}{t^3}$ are correct simplified forms.

Answer

\frac{1}{t^3}

Final answer

1/t^3

Detailed explanation

Walkthrough

This question asks us to simplify an algebraic fraction involving powers of the same base variable $t$ . We rely on two fundamental rules of indices (exponents):

Product Law: When multiplying terms with the same base, add the exponents. Here, we have $t^4$ multiplied by $t^3$ . So, $4 + 3 = 7$ , giving us $t^7$ in the numerator.
Quotient Law: When dividing terms with the same base, subtract the exponent of the denominator from the exponent of the numerator. Here, we have $t^7$ divided by $t^{10}$ . So, $7 - 10 = -3$ , giving us $t^{-3}$ .

A negative exponent indicates a reciprocal. Therefore, $t^{-3}$ is equivalent to $\frac{1}{t^3}$ . This is the standard form for a simplified answer unless a negative index is explicitly allowed.

Key Takeaways

Always look for opportunities to apply index laws before attempting other methods.
Remember that $a^m \times a^n = a^{m+n}$ and $\frac{a^m}{a^n} = a^{m-n}$ .
A negative index means "take the reciprocal": $a^{-n} = \frac{1}{a^n}$ .

Common Mistakes

Adding instead of subtracting: Students might incorrectly calculate $7 + 10 = 17$ in the denominator or numerator.
Subtracting in the wrong order: Calculating $10 - 7 = 3$ and writing $t^3$ instead of $t^{-3}$ . It is crucial to remember that the top exponent minus the bottom exponent determines the sign.
Leaving the answer as $t^{-3}$ : While mathematically correct, exam boards often prefer positive indices for final answers. Both are usually accepted, but $\frac{1}{t^3}$ is safer.

Things to Be Careful About

Ensure you are applying the correct law for multiplication vs. division.
Check the signs carefully when subtracting exponents ( $7 - 10$ results in a negative number).
The mark scheme accepts both $\frac{1}{t^3}$ and $t^{-3}$ , so either form is valid.

Techniques used

apply the product law of indicesapply the quotient law of indicesexpress the result using a negative index or reciprocal

(b)

$(\sqrt{6})^2$

______

1M Similar

DifficultyEasy

Worked solution

Approach

Understand the relationship between square roots and squaring. Squaring a square root cancels out the operations, returning the original number under the radical.

Working

The expression is:

(\sqrt{6})^2

By definition, $\sqrt{x}$ is the number which, when squared, gives $x$ . Therefore:

(\sqrt{x})^2 = x

Applying this to our specific case where $x = 6$ :

(\sqrt{6})^2 = 6

Alternatively, you can think of it as:

\sqrt{6} \times \sqrt{6} = \sqrt{6 \times 6} = \sqrt{36} = 6

Answer

6

Final answer

6

Detailed explanation

Walkthrough

This question tests the fundamental definition of a square root. The symbol $\sqrt{}$ denotes the principal (positive) square root. If you take a number, find its square root, and then square that result, you end up back where you started. Mathematically, the square and the square root are inverse operations. Thus, $(\sqrt{6})^2$ simply removes the square root sign, leaving the number inside, which is 6.

Key Takeaways

Squaring and taking a square root are inverse operations.
$(\sqrt{n})^2 = n$ for any non-negative number $n$ .
$\sqrt{n} \times \sqrt{n} = n$ .

Common Mistakes

Confusing with multiplication: Thinking the answer is $2\sqrt{6}$ (treating the square as a coefficient).
Calculating $6^2$ incorrectly: Some might multiply $6 \times 2$ instead of squaring the root.
Sign errors: Forgetting that the principal square root is always positive, so the answer must be positive 6, not $\pm 6$ .

Things to Be Careful About

The notation $(\sqrt{6})^2$ means "the square of the square root of 6". Do not confuse this with $\sqrt{6^2}$ (which is also 6) or $\sqrt{6 \times 2}$ .
The mark scheme specifies "cao" (correct answer only), meaning no working is strictly required for marks, but understanding the concept prevents simple slips.

Techniques used

square a square root

Q5Medium-EasyStatistics

A group of people are asked what type of holiday they prefer.
The table gives information about the results.

Type of holiday	Number of people	Pie chart angle
Camping	15	$60°$
Beach	45
Cruise	20
Hiking	10

(a)

Complete the table.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Find the total number of people surveyed, then use the given Camping data to find the degrees per person. Multiply this by the number of people for each holiday type to find the missing pie chart angles.

Working

Total number of people = $15 + 45 + 20 + 10 = 90$ .

Degrees per person = $\frac{60^\circ}{15} = 4^\circ$ .

Beach angle = $45 \times 4^\circ = 180^\circ$ .
Cruise angle = $20 \times 4^\circ = 80^\circ$ .
Hiking angle = $10 \times 4^\circ = 40^\circ$ .

Check: $60^\circ + 180^\circ + 80^\circ + 40^\circ = 360^\circ$ .

Answer

The missing pie chart angles are $180^\circ$ , $80^\circ$ , and $40^\circ$ .

Final answer

180°, 80°, 40°

Detailed explanation

Walkthrough

First, we calculate the total number of people surveyed by adding up the numbers for each holiday type: $15 + 45 + 20 + 10 = 90$ . Next, we use the information given for Camping: 15 people correspond to a pie chart angle of $60^\circ$ . This allows us to find the angle for 1 person, which is $\frac{60^\circ}{15} = 4^\circ$ . We then multiply this value by the number of people for each of the other holidays to find their respective angles: Beach is $45 \times 4^\circ = 180^\circ$ , Cruise is $20 \times 4^\circ = 80^\circ$ , and Hiking is $10 \times 4^\circ = 40^\circ$ . Finally, we verify that all angles add up to $360^\circ$ : $60^\circ + 180^\circ + 80^\circ + 40^\circ = 360^\circ$ .

Key Takeaways

When working with pie charts, the total angle is always $360^\circ$ . You can find the angle per unit (or per person) by dividing a known angle by its corresponding quantity, then multiply this by the other quantities to find their angles. Always check that the angles sum to $360^\circ$ .

Common Mistakes

Forgetting to calculate the total number of people first and trying to use the angles directly without finding the degrees per person.
Not checking that the final angles add up to $360^\circ$ .
Writing the angles without the degree symbol $^\circ$ .

Things to Be Careful About

Ensure the angles are in degrees and include the $^\circ$ symbol.
The mark scheme accepts alternative methods such as finding the fraction of the total (e.g., $\frac{45}{90} \times 360^\circ$ ) or using the proportion $\frac{60}{15} \times k$ where $k$ is the number of people for each holiday.

Techniques used

calculate total number of peoplefind degrees per personcalculate missing angles

(b)

Complete the pie chart to show this information.

Fig. 1

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Using the angles calculated in part (a), draw the remaining sectors in the pie chart. Ensure the angles add up to $360^\circ$ and label each sector with the correct holiday type.

Working

Camping is already drawn with $60^\circ$ .
Draw Beach as a semicircle ( $180^\circ$ ) adjacent to Camping.
Draw Cruise as an $80^\circ$ sector adjacent to Beach.
Draw Hiking as a $40^\circ$ sector adjacent to Cruise, which closes the circle ( $60^\circ + 180^\circ + 80^\circ + 40^\circ = 360^\circ$ ).

Answer

A completed pie chart with sectors Camping ( $60^\circ$ ), Beach ( $180^\circ$ ), Cruise ( $80^\circ$ ), and Hiking ( $40^\circ$ ).

Final answer

Pie chart with sectors Camping 60°, Beach 180°, Cruise 80°, Hiking 40°

Detailed explanation

Walkthrough

The pie chart already has the Camping sector drawn with an angle of $60^\circ$ . Using the angles calculated in part (a), we add the remaining sectors in sequence. Beach has an angle of $180^\circ$ , which is exactly half the circle, so we draw a diameter to create this semicircle. Next, we add the Cruise sector with an angle of $80^\circ$ adjacent to Beach. Finally, we add the Hiking sector with an angle of $40^\circ$ , which should perfectly close the circle since $60^\circ + 180^\circ + 80^\circ + 40^\circ = 360^\circ$ . Each sector must be clearly labelled with its corresponding holiday type.

Key Takeaways

A completed pie chart must have all sectors labelled and their angles must sum to exactly $360^\circ$ . When drawing, it is often easiest to start with the largest sector (Beach, $180^\circ$ ) to create a straight line, then add the remaining sectors sequentially.

Common Mistakes

Drawing sectors with incorrect angles that do not match the calculated values.
Forgetting to label the sectors with the holiday names.
Not ensuring the sectors fit together perfectly to form a complete circle ( $360^\circ$ ).

Things to Be Careful About

The mark scheme gives follow-through marks if the angles in the table add up to $360^\circ$ , even if there was an earlier error.
Ensure sectors are drawn accurately and labelled clearly to avoid losing marks.

Techniques used

draw sectors with calculated angleslabel sectors with holiday types

Q62MMedium-EasyNumber

A laptop costs $800.
In a sale, the cost is reduced by 15%.

Work out the cost of the laptop in the sale.

$ ______

DifficultyMedium-Easy

Worked solution

Approach

The laptop's original cost is $800. The sale reduces this price by 15%. We can find the sale price by calculating 15% of $800 and subtracting it from the original price, or by directly calculating 85% of the original price.

Working

Method 1: Calculate the discount first
Calculate 15% of $800:

\frac{15}{100} \times 800 = 15 \times 8 = 120

The discount is $120.
Subtract the discount from the original price:

800 - 120 = 680

Method 2: Calculate the remaining percentage
If the price is reduced by 15%, the customer pays:

100\% - 15\% = 85\%

Calculate 85% of $800:

\frac{85}{100} \times 800 = 85 \times 8

85 \times 8 = 680

Answer

The cost of the laptop in the sale is $680.

680

Final answer

680

Detailed explanation

Walkthrough

We are given an original price of $800 and a discount rate of 15%. The goal is to find the final sale price.

Step 1: Understand the reduction. A 15% reduction means the new price is 15% less than the old price. This leaves 85% of the original value (since $100\% - 15\% = 85\%$ ).

Step 2: Perform the calculation. There are two common ways to do this:

Find the discount amount ( $15\%$ of $800$ ) and subtract it from $800$ .
$15\%$ of $800$ is calculated as $\frac{15}{100} \times 800$ . Since $800 \div 100 = 8$ , this becomes $15 \times 8 = 120$ . Then $800 - 120 = 680$ .
Find the remaining percentage ( $85\%$ ) and multiply by the original price.
$85\%$ of $800$ is calculated as $\frac{85}{100} \times 800$ . Since $800 \div 100 = 8$ , this becomes $85 \times 8 = 680$ .

Both methods yield the same result. The mark scheme accepts either the working for $800 - \frac{15}{100} \times 800$ or $\frac{85}{100} \times 800$ .

Key Takeaways

To reduce a quantity by a percentage, you can either calculate the percentage amount and subtract, or calculate the remaining percentage and multiply.
Calculating percentages of round numbers like 800 is often easier by dividing by 100 first to get 1%, then multiplying by the required percentage.

Common Mistakes

Adding the percentage instead of subtracting (calculating $800 + 120 = 920$ ).
Calculating only the discount amount ($120) and forgetting to subtract it from the original price.
Arithmetic errors when multiplying $15 \times 8$ or $85 \times 8$ .

Things to Be Careful About

Ensure you are calculating a reduction, not an increase.
Check that your final answer makes sense (it should be less than $800).

Techniques used

calculate the percentage reduction amountsubtract the reduction from the original pricemultiply by the remaining percentage

Similar questions

Q72MMedium-EasyNumber

Work out $\frac{3}{4} + \frac{5}{6}$ .

Give your answer as a mixed number in its simplest form.

______

DifficultyMedium-Easy

Worked solution

Approach

To add $\frac{3}{4}$ and $\frac{5}{6}$ , we must first find a common denominator so that the fractions represent parts of the same whole size. The least common multiple (LCM) of the denominators 4 and 6 will serve as this common denominator.

Working

First, determine the LCM of 4 and 6:

\begin{aligned} 4 &= 2 \times 2 \\ 6 &= 2 \times 3 \\ \text{LCM}(4, 6) &= 2 \times 2 \times 3 = 12 \end{aligned}

Next, convert each fraction to an equivalent fraction with a denominator of 12:
For $\frac{3}{4}$ , multiply both numerator and denominator by 3:

\frac{3 \times 3}{4 \times 3} = \frac{9}{12}

For $\frac{5}{6}$ , multiply both numerator and denominator by 2:

\frac{5 \times 2}{6 \times 2} = \frac{10}{12}

Now add the two fractions:

\frac{9}{12} + \frac{10}{12} = \frac{9 + 10}{12} = \frac{19}{12}

The question requires the answer as a mixed number in simplest form. Divide the numerator by the denominator:

19 \div 12 = 1 \text{ remainder } 7

So,

\frac{19}{12} = 1 \frac{7}{12}

The fraction $\frac{7}{12}$ is already in its simplest form because 7 is prime and does not divide 12.

Answer

1 \frac{7}{12}

Final answer

1 7/12

Detailed explanation

Walkthrough

Adding fractions with different denominators requires making the denominators the same. This is done by finding a common multiple of the original denominators. Here, the smallest such number (the Least Common Multiple) for 4 and 6 is 12.

We then adjust the numerators proportionally. Since we multiplied the denominator 4 by 3 to get 12, we must also multiply the numerator 3 by 3 to keep the value unchanged ( $\frac{3}{4} = \frac{9}{12}$ ). Similarly, since we multiplied the denominator 6 by 2 to get 12, we multiply the numerator 5 by 2 ( $\frac{5}{6} = \frac{10}{12}$ ).

With the denominators now identical, we simply add the top numbers (numerators): $9 + 10 = 19$ . The result is $\frac{19}{12}$ . Because the numerator is larger than the denominator, this is an improper fraction. To convert it to a mixed number, we ask how many times 12 goes into 19. It goes in 1 time, with 7 left over. Thus, the integer part is 1 and the fractional part is $\frac{7}{12}$ .

Key Takeaways

Always find a common denominator before adding or subtracting fractions.
Multiplying the numerator and denominator by the same number creates an equivalent fraction without changing its value.
An improper fraction (numerator > denominator) should be converted to a mixed number if requested.

Common Mistakes

Adding numerators and denominators directly (e.g., $\frac{3+5}{4+6} = \frac{8}{10}$ ), which is mathematically incorrect.
Using a common denominator that is not the LCM (e.g., 24) but failing to simplify the final answer correctly.
Forgetting to convert the final improper fraction into a mixed number when explicitly asked.

Things to Be Careful About

Ensure the final mixed number is in simplest form. In this case, check if the fractional part $\frac{7}{12}$ can be reduced further (it cannot).

Techniques used

find a common denominator for fractionsadd numerators while keeping the denominator constantconvert an improper fraction to a mixed numbersimplify a fraction to its lowest terms

Similar questions

Q82MMedium-EasyNumber

Sophia walks at an average speed of $4\,\text{km/h}$ .

Work out the time Sophia takes to walk $13\,\text{km}$ .
Give your answer in hours and minutes.

______ hours ______ minutes

DifficultyMedium-Easy

Worked solution

Approach

We are given the distance Sophia walks and her average speed. We need to find the time taken using the relationship between distance, speed, and time. The answer must be in hours and minutes.

Working

First, recall the formula relating distance ( $d$ ), speed ( $s$ ), and time ( $t$ ):

d = s \times t

Rearranging for time:

t = \frac{d}{s}

Substitute the given values: $d = 13\,\text{km}$ and $s = 4\,\text{km/h}$ .

t = \frac{13}{4}\,\text{hours}

Calculate the value of this fraction:

\frac{13}{4} = 3.25\,\text{hours}

This means Sophia takes 3 full hours plus $0.25$ of an hour. To convert the fractional part into minutes, multiply by 60 (since there are 60 minutes in 1 hour):

0.25 \times 60 = 15\,\text{minutes}

Alternatively, converting the total time directly:

\frac{13}{4} \times 60 = 13 \times 15 = 195\,\text{minutes}

Since $195 = 3 \times 60 + 15$ , this is 3 hours and 15 minutes.

Answer

3 \text{ hours } 15 \text{ minutes}

Final answer

3 hours 15 minutes

Detailed explanation

Walkthrough

Identify the knowns: Sophia walks a distance of $13\,\text{km}$ at a speed of $4\,\text{km/h}$ .
Choose the correct formula: The fundamental relationship is $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$ . Since we want to find the time, we rearrange it to $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$ .
Perform the division: $\text{Time} = \frac{13}{4} \text{ hours}$ Performing the division gives $3.25$ hours.
Convert units: The question asks for the answer in hours and minutes.
- The integer part is $3$ , so that is 3 hours.
- The decimal part is $0.25$ hours. To convert this to minutes, multiply by 60 (the number of minutes in an hour): $0.25 \times 60 = 15 \text{ minutes}$
- Combine them: 3 hours 15 minutes.

Key Takeaways

Always check the required units for your final answer. If the input is in km/h but the output needs minutes, a conversion step is essential.
Remember the formula triangle: Time = Distance / Speed.
To convert decimal hours to minutes, always multiply the decimal part by 60, not 100.

Common Mistakes

Division Error: Dividing $4$ by $13$ instead of $13$ by $4$ .
Unit Conversion: Multiplying the decimal part by 100 (treating it like a percentage) instead of 60.
Ignoring Units: Forgetting to include "hours" or "minutes" in the final statement if explicit writing is required, though the blanks here guide the format.

Things to Be Careful About

Ensure you divide distance by speed, not speed by distance.
When converting $0.25$ hours, remember that an hour has 60 minutes, so $0.25 \times 60 = 15$ , not $25$ .
The mark scheme accepts $\frac{13}{4} \times 60$ as a method step (M1), showing that converting total minutes first is also a valid path.

Techniques used

apply the formula $\text{time} = \frac{\text{distance}}{\text{speed}}$convert decimal hours to hours and minutes

Similar questions

Q9MediumAlgebra and Graphs

A sequence of patterns is made using crosses and circles.

Fig. 2

(a)

Draw Pattern 4.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

Observe the dimensions of the circle grid and the arrangement of crosses in the first three patterns, then extend this to Pattern 4.

Working

Pattern 1 has a $1 \times 2$ grid of circles and $4$ crosses.
Pattern 2 has a $2 \times 3$ grid of circles and $6$ crosses.
Pattern 3 has a $3 \times 4$ grid of circles and $8$ crosses.

Following this rule, Pattern 4 must have a $4 \times 5$ grid of circles (4 rows and 5 columns) and $10$ crosses.
The crosses form an L-shape along the right and bottom edges of the circle grid: $4$ crosses to the right of the circles and $6$ crosses below the circles.

Answer

A $4 \times 5$ grid of circles with $10$ crosses arranged in an L-shape along the right and bottom edges.

Final answer

A 4x5 grid of circles with 10 crosses arranged in an L-shape along the right and bottom edges.

Detailed explanation

Walkthrough

By examining the first three patterns, we can see that Pattern $n$ has a grid of circles with dimensions $n$ rows by $(n+1)$ columns. The number of circles is therefore $n(n+1)$ . The crosses are arranged in an L-shape around the right and bottom edges of this grid. For Pattern $n$ , there are $n$ crosses to the right of the circles and $(n+2)$ crosses below the circles, giving a total of $n + n + 2 = 2n + 2$ crosses.

For Pattern 4, we substitute $n = 4$ : the grid of circles is $4 \times 5$ (20 circles), and the number of crosses is $2(4) + 2 = 10$ . The drawing shows 4 rows of 5 circles, with 4 crosses to the right and 6 crosses along the bottom row.

Key Takeaways

Patterns in sequences can often be understood by breaking them into geometric components. Identifying the dimensions of sub-shapes (like the circle grid here) helps predict future terms.

Common Mistakes

Drawing the wrong number of rows or columns for the circle grid (e.g., a $4 \times 4$ or $5 \times 5$ grid instead of $4 \times 5$ ).
Miscounting the crosses, forgetting that the bottom row has one more cross than the number of rows.

Things to Be Careful About

When drawing pattern sequence questions, ensure the shape is drawn clearly and the counts of each element match the expected values. The mark scheme accepts a correct drawing even if the exact style varies slightly, as long as the dimensions and counts are right.

Techniques used

observe the geometric growth of the patterndetermine the dimensions of the grid for the next termdraw the corresponding arrangement

(b)

Complete the table for Pattern 4 and Pattern 5.

Pattern number ( $n$ )	1	2	3	4	5
Number of crosses	4	6	8
Number of circles	2	6	12

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the rules established from the first three patterns to find the number of crosses and circles for Patterns 4 and 5.

Working

Crosses: The number of crosses follows the linear pattern $2n + 2$ .
For Pattern 4 ( $n = 4$ ): $2(4) + 2 = 10$ .
For Pattern 5 ( $n = 5$ ): $2(5) + 2 = 12$ .

Circles: The number of circles follows the pattern $n(n + 1)$ .
For Pattern 4 ( $n = 4$ ): $4 \times 5 = 20$ .
For Pattern 5 ( $n = 5$ ): $5 \times 6 = 30$ .

Answer

Pattern number ( $n$ )	1	2	3	4	5
Number of crosses	4	6	8	10	12
Number of circles	2	6	12	20	30

Final answer

Pattern 4: 10 crosses, 20 circles; Pattern 5: 12 crosses, 30 circles

Detailed explanation

Walkthrough

The number of crosses increases by 2 each time (4, 6, 8...), which is a linear sequence with a common difference of 2. The formula is $2n + 2$ . Substituting $n = 4$ and $n = 5$ gives 10 and 12.

The number of circles is 2, 6, 12... The differences between terms are 4, 6... The second differences are constant at 2, indicating a quadratic sequence. The pattern is $n \times (n+1)$ . For $n = 4$ , this is $4 \times 5 = 20$ . For $n = 5$ , this is $5 \times 6 = 30$ .

Key Takeaways

Linear sequences have a constant first difference, while quadratic sequences have a constant second difference. Recognizing these patterns allows you to fill in missing terms in a table.

Common Mistakes

Assuming the number of circles increases by a constant amount (e.g., adding 4 or 6 each time) instead of recognizing the quadratic growth.
Miscalculating $4 \times 5$ or $5 \times 6$ .

Things to Be Careful About

Ensure the values are placed in the correct columns for Pattern 4 and Pattern 5. The table requires both crosses and circles for both patterns.

Techniques used

extend the linear pattern for crossesextend the quadratic pattern for circles

(c)

The expression for the number of crosses in Pattern $n$ is $2n + 2$ .

Find the number of crosses in Pattern 35.

______

1M Similar

(d)

Find an expression, in terms of $n$ , for the number of circles in Pattern $n$ .

______

2M Similar

DifficultyMedium

Worked solution

Approach

Examine the sequence for the number of circles: $2, 6, 12, 20, 30, \dots$ and find a relationship with the pattern number $n$ .

Working

List the number of circles for each $n$ :

For $n = 1$ , circles $= 2 = 1 \times 2$
For $n = 2$ , circles $= 6 = 2 \times 3$
For $n = 3$ , circles $= 12 = 3 \times 4$
For $n = 4$ , circles $= 20 = 4 \times 5$
For $n = 5$ , circles $= 30 = 5 \times 6$

In each case, the number of circles is the product of $n$ and $(n + 1)$ .

Alternatively, using the method of differences:
First differences: $6 - 2 = 4$ , $12 - 6 = 6$ , $20 - 12 = 8$ , $30 - 20 = 10$
Second differences: $6 - 4 = 2$ , $8 - 6 = 2$ , $10 - 8 = 2$

Since the second difference is constant ( $2$ ), the sequence is quadratic and has the form $an^2 + bn + c$ where $2a = 2$ , so $a = 1$ .
The expression is $n^2 + bn + c$ . Substituting $n = 1$ gives $1 + b + c = 2$ , and $n = 2$ gives $4 + 2b + c = 6$ . Solving these gives $b = 1$ and $c = 0$ .

Thus, the expression is $n^2 + n$ , which can be factored as $n(n + 1)$ .

Answer

n(n + 1) \quad \text{or} \quad n^2 + n

Final answer

n(n + 1)

Detailed explanation

Walkthrough

The number of circles is $2, 6, 12, 20, 30$ for $n = 1, 2, 3, 4, 5$ . We can factor each of these numbers to find a pattern:
$2 = 1 \times 2$
$6 = 2 \times 3$
$12 = 3 \times 4$
$20 = 4 \times 5$
$30 = 5 \times 6$

The first factor is always $n$ , and the second factor is always $n + 1$ . Therefore, the expression for the number of circles in Pattern $n$ is $n(n + 1)$ .

Using the method of differences confirms this: the constant second difference of $2$ indicates a quadratic sequence $n^2 + n$ .

Key Takeaways

Factoring the terms of a sequence can reveal a simple product pattern. If factoring is not immediately obvious, the method of differences (finding constant second differences) is a reliable way to derive the quadratic nth term.

Common Mistakes

Assuming the sequence is linear and adding a constant difference.
Writing $n^2 + n$ but not accepting it if the mark scheme expects the factored form $n(n + 1)$ (though both are generally accepted in 4024 unless specified).
Forgetting that $n$ starts at 1, not 0.

Things to Be Careful About

The question asks for an expression in terms of $n$ . Both $n(n + 1)$ and $n^2 + n$ are correct. Ensure the expression is fully simplified and clearly written.

Techniques used

identify the pattern in the number of circlesformulate a quadratic expression for the nth term

Q10Medium-EasyGeometryNumber

The scale drawing shows a field $ABCDE$ .
The field contains a stage that is a sector of a circle, centre $C$ .

The scale is $1\,\text{cm}$ to $2.5\,\text{m}$ .

Fig. 3

(a)

Find the actual radius of the stage.

______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Measure the radius of the stage sector (the distance from centre $C$ to the circular arc) on the printed diagram using a ruler in centimetres. Then, multiply this measurement by the scale factor of $2.5\,\text{m/cm}$ to obtain the actual radius.

Working

Measuring the radius on the scale diagram gives:

\text{Measured radius} = 3.0\,\text{cm} \quad (\text{acceptable range: } 2.8\,\text{cm to } 3.2\,\text{cm})

Using the scale $1\,\text{cm} = 2.5\,\text{m}$ :

\text{Actual radius} = 3.0 \times 2.5 = 7.5\,\text{m}

Answer

7.5

Final answer

7.5

Detailed explanation

Walkthrough

Use a ruler to measure the distance from the centre $C$ to any point on the curved arc of the stage. The measurement should lie between $2.8\,\text{cm}$ and $3.2\,\text{cm}$ (nominal measurement is $3.0\,\text{cm}$ ).
The scale is given as $1\,\text{cm}$ to $2.5\,\text{m}$ . To convert the measured diagram length into the actual length in metres, multiply the measurement by $2.5$ .
Multiplying $3.0$ by $2.5$ gives $7.5\,\text{m}$ . Any answer in the range $7$ to $8\,\text{m}$ (corresponding to measurements in the range $2.8$ to $3.2\,\text{cm}$ ) is accepted.

Key Takeaways

For scale drawings, always measure accurately with a ruler to the nearest millimetre.
Convert diagram lengths to actual lengths by multiplying by the unit scale factor.

Common Mistakes

Measuring the distance from $A$ , $B$ , $D$ , or $E$ instead of the centre $C$ .
Dividing by $2.5$ instead of multiplying by $2.5$ .
Forgetting to convert the measurement to actual metres using the scale.

Things to Be Careful About

Check that your measurement starts precisely at the vertex $C$ and goes directly to the arc along a straight line (such as along edge $CB$ or $CD$ ).

Techniques used

measure a length with a rulermultiply by a scale factor to find an actual distance

(b)

The rest of the field is split into two zones, zone 1 and zone 2.
Zone 1 and zone 2 do not include the stage.
Zone 1 is the region that is nearer to $EA$ than to $ED$ .

3M

(i)

Using compasses and a straight edge only, construct the boundary between zone 1 and zone 2.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The boundary between points nearer to line $EA$ than to line $ED$ is the angle bisector of $\angle AED$ . Construct this bisector using a compass and straight edge, ensuring construction arcs are visible, and extend the line until it meets the boundary of the stage.

Working

Place the compass point at vertex $E$ and draw an arc that cuts the line segment $EA$ and the line segment $ED$ .
From each of the two intersection points, draw two intersecting arcs inside the angle $\angle AED$ using an equal compass radius.
Using a straight edge, draw a straight line from vertex $E$ through the intersection of these two arcs, continuing until it reaches the arc of the stage.

Answer

Straight line bisecting angle $AED$ constructed with correct arcs and ending at the stage boundary.

Final answer

Straight line bisecting angle AED constructed with correct arcs and ending at the stage boundary

Detailed explanation

Walkthrough

The locus of points equidistant from two intersecting lines (here $EA$ and $ED$ ) is the angle bisector of the angle between them ( $\angle AED$ ). Points on one side of this bisector are closer to $EA$ , while points on the other side are closer to $ED$ .
To construct the bisector:
1. Place the compass point at $E$ and draw arcs crossing both $EA$ and $ED$ .
2. From these intersection points, use compasses to make intersecting arcs within the angle.
3. Draw a straight line from $E$ through the intersection point of the arcs.
4. Stop or continue the line where it meets the boundary of the stage, as the stage is not part of zone 1 or zone 2.

Key Takeaways

The locus of points equidistant from two straight lines meeting at a vertex is the bisector of the angle formed by those lines.
Construction marks (arcs) must be clearly visible to earn full method marks.

Common Mistakes

Drawing the bisector by eye using a protractor without showing compass construction arcs (loses the construction mark).
Constructing the perpendicular bisector of a segment instead of the angle bisector of $\angle AED$ .

Things to Be Careful About

Ensure construction arcs are drawn neatly and clearly.
Do not erase construction arcs.

Techniques used

construct an angle bisector using compasses and straight edge

(ii)

Shade the region that represents zone 1.

1M Similar

(c)

The field is used for a concert.
Tickets for the concert cost $30.75 each.

Work out the cost of 8 tickets.

$ ______

1M Similar

Q112MMedium-EasyAlgebra and Graphs

Factorise.

4m^2 - 14m

______

DifficultyMedium-Easy

Worked solution

Approach

To factorise the expression $4m^2 - 14m$ , we look for the highest common factor (HCF) shared by both terms. We then place this HCF outside a pair of parentheses and divide each term inside the original expression by this HCF to determine what remains inside.

Working

First, examine the coefficients: $4$ and $14$ . Both numbers are divisible by $2$ .

Next, examine the variables: $m^2$ and $m$ . Both terms contain at least one $m$ .

Therefore, the highest common factor is $2m$ .

Now, factor out $2m$ :

4m^2 \div 2m = 2m

-14m \div 2m = -7

Write the HCF outside the brackets and the results of the division inside:

4m^2 - 14m = 2m(2m - 7)

Answer

The fully factorised form is:

2m(2m - 7)

Final answer

2m(2m - 7)

Detailed explanation

Walkthrough

Factorising is the reverse process of expanding brackets. Instead of multiplying terms out, we are looking for a quantity that divides into every term in the expression evenly.

Identify the numerical factor: Look at the numbers $4$ and $14$ . The largest number that divides both of them is $2$ .
Identify the variable factor: Look at the algebraic parts $m^2$ and $m$ . Since $m^2 = m \times m$ , the lowest power of $m$ present in all terms is $m$ .
Combine to find the HCF: The Highest Common Factor is $2 \times m = 2m$ .
Divide and place in brackets: Divide the original terms by $2m$ $2 m$ .
- $4m^2 / 2m = 2m$
- $-14m / 2m = -7$
  Place these results inside the brackets next to the HCF: $2m(2m - 7)$ .

Key Takeaways

Always check both the numerical coefficient and the variable part when finding the HCF.
Ensure that when you expand your answer ( $2m \times 2m - 2m \times 7$ ), you get back the original question ( $4m^2 - 14m$ ). This is a good way to verify your work.

Common Mistakes

Partial factorisation: Forgetting to include the variable $m$ in the HCF (writing just $2$ instead of $2m$ ).
Arithmetic errors: Incorrectly dividing $14$ by $2$ or mismanaging the negative sign (e.g., writing $+7$ instead of $-7$ ).
Leaving terms behind: Writing $2m(m - 7)$ which expands to $2m^2 - 14m$ , missing the factor of $2$ on the first term.

Things to Be Careful About

The mark scheme awards partial credit (B1) for incomplete factorisations such as $2(2m^2 - 7m)$ or $m(4m - 14)$ . However, the final answer must be fully factorised to receive full marks (A1). Ensure every term inside the bracket cannot be divided by any further common factor.

Techniques used

identify the highest common factorfactorise by grouping terms

Similar questions

Q121MEasyNumber

Here is a list of numbers.

\frac{1}{3} \quad \sqrt{4} \quad 2^0 \quad \sqrt{5} \quad \frac{10}{8} \quad 2^{-1}

Write down the number in the list that is irrational.

______

DifficultyEasy

Worked solution

Approach

An irrational number cannot be written as a fraction of two integers. Simplify each expression and check whether it can be written as a rational number. The only one that cannot is the square root of a non-perfect square.

Working

Evaluate the powers and root:

\sqrt{4} = 2, \quad 2^0 = 1, \quad 2^{-1} = \frac{1}{2}

Now classify each number:

\frac{1}{3} \text{ is rational}, \quad \sqrt{4}=2 \text{ is rational}, \quad 2^0=1 \text{ is rational}

\sqrt{5} \text{ is irrational}, \quad \frac{10}{8}=\frac{5}{4} \text{ is rational}, \quad 2^{-1}=\frac{1}{2} \text{ is rational}

Since $5$ is not a perfect square, $\sqrt{5}$ cannot be written as an exact fraction.

Answer

\sqrt{5}

Final answer

√5

Detailed explanation

Walkthrough

We need to decide which of the six numbers is irrational. An irrational number cannot be written as a fraction $\frac{a}{b}$ of two integers. Start by simplifying the expressions that look like powers or roots. $\sqrt{4}=2$ , so it is rational. $2^0=1$ , so it is rational. $2^{-1}=\frac{1}{2}$ , so it is rational. $\frac{10}{8}$ simplifies to $\frac{5}{4}$ , so it is rational. $\frac{1}{3}$ is already a fraction, so it is rational. That leaves $\sqrt{5}$ . Since $5$ is not a perfect square, $\sqrt{5}$ cannot be written as a fraction, so it is irrational.

Key Takeaways

A rational number can be written as a fraction of two integers.
Square roots of non-perfect squares are irrational.
Powers with exponent $0$ equal $1$ (for a non-zero base), and negative exponents give reciprocals.

Common Mistakes

Thinking $\sqrt{4}$ is irrational because it contains a square root sign; it simplifies to $2$ .
Thinking $2^0 = 0$ ; it equals $1$ .
Thinking $2^{-1} = -2$ ; it equals $\frac{1}{2}$ .
Confusing $\frac{10}{8}$ with a non-terminating decimal; it simplifies to $\frac{5}{4}$ .

Things to Be Careful About

The answer must be exactly $\sqrt{5}$ (cao); no equivalent decimal is needed.
This is a non-calculator paper, but no calculator is needed here — just simplify each item by hand.
An irrational number is not just “a number with a square root sign”; only roots of non-perfect squares are irrational.

Techniques used

simplify powers and rootsrecognise square roots of non-perfect squares as irrationalclassify numbers as rational or irrational

Similar questions

Q132MMedium-EasyNumber

Evaluate.

5 \times 10^7 - 8 \times 10^6

Give your answer in standard form.

______

DifficultyMedium-Easy

Worked solution

Approach

Both terms are powers of $10$ . To subtract them, rewrite both with the same power of $10$ (or as ordinary numbers), then convert the result back to standard form.

Working

Rewrite $5 \times 10^7$ using $10^6$ as the common power:

5 \times 10^7 = 50 \times 10^6

Subtract the two numbers:

50 \times 10^6 - 8 \times 10^6 = 42 \times 10^6

Convert to standard form by writing $42 = 4.2 \times 10$ :

42 \times 10^6 = 4.2 \times 10 \times 10^6 = 4.2 \times 10^7

Answer

4.2 \times 10^7

Final answer

4.2 × 10^7

Detailed explanation

Walkthrough

The two terms are not in the same form because one is multiplied by $10^7$ and the other by $10^6$ . Before subtracting, make the powers of $10$ match.

Rewrite $5 \times 10^7$ as $50 \times 10^6$ , since $10^7 = 10 \times 10^6$ . Now both terms have the same power of $10$ , so subtract the coefficients directly:

50 \times 10^6 - 8 \times 10^6 = 42 \times 10^6

The result $42 \times 10^6$ is not yet in standard form, because the coefficient $42$ is not between $1$ and $10$ . Convert it by writing $42 = 4.2 \times 10$ :

42 \times 10^6 = 4.2 \times 10 \times 10^6 = 4.2 \times 10^7

An equivalent route is to rewrite $8 \times 10^6$ as $0.8 \times 10^7$ , giving $5 \times 10^7 - 0.8 \times 10^7 = 4.2 \times 10^7$ .

The answer is $4.2 \times 10^7$ .

Key Takeaways

To add or subtract numbers in standard form, first adjust them to the same power of $10$ .
Standard form requires a coefficient $A$ with $1 \leq A < 10$ , multiplied by an integer power of $10$ .
Moving the decimal point one place to the left increases the power of $10$ by $1$ .

Common Mistakes

Subtracting coefficients without aligning powers of $10$ , e.g. treating $5 \times 10^7 - 8 \times 10^6$ as $(5-8) \times 10^6$ or $(5-8) \times 10^7$ .
Leaving the answer as $42 \times 10^6$ , which is not standard form.
Misplacing the decimal point when converting, such as writing $4.2 \times 10^6$ instead of $4.2 \times 10^7$ .
Giving $42 \times 10^6$ or $420 \times 10^5$ without converting to standard form; these may earn partial credit but not the final answer.

Things to Be Careful About

This is the non-calculator component, so show the adjustment by hand: $5 \times 10^7 = 50 \times 10^6$ .
The final answer is cao: it must be exactly $4.2 \times 10^7$ . The mark scheme allows B1 for seeing the figures $42$ or for an answer of the form $A \times 10^7$ with $1 \leq A < 10$ , and M1 for adjusting both numbers to the same power of $10$ (or writing $50\,000\,000$ and $8\,000\,000$ ).
Make sure the coefficient in standard form satisfies $1 \leq A < 10$ ; here it is $4.2$ .

Techniques used

rewrite numbers to a common power of tensubtract coefficientsconvert to standard form

Similar questions

Q141MEasyNumber

$120 = 2^3 \times 3 \times 5$
$126 = 2 \times 3^2 \times 7$

The lowest common multiple (LCM) of 120 and 126 is 2520.

Write 2520 as a product of its prime factors.

______

DifficultyEasy

Worked solution

Approach

To write the Lowest Common Multiple (LCM) as a product of its prime factors, we need to include every prime factor present in the original numbers, raised to the highest power that appears in any of those numbers.

Working

We are given:
$120 = 2^3 \times 3^1 \times 5^1$
$126 = 2^1 \times 3^2 \times 7^1$

The LCM is formed by taking the highest power of each prime base found in these factorisations:

For the prime $2$ , the powers are $2^3$ and $2^1$ . The highest is $2^3$ .
For the prime $3$ , the powers are $3^1$ and $3^2$ . The highest is $3^2$ .
For the prime $5$ , it appears only in $120$ with power $5^1$ . So we take $5^1$ (or just $5$ ).
For the prime $7$ , it appears only in $126$ with power $7^1$ . So we take $7^1$ (or just $7$ ).

Multiplying these together gives the prime factorisation of the LCM:
$\text{LCM} = 2^3 \times 3^2 \times 5 \times 7$

We can verify this: $8 \times 9 \times 5 \times 7 = 72 \times 35 = 2520$ . This matches the given LCM.

Answer

$2^3 \times 3^2 \times 5 \times 7$

Final answer

2^3 x 3^2 x 5 x 7

Detailed explanation

Walkthrough

The problem asks us to express the number 2520 (which is the LCM of 120 and 126) as a product of its prime factors. We are already provided with the prime factorisations of 120 and 126.

The rule for finding the LCM from prime factorisations is to take each unique prime factor involved and raise it to the highest exponent that appears in the factorisations of the individual numbers.

Identify all unique primes: Looking at $120 = 2^3 \times 3 \times 5$ and $126 = 2 \times 3^2 \times 7$ , the primes involved are 2, 3, 5, and 7.
Find the highest power for each:
- Prime 2: Appears as $2^3$ in 120 and $2^1$ in 126. Max power is 3. So we use $2^3$ .
- Prime 3: Appears as $3^1$ in 120 and $3^2$ in 126. Max power is 2. So we use $3^2$ .
- Prime 5: Appears as $5^1$ in 120 and not in 126. Max power is 1. So we use $5$ .
- Prime 7: Appears as $7^1$ in 126 and not in 120. Max power is 1. So we use $7$ .
Combine them: Multiply these selected factors together.

Result: $2^3 \times 3^2 \times 5 \times 7$ .

Key Takeaways

The LCM contains all prime factors present in the numbers.
The exponent for each prime in the LCM is the maximum of its exponents in the individual numbers.
Index notation ( $2^3$ ) is preferred over repeated multiplication ( $2 \times 2 \times 2$ ) unless specified otherwise, though both are mathematically correct.

Common Mistakes

Listing the lowest power instead of the highest (this would give the HCF, not the LCM).
Omitting a prime factor that appears in only one of the numbers (e.g., forgetting the 5 or the 7).
Multiplying the numbers together ( $120 \times 126$ ) instead of combining their factors correctly.

Things to Be Careful About

Ensure you identify the highest power. A common error is to just list the factors once without checking the exponents.
The mark scheme accepts both index form ( $2^3 \times 3^2 \times 5 \times 7$ ) and expanded form ( $2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7$ ). Index form is generally more concise and standard for this type of question.

Techniques used

combine prime factors from multiple numbersselect the highest power of each prime for LCMexpress the result in index form

Similar questions

Q152MMedium-EasyGeometry

Each interior angle of a regular polygon is $160°$ .

Find the number of sides of the polygon.

______

DifficultyMedium-Easy

Worked solution

Approach

The easiest way to find the number of sides is to use the exterior angle. The interior and exterior angles of any polygon lie on a straight line, so they add up to $180^\circ$ . Once we have the exterior angle, we can divide the total sum of exterior angles ( $360^\circ$ ) by it to get the number of sides.

Working

First, calculate the size of one exterior angle:

\text{Exterior angle} = 180^\circ - \text{Interior angle}

\text{Exterior angle} = 180^\circ - 160^\circ = 20^\circ

For any convex polygon, the sum of the exterior angles (taken one at each vertex) is always $360^\circ$ . For a regular polygon with $n$ sides, all exterior angles are equal. Therefore:

n = \frac{360^\circ}{\text{Exterior angle}}

Substitute the value we found:

n = \frac{360^\circ}{20^\circ}

n = 18

Alternatively, using the interior angle formula $\frac{180(n-2)}{n}$ :

\frac{180(n - 2)}{n} = 160

Multiply both sides by $n$ :

180(n - 2) = 160n

Expand the brackets:

180n - 360 = 160n

Rearrange to group terms:

180n - 160n = 360

20n = 360

Divide by 20:

n = 18

Answer

The polygon has 18 sides.

18

Final answer

18

Detailed explanation

Walkthrough

A regular polygon has sides and angles that are all equal. To find the number of sides given an interior angle, we can look at the relationship between the interior and exterior angles.

At any corner (vertex) of a polygon, the interior angle and the exterior angle form a straight line. A straight line measures $180^\circ$ . So, if the interior angle is $160^\circ$ , the exterior angle must be:

180^\circ - 160^\circ = 20^\circ

Another key property of polygons is that if you walk around the outside of any polygon, turning at each corner, you make exactly one full turn when you return to your starting point. This means the sum of the exterior angles of any polygon is always $360^\circ$ .

Since the polygon is regular, every exterior angle is the same size ( $20^\circ$ ). To find out how many corners (and therefore sides) there are, we simply ask: "How many times does $20^\circ$ fit into $360^\circ$ ?"

\frac{360}{20} = 18

So, there are 18 sides.

Key Takeaways

Interior + Exterior = $180^\circ$ : These two angles are supplementary.
Sum of Exterior Angles: The sum of the exterior angles of any polygon is always $360^\circ$ .
Regular Polygons: In a regular polygon, the number of sides $n$ can be found by $n = \frac{360}{\text{exterior angle}}$ .

Common Mistakes

Subtracting from $90^\circ$ : Some students confuse complementary angles with supplementary ones. Remember that interior and exterior angles are on a straight line, so they add to $180^\circ$ , not $90^\circ$ .
Using the wrong sum formula: Confusing the sum of interior angles ( $180(n-2)$ ) with the sum of exterior angles ( $360$ ).
Calculation errors: Dividing 360 by the interior angle ( $160$ ) instead of the exterior angle ( $20$ ).

Things to Be Careful About

Make sure to state the final answer clearly as a whole number representing the count of sides.
When using the interior angle formula method, be careful with algebraic manipulation (expanding brackets and collecting like terms).

Techniques used

calculate the exterior angle of a regular polygondetermine the number of sides from the exterior angle

Similar questions

Q16MediumTransformations and Vectors

Fig. 4

(a)

Draw the image of shape $A$ after a translation by vector $\begin{pmatrix} 3 \\ -5 \end{pmatrix}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

To translate shape $A$ by the vector $\begin{pmatrix} 3 \\ -5 \end{pmatrix}$ , add $3$ to the $x$ -coordinate and $-5$ to the $y$ -coordinate of each vertex.

Working

The vertices of shape $A$ are $(-2, 2)$ , $(2, 2)$ , $(2, 3)$ , $(-1, 3)$ , $(-1, 4)$ , and $(-2, 4)$ .

Applying the translation $(x, y) \to (x+3, y-5)$ :

\begin{aligned} (-2, 2) &\to (-2+3, 2-5) = (1, -3) \\ (2, 2) &\to (2+3, 2-5) = (5, -3) \\ (2, 3) &\to (2+3, 3-5) = (5, -2) \\ (-1, 3) &\to (-1+3, 3-5) = (2, -2) \\ (-1, 4) &\to (-1+3, 4-5) = (2, -1) \\ (-2, 4) &\to (-2+3, 4-5) = (1, -1) \end{aligned}

The new vertices are $(1, -3)$ , $(5, -3)$ , $(5, -2)$ , $(2, -2)$ , $(2, -1)$ , and $(1, -1)$ .

Answer

Draw the translated shape with vertices at $(1, -1)$ , $(2, -1)$ , $(2, -2)$ , $(5, -2)$ , $(5, -3)$ , and $(1, -3)$ .

Final answer

Shape drawn with vertices at (1, -1), (2, -1), (2, -2), (5, -2), (5, -3), (1, -3)

Detailed explanation

Walkthrough

The problem asks for the image of shape $A$ after a translation by the vector $\begin{pmatrix} 3 \\ -5 \end{pmatrix}$ . This means every point on the shape moves 3 units to the right and 5 units down. We take each vertex of shape $A$ and apply this change: add 3 to the $x$ -coordinate and subtract 5 from the $y$ -coordinate. For example, the vertex $(-2, 2)$ becomes $(-2+3, 2-5) = (1, -3)$ . We do this for all six vertices to get the new coordinates: $(1, -3)$ , $(5, -3)$ , $(5, -2)$ , $(2, -2)$ , $(2, -1)$ , and $(1, -1)$ . Plotting these points and joining them in the same order as the original shape gives the translated image.

Key Takeaways

A translation vector $\begin{pmatrix} a \\ b \end{pmatrix}$ adds $a$ to the $x$ -coordinates and $b$ to the $y$ -coordinates of all points.
The shape and orientation remain unchanged; only the position changes.

Common Mistakes

Forgetting to change the sign of the $y$ -component (e.g., adding 5 instead of subtracting 5).
Adding the vector components to the wrong coordinates (e.g., adding 3 to $y$ and -5 to $x$ ).
Not drawing the complete shape or connecting the vertices incorrectly.

Things to Be Careful About

Ensure the coordinates are calculated correctly; a small arithmetic error shifts the entire shape.
The final answer is a drawing, so accuracy in plotting is essential for full marks.

Techniques used

add vector components to vertex coordinatesplot the translated verticesdraw the translated shape

(b)

Describe fully the single transformation that maps shape $A$ onto shape $B$ .

______

3M Similar

DifficultyMedium

Worked solution

Approach

Compare shape $A$ and shape $B$ . They are congruent but have different orientations. Shape $A$ is oriented horizontally and shape $B$ vertically, suggesting a rotation. To find the centre and angle, we can test a point and its image, or use the property that the centre lies on the perpendicular bisector of the line joining a point and its image.

Working

Let's test a rotation of $90^{\circ}$ clockwise. Pick a vertex on $A$ , say $(-2, 2)$ . Its corresponding vertex on $B$ (top-left of the long vertical side) is $(5, 7)$ .

If the centre is $(h, k)$ , rotating $(-2, 2)$ by $90^{\circ}$ clockwise about $(h, k)$ gives $(h + (2-k), k - (-2-h)) = (h+2-k, k+2+h)$ .

Set this equal to $(5, 7)$ :

h + 2 - k = 5 \implies h - k = 3

k + 2 + h = 7 \implies h + k = 5

Adding the equations: $2h = 8 \implies h = 4$ . Substituting back: $4 + k = 5 \implies k = 1$ .

So the centre is $(4, 1)$ . Let's verify with another point. Vertex $(2, 2)$ on $A$ should map to $(5, 3)$ on $B$ .
Vector from $(4, 1)$ to $(2, 2)$ is $(-2, 1)$ . Rotating $90^{\circ}$ clockwise gives $(1, 2)$ . Adding to centre: $(4+1, 1+2) = (5, 3)$ . This matches.

Answer

Rotation $90^{\circ}$ clockwise about the centre $(4, 1)$ .

Final answer

Rotation 90 degrees clockwise about centre (4, 1)

Detailed explanation

Walkthrough

First, observe the shapes. Shape $A$ is an L-shape lying mostly flat, while shape $B$ is an L-shape standing up. This change in orientation indicates a rotation. To confirm, we can check if a rotation maps the vertices. Let's guess a $90^{\circ}$ clockwise rotation. We need to find the centre $(h, k)$ .

Take a vertex from $A$ , for example $(-2, 2)$ . Looking at the corresponding position on $B$ (the top-left corner of the main vertical block), the point is $(5, 7)$ . For a $90^{\circ}$ clockwise rotation about $(h, k)$ , the point $(x, y)$ maps to $(h + y - k, k - x + h)$ . Substituting $(-2, 2)$ and $(5, 7)$ :

h + 2 - k = 5 \implies h - k = 3

k - (-2) + h = 7 \implies h + k = 5

Solving these simultaneous equations gives $h = 4$ and $k = 1$ . So the centre is $(4, 1)$ .

We verify with another point, say $(2, 2)$ on $A$ . The vector from $(4, 1)$ to $(2, 2)$ is $\begin{pmatrix} -2 \\ 1 \end{pmatrix}$ . Rotating this vector $90^{\circ}$ clockwise gives $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ . Adding this to the centre $(4, 1)$ gives $(5, 3)$ , which is indeed a vertex on $B$ .

Key Takeaways

To describe a rotation fully, you must state the type (rotation), the angle and direction ( $90^{\circ}$ clockwise), and the centre coordinates.
The centre of rotation can be found by solving equations derived from the rotation formula or by constructing perpendicular bisectors of lines joining corresponding points.

Common Mistakes

Forgetting to state the centre coordinates (a partial description scores no marks).
Getting the direction wrong (e.g., saying anticlockwise when it is clockwise).
Not verifying the centre with a second point.

Things to Be Careful About

A full description requires three pieces of information: type, angle/direction, and centre. Missing any one costs a mark.
Ensure the angle is specified as $90^{\circ}$ and the direction clearly as clockwise.

Techniques used

identify transformation type by comparing shapestest a point to find the centre of rotationdetermine angle and direction of rotation

Q17Medium-EasyAlgebra and GraphsMensuration

Fig. 5

The diagram shows a shaded triangle, $ABC$ , drawn on a $1\,\text{cm}$ square grid.
The equation of the line $AC$ is $y = -\frac{3}{2}x + 1$ .

(a)

The shaded region inside triangle $ABC$ is defined by three inequalities.
One of these inequalities is $y \geq -\frac{3}{2}x + 1$ .

Find the other two inequalities.

______
______

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Identify the equations of the lines forming the other two boundaries of the triangle, then determine the correct inequality sign based on the shaded region.

Working

Line AB:
Passes through $A(0, 1)$ and $B(4, 5)$ .

\text{Gradient} = \frac{5 - 1}{4 - 0} = \frac{4}{4} = 1

The y-intercept is $1$ , so the equation is $y = x + 1$ .
The shaded region is below this line, so the inequality is $y \leq x + 1$ .

Line BC:
Passes through $B(4, 5)$ and $C(4, -5)$ .
This is a vertical line with equation $x = 4$ .
The shaded region is to the left of this line, so $x \leq 4$ .
Since the triangle is also bounded by the y-axis ( $x = 0$ ) at vertex $A$ , the full inequality is $0 \leq x \leq 4$ .

Answer

y \leq x + 1, \quad 0 \leq x \leq 4

Final answer

y <= x + 1, 0 <= x <= 4

Detailed explanation

Walkthrough

The shaded region is bounded by three lines. One is given as $y \geq -\frac{3}{2}x + 1$ , which is the line $AC$ . We need to find the inequalities for lines $AB$ and $BC$ .

First, consider line $AB$ . It passes through $A(0, 1)$ and $B(4, 5)$ . The gradient is $\frac{5 - 1}{4 - 0} = 1$ . The y-intercept is $1$ , giving the equation $y = x + 1$ . Since the shaded region lies below this line, the inequality is $y \leq x + 1$ .

Next, consider line $BC$ . It passes through $B(4, 5)$ and $C(4, -5)$ . Both points have an x-coordinate of $4$ , so this is the vertical line $x = 4$ . The shaded region is to the left of this line, meaning $x$ values are less than or equal to $4$ . Additionally, the leftmost point of the triangle is $A$ on the y-axis where $x = 0$ , so $x \geq 0$ . Combining these gives $0 \leq x \leq 4$ .

Key Takeaways

The equation of a line can be found using two points: $y = mx + c$ where $m$ is the gradient and $c$ is the y-intercept.
Vertical lines have equations of the form $x = k$ .
The direction of an inequality is determined by which side of the boundary line the shaded region lies.

Common Mistakes

Calculating the gradient incorrectly (e.g., $\frac{4-5}{4-1}$ instead of $\frac{5-1}{4-0}$ ).
Forgetting the $0 \leq$ part of the inequality for the vertical boundary, writing only $x \leq 4$ . The mark scheme awards a follow-through mark (SC1) if both are correct but one is missing, but full marks require the full range.
Writing the inequality for line $AB$ as $y \geq x + 1$ by misinterpreting the shaded side.

Things to Be Careful About

Ensure the inequality signs match the boundary lines (solid lines mean $\leq$ or $\geq$ , dashed lines mean $<$ or $>$ ). Here all boundaries are solid.
The mark scheme accepts $x \leq 4$ on its own as a B1, but $0 \leq x \leq 4$ is the complete and correct description of the region's x-bounds.
Always verify your inequality by testing a point inside the shaded region, such as $(2, 0)$ : $0 \leq 2 + 1$ is true, and $0 \leq 2 \leq 4$ is true.

Techniques used

find the equation of a line from two pointsdetermine the inequality direction from a shaded region

(b)

Work out the area of triangle $ABC$ .

______ $\text{cm}^2$

2M Similar

DifficultyEasy

Worked solution

Approach

Use the formula for the area of a triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ . Choose the vertical side $BC$ as the base to make calculation straightforward.

Working

Base:
Side $BC$ is vertical along the line $x = 4$ , from $y = -5$ to $y = 5$ .

\text{Length of } BC = 5 - (-5) = 10 \text{ cm}

Height:
The perpendicular distance from vertex $A(0, 1)$ to the line $BC$ ( $x = 4$ ) is the horizontal distance.

\text{Height} = 4 - 0 = 4 \text{ cm}

Area:

\text{Area} = \frac{1}{2} \times 10 \times 4 = 20 \text{ cm}^2

Answer

20

Final answer

20

Detailed explanation

Walkthrough

To find the area of triangle $ABC$ , we identify a base and the corresponding perpendicular height. The side $BC$ is vertical, making it an excellent choice for the base.

The coordinates of $B$ are $(4, 5)$ and $C$ are $(4, -5)$ . The length of $BC$ is the difference in y-coordinates: $5 - (-5) = 10$ cm. This is our base.

The height is the perpendicular distance from the opposite vertex $A(0, 1)$ to the line containing the base. Since $BC$ lies on the vertical line $x = 4$ , the perpendicular distance is the horizontal distance from $x = 0$ to $x = 4$ , which is $4$ cm.

Applying the area formula: $\frac{1}{2} \times 10 \times 4 = 20$ cm $^2$ .

Key Takeaways

When a side of a triangle is horizontal or vertical on a grid, it is convenient to use it as the base.
The height is always the perpendicular distance from the opposite vertex to the line containing the base.

Common Mistakes

Using the wrong height (e.g., using the y-coordinate of $A$ instead of the horizontal distance to $BC$ ).
Forgetting to multiply by $\frac{1}{2}$ in the area formula.
Counting grid squares incorrectly.

Things to Be Careful About

The grid is in cm, so the area is in cm $^2$ . The question asks for the answer in cm $^2$ .
Ensure the height is perpendicular to the base. If using $BC$ as the base, the height must be horizontal.

Techniques used

calculate the area of a triangle using base and height

Q18Medium-EasyStatistics

Lin records the masses, in grams, of 60 onions.
The cumulative frequency diagram shows her results.

Fig. 6

(a)

Use the diagram to find an estimate for the interquartile range.

______ $\text{g}$

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The interquartile range is the difference between the upper quartile (UQ) and the lower quartile (LQ). For a total frequency of 60, the LQ is at cumulative frequency $60 \div 4 = 15$ and the UQ is at cumulative frequency $3 \times 60 \div 4 = 45$ . Read these masses from the cumulative frequency diagram and subtract.

Working

Total frequency = 60.

Lower quartile position:

\frac{60}{4} = 15

Upper quartile position:

\frac{3 \times 60}{4} = 45

From the cumulative frequency diagram:

At cumulative frequency 15, the mass is approximately 205 g. So LQ = 205 g.
At cumulative frequency 45, the mass is 300 g. So UQ = 300 g.

Interquartile range = UQ - LQ

300 - 205 = 95

Answer

95

Final answer

95

Detailed explanation

Walkthrough

The interquartile range (IQR) measures the spread of the middle 50% of the data. To find it from a cumulative frequency diagram, we first locate the positions of the lower quartile (LQ) and upper quartile (UQ) on the vertical cumulative frequency axis.

For a dataset of $n = 60$ onions:

The LQ is at the $\frac{1}{4}$ position: $60 \div 4 = 15$ .
The UQ is at the $\frac{3}{4}$ position: $60 \times \frac{3}{4} = 45$ .

Next, draw horizontal lines from cumulative frequencies 15 and 45 across to the curve, then drop vertical lines down to the horizontal mass axis to read the corresponding values. From the diagram, the mass at CF = 15 is about 205 g, and the mass at CF = 45 is 300 g.

Finally, subtract the LQ from the UQ to get the IQR: $300 - 205 = 95$ g.

Key Takeaways

Quartile positions on a cumulative frequency graph are found at $\frac{n}{4}$ and $\frac{3n}{4}$ on the cumulative frequency axis.
The interquartile range is always calculated as Upper Quartile $-$ Lower Quartile.
Reading values from a graph requires drawing horizontal lines to the curve first, then vertical lines to the relevant axis.

Common Mistakes

Using $\frac{n}{2}$ (which gives 30) for the upper quartile instead of $\frac{3n}{4}$ (which gives 45).
Reading the cumulative frequency axis as ordinary frequency.
Forgetting to subtract the LQ from the UQ to get the IQR, and instead just reporting one of the quartile values.

Things to Be Careful About

Accuracy when reading from the graph is essential. The mark scheme accepts 205 for the LQ and 300 for the UQ, giving an IQR of 95. Small reading errors (e.g., 200 or 210) will lose the accuracy mark.
Always show the subtraction step ( $300 - 205$ ) to earn the method mark for the IQR calculation.

Techniques used

locate quartile positions on the cumulative frequency axisread off corresponding masses from the cumulative frequency curvecalculate interquartile range as upper quartile minus lower quartile

(b)

An onion is large if its mass is at least $N$ grams.
24 of the 60 onions are large.

Find the value of $N$ .

$N$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

An onion is large if its mass is at least $N$ grams. This means 24 onions have mass $\geq N$ . The remaining onions have mass $< N$ . Find the number of onions with mass less than $N$ , then read the corresponding mass from the cumulative frequency diagram.

Working

Total number of onions = 60.
Number of large onions (mass $\geq N$ ) = 24.

Number of onions with mass less than $N$ :

60 - 24 = 36

From the cumulative frequency diagram, find the mass at cumulative frequency 36.
At cumulative frequency 36, the mass is approximately 260 g.

Therefore, $N = 260$ .

Answer

260

Final answer

260

Detailed explanation

Walkthrough

The question states that 24 of the 60 onions are large, meaning their mass is at least $N$ grams. We need to find the threshold mass $N$ .

The cumulative frequency diagram shows how many onions have a mass less than or equal to a given value. If 24 onions have mass $\geq N$ , then the number of onions with mass $< N$ is the total minus the large ones:

60 - 24 = 36

Now, locate cumulative frequency 36 on the vertical axis. Draw a horizontal line to the curve and then a vertical line down to the horizontal mass axis. The value read is approximately 260 g.

This means 36 onions have a mass less than 260 g, and the remaining 24 onions have a mass of at least 260 g. Thus, $N = 260$ .

Key Takeaways

When a problem gives the number of items above a threshold, use the complementary frequency (total $-$ above) to find the cumulative frequency to read from the graph.
Cumulative frequency always represents the number of items below (or equal to) a given value.

Common Mistakes

Trying to find the mass corresponding to cumulative frequency 24 directly, instead of using the complementary frequency $60 - 24 = 36$ .
Misreading the graph by dropping a vertical line to the cumulative frequency axis instead of the mass axis.

Things to Be Careful About

The mark scheme awards a method mark (B1) for showing the calculation $60 - 24 = 36$ or simply seeing 36. Always show this working.
Read the mass axis carefully; at CF = 36, the value is 260, not 250 or 270.

Techniques used

use complementary frequency to find the threshold valueread off the corresponding mass from the cumulative frequency curve

Q192MMedium-EasyGeometryMensuration

The diagram shows two mathematically similar mugs.

Fig. 7

The small mug has width $w\,\text{cm}$ and holds $270\,\text{ml}$ when full.
The large mug has width $8\,\text{cm}$ and holds $640\,\text{ml}$ when full.

Find the value of $w$ .

$w$ = ______

DifficultyMedium-Easy

Worked solution

Approach

For two mathematically similar solids, the ratio of their volumes is the cube of the ratio of their corresponding linear dimensions. We can set up an equation using the given volumes and widths to find $w$ .

Working

The volume ratio of the small mug to the large mug is $\frac{270}{640}$ .

The linear scale factor from the large mug to the small mug is $\frac{w}{8}$ .

Since volume scales as the cube of the linear dimension:

\left(\frac{w}{8}\right)^3 = \frac{270}{640}

Simplify the volume fraction by dividing the numerator and denominator by $10$ :

\frac{270}{640} = \frac{27}{64}

Take the cube root of both sides:

\frac{w}{8} = \sqrt[3]{\frac{27}{64}}

\frac{w}{8} = \frac{3}{4}

Multiply both sides by $8$ to solve for $w$ :

w = 8 \times \frac{3}{4}

w = 6

Answer

w = 6

Final answer

6

Detailed explanation

Walkthrough

The question involves two mathematically similar mugs. A key property of similar solids is that the ratio of their volumes is equal to the cube of the ratio of their corresponding linear dimensions (such as width, height, or radius).

First, we identify the volume ratio. The small mug holds $270\,\text{ml}$ and the large mug holds $640\,\text{ml}$ . The ratio of the small volume to the large volume is $\frac{270}{640}$ .

Next, we identify the linear scale factor. The width of the small mug is $w\,\text{cm}$ and the width of the large mug is $8\,\text{cm}$ . The ratio of the small width to the large width is $\frac{w}{8}$ .

We equate the cube of the linear scale factor to the volume ratio:

\left(\frac{w}{8}\right)^3 = \frac{270}{640}

To make the arithmetic easier, we simplify the fraction $\frac{270}{640}$ by cancelling the common factor of $10$ , giving $\frac{27}{64}$ .

Taking the cube root of both sides gives:

\frac{w}{8} = \sqrt[3]{\frac{27}{64}} = \frac{3}{4}

Finally, we multiply both sides by $8$ to isolate $w$ :

w = 8 \times \frac{3}{4} = 6

Key Takeaways

For similar solids, the volume ratio is the cube of the linear scale factor: $\frac{V_1}{V_2} = \left(\frac{L_1}{L_2}\right)^3$ .
To find a linear dimension from a volume ratio, always take the cube root of the volume ratio.
Simplifying fractions before taking roots makes the calculation much easier and reduces the chance of arithmetic errors.

Common Mistakes

Forgetting to take the cube root and using the volume ratio $\frac{27}{64}$ directly as the linear scale factor, leading to an incorrect answer of $w = \frac{27}{8}$ .
Attempting to take the cube root of $\frac{270}{640}$ without simplifying first, which can lead to arithmetic mistakes.
Mixing up the ratio order (e.g., using $\frac{8}{w}$ instead of $\frac{w}{8}$ ), though this will still yield the correct final answer if applied consistently.

Things to Be Careful About

Units: The volumes are given in millilitres ( $\text{ml}$ ) and the widths in centimetres ( $\text{cm}$ ). Since $1\,\text{ml} = 1\,\text{cm}^3$ , the units are compatible and no conversion is needed for the ratio.
Exact form: The answer is an exact integer ( $6$ ), so no rounding is required. If the answer had been a non-integer, it would need to be given to an appropriate number of significant figures or as an exact surd/fraction depending on the question's instructions.
The diagram is marked 'NOT TO SCALE', so candidates must rely entirely on the mathematical relationships rather than measuring the diagram.

Techniques used

identify the linear scale factor from the volume ratiotake the cube root of the volume ratioset up an equation relating the widths and the scale factor

Similar questions

Q20EasyNumberAlgebra and Graphs

(a)

Simplify.

(16a^{20})^{\frac{3}{4}}

______

2M Similar

DifficultyEasy

Worked solution

Approach

Use the index law for powers raised to another power: $(x^a)^b = x^{a \times b}$ , and apply the exponent to both the numerical coefficient and the variable.

Working

We are given:

(16a^{20})^{\frac{3}{4}}

Apply the outer exponent $\frac{3}{4}$ to each factor inside the bracket:

= 16^{\frac{3}{4}} \times (a^{20})^{\frac{3}{4}}

First, evaluate $16^{\frac{3}{4}}$ . We know that $16 = 2^4$ , so:

16^{\frac{3}{4}} = (2^4)^{\frac{3}{4}} = 2^{4 \times \frac{3}{4}} = 2^3 = 8

Next, simplify the variable part by multiplying the exponents:

(a^{20})^{\frac{3}{4}} = a^{20 \times \frac{3}{4}}

Calculate the new exponent:

20 \times \frac{3}{4} = \frac{60}{4} = 15

So:

(a^{20})^{\frac{3}{4}} = a^{15}

Combine the results:

8a^{15}

Answer

8a^{15}

Final answer

8a^{15}

Detailed explanation

Walkthrough

The question asks us to simplify an expression with a fractional exponent applied to a term containing both a number and a variable. The key is to use two main index laws:

Power of a Product: $(xy)^n = x^n y^n$ . This allows us to take the exponent $\frac{3}{4}$ to both the $16$ and the $a^{20}$ separately.
Power of a Power: $(x^a)^b = x^{ab}$ . This tells us to multiply the existing exponent by the new one.

For the number $16$ , we need to find $16^{\frac{3}{4}}$ . A fractional exponent $\frac{p}{q}$ means taking the $q$ -th root and then raising to the power of $p$ . So, $16^{\frac{3}{4}} = (\sqrt[4]{16})^3$ . Since $2^4 = 16$ , the fourth root of $16$ is $2$ . Then $2^3 = 8$ . Alternatively, write $16$ as $2^4$ : $(2^4)^{3/4} = 2^{4 \cdot 3/4} = 2^3 = 8$ .

For the variable $a^{20}$ , we simply multiply the exponents: $20 \times \frac{3}{4} = 15$ . Thus, the term becomes $a^{15}$ .

Combining these gives $8a^{15}$ .

Key Takeaways

When a bracket contains multiple factors raised to a power, apply the power to EACH factor.
To evaluate a fractional power $x^{m/n}$ , you can find the $n$ -th root first, then raise to the $m$ -th power.
Always check if the base number is a perfect power (like $16=2^4$ ) to make fractional exponent calculations easier.

Common Mistakes

Forgetting to apply the exponent to the coefficient $16$ .
Adding the exponents instead of multiplying them (e.g., writing $a^{20 + 3/4}$ ).
Incorrectly calculating $16^{3/4}$ (e.g., doing $16 \times 3 / 4$ or $16^{0.75}$ incorrectly).
Not simplifying the final answer to the form $ka^n$ .

Things to Be Careful About

Ensure the final answer is in its simplest form ( $8a^{15}$ rather than $2^3 a^{15}$ ).
Remember that the mark scheme accepts intermediate forms like $8a^k$ (where $k \neq 0$ ) or just seeing $8a^{15}$ not as the final answer, but you should always provide the fully simplified result.

Techniques used

apply the power of a product rulemultiply indices using the power law

(b)

Expand and simplify.

(2c + 9d)(4c - 3d)

______

2M Similar

DifficultyEasy

Worked solution

Approach

Expand the product of two binomials $(2c + 9d)(4c - 3d)$ by multiplying every term in the first bracket by every term in the second bracket (often called the FOIL method). Then, combine any like terms.

Working

Multiply the terms:

First: $2c \times 4c = 8c^2$
Outer: $2c \times (-3d) = -6cd$
Inner: $9d \times 4c = 36cd$
Last: $9d \times (-3d) = -27d^2$

Write out the expanded expression:

8c^2 - 6cd + 36cd - 27d^2

Identify and collect the like terms. The middle terms $-6cd$ and $+36cd$ are like terms because they both contain $cd$ :

-6cd + 36cd = 30cd

Substitute this back into the expression:

8c^2 + 30cd - 27d^2

There are no other like terms to combine.

Answer

8c^2 + 30cd - 27d^2

Final answer

8c^2 + 30cd - 27d^2

Detailed explanation

Walkthrough

This question requires expanding two brackets containing variables. The most reliable method is to multiply each term in the first bracket by each term in the second bracket.

Let's break it down step-by-step:

Multiply $2c$ by $4c$ to get $8c^2$ .
Multiply $2c$ by $-3d$ to get $-6cd$ .
Multiply $9d$ by $4c$ to get $36cd$ .
Multiply $9d$ by $-3d$ to get $-27d^2$ .

After expanding, we have four terms: $8c^2 - 6cd + 36cd - 27d^2$ .

The next step is simplification. We look for 'like terms', which are terms that have exactly the same variable parts. Here, $-6cd$ and $36cd$ are like terms. We add their coefficients: $-6 + 36 = 30$ . So, they become $+30cd$ .

The terms $8c^2$ and $-27d^2$ do not have matching variable partners, so they remain as they are.

The final simplified expression is $8c^2 + 30cd - 27d^2$ .

Key Takeaways

Expansion of $(A+B)(C+D)$ results in $AC + AD + BC + BD$ .
Pay close attention to signs (positive and negative) during multiplication.
Only terms with identical variables and powers can be combined (collected).

Common Mistakes

Sign errors: e.g., forgetting that $9d \times -3d$ is negative, or $-6cd + 36cd$ is positive.
Forgetting to multiply the coefficients (e.g., writing $2c \times 4c$ as $8c$ or $24c^2$ ).
Failing to collect like terms at the end.
Creating extra terms or missing terms during expansion.

Things to Be Careful About

The mark scheme awards a method mark (M1) for getting three correct terms from the initial expansion ( $8c^2, -6cd, 36cd, -27d^2$ ), even if the final combination is wrong. However, the final answer must be fully simplified.
Ensure the order of terms does not matter, but standard practice is descending powers of the first variable ( $c$ ).

Techniques used

expand two binomials using FOILcollect like terms

Q213MMedium[Legacy] Matrices

The inverse of a matrix $\mathbf{A}$ is given by

\mathbf{A}^{-1} = \frac{1}{20} \begin{pmatrix} m & 7 \\ -1 & k \end{pmatrix}

$m$ and $k$ are positive integers and $m < k$ .
The determinant of matrix $\mathbf{A}$ is 20.

Find $\mathbf{A}$ .

$\mathbf{A}$ = ______

DifficultyMedium

Worked solution

Approach

For a $2 \times 2$ matrix, the inverse swaps the leading diagonal, negates the other diagonal, and divides by the determinant. Since the determinant is given as 20 and the inverse carries a factor of $\frac{1}{20}$ , we equate the two expressions entry by entry to recover $\mathbf{A}$ in terms of $m$ and $k$ . Then the determinant condition gives an equation for $m$ and $k$ , which we solve using the fact that they are positive integers with $m < k$ .

Working

Consider a general $2 \times 2$ matrix:

\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

Its inverse is:

\mathbf{A}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Since $\det(\mathbf{A}) = 20$ , the scalar factor is $\frac{1}{20}$ . The given inverse is:

\mathbf{A}^{-1} = \frac{1}{20} \begin{pmatrix} m & 7 \\ -1 & k \end{pmatrix}

Equating the two expressions:

\frac{1}{20} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \frac{1}{20} \begin{pmatrix} m & 7 \\ -1 & k \end{pmatrix}

Comparing entries:

$d = m$
$-b = 7 \Rightarrow b = -7$
$-c = -1 \Rightarrow c = 1$
$a = k$

So:

\mathbf{A} = \begin{pmatrix} k & -7 \\ 1 & m \end{pmatrix}

Now use $\det(\mathbf{A}) = 20$ :

\det(\mathbf{A}) = k \cdot m - (-7)(1) = km + 7 = 20

km = 13

Since $m$ and $k$ are positive integers with $m < k$ , and 13 is prime, the only factor pair is $m = 1$ , $k = 13$ .

Therefore:

\mathbf{A} = \begin{pmatrix} 13 & -7 \\ 1 & 1 \end{pmatrix}

Answer

\mathbf{A} = \begin{pmatrix} 13 & -7 \\ 1 & 1 \end{pmatrix}

Final answer

A = [[13, -7], [1, 1]]

Detailed explanation

Walkthrough

The key to this problem is knowing the formula for the inverse of a $2 \times 2$ matrix. For a matrix

\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

the inverse is obtained by swapping $a$ and $d$ , negating $b$ and $c$ , and dividing by the determinant $ad - bc$ :

\mathbf{A}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Since the determinant of $\mathbf{A}$ is given as 20, and the inverse is written with a factor of $\frac{1}{20}$ , the matrix inside the brackets must be exactly the swapped-and-negated version of $\mathbf{A}$ . By comparing the four entries, we recover $\mathbf{A}$ in terms of $m$ and $k$ : the top-left entry of $\mathbf{A}$ is $k$ , the top-right is $-7$ , the bottom-left is $1$ , and the bottom-right is $m$ .

Next, we use the determinant condition. The determinant of $\mathbf{A}$ is $k \cdot m - (-7)(1) = km + 7$ , and this must equal 20. So $km = 13$ . Since $m$ and $k$ are positive integers with $m < k$ , and 13 is prime, the only factor pair is $m = 1$ and $k = 13$ . Substituting gives the final answer.

Key Takeaways

The inverse of a $2 \times 2$ matrix swaps the leading diagonal, negates the other diagonal, and divides by the determinant.
The determinant of a $2 \times 2$ matrix is the product of the leading diagonal minus the product of the other diagonal.
When a product of two positive integers is a prime number, the only factor pair is 1 and the prime itself.

Common Mistakes

Forgetting to negate both off-diagonal entries when forming the inverse — this would give the wrong signs in $\mathbf{A}$ .
Swapping $m$ and $k$ : the condition $m < k$ forces $m = 1$ and $k = 13$ , not the other way round.
Forgetting that the determinant of $\mathbf{A}$ appears as the denominator in the inverse formula — since it is 20, the factor $\frac{1}{20}$ matches.

Things to Be Careful About

The condition $m < k$ is essential: it tells you which factor of 13 is $m$ and which is $k$ .
The determinant of $\mathbf{A}$ is 20, which is why the inverse carries the factor $\frac{1}{20}$ .
The determinant equation $km + 7 = 20$ may be written in equivalent forms (e.g. $20 = km + 7$ ); the mark scheme accepts any equivalent form, seen or implied.
This is a non-calculator question; the arithmetic is simple (factorising 13), so no calculator is needed.
The answer must be given as a $2 \times 2$ matrix with the correct entries in the correct positions.

Techniques used

recall the inverse formula for a 2x2 matrixequate corresponding entries of the inverseform the determinant equationsolve for positive integer factor pairs

Similar questions

Q22MediumAlgebra and GraphsNumber

$f(x) = \frac{3x - 1}{2} \quad g(x) = 5^x$

(a)

Find $f(-7)$ .

______

1M Similar

(b)

Find $f^{-1}(x)$ .

$f^{-1}(x)$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Write $y = f(x)$ , interchange $x$ and $y$ , then rearrange to make $y$ the subject. Rename $y$ as $f^{-1}(x)$ .

Working

Let

y = \frac{3x - 1}{2}

For the inverse, swap the roles of $x$ and $y$ :

x = \frac{3y - 1}{2}

Multiply both sides by $2$ :

2x = 3y - 1

Add $1$ to both sides:

2x + 1 = 3y

Divide by $3$ :

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

So

f^{-1}(x) = \frac{2x + 1}{3}

Answer

f^{-1}(x) = \frac{2x + 1}{3}

Final answer

f^{-1}(x) = (2x + 1)/3

Detailed explanation

Walkthrough

To find the inverse of a function, we reverse the operations. Write the function as $y = \frac{3x - 1}{2}$ . The inverse is found by swapping $x$ and $y$ , because the inverse maps outputs back to inputs. This gives $x = \frac{3y - 1}{2}$ . Now solve for $y$ : multiply by $2$ , add $1$ , then divide by $3$ . This yields $y = \frac{2x + 1}{3}$ , so $f^{-1}(x) = \frac{2x + 1}{3}$ .

Key Takeaways

The inverse function undoes the original function. The method of swapping variables and rearranging is standard for linear functions. It also reinforces changing the subject of a formula.

Common Mistakes

A common mistake is to stop after swapping variables without solving for $y$ . Another is to rearrange incorrectly, for example writing $y = \frac{2x - 1}{3}$ instead of $y = \frac{2x + 1}{3}$ . The mark scheme awards a method mark for the correct first step, so showing $2y = 3x - 1$ or $x = \frac{3y - 1}{2}$ is important.

Things to Be Careful About

The answer is an equivalent form, so $\frac{2x + 1}{3}$ is accepted; do not simplify incorrectly to $\frac{2x}{3} + 1$ . The variable in the final inverse is $x$ , not $y$ . On the non-calculator paper, the rearrangement must be shown by hand.

Techniques used

write y in place of f(x)swap x and y to find the inverserearrange to make y the subjectreplace y by f^{-1}(x)

(c)

$f\left(\frac{9}{25}\right) = g(x)$

Find $x$ .

$x$ = ______

3M Similar

DifficultyMedium

Worked solution

Approach

First evaluate $f\left(\frac{9}{25}\right)$ . Then write the resulting value as a power of $5$ and compare with $g(x) = 5^x$ to find $x$ .

Working

f\left(\frac{9}{25}\right) = \frac{3\left(\frac{9}{25}\right) - 1}{2}

Simplify the numerator:

3 \times \frac{9}{25} - 1 = \frac{27}{25} - \frac{25}{25} = \frac{2}{25}

So

f\left(\frac{9}{25}\right) = \frac{\frac{2}{25}}{2} = \frac{2}{25} \times \frac{1}{2} = \frac{1}{25}

Since $g(x) = 5^x$ , we need

5^x = \frac{1}{25}

Write $\frac{1}{25}$ as a power of $5$ :

\frac{1}{25} = \frac{1}{5^2} = 5^{-2}

Therefore

5^x = 5^{-2}

and so

x = -2

Answer

x = -2

Final answer

-2

Detailed explanation

Walkthrough

Start by substituting $\frac{9}{25}$ into $f$ . The numerator becomes $3 \times \frac{9}{25} - 1 = \frac{27}{25} - \frac{25}{25} = \frac{2}{25}$ . Dividing this by $2$ gives $\frac{1}{25}$ . The equation is now $5^x = \frac{1}{25}$ . Since $\frac{1}{25} = 5^{-2}$ , the bases match and the exponents must be equal, so $x = -2$ .

Key Takeaways

This question combines function notation with index laws. It shows that evaluating a function can produce a fraction, and that solving an equation like $5^x = \frac{1}{25}$ requires writing both sides with the same base.

Common Mistakes

A common error is to compute $3 \times \frac{9}{25} - 1$ as $\frac{27}{25} - 1 = \frac{2}{25}$ correctly but then forget to divide by $2$ , giving $\frac{2}{25}$ instead of $\frac{1}{25}$ . Another is to write $\frac{1}{25}$ as $5^2$ instead of $5^{-2}$ . The mark scheme gives partial credit for reaching $f\left(\frac{9}{25}\right) = \frac{1}{25}$ or $5^x = \frac{1}{25}$ .

Things to Be Careful About

Remember that $\frac{1}{25} = 5^{-2}$ , not $5^2$ . Also, the division by $2$ in $f$ applies to the whole numerator, so $\frac{\frac{2}{25}}{2} = \frac{1}{25}$ . On the non-calculator paper, the fraction arithmetic must be shown by hand.

Techniques used

evaluate a function at a fractional inputsimplify a compound fractionexpress a fraction as a power of a given baseequate exponents when bases are equal

Q23MediumNumberAlgebra and Graphs

A theatre offers a singing lesson ( $S$ ), a dancing lesson ( $D$ ) and an acting lesson ( $A$ ).
A group of 40 people are asked which lessons they take part in.
Some of the results are shown in the Venn diagram.

Fig. 8

(a)

All 40 people take part in at least one lesson.
3 people take part in a singing lesson and an acting lesson but not a dancing lesson.
7 people take part in a dancing lesson only.
19 people take part in a singing lesson.
4 times as many people take part in a singing lesson only as those who take part in all three lessons.

Use this information to complete the Venn diagram.

3M Similar

DifficultyMedium

Worked solution

Approach

Let $x$ represent the number of people taking all three lessons ( $S \cap D \cap A$ ). The number taking only a singing lesson is given as $4x$ . We can form an equation using the total number of people in the singing lesson ( $S$ ), which is 19.

Working

The total in $S$ is the sum of its four disjoint regions:

\begin{aligned} (S \text{ only}) + (S \cap D \text{ only}) + (S \cap A \text{ only}) + (S \cap D \cap A) &= 19 \\ 4x + 6 + 3 + x &= 19 \\ 5x + 9 &= 19 \\ 5x &= 10 \\ x &= 2 \end{aligned}

So the number taking all three lessons is $2$ , and the number taking only a singing lesson is $4 \times 2 = 8$ .

The total number of people is 40, and everyone takes at least one lesson (so the outside region is 0). Summing the known regions:

8 + 7 + 10 + 6 + 3 + 2 = 36

The remaining region is $D \cap A$ only:

40 - 36 = 4

Answer

The completed Venn diagram has the following values:

$S$ only: 8
$D$ only: 7
$A$ only: 10
$S \cap D$ only: 6
$S \cap A$ only: 3
$D \cap A$ only: 4
$S \cap D \cap A$ : 2
Outside: 0

Final answer

S only: 8, D only: 7, A only: 10, S∩D only: 6, S∩A only: 3, D∩A only: 4, S∩D∩A: 2, Outside: 0

Detailed explanation

Walkthrough

The problem gives us a Venn diagram with three sets (S, D, A) and several known values, but leaves some regions blank. We are told that 19 people take a singing lesson, and that the number taking only singing is 4 times the number taking all three. By letting the number taking all three be $x$ , we can express the number taking only singing as $4x$ . Adding up all the disjoint regions inside the S circle gives the total for S, which allows us to form the equation $4x + 6 + 3 + x = 19$ . Solving this gives $x = 2$ , so the all-three region is 2 and the S-only region is 8. Finally, since all 40 people are inside the circles, we sum the known regions (8 + 7 + 10 + 6 + 3 + 2 = 36) and subtract from 40 to find the missing D∩A region, which is 4.

Key Takeaways

When a Venn diagram has a relationship between two unknown regions (e.g., one is a multiple of another), define a variable for one and express the other in terms of it.
The total for a set is the sum of all its disjoint internal regions.
The universal set total is the sum of all disjoint regions across the entire diagram, including outside the sets.

Common Mistakes

Forgetting that the total in a set (like S = 19) includes the intersections, not just the 'only' region.
Adding the intersection values incorrectly or missing a region when summing to the universal total.
Not realising that 'all 40 people take part in at least one lesson' means the region outside all circles is 0.

Things to Be Careful About

Ensure all regions inside the Venn diagram sum exactly to the universal total (40).
When writing the final answer for a diagram completion, clearly state the value for each region so the marker can verify the work.
The mark scheme awards method marks for forming the equation and finding the values, so show the algebraic steps clearly.

Techniques used

define a variable for the unknown central regionform a linear equation from the total in one setsolve the equation to find the unknown valuesuse the universal set total to find the remaining region

(b)

Use set notation to describe the subset with 10 people.

______

1M Similar

Q24MediumCoordinate Geometry

$P$ is the point $(-1, 4)$ and $Q$ is the point $(-3, -2)$ .

(a)

Find the coordinates of the midpoint of the line $PQ$ .

( ______ , ______ )

1M Similar

(b)

Find the equation of the line perpendicular to $PQ$ which passes through the point $P$ .

______

4M Similar

DifficultyMedium

Worked solution

Approach

First, determine the gradient (slope) of the line passing through $P$ and $Q$ . Then, use the property that perpendicular lines have gradients whose product is $-1$ (negative reciprocals) to find the gradient of the required line. Finally, use the point-slope form or substitute the known point and new gradient into $y = mx + c$ to find the y-intercept and write the full equation.

Working

Step 1: Find the gradient of $PQ$ ( $m_{PQ}$ )
Using $P(-1, 4)$ and $Q(-3, -2)$ :

m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 4}{-3 - (-1)}

m_{PQ} = \frac{-6}{-3 + 1} = \frac{-6}{-2} = 3

Step 2: Find the gradient of the perpendicular line ( $m_{\perp}$ )
The gradient of a line perpendicular to another is the negative reciprocal of the original gradient.

m_{\perp} = -\frac{1}{m_{PQ}} = -\frac{1}{3}

Step 3: Find the equation of the line
The line passes through point $P(-1, 4)$ and has gradient $m = -\frac{1}{3}$ . We can use the equation $y = mx + c$ .
Substitute $x = -1$ , $y = 4$ , and $m = -\frac{1}{3}$ :

4 = \left(-\frac{1}{3}\right)(-1) + c

4 = \frac{1}{3} + c

Solve for $c$ :

c = 4 - \frac{1}{3}

To subtract, write $4$ as $\frac{12}{3}$ :

c = \frac{12}{3} - \frac{1}{3} = \frac{11}{3}

Now substitute $m$ and $c$ back into $y = mx + c$ :

y = -\frac{1}{3}x + \frac{11}{3}

Alternatively, using point-slope form $y - y_1 = m(x - x_1)$ :

y - 4 = -\frac{1}{3}(x - (-1))

y - 4 = -\frac{1}{3}(x + 1)

y = -\frac{1}{3}x - \frac{1}{3} + 4

y = -\frac{1}{3}x + \frac{11}{3}

Answer

$y = -\frac{1}{3}x + \frac{11}{3}$

Final answer

y = -1/3x + 11/3

Detailed explanation

Walkthrough

Calculate Gradient of PQ: The gradient formula is $\frac{\text{change in } y}{\text{change in } x}$ . Taking $P$ as $(x_1, y_1)$ and $Q$ as $(x_2, y_2)$ , we compute $\frac{-2 - 4}{-3 - (-1)}$ . This simplifies to $\frac{-6}{-2} = 3$ . Note that the order must be consistent (either $Q-P$ or $P-Q$ ).
Perpendicular Gradient: Two lines are perpendicular if their gradients multiply to $-1$ . If $m_{PQ} = 3$ , then $m_{\perp} \times 3 = -1$ , so $m_{\perp} = -\frac{1}{3}$ . This is the "negative reciprocal" rule.
Find Equation: Use $y = mx + c$ . We know $m = -\frac{1}{3}$ . We also know the line passes through $P(-1, 4)$ , meaning when $x = -1$ , $y = 4$ . Substitute these into the equation: $4 = -\frac{1}{3}(-1) + c$ . This becomes $4 = \frac{1}{3} + c$ . Subtracting $\frac{1}{3}$ from $4$ gives $\frac{11}{3}$ . Thus, $c = \frac{11}{3}$ . The final equation is $y = -\frac{1}{3}x + \frac{11}{3}$ .

Key Takeaways

Gradient formula: $m = \frac{y_2-y_1}{x_2-x_1}$ .
Perpendicular gradient relationship: $m_1 m_2 = -1$ .
Finding $c$ : Substitute a known point $(x, y)$ and the calculated $m$ into $y=mx+c$ .

Common Mistakes

Swapping $x$ and $y$ in the gradient calculation (calculating $\Delta x / \Delta y$ ).
Sign errors when subtracting negative coordinates (e.g., $-3 - (-1)$ becoming $-4$ instead of $-2$ ).
Forgetting the negative sign when taking the reciprocal (writing $1/3$ instead of $-1/3$ ).
Arithmetic errors when solving for $c$ (e.g., adding instead of subtracting fractions).

Things to Be Careful About

Show working for the gradient calculation clearly to secure method marks.
Ensure the final answer is in the standard linear form $y = mx + c$ unless otherwise specified.
On Component 1 (Non-calculator), keep answers in fraction form rather than decimal approximations.

Techniques used

calculate the gradient of a line from coordinatesdetermine the gradient of a perpendicular linefind the equation of a line given a point and gradient

Q253MMedium-EasyStatistics

The table shows the times each of 110 students take to travel to school one day.

Time ( $t$ minutes)	$0 < t \leq 5$	$5 < t \leq 10$	$10 < t \leq 20$	$20 < t \leq 40$
Frequency	30	25	35	20

Complete the histogram to show this information.

Fig. 9

DifficultyMedium-Easy

Worked solution

Approach

To complete the histogram, we need to calculate the frequency density for each class interval using the formula $\text{Frequency density} = \frac{\text{Frequency}}{\text{Class width}}$ , then draw rectangles with the corresponding widths and heights on the grid.

Working

For $5 < t \leq 10$ :

\begin{aligned} \text{Class width} &= 10 - 5 = 5 \\ \text{Frequency density} &= \frac{25}{5} = 5 \end{aligned}

For $10 < t \leq 20$ :

\begin{aligned} \text{Class width} &= 20 - 10 = 10 \\ \text{Frequency density} &= \frac{35}{10} = 3.5 \end{aligned}

For $20 < t \leq 40$ :

\begin{aligned} \text{Class width} &= 40 - 20 = 20 \\ \text{Frequency density} &= \frac{20}{20} = 1 \end{aligned}

Answer

The histogram is completed by drawing three additional rectangles:

From $t = 5$ to $t = 10$ with height $5$ .
From $t = 10$ to $t = 20$ with height $3.5$ .
From $t = 20$ to $t = 40$ with height $1$ .

Final answer

Histogram with bars for 5<t<=10 at height 5, 10<t<=20 at height 3.5, and 20<t<=40 at height 1.

Detailed explanation

Walkthrough

A histogram uses frequency density on the vertical axis, not frequency. Frequency density is calculated by dividing the frequency by the class width. The first bar ( $0 < t \leq 5$ ) is already drawn with a height of 6, which matches $\frac{30}{5} = 6$ . We apply the same formula to the remaining intervals. For $5 < t \leq 10$ , the width is 5 and frequency is 25, so the density is 5. For $10 < t \leq 20$ , the width is 10 and frequency is 35, giving a density of 3.5. For $20 < t \leq 40$ , the width is 20 and frequency is 20, giving a density of 1. These values are then plotted as rectangles on the grid with their respective widths and heights.

Key Takeaways

Always use frequency density, not frequency, for the vertical axis of a histogram with unequal class widths.
Frequency density is found by dividing frequency by class width: $\text{FD} = \frac{f}{w}$ .
The width of each rectangle must match the class interval, and the height must match the calculated frequency density.

Common Mistakes

Using the frequency (25, 35, 20) as the height of the bars instead of the frequency density.
Miscalculating the class width, especially for the interval $20 < t \leq 40$ where the width is 20, not 2.
Drawing rectangles that do not touch each other (histogram bars for continuous data must be adjacent with no gaps).
Not extending the final bar to $t = 40$ correctly, stopping at $t = 20$ or misreading the axis.

Things to Be Careful About

The vertical axis is labelled 'Frequency density', not 'Frequency'. Check the axis labels before plotting.
The class intervals are continuous and adjacent, so the rectangles must share their vertical boundaries (e.g., the bar for $5 < t \leq 10$ must start exactly where the bar for $0 < t \leq 5$ ends, at $t = 5$ ).
Accuracy in drawing is important; heights of 3.5 and 1 must be read precisely from the grid lines.

Techniques used

calculate frequency density from frequency and class widthdraw rectangles on a histogram grid with correct widths and heights

Similar questions

Q26MediumTransformations and Vectors

Fig. 10

In the diagram,
$\overrightarrow{OA} = \mathbf{a}$ , $\overrightarrow{OB} = 2\mathbf{b}$ and $AB : BC = 1 : 3$ .
$OBD$ is a straight line.

(a)

Express $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .

$\overrightarrow{AB}$ = ______

1M Similar

(b)

Show that $\overrightarrow{OC} = 8\mathbf{b} - 3\mathbf{a}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the given ratio $AB : BC = 1 : 3$ to express $\overrightarrow{BC}$ as a multiple of $\overrightarrow{AB}$ . Then, find $\overrightarrow{OC}$ by adding vectors along the path $O \to B \to C$ .

Working

Since $A$ , $B$ , and $C$ are collinear and $AB : BC = 1 : 3$ , the vector $\overrightarrow{BC}$ is three times the vector $\overrightarrow{AB}$ :

\overrightarrow{BC} = 3 \overrightarrow{AB}

Substitute $\overrightarrow{AB} = 2\mathbf{b} - \mathbf{a}$ from part (a):

\overrightarrow{BC} = 3(2\mathbf{b} - \mathbf{a}) = 6\mathbf{b} - 3\mathbf{a}

Now, find $\overrightarrow{OC}$ using the path $O \to B \to C$ :

\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC}

Substitute $\overrightarrow{OB} = 2\mathbf{b}$ and $\overrightarrow{BC} = 6\mathbf{b} - 3\mathbf{a}$ :

\overrightarrow{OC} = 2\mathbf{b} + (6\mathbf{b} - 3\mathbf{a})

Combine like terms:

\overrightarrow{OC} = 8\mathbf{b} - 3\mathbf{a}

Answer

\overrightarrow{OC} = 8\mathbf{b} - 3\mathbf{a}

Final answer

8b - 3a

Detailed explanation

Walkthrough

The problem gives the ratio $AB : BC = 1 : 3$ . Since the points are on the same line and in that order, the vector from $B$ to $C$ is three times the vector from $A$ to $B$ . We calculated $\overrightarrow{AB} = 2\mathbf{b} - \mathbf{a}$ in part (a), so $\overrightarrow{BC} = 3(2\mathbf{b} - \mathbf{a}) = 6\mathbf{b} - 3\mathbf{a}$ . To find $\overrightarrow{OC}$ , we can travel from $O$ to $B$ and then from $B$ to $C$ . Adding $\overrightarrow{OB} = 2\mathbf{b}$ and $\overrightarrow{BC} = 6\mathbf{b} - 3\mathbf{a}$ gives $8\mathbf{b} - 3\mathbf{a}$ . An alternative valid route is to use $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC}$ , where $\overrightarrow{AC} = 4\overrightarrow{AB} = 8\mathbf{b} - 4\mathbf{a}$ , leading to $\mathbf{a} + 8\mathbf{b} - 4\mathbf{a} = 8\mathbf{b} - 3\mathbf{a}$ .

Key Takeaways

When points are collinear and given in a ratio, you can scale the vector between them to find other vectors on the same line. Vector addition is path-independent, so you can choose any route (like $O \to B \to C$ or $O \to A \to C$ ) to reach the destination.

Common Mistakes

Assuming $\overrightarrow{AC} = 3\overrightarrow{AB}$ instead of $4\overrightarrow{AB}$ (forgetting that $AC = AB + BC$ ).
Making sign errors when distributing the scalar 3 into $(2\mathbf{b} - \mathbf{a})$ .
Incorrectly adding the vectors, such as $2\mathbf{b} + 6\mathbf{b} - 3\mathbf{a} = 8\mathbf{a} - 3\mathbf{b}$ (mixing up the coefficients of $\mathbf{a}$ and $\mathbf{b}$ ).

Things to Be Careful About

The question asks to "show that", so you must display the working that leads to $8\mathbf{b} - 3\mathbf{a}$ . Just writing the final answer will not earn full marks. Ensure all vector symbols ( $\mathbf{a}$ , $\mathbf{b}$ ) are bold or have arrows, depending on the required notation, and that the final expression matches the target exactly.

Techniques used

use the given ratio to find a multiple of the vector ABadd vectors along a path from O to C

(c)

$\overrightarrow{AD} = k \overrightarrow{OC}$

Find the value of $k$ .

$k$ = ______

1M Similar

DifficultyMedium

Worked solution

Approach

Since $OBD$ is a straight line, $\overrightarrow{OD}$ must be a scalar multiple of $\overrightarrow{OB}$ . Express $\overrightarrow{AD}$ in terms of this unknown scalar and the given vectors, then equate it to $k \overrightarrow{OC}$ to solve for $k$ by comparing coefficients.

Working

Since $O$ , $B$ , and $D$ are collinear, there exists a scalar $\lambda$ such that:

\overrightarrow{OD} = \lambda \overrightarrow{OB} = \lambda(2\mathbf{b}) = 2\lambda\mathbf{b}

Now, express $\overrightarrow{AD}$ using the path $A \to O \to D$ :

\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA}

Substitute $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OD} = 2\lambda\mathbf{b}$ :

\overrightarrow{AD} = 2\lambda\mathbf{b} - \mathbf{a}

We are given that $\overrightarrow{AD} = k \overrightarrow{OC}$ . Substitute $\overrightarrow{OC} = 8\mathbf{b} - 3\mathbf{a}$ from part (b):

\overrightarrow{AD} = k(8\mathbf{b} - 3\mathbf{a}) = 8k\mathbf{b} - 3k\mathbf{a}

Equate the two expressions for $\overrightarrow{AD}$ :

2\lambda\mathbf{b} - \mathbf{a} = 8k\mathbf{b} - 3k\mathbf{a}

Since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel vectors, we can equate their coefficients on both sides.

Equating the coefficients of $\mathbf{a}$ :

-1 = -3k

Solve for $k$ :

k = \frac{1}{3}

Answer

k = \frac{1}{3}

Final answer

1/3

Detailed explanation

Walkthrough

The key to this part is using the fact that $OBD$ is a straight line. This means $\overrightarrow{OD}$ is parallel to $\overrightarrow{OB}$ , so we can write $\overrightarrow{OD} = \lambda \overrightarrow{OB}$ for some number $\lambda$ . This gives $\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = 2\lambda\mathbf{b} - \mathbf{a}$ . We are also told $\overrightarrow{AD} = k\overrightarrow{OC} = k(8\mathbf{b} - 3\mathbf{a}) = 8k\mathbf{b} - 3k\mathbf{a}$ . Because $\mathbf{a}$ and $\mathbf{b}$ are not parallel, the only way these two expressions can be equal is if the coefficients of $\mathbf{a}$ match and the coefficients of $\mathbf{b}$ match. Looking at the $\mathbf{a}$ terms: $-1 = -3k$ , which immediately gives $k = 1/3$ . (We could also find $\lambda$ by matching the $\mathbf{b}$ terms: $2\lambda = 8k = 8/3$ , so $\lambda = 4/3$ , meaning $D$ is further along the line than $B$ ).

Key Takeaways

When a vector is expressed as a linear combination of two non-parallel vectors in two different ways, you can equate the corresponding coefficients to form simultaneous equations. Collinearity with the origin allows you to introduce a scalar multiplier for the position vector.

Common Mistakes

Assuming $\overrightarrow{AD}$ is parallel to $\overrightarrow{OB}$ instead of $\overrightarrow{OC}$ .
Forgetting that $\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA}$ and trying to add them.
Not realizing that equating coefficients is valid because $\mathbf{a}$ and $\mathbf{b}$ are non-parallel; trying to solve for both $\lambda$ and $k$ without separating the components.

Things to Be Careful About

The question asks for the value of $k$ , not $\lambda$ or the position of $D$ . Ensure you only solve for what is asked. Also, remember that $\mathbf{a}$ and $\mathbf{b}$ must be non-parallel for the coefficient comparison to be valid; the diagram confirms they are not parallel.

Techniques used

express vector AD in terms of a and b using the collinear points O, B, Dequate coefficients of linearly independent vectors to find k

Mathematics (Syllabus D) 4024/12 — October/November 2024

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Key Takeaways

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Key Takeaways

Common Mistakes

Things to Be Careful About

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Working