ISEE Middle Level Quiz: Calculating Averages
20 questions · exam conditions
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Calculating AveragesQuestion 1 of 20

The average of three numbers, a, b, and c, is 16. Number b is twice number a. Number c is 8 more than number b. What is the value of number a?

8
12
16
24
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ISEE Middle Level Quiz

ISEE Middle Level Quiz: Calculating Averages

Practice Calculating Averages in ISEE Middle Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Calculating Averages, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Middle Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The average of three numbers, a, b, and c, is 16. Number b is twice number a. Number c is 8 more than number b. What is the value of number a?

  1. 8 (correct answer)
  2. 12
  3. 16
  4. 24
Explanation: The sum of the three numbers is 3×16=483 \times 16 = 48, so a+b+c=48a + b + c = 48. We are given relationships between the numbers: b=2ab = 2a and c=b+8c = b + 8. We can express cc in terms of aa: c=(2a)+8c = (2a) + 8. Now substitute these expressions into the sum equation: a+(2a)+(2a+8)=48a + (2a) + (2a + 8) = 48. Combine like terms: 5a+8=485a + 8 = 48. Subtract 8 from both sides: 5a=405a = 40. Divide by 5: a=8a = 8.

Question 2

What is the mean daily temperature from 61, 63, 62, 60, 61, 47, 62?

  1. 59 (correct answer)
  2. 62
  3. 416
  4. 61
Explanation: This question tests middle school quantitative reasoning skills, specifically calculating the average from a data set. The average, or mean, is found by adding up all the numbers in a data set and then dividing by the number of values. In this scenario, you need to calculate the average of seven daily temperatures: 61, 63, 62, 60, 61, 47, 62. Adding these temperatures gives 416, and dividing by 7 yields approximately 59.43, which rounds to 59. Choice A is correct because it represents the properly rounded average temperature. The outlier of 47 degrees significantly lowers the average compared to the other temperatures clustering around 61-63. To help students, emphasize the importance of including all data points and practice recognizing how outliers affect the mean in real-world contexts like weather data.

Question 3

At a carnival game, 5 people won a prize of $10, 3 people won a prize of $20, and 2 people won a prize of $50. What was the average prize value won by these 10 people?

  1. $21 (correct answer)
  2. $25
  3. $26.67
  4. $30
Explanation: To find the average prize value, calculate the total value of all prizes and divide by the number of people. Total value = (5 \times \10) + (3 \times $20) + (2 \times $50) = $50 + $60 + $100 = $210.Thetotalnumberofpeopleis. The total number of people is 5 + 3 + 2 = 10.Theaverageprizeis. The average prize is $210 / 10 = $21$.

Question 4

Over 4 weeks, a food blogger rated a new restaurant each week on a scale of 1 to 10. Her average rating for the first 3 weeks was 7. Her rating in the fourth week was 9. What was her average rating for all 4 weeks?

  1. 7.0
  2. 7.5 (correct answer)
  3. 8.0
  4. 8.5
Explanation: The sum of the ratings for the first 3 weeks is 3×7=213 \times 7 = 21. The rating for the fourth week is 9. The total sum of the ratings for all 4 weeks is 21+9=3021 + 9 = 30. The average rating for the 4 weeks is the total sum divided by the number of weeks: 30/4=7.530 / 4 = 7.5.

Question 5

In a certain class, 12 students have an average height of 60 inches. The remaining 8 students have an average height of 55 inches. What is the average height, in inches, of all 20 students in the class?

  1. 57.5
  2. 58 (correct answer)
  3. 58.5
  4. 59
Explanation: This is a weighted average problem. First, find the total height of each group. For the first group: 12 students×60 inches/student=720 inches12 \text{ students} \times 60 \text{ inches/student} = 720 \text{ inches}. For the second group: 8 students×55 inches/student=440 inches8 \text{ students} \times 55 \text{ inches/student} = 440 \text{ inches}. The total height of all students is 720+440=1160 inches720 + 440 = 1160 \text{ inches}. The total number of students is 12+8=2012 + 8 = 20. The overall average height is 1160/20=58 inches1160 / 20 = 58 \text{ inches}.

Question 6

A cyclist travels 30 miles in 2 hours and then travels another 60 miles in 3 hours. What is the cyclist's average speed in miles per hour for the entire journey?

  1. 17.5
  2. 18 (correct answer)
  3. 19
  4. 20
Explanation: Average speed is calculated as total distance divided by total time. The total distance is 30+60=9030 + 60 = 90 miles. The total time is 2+3=52 + 3 = 5 hours. The average speed is 90 miles/5 hours=1890 \text{ miles} / 5 \text{ hours} = 18 miles per hour. Averaging the individual speeds (15 mph and 20 mph) would be incorrect.

Question 7

The average of a list of 5 numbers is 18. If one of the numbers in the list, 10, is removed and replaced with the number 30, what is the new average of the list?

  1. 20
  2. 22 (correct answer)
  3. 24
  4. 26
Explanation: The original sum of the 5 numbers is 5×18=905 \times 18 = 90. When 10 is removed, the sum becomes 90−10=8090 - 10 = 80. When 30 is then added, the new sum is 80+30=11080 + 30 = 110. The number of items in the list remains 5. The new average is 110/5=22110 / 5 = 22. Alternatively, the sum increased by 30−10=2030 - 10 = 20. This increase is distributed over the 5 numbers, so the average increases by 20/5=420 / 5 = 4. The new average is 18+4=2218 + 4 = 22.

Question 8

What is the average (arithmetic mean) of the set of all integers from –8 to 4, inclusive?

  1. -4
  2. -2.5
  3. -2 (correct answer)
  4. 2
Explanation: The set of integers is an evenly spaced sequence. For any evenly spaced sequence, the average is equal to the average of the first and last terms. The average is (−8+4)/2=−4/2=−2(-8 + 4) / 2 = -4 / 2 = -2. Alternatively, one could sum all the integers (−8,−7,...,3,4-8, -7, ..., 3, 4), which is -26, and divide by the count of the integers (4−(−8)+1=134 - (-8) + 1 = 13). The result is −26/13=−2-26 / 13 = -2.

Question 9

What is the average of the first five positive even integers that are multiples of 4?

  1. 10
  2. 12 (correct answer)
  3. 15
  4. 20
Explanation: The first five positive even integers that are multiples of 4 are 4, 8, 12, 16, and 20. Since this is an evenly spaced set of numbers, the average is the median, which is the middle number, 12. Alternatively, sum the numbers 4+8+12+16+20=604+8+12+16+20=60 and divide by the count, 5. The average is 60/5=1260/5 = 12.

Question 10

Jordan's scores on his first four science tests are 88, 92, 85, and 91. What score must he earn on his fifth test to have an average (arithmetic mean) score of exactly 90 for all five tests?

  1. 89
  2. 90
  3. 94 (correct answer)
  4. 96
Explanation: To find the required score on the fifth test, first calculate the total score needed for an average of 90 over five tests. This is 90×5=45090 \times 5 = 450. Next, find the sum of Jordan's first four scores: 88+92+85+91=35688 + 92 + 85 + 91 = 356. The score needed on the fifth test is the difference between the required total and the current total: 450−356=94450 - 356 = 94.

Question 11

The average of a set of four numbers is 15. A second set of six numbers has an average of 25. What is the average of the combined set of all ten numbers?

  1. 19
  2. 20
  3. 21 (correct answer)
  4. 22
Explanation: First, find the sum of the numbers in each set. The sum of the first set is 4×15=604 \times 15 = 60. The sum of the second set is 6×25=1506 \times 25 = 150. The sum of the combined set of ten numbers is 60+150=21060 + 150 = 210. The average of the combined set is the total sum divided by the total count: 210/10=21210 / 10 = 21. Simply averaging the two averages (15 and 25) is incorrect because the sets are of different sizes.

Question 12

What would the average be without the outlier 2 in 18, 19, 20, 18, 2, 19?

  1. 16
  2. 19 (correct answer)
  3. 18
  4. 96
Explanation: This question tests middle school quantitative reasoning skills, specifically calculating the average after removing an outlier from a data set. The average, or mean, is found by adding up all the numbers in a data set and then dividing by the number of values. In this scenario, you need to calculate the average of 18, 19, 20, 18, 19 after removing the outlier 2. Adding these five remaining values gives 94, and dividing by 5 yields 18.8, which rounds to 19. Choice B is correct because it accurately reflects the mean without the outlier. With the outlier included, the average would drop to 16, demonstrating the significant impact of extreme values. To help students master this concept, practice identifying outliers using visual representations like dot plots and calculating averages both with and without outliers.

Question 13

What would the average be without the outlier 95 in 68, 70, 72, 69, 71, 95?

  1. 74
  2. 70 (correct answer)
  3. 72
  4. 69
Explanation: This question tests middle school quantitative reasoning skills, specifically calculating the average after removing an outlier from a data set. The average, or mean, is found by adding up all the numbers in a data set and then dividing by the number of values. In this scenario, you need to calculate the average of the data set 68, 70, 72, 69, 71 after removing the outlier 95. Adding these five remaining values gives 350, and dividing by 5 yields exactly 70. Choice B is correct because it accurately reflects the mean without the outlier. This demonstrates how outliers can significantly affect averages - with the outlier included, the average would be about 74.2. To help students, practice identifying outliers and understanding their impact on statistical measures, using real-world examples like unusually high or low test scores.

Question 14

The measures of three angles in a quadrilateral are 80°, 100°, and 110°. What is the average (arithmetic mean) of the measures of all four angles in the quadrilateral?

  1. 85°
  2. 96.7°
  3. 95°
  4. 90° (correct answer)
Explanation: When you encounter angle problems in quadrilaterals, remember that the sum of all interior angles in any quadrilateral is always 360°. This is a fundamental property you can rely on to solve missing angle problems. Given three angles of 80°, 100°, and 110°, you first need to find the fourth angle. Add the known angles: 80°+100°+110°=290°80° + 100° + 110° = 290°. Since all four angles must sum to 360°, the fourth angle is 360°−290°=70°360° - 290° = 70°. Now you can find the average of all four angles: 80°+100°+110°+70°4=360°4=90°\frac{80° + 100° + 110° + 70°}{4} = \frac{360°}{4} = 90°. This confirms answer choice D is correct. Looking at the wrong answers: Choice A (85°) likely comes from averaging only the three given angles and getting confused about what to do next. Choice B (96.7°) results from incorrectly averaging just the three given angles: 290°3≈96.7°\frac{290°}{3} ≈ 96.7°. Choice C (95°) might come from estimating or making an arithmetic error when trying to include the fourth angle. Here's a powerful shortcut to remember: the average of all angles in any quadrilateral will always be 90° because 360°4=90°\frac{360°}{4} = 90°. This means you don't actually need to find the fourth angle to answer this type of question. Whenever you're asked for the average of all angles in a quadrilateral, the answer is automatically 90°, regardless of the individual angle measures given.

Question 15

After taking his third test, Leo's average score increased from 84 to 86. What was Leo's score on his third test?

  1. 85
  2. 88
  3. 90 (correct answer)
  4. 92
Explanation: Before the third test, Leo had taken two tests with an average of 84. The sum of his scores on the first two tests was 2×84=1682 \times 84 = 168. After the third test, he had taken three tests with an average of 86. The sum of his scores on the three tests was 3×86=2583 \times 86 = 258. The score on his third test is the difference between the new total and the old total: 258−168=90258 - 168 = 90.

Question 16

The average monthly rainfall in a city over a 6-month period was 3.5 inches. What was the total rainfall in that city during this period?

  1. 3.5 inches
  2. 9.5 inches
  3. 21.0 inches (correct answer)
  4. 24.5 inches
Explanation: The average is the total sum divided by the count. To find the total sum, multiply the average by the count. Total rainfall = Average rainfall × Number of months. Total rainfall = 3.5 inches/month×6 months=21.0 inches3.5 \text{ inches/month} \times 6 \text{ months} = 21.0 \text{ inches}.

Question 17

The average (arithmetic mean) of three distinct positive integers is 8. The median of these three integers is 9. What is the largest possible value for one of these integers?

  1. 11
  2. 13
  3. 14 (correct answer)
  4. 15
Explanation: If the average of three integers is 8, their sum is 3×8=243 \times 8 = 24. The median is 9, so the middle integer is 9. Let the three integers in increasing order be a,9,ba, 9, b. Their sum is a+9+b=24a + 9 + b = 24, which means a+b=15a + b = 15. To make bb (the largest integer) as large as possible, aa (the smallest integer) must be as small as possible. Since the integers are distinct and positive, the smallest possible value for aa is 1. If a=1a=1, then 1+b=151 + b = 15, which gives b=14b = 14. The set {1, 9, 14} satisfies all conditions.

Question 18

The average of nine numbers is 20. If a new number, 50, is added to the set, what is the new average of the ten numbers?

  1. 20
  2. 23 (correct answer)
  3. 25
  4. 35
Explanation: First, find the sum of the original nine numbers: 9×20=1809 \times 20 = 180. Next, add the new number to the sum: 180+50=230180 + 50 = 230. This is the sum of the new set of ten numbers. To find the new average, divide the new sum by the new count: 230/10=23230 / 10 = 23.

Question 19

The average of a set of 10 distinct positive integers is 15. If a new integer, 15, is added to the set, what is the new average of the 11 integers?

  1. Less than 15
  2. Cannot be determined
  3. Greater than 15
  4. Exactly 15 (correct answer)
Explanation: When you encounter average problems involving adding new numbers to a set, focus on how the new number compares to the existing average. Start with what you know: 10 distinct positive integers have an average of 15, so their sum is 10×15=15010 \times 15 = 150. When you add 15 to this set, you now have 11 integers with a total sum of 150+15=165150 + 15 = 165. The new average becomes 16511=15\frac{165}{11} = 15. Since you're adding a number (15) that equals the original average (15), the average remains unchanged at exactly 15. Here's why the other choices miss the mark: Choice (A) suggests the average decreases, but this only happens when you add a number smaller than the current average. Choice (C) claims the average increases, which occurs only when you add a number larger than the current average. Choice (B) states the answer cannot be determined, but you have all the information needed—the original average and the value being added. The key insight is that adding a number equal to the current average always preserves that average. Think of it this way: if everyone in a group has the same "weight" in terms of the average, adding another person with that exact same "weight" keeps the balance unchanged. Remember this pattern: when adding to a set, if the new number equals the current average, the average stays the same. If it's higher than the average, the new average increases. If it's lower, the new average decreases.

Question 20

The average of five consecutive odd integers is 21. What is the value of the largest of these integers?

  1. 21
  2. 23
  3. 25 (correct answer)
  4. 27
Explanation: For any set of consecutive odd integers, the average is equal to the median (the middle number). Since there are five integers, the middle number is the third one. Therefore, 21 is the third integer in the sequence. The five consecutive odd integers are 17, 19, 21, 23, and 25. The largest of these integers is 25.