Past Exam Questions

To find the intersection points, equate the equations of the curve and the line:

\(x^2 + 2cx + 4 = 4x + c\)

Rearrange to form a quadratic equation:

\(x^2 + 2cx - 4x + 4 - c = 0\)

This simplifies to:

\(x^2 + (2c - 4)x + (4 - c) = 0\)

For the curve and line to intersect at two distinct points, the discriminant \(b^2 - 4ac\) must be greater than zero.

Here, \(a = 1\), \(b = 2c - 4\), and \(c = 4 - c\).

Calculate the discriminant:

\((2c - 4)^2 - 4(1)(4 - c) > 0\)

Expand and simplify:

\(4c^2 - 16c + 16 - 16 + 4c > 0\)

\(4c^2 - 12c > 0\)

Factorize:

\(4c(c - 3) > 0\)

This inequality holds when \(c < 0\) or \(c > 3\).

To find the set of values of \(k\) for which the curve and line do not intersect, we set the equations equal to each other:

\(kx^2 + 2x - k = kx - 2\)

Rearrange to form a quadratic equation:

\(kx^2 + (2-k)x + (2-k) = 0\)

For the curve and line not to intersect, the discriminant of this quadratic must be less than zero:

\(b^2 - 4ac < 0\)

Here, \(a = k\), \(b = 2-k\), \(c = 2-k\).

Calculate the discriminant:

\((2-k)^2 - 4k(2-k) < 0\)

\(4 - 4k + k^2 - 8k + 4k^2 < 0\)

\(5k^2 - 12k + 4 < 0\)

Factor the quadratic:

\((k-2)(5k-2) < 0\)

Find the critical points by setting each factor to zero:

\(k - 2 = 0 \Rightarrow k = 2\)

\(5k - 2 = 0 \Rightarrow k = \frac{2}{5}\)

The inequality \((k-2)(5k-2) < 0\) holds between the roots:

\(\frac{2}{5} < k < 2\)

For the quadratic equation \(ax^2 + bx + c = 0\) to have exactly one root, the discriminant must be zero. The discriminant \(\Delta\) is given by \(b^2 - 4ac\).

Here, \(a = 16\), \(b = -24\), and \(c = 10 - k\).

Set the discriminant to zero:

\((-24)^2 - 4 \times 16 \times (10 - k) = 0\)

\(576 - 64(10 - k) = 0\)

\(576 - 640 + 64k = 0\)

\(-64 + 64k = 0\)

\(64k = 64\)

\(k = 1\)

Substitute \(k = 1\) back into the equation:

\(16x^2 - 24x + 9 = 0\)

Factor the quadratic:

\((4x - 3)^2 = 0\)

Thus, \(4x - 3 = 0\)

\(4x = 3\)

\(x = \frac{3}{4}\)

Solution

(a) Back-to-back stem-and-leaf diagram:

Key: \( 1 \mid 17 \mid 0 \) means \(171\) cm (Aces) and \(170\) cm (Jets).

(b) Aces ordered: 164, 166, 168, 169, 171, 173, 174, 180, 181, 182, 182.

Median \(= 173\ \text{cm}\).

Lower quartile \(= 168\ \text{cm}\), upper quartile \(= 181\ \text{cm}\).

\( \text{IQR} = 181 - 168 = 13\ \text{cm} \).

(c) Spread comparison (ranges):

Aces: \( 182 - 164 = 18 \) cm; Jets: \( 190 - 166 = 24 \) cm. Jets have the wider spread.

Solution

(i) The back-to-back stem-and-leaf diagram is constructed as follows:

Key: \( 5 \mid 1 \mid 6 \) means \(165\) cm (Anvils) and \(166\) cm (Brecons).

(ii) To find the median and interquartile range for the Anvils:

Ordered heights: 158, 165, 169, 170, 172, 173, 175, 180, 181, 184, 196.

Median \(= 173\).

Lower quartile \(Q_1 = 169\), upper quartile \(Q_3 = 181\).

\( \text{IQR} = Q_3 - Q_1 = 181 - 169 = 12 \).

To create a stem-and-leaf diagram, separate each number into a stem and a leaf, then order the leaves.

Key: \( 2 \mid 8 \) means \(28\) medals.

Solution

(i) The back-to-back stem-and-leaf diagram is constructed as follows:

Key: \( 9 \mid 4 \mid 2 \) represents \(0.049\ \text{g}\) (Factory A) and \(0.042\ \text{g}\) (Factory B).

(ii) To find the median and interquartile range for factory B:

Ordered masses: \( 0.031,\ 0.035,\ 0.038,\ 0.042,\ 0.044,\ 0.047,\ 0.048,\ 0.049,\ 0.051,\ 0.054,\ 0.056,\ 0.058,\ 0.064 \).

Median \(= 0.048\ \text{g}\).

Lower quartile \(Q_1 = 0.040\ \text{g}\), upper quartile \(Q_3 = 0.055\ \text{g}\).

\( \text{IQR} = Q_3 - Q_1 = 0.055 - 0.040 = 0.015\ \text{g} \).

(iii) Comparisons:

• The masses are generally heavier in factory A.
• The masses are more spread out in factory B.

Solution

(i) The back-to-back stem-and-leaf diagram is constructed as follows:

Key: \( 3 \mid 1 \mid 5 \) represents \(13\) kph for Bronlea and \(15\) kph for Rogate.

(ii) Median for Bronlea (ordered): \(3, 6, 13, 14, 17, 21, 22, 24, 25, 27, 28, 32, 33, 45\). Median \(=\) 8th value \(= 23\ \text{kph}\).

Rogate (ordered): \(4, 5, 7, 7, 10, 11, 13, 15, 16, 18, 23, 23, 26, 34\). \(Q_1 = 7,\ Q_3 = 23\). IQR \(= 23 - 7 = 16\).

(iii) From the diagram, Rogate is less windy than Bronlea.

Solution

(i) Back-to-back stem-and-leaf diagram:

Key: \( 1 \mid 9 \mid 4 \) means \(91\ \text{kg}\) (Team A) and \(94\ \text{kg}\) (Team B).

(ii) Interquartile range for Team A:

Ordered weights: 82, 84, 84, 91, 96, 97, 98, 99, 100, 104, 107, 109, 115, 116, 122.

\( Q_1 = 91,\quad Q_3 = 109 \)

\( \text{IQR} = Q_3 - Q_1 = 109 - 91 = 18\ \text{kg}. \)

(iii) New player in Team B:

\( \sum_{15} = 1399 \)

\( \dfrac{1399 + x}{16} = 93.9 \)

\( 1399 + x = 16 \times 93.9 = 1502.4 \)

\( x = 1502.4 - 1399 = 103.4 \ \text{kg} \approx 103\ \text{kg}. \)

To create a stem-and-leaf diagram, separate each number into a stem (the leading digit) and a leaf (the trailing digit). For example, the number 23 is split into a stem of 2 and a leaf of 3.

Step 1: Organize the data

• Stems: 1, 2, 3, 4
• Leaves for stem 1: 4, 5, 7, 8, 9, 9
• Leaves for stem 2: 1, 2, 2, 3, 4, 5, 6, 6, 8, 8
• Leaves for stem 3: 0, 2, 6, 8
• Leaves for stem 4: 1, 2, 5, 6, 7

Step 2: Construct the diagram

Step 3: Key

Key: \( 1 \mid 4 \) represents \( 14 \) glasses of water.

Summary:

• The lowest number of glasses is 14 and the highest is 47.
• Most values are clustered in the 20s, indicating a typical weekly consumption between 20–30 glasses.
• The stem-and-leaf diagram provides a clear visualization of the distribution and helps identify patterns quickly.

Solution

(i) The back-to-back stem-and-leaf diagram is constructed as follows:

Key: \( 3 \mid 5 \mid 4 \) represents \( 53 \) seconds for adults and \( 54 \) seconds for children.

(ii) Comparisons:

• Children's estimates are more spread out than adults'.
• Adults' estimates are generally lower than children's.
• Adults' estimates are roughly symmetrical, whereas children's estimates are skewed to the right.

Solution

(i) Construct the stem-and-leaf diagram:

Key: \(1 \mid 4\) represents \$140.

(ii) Find the median and interquartile range:

Ordered data: 10, 40, 60, 80, 100, 130, 140, 140, 140, 150, 150, 150, 160, 160, 160, 160, 170, 180, 180, 200, 210, 250, 270, 280, 310, 450, 570

\(\text{Median} = 160\) (14th value)

\(\text{LQ} = 140\) (7th value), \(\text{UQ} = 210\) (21st value)

\(\text{IQR} = \text{UQ} - \text{LQ} = 210 - 140 = 70\)

(iii) Identify outliers:

\(1.5 \times \text{IQR} = 1.5 \times 70 = 105\)

\(\text{Lower limit} = 140 - 105 = 35\)

\(\text{Upper limit} = 210 + 105 = 315\)

Outliers: 10, 450, 570

(i) To create a stem-and-leaf diagram, separate the thousands and hundreds (stem) from the tens and units (leaf) for each price.

Sort the prices: 14 300, 15 300, 15 400, 15 500, 16 100, 16 400, 16 800, 16 800, 17 300, 17 700, 18 500.

Stem-and-leaf diagram:

Key: \(14 \mid 3\) represents \(14300\) dollars.

(ii) To find the lower quartile (Q1), arrange the prices in ascending order and find the median of the first half.

Sorted prices: 14 300, 15 300, 15 400, 15 500, 16 100, 16 400, 16 800, 16 800, 17 300, 17 700, 18 500.

The first half: 14 300, 15 300, 15 400, 15 500, 16 100.

The median of the first half (lower quartile) is 15 400.

(i) The back-to-back stem-and-leaf diagram is constructed as follows:

Key: \( 5 \mid 8 \mid 4 \) means \( 0.85 \) m for flat screen.

(ii) To find the median of the conventional TVs, order the data: 0.65, 0.67, 0.69, 0.71, 0.74, 0.75, 0.77, 0.85, 0.86. The median is the middle value, 0.74.

The interquartile range (IQR) is calculated as the difference between the upper quartile (UQ) and the lower quartile (LQ). LQ = 0.68, UQ = 0.81, so IQR = 0.81 - 0.68 = 0.13.

(iii) For the flat screen TVs, calculate the mean:

Mean = \(\frac{0.85 + 0.94 + 0.91 + 0.96 + 1.04 + 0.89 + 1.07 + 0.92 + 0.76}{9} = 0.927\).

Calculate the standard deviation (SD):

SD = \(\sqrt{\frac{(0.85-0.927)^2 + (0.94-0.927)^2 + (0.91-0.927)^2 + (0.96-0.927)^2 + (1.04-0.927)^2 + (0.89-0.927)^2 + (1.07-0.927)^2 + (0.92-0.927)^2 + (0.76-0.927)^2}{9}}\)\( \approx 0.0882\).

(a) To find the median for the Cheetahs, list the times in order: 61, 63, 65, 78, 79, 82, 87, 88, 89, 90, 97, 98, 99, 100, 101, 102, 103, 104, 105. The median is the 10th value: 99 minutes.

For the interquartile range (IQR), find the lower quartile (LQ) and upper quartile (UQ). LQ is the 5th value: 83 minutes. UQ is the 15th value: 106 minutes. IQR = UQ - LQ = 106 - 83 = 23 minutes.

(b) The median time for the Cheetahs (99 minutes) is less than the median time for the Panthers (103 minutes), indicating that the Cheetahs are generally faster. The IQR for the Cheetahs (23 minutes) is greater than the IQR for the Panthers (14 minutes), indicating that the Cheetahs' times are more spread out.

(c) The total time for 20 Cheetahs is 99 minutes × 20 = 1980 minutes. The total time for the 19 Cheetahs is 1862 minutes. Kenny's time = 1980 - 1862 = 118 minutes.

(i) To create a back-to-back stem-and-leaf diagram, we use the integer part of the weights as the stem and the decimal part as the leaves.

Key: \( 1 \mid 196 \mid 2 \) means \( 1.961 \) kg for sugar and \( 1.962 \) kg for flour.

(ii) To find the median of the sugar weights, we arrange them in order: 1.961, 1.968, 1.977, 1.983, 1.984, 1.989, 1.994, 2.008, 2.011, 2.014, 2.017.

The median is the middle value: 1.989 kg.

To find the interquartile range, we find the lower quartile (LQ) and upper quartile (UQ).

LQ is the median of the first half: 1.977 kg.

UQ is the median of the second half: 2.011 kg.

\(Interquartile Range = UQ - LQ = 2.011 - 1.977 = 0.034 kg.\)

(i) To draw a stem-and-leaf diagram, separate each number into a 'stem' (the tens digit) and a 'leaf' (the units digit). Arrange the leaves in ascending order for each stem.

Key: 1 | 2 represents 12 people.

(ii) To find the median, order the data: 2, 5, 6, 8, 8, 12, 14, 16, 17, 17, 19, 21, 22, 23, 23, 23, 25, 26, 27, 31, 35. The median is the middle value, which is 19.

The lower quartile (LQ) is the median of the first half: 2, 5, 6, 8, 8, 12, 14, 16, 17, 17. LQ = 10.

The upper quartile (UQ) is the median of the second half: 21, 22, 23, 23, 23, 25, 26, 27, 31, 35. UQ = 24

The interquartile range (IQR) is UQ - LQ = 24 - 10 = 14.

(iii) The median is preferable because the mode could be any number which is duplicated more than twice, making it less reliable as a measure of central tendency.

To create a stem-and-leaf diagram, we first identify the stems and leaves from the data set. The stems are the tens digits, and the leaves are the units digits.

1. List the stems: 10, 11, 12, 13, 14, 15, 16.

2. Assign the leaves to each stem:

Key: \( 10 \mid 4 \) represents \( 104 \).

(i) To find the median, we need to arrange the data in order and find the middle value. There are 31 data points (including \(x\)). The median is the 16th value, which is 24.

The lower quartile (LQ) is the median of the first half of the data. The first 15 values are considered, and the 8th value is 16.

(ii) The interquartile range (IQR) is given as 19. The upper quartile (UQ) is calculated as \(\text{LQ} + 19 = 35\). The value of \(x\) must be 5 to maintain the order and achieve the correct UQ.

(i) The back-to-back stem-and-leaf diagram is constructed as follows:

Key: \( 7 \mid 3 \mid 2 \) means \( 13.7 \) minutes (16 yr olds) and \( 13.2 \) minutes (9 yr olds).

(ii) The sum of the times for the original 8 pupils is 106.8 minutes.

The total time for 9 pupils is given by the mean:

\(13.6 \times 9 = 122.4 \text{ minutes}\)

The new pupil's time is the difference between the total time for 9 pupils and the sum for 8 pupils:

\(122.4 - 106.8 = 15.6 \text{ minutes}\)