§ 1.1 · Reading Graphs & Charts
Data Analysis
Core Concept: Reading the visual story behind data
Data-analysis questions test whether you can extract numbers from pictures: bar graphs, line graphs, and pie charts. The math is light — the trap is misreading the axis, units, or percentage.
- Bar graph: compare heights; tallest = largest count.
- Line graph: read where the curve crosses a value on the y-axis, then find the x-value.
- Pie chart: each slice's percent × total = count for that category. All slices must sum to 100%.
The graph shows the number of apples that each of five students ate during a week. Which student ate the greatest number of apples?
- A. Dae
- B. Joe
- C. Etta
- D. Susie
- E. Vinh
Concept
On a bar graph, the tallest bar represents the largest quantity.
Strategy
Read each student's bar height against the y-axis; pick the maximum.
Steps
- Dae = 6, Joe = 3, Etta = 5, Susie = 4, Vinh = 1.
- The largest value is 6.
Pitfall
Don't confuse “greatest” with “least” — Vinh has the smallest bar.
The line graph shows the distance Andrew walked over time. How long did it take Andrew to walk the first 2 km?
- A. 15 minutes
- B. 1 hour, 15 min
- C. 1 hour, 45 min
- D. 2 hours
- E. 45 minutes
Concept
On a distance-vs-time graph, to find the time for a given distance, draw a horizontal line at that distance and read where it meets the curve.
Steps
- Draw a horizontal line at y = 2 km.
- It meets the curve at x = 1.25 hours.
- 0.25 hours = 15 minutes, so total = 1 hour 15 minutes.
Pitfall
Don't read off “2 hours” from the x-axis. Always go from distance → time.
Answer
B — 1 hour, 15 minutes
The pie chart shows results of a survey asking students their favourite fruit. 100 students were surveyed. How many chose banana? Orange 40%, Apple 40%, Banana 20%.
- A. 40
- B. 80
- C. 100
- D. 20
- E. 60
Concept
Pie chart count = percentage × total respondents.
Steps
20% of 100 = 0.20 × 100 = 20 students.
In the pie chart shown, 80 students chose juice. The juice slice has a right angle. How many students chose milk?
- A. 120
- B. 160
- C. 240
- D. 180
- E. 80
Concept
A right angle in a pie chart equals 90° / 360° = 1/4 of the total.
Strategy
Use the juice fraction and its count to find the total, then identify the milk fraction from the chart.
Steps
- Juice = 1/4 of total, so total = 80 × 4 = 320.
- From the chart, milk represents 1/2 of the circle.
- Milk = 320 × 1/2 = 160.
Pitfall
Always identify all slices before subtracting. The “rest of the circle” may include a third category.
§ 1.2 · Measures of Center & Spread
Mean, Median, Mode, Range
Core Concept: Four ways to summarize a list of numbers
- Mean: sum ÷ count. Sensitive to outliers.
- Median: the middle value after sorting. For an even count, average the two middle values.
- Mode: the value that appears most often. May have no mode, one mode, or multiple modes.
- Range: max − min. Measures spread, not center.
- Reversible identity: sum = mean × count.
Alexis took a total of 243,000 steps during the 30 days of April. What was her mean number of steps per day?
- A. 7900
- B. 8100
- C. 8000
- D. 7100
- E. 8200
Steps
Mean = 243000 ÷ 30 = 8100.
Quick Check
Estimate: 240,000 ÷ 30 = 8,000. The exact answer must be slightly larger.
Answer
B — 8,100 steps/day
Five different integers in a list have a median of 10 and a range of 7. What is the smallest possible integer in the list?
Concept
For 5 sorted integers a < b < c < d < e, the median is c and the range is e − a.
Strategy
Fix c = 10 and e − a = 7. Minimize a by making e as small as possible while keeping d strictly between 10 and e.
Steps
- We need 10 < d < e, so e ≥ 12.
- Smallest e = 12, so smallest a = 12 − 7 = 5.
- Verify: {5, 6, 10, 11, 12} works.
- If a = 4, then e = 11, but no integer d can fit between 10 and 11.
Pitfall
“Different integers” rules out repeats.
Heights of 4 athletes are 135, 160, 170, 175 cm. Laurissa joins. On the new team of 5, mode = median = mean. How tall is Laurissa?
- A. 135
- B. 160
- C. 170
- D. 175
- E. 148
Concept
A mode requires at least one repeated value. Since the original four are all different, Laurissa must equal one of them.
Strategy
Test each candidate that matches an existing height.
Steps
- Try L = 160: sorted list {135, 160, 160, 170, 175}.
- Mode = 160, median = 160, mean = 800 ÷ 5 = 160.
- All three measures equal 160.
The mean of six numbers is 10. When 25 is removed, what is the mean of the remaining numbers?
Strategy
Use sum = mean × count twice.
Steps
- Original sum = 10 × 6 = 60.
- After removing 25: new sum = 60 − 25 = 35, new count = 5.
- New mean = 35 ÷ 5 = 7.
§ 1.3 · Foundations
Probability Basics
Core Concept: The favorable / total framework
P(A) = favorable outcomes / total outcomes
- P(not A) = 1 − P(A).
- Mutually exclusive: P(A or B) = P(A) + P(B).
- Independent events: P(A and B) = P(A) × P(B).
- Overlap: P(A or B) = P(A) + P(B) − P(A and B).
A circular spinner has 12 unshaded sections and 3 shaded sections. Each unshaded section is 3× the size of each shaded section. The arrow is spun once. What is the probability it lands in a shaded section?
- A. 1/15
- B. 1/5
- C. 1/12
- D. 1/13
- E. 1/4
Key Pitfall
Sections are not equal sized. You cannot just write 3/15.
Strategy
Let one shaded section = 1 unit of area. Then each unshaded section = 3 units.
Steps
- Shaded area = 3 × 1 = 3 units.
- Unshaded area = 12 × 3 = 36 units.
- Total area = 39 units.
- P(shaded) = 3/39 = 1/13.
A bag contains some candies. P(choosing a red candy) = 5/6. The total number of candies could be:
Concept
If P = 5/6 in lowest terms, the total candies must be a multiple of 6.
Steps
From the options, only 6 is a multiple of 6.
A number is chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. What is the probability the chosen number is divisible by 2, or by 3, or by both?
- A. 4/9
- B. 5/9
- C. 6/9
- D. 7/9
- E. 8/9
Concept
Inclusion–Exclusion: |A ∪ B| = |A| + |B| − |A ∩ B|.
Steps
- Divisible by 2: {2, 4, 6, 8} → 4 numbers.
- Divisible by 3: {3, 6, 9} → 3 numbers.
- Divisible by both: {6} → 1 number.
- Union = 4 + 3 − 1 = 6.
- P = 6/9.
Pitfall
Forgetting to subtract the overlap double-counts 6.
A standard fair coin is tossed three times. What is the probability that all three outcomes are the same?
- A. 1/2
- B. 3/16
- C. 1/4
- D. 5/16
- E. 1/8
Strategy
List the sample space. There are 2³ = 8 outcomes.
Steps
- Total outcomes = 8.
- All same outcomes: HHH and TTT → 2 outcomes.
- P = 2/8 = 1/4.
§ 1.4 · Two-Dice Sample Spaces
Probability in Dice Problems
Core Concept: Master the 6 × 6 grid
For two standard dice there are exactly 36 equally likely outcomes. Sum 7 is the most common; the extremes 2 and 12 are rarest.
Ways by sum: 2:1, 3:2, 4:3, 5:4, 6:5, 7:6, 8:5, 9:4, 10:3, 11:2, 12:1.
Two standard six-sided dice are rolled. If the two top-face numbers are added, which sum is least likely?
Strategy
Sums closer to the extremes have fewer combinations. From the choices, 11 is farthest from the center.
Steps
Sum 7: 6 ways. Sum 8: 5 ways. Sum 9: 4 ways. Sum 10: 3 ways. Sum 11: 2 ways.
Two standard dice are rolled. What is the probability the sum is a prime number?
- A. 5/12
- B. 7/12
- C. 1/2
- D. 5/6
- E. 1/3
Concept
Prime sums possible: 2, 3, 5, 7, 11.
Steps
- Sum 2: 1 way. Sum 3: 2. Sum 5: 4. Sum 7: 6. Sum 11: 2.
- Total favorable = 15.
- P = 15/36 = 5/12.
Two standard dice are rolled. The product of the two numbers is computed. What is the probability that the ones digit of the product is 0?
- A. 11/36
- B. 2/9
- C. 1/36
- D. 1/6
- E. 5/36
Concept
Ones digit is 0 ⇔ product is a multiple of 10 ⇔ product has both a factor of 2 and a factor of 5.
Strategy
Since 5 only appears on die face 5, at least one die must show 5. The other die must show 2, 4, or 6.
Steps
- (5,2), (5,4), (5,6) → 3 ways.
- (2,5), (4,5), (6,5) → 3 ways.
- Total = 6 ways. P = 6/36 = 1/6.
Pitfall
Order matters: (5,2) and (2,5) are separate outcomes.
A special 6-sided die is rolled. P(multiple of 3) = 1/2, P(even) = 1/3. Which face-set works?
- A. {1,2,3,5,5,6}
- B. {1,2,3,3,5,6}
- C. {1,2,3,4,6,6}
- D. {1,2,3,3,4,6}
- E. {2,3,3,3,5,6}
Strategy
Out of 6 faces, need exactly 3 multiples of 3 and exactly 2 even faces.
Steps
Option B = {1, 2, 3, 3, 5, 6}. Multiples of 3: 3, 3, 6 → 3 faces. Even numbers: 2, 6 → 2 faces.
Pitfall
Each face is counted separately, even if the same number appears twice.
Answer
B — {1, 2, 3, 3, 5, 6}
§ 1.6 · Mixed Review · 10 Problems
Homework
Based on the bar graph, which sport is played by the most students? Hockey 40, Basketball 30, Soccer 50, Volleyball 20, Badminton 10.
- A. hockey
- B. basketball
- C. soccer
- D. volleyball
- E. badminton
Steps
Soccer = 50 is the tallest bar.
480 students voted for a favourite subject: Math 40%, Science 30%, Music 20%, Art 10%. How many voted for math?
- A. 184
- B. 192
- C. 96
- D. 144
- E. 288
Steps
Math voters = 0.40 × 480 = 192.
The mean of the four integers 78, 83, 82, and x is 80. Which statement is true?
- A. x is 2 greater
- B. x is 1 less
- C. x is 2 less
- D. x is 3 less
- E. x equals mean
Steps
- Sum = 80 × 4 = 320.
- x = 320 − 78 − 83 − 82 = 77.
- 80 − 77 = 3, so x is 3 less than the mean.
Answer
D — x is 3 less than the mean
The mean of the list 2, 9, 4, n, 2n is 6. What is n?
Steps
- Sum = 6 × 5 = 30.
- 2 + 9 + 4 + n + 2n = 30.
- 15 + 3n = 30, so n = 5.
The list of seven numbers 3, 15, 8, 8, 9, 9, n has exactly one mode, which is 8. What is n?
Strategy
Currently 8 appears twice and 9 appears twice. To make 8 the unique mode, 8 must appear more times.
Steps
- If n = 8: counts become 8 ×3 and 9 ×2. Unique mode = 8.
- If n = 9: mode becomes 9.
- If n = anything else: 8 and 9 stay tied.
Eloise bought water pumps for charity. Mean price was $85 per pump. Total spent was $765. How many pumps did she buy?
Steps
Count = total ÷ mean = 765 ÷ 85 = 9.
An integer is randomly chosen from {10, 11, 12, …, 19}. What is the probability the chosen integer is even?
- A. 3/10
- B. 4/10
- C. 5/10
- D. 6/10
- E. 7/10
Steps
Even numbers: 10, 12, 14, 16, 18 → 5 numbers. P = 5/10.
A spinner has 20 equal sections numbered 1–20. What is the probability the arrow stops in a section containing a divisor of 20?
- A. 12/20
- B. 14/20
- C. 15/20
- D. 7/20
- E. 6/20
Concept
Divisors of 20 divide 20 evenly.
Steps
- Divisors of 20: 1, 2, 4, 5, 10, 20.
- There are 6 divisors.
- P = 6/20.
Pitfall
Do not include numbers like 3, 6, 7, 8, or 9.
A box contains 15 red, 20 blue, and 16 green jelly beans. Jack first picks and eats a green, then picks and eats a blue. What is the probability his next pick is red?
- A. 15/31
- B. 34/49
- C. 15/49
- D. 2/7
- E. 1/15
Strategy
Conditional probability — update the counts after each removal.
Steps
- Start: 15R, 20B, 16G, total 51.
- After eating 1 green: 15R, 20B, 15G, total 50.
- After eating 1 blue: 15R, 19B, 15G, total 49.
- P(red next) = 15/49.
Pitfall
The red count does not change; only the total changes.
A dime = $0.10, a quarter = $0.25. Terry's jar contains $1.00 in dimes and $1.00 in quarters. He removes one coin at random. What is the probability it is a dime?
- A. 1/10
- B. 2/7
- C. 10/11
- D. 2/5
- E. 5/7
Key Pitfall
Equal dollar amounts do not mean equal coin counts.
Steps
- $1.00 in dimes = $1.00 ÷ $0.10 = 10 dimes.
- $1.00 in quarters = $1.00 ÷ $0.25 = 4 quarters.
- Total coins = 14.
- P(dime) = 10/14 = 5/7.
Variation
If the jar had $1 in nickels and $1 in quarters: 20 nickels and 4 quarters → P(nickel) = 20/24 = 5/6.