Gauss 7 · Past Contest Questions

Statistics & Probability

Module 01 of 06 · Full Teaching Edition

Module 01 of 06
Problems 27 total
Source Gauss 7 Past Questions
Format Teaching Edition

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%.
1
2025 Gauss 7 · Q3
Explorer

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

  1. Dae = 6, Joe = 3, Etta = 5, Susie = 4, Vinh = 1.
  2. The largest value is 6.

Pitfall

Don't confuse “greatest” with “least” — Vinh has the smallest bar.

Answer

A — Dae (6 apples)

2
2021 Gauss 7 · Q8
Explorer

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

  1. Draw a horizontal line at y = 2 km.
  2. It meets the curve at x = 1.25 hours.
  3. 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

3
2018 Gauss 7 · Q2
Explorer

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.

Answer

D — 20 students

4
2020 Gauss 7 · Q6
Practice

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

  1. Juice = 1/4 of total, so total = 80 × 4 = 320.
  2. From the chart, milk represents 1/2 of the circle.
  3. Milk = 320 × 1/2 = 160.

Pitfall

Always identify all slices before subtracting. The “rest of the circle” may include a third category.

Answer

B — 160 students

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.
5
2020 Gauss 7 · Q5
Explorer

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

6
2024 Gauss 7 · Q19
Explorer

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?

  • A. 4
  • B. 5
  • C. 6
  • D. 7
  • E. 8

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

  1. We need 10 < d < e, so e ≥ 12.
  2. Smallest e = 12, so smallest a = 12 − 7 = 5.
  3. Verify: {5, 6, 10, 11, 12} works.
  4. If a = 4, then e = 11, but no integer d can fit between 10 and 11.

Pitfall

“Different integers” rules out repeats.

Answer

B — 5

7
2018 Gauss 7 · Q19
Explorer

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

  1. Try L = 160: sorted list {135, 160, 160, 170, 175}.
  2. Mode = 160, median = 160, mean = 800 ÷ 5 = 160.
  3. All three measures equal 160.

Answer

B — 160 cm

8
2016 Gauss 7 · Q18
Practice

The mean of six numbers is 10. When 25 is removed, what is the mean of the remaining numbers?

  • A. 6
  • B. 7
  • C. 8
  • D. 9
  • E. 10

Strategy

Use sum = mean × count twice.

Steps

  1. Original sum = 10 × 6 = 60.
  2. After removing 25: new sum = 60 − 25 = 35, new count = 5.
  3. New mean = 35 ÷ 5 = 7.

Answer

B — 7

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).
9
2024 Gauss 7 · Q17
Explorer

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

  1. Shaded area = 3 × 1 = 3 units.
  2. Unshaded area = 12 × 3 = 36 units.
  3. Total area = 39 units.
  4. P(shaded) = 3/39 = 1/13.

Answer

D — 1/13

10
2021 Gauss 7 · Q6
Explorer

A bag contains some candies. P(choosing a red candy) = 5/6. The total number of candies could be:

  • A. 3
  • B. 10
  • C. 17
  • D. 6
  • E. 7

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.

Answer

D — 6

11
2024 Gauss 7 · Q10
Explorer

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

  1. Divisible by 2: {2, 4, 6, 8} → 4 numbers.
  2. Divisible by 3: {3, 6, 9} → 3 numbers.
  3. Divisible by both: {6} → 1 number.
  4. Union = 4 + 3 − 1 = 6.
  5. P = 6/9.

Pitfall

Forgetting to subtract the overlap double-counts 6.

Answer

C — 6/9

12
2022 Gauss 7 · Q16
Practice

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

  1. Total outcomes = 8.
  2. All same outcomes: HHH and TTT → 2 outcomes.
  3. P = 2/8 = 1/4.

Answer

C — 1/4

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.

13
2025 Gauss 7 · Q15
Explorer

Two standard six-sided dice are rolled. If the two top-face numbers are added, which sum is least likely?

  • A. 7
  • B. 8
  • C. 9
  • D. 10
  • E. 11

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.

Answer

E — sum of 11

14
2021 Gauss 7 · Q20
Explorer

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

  1. Sum 2: 1 way. Sum 3: 2. Sum 5: 4. Sum 7: 6. Sum 11: 2.
  2. Total favorable = 15.
  3. P = 15/36 = 5/12.

Pitfall

9 is not prime.

Answer

A — 5/12

15
2023 Gauss 7 · Q19
Explorer

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

  1. (5,2), (5,4), (5,6) → 3 ways.
  2. (2,5), (4,5), (6,5) → 3 ways.
  3. Total = 6 ways. P = 6/36 = 1/6.

Pitfall

Order matters: (5,2) and (2,5) are separate outcomes.

Answer

D — 1/6

16
2013 Gauss 7 · Q19
Practice

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}

Complicated Probability

17
2016 Gauss 7 · Q21
Challenge

A 10 × 10 grid is created using 100 points. Point P is given. One of the other 99 points is randomly chosen to be Q. What is the probability that segment PQ is vertical or horizontal?

  • A. 2/11
  • B. 1/5
  • C. 1/10
  • D. 4/25
  • E. 5/33

Concept

PQ is horizontal ⇔ P and Q are in the same row. PQ is vertical ⇔ same column.

Strategy

Count points in the same row as P plus points in the same column as P, excluding P itself.

Steps

  1. Same row as P: 10 − 1 = 9 other points.
  2. Same column as P: 10 − 1 = 9 other points.
  3. No overlap, since P itself is excluded.
  4. Favorable Q's = 18.
  5. P(vertical or horizontal) = 18/99 = 2/11.

Pitfall

This answer does not depend on where P sits in the grid.

Variation

For an n × n grid: P = 2(n−1)/(n²−1) = 2/(n+1). For n = 10, this gives 2/11.

Answer

A — 2/11

Homework

18
2017 Gauss 7 · Q2
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.

Answer

C — Soccer

19
2015 Gauss 7 · Q11
Homework

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.

Answer

B — 192

20
2017 Gauss 7 · Q17
Homework

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

  1. Sum = 80 × 4 = 320.
  2. x = 320 − 78 − 83 − 82 = 77.
  3. 80 − 77 = 3, so x is 3 less than the mean.

Answer

D — x is 3 less than the mean

21
2023 Gauss 7 · Q15
Homework

The mean of the list 2, 9, 4, n, 2n is 6. What is n?

  • A. 9
  • B. 12
  • C. 10
  • D. 5
  • E. 6

Steps

  1. Sum = 6 × 5 = 30.
  2. 2 + 9 + 4 + n + 2n = 30.
  3. 15 + 3n = 30, so n = 5.

Answer

D — n = 5

22
2025 Gauss 7 · Q7
Homework

The list of seven numbers 3, 15, 8, 8, 9, 9, n has exactly one mode, which is 8. What is n?

  • A. 15
  • B. 9
  • C. 3
  • D. 8
  • E. 10

Strategy

Currently 8 appears twice and 9 appears twice. To make 8 the unique mode, 8 must appear more times.

Steps

  1. If n = 8: counts become 8 ×3 and 9 ×2. Unique mode = 8.
  2. If n = 9: mode becomes 9.
  3. If n = anything else: 8 and 9 stay tied.

Answer

D — n = 8

23
2024 Gauss 7 · Q13
Homework

Eloise bought water pumps for charity. Mean price was $85 per pump. Total spent was $765. How many pumps did she buy?

  • A. 7
  • B. 8
  • C. 9
  • D. 10
  • E. 6

Steps

Count = total ÷ mean = 765 ÷ 85 = 9.

Answer

C — 9 pumps

24
2023 Gauss 7 · Q9
Homework

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.

Answer

C — 5/10

25
2017 Gauss 7 · Q16
Homework

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

  1. Divisors of 20: 1, 2, 4, 5, 10, 20.
  2. There are 6 divisors.
  3. P = 6/20.

Pitfall

Do not include numbers like 3, 6, 7, 8, or 9.

Answer

E — 6/20

26
2020 Gauss 7 · Q14
Homework

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

  1. Start: 15R, 20B, 16G, total 51.
  2. After eating 1 green: 15R, 20B, 15G, total 50.
  3. After eating 1 blue: 15R, 19B, 15G, total 49.
  4. P(red next) = 15/49.

Pitfall

The red count does not change; only the total changes.

Answer

C — 15/49

27
2022 Gauss 7 · Q10
Homework

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. $1.00 in dimes = $1.00 ÷ $0.10 = 10 dimes.
  2. $1.00 in quarters = $1.00 ÷ $0.25 = 4 quarters.
  3. Total coins = 14.
  4. 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.

Answer

E — 5/7