Calculating Averages - ISEE Middle Level Quantitative Reasoning

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?
Answer: 7.5 The sum of the ratings for the first 3 weeks is $3 \times 7 = 21$ . The rating for the fourth week is 9. The total sum of the ratings for all 4 weeks is $21 + 9 = 30$ . The average rating for the 4 weeks is the total sum divided by the number of weeks: $30 / 4 = 7.5$ .
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?
Answer: 18 Average speed is calculated as total distance divided by total time. The total distance is $30 + 60 = 90$ miles. The total time is $2 + 3 = 5$ hours. The average speed is $90 \text{ miles} / 5 \text{ hours} = 18$ miles per hour. Averaging the individual speeds (15 mph and 20 mph) would be incorrect.
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?
Answer: 14 If the average of three integers is 8, their sum is $3 \times 8 = 24$ . The median is 9, so the middle integer is 9. Let the three integers in increasing order be $a, 9, b$ . Their sum is $a + 9 + b = 24$ , which means $a + b = 15$ . To make $b$ (the largest integer) as large as possible, $a$ (the smallest integer) must be as small as possible. Since the integers are distinct and positive, the smallest possible value for $a$ is 1. If $a=1$ , then $1 + b = 15$ , which gives $b = 14$ . The set {1, 9, 14} satisfies all conditions.
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?
Answer: Exactly 15 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 $10 \times 15 = 150$ . When you add 15 to this set, you now have 11 integers with a total sum of $150 + 15 = 165$ . The new average becomes $\frac{165}{11} = 15$ . 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.
After taking his third test, Leo's average score increased from 84 to 86. What was Leo's score on his third test?
Answer: 90 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 $2 \times 84 = 168$ . After the third test, he had taken three tests with an average of 86. The sum of his scores on the three tests was $3 \times 86 = 258$ . The score on his third test is the difference between the new total and the old total: $258 - 168 = 90$ .
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?
Answer: 21 First, find the sum of the numbers in each set. The sum of the first set is $4 \times 15 = 60$ . The sum of the second set is $6 \times 25 = 150$ . The sum of the combined set of ten numbers is $60 + 150 = 210$ . The average of the combined set is the total sum divided by the total count: $210 / 10 = 21$ . Simply averaging the two averages (15 and 25) is incorrect because the sets are of different sizes.
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?
Answer: 21.0 inches 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 = $3.5 \text{ inches/month} \times 6 \text{ months} = 21.0 \text{ inches}$ .
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?
Answer: 58 This is a weighted average problem. First, find the total height of each group. For the first group: $12 \text{ students} \times 60 \text{ inches/student} = 720 \text{ inches}$ . For the second group: $8 \text{ students} \times 55 \text{ inches/student} = 440 \text{ inches}$ . The total height of all students is $720 + 440 = 1160 \text{ inches}$ . The total number of students is $12 + 8 = 20$ . The overall average height is $1160 / 20 = 58 \text{ inches}$ .
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?
Answer: 90° 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: $80° + 100° + 110° = 290°$ . Since all four angles must sum to 360°, the fourth angle is $360° - 290° = 70°$ . Now you can find the average of all four angles: $\frac{80° + 100° + 110° + 70°}{4} = \frac{360°}{4} = 90°$ . 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: $\frac{290°}{3} ≈ 96.7°$ . 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 $\frac{360°}{4} = 90°$ . 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.
The average weight of 6 packages is 24 pounds. If one package weighing 30 pounds is removed, what is the average weight of the remaining 5 packages?
Answer: 22.8 pounds When you encounter average problems involving the removal of items, you need to work with totals rather than just the averages themselves. The key insight is that removing an item changes both the total and the count. Start by finding the total weight of all 6 packages. Since the average weight is 24 pounds, the total weight is $6 \times 24 = 144$ pounds. When you remove the 30-pound package, the remaining total becomes $144 - 30 = 114$ pounds. Now you have 5 packages with a total weight of 114 pounds, so the new average is $114 \div 5 = 22.8$ pounds. Looking at the wrong answers: Choice A (19.0 pounds) represents a common error where students might subtract the removed package's weight from the original average (24 - 5 = 19), but this doesn't account for how averages actually work. Choice B (25.2 pounds) could result from incorrectly adding instead of subtracting, or from calculation errors with the totals. Choice C (24.0 pounds) suggests the misconception that removing any package won't change the average, but since the removed package (30 pounds) weighs more than the original average (24 pounds), the remaining packages must have a lower average. Remember this pattern: when you remove an item that's above the average, the new average decreases. When you remove an item below the average, the new average increases. Always work with totals first, then recalculate the average with the new count.
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?
Answer: 22 The original sum of the 5 numbers is $5 \times 18 = 90$ . When 10 is removed, the sum becomes $90 - 10 = 80$ . When 30 is then added, the new sum is $80 + 30 = 110$ . The number of items in the list remains 5. The new average is $110 / 5 = 22$ . Alternatively, the sum increased by $30 - 10 = 20$ . This increase is distributed over the 5 numbers, so the average increases by $20 / 5 = 4$ . The new average is $18 + 4 = 22$ .
What is the mean daily temperature from 61, 63, 62, 60, 61, 47, 62?
Answer: 59 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.
What is the average (arithmetic mean) of the set of all integers from –8 to 4, inclusive?
Answer: -2 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 $(-8 + 4) / 2 = -4 / 2 = -2$ . Alternatively, one could sum all the integers ( $-8, -7, ..., 3, 4$ ), which is -26, and divide by the count of the integers ( $4 - (-8) + 1 = 13$ ). The result is $-26 / 13 = -2$ .
What would the average be without the outlier 95 in 68, 70, 72, 69, 71, 95?
Answer: 70 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.
The average of five consecutive odd integers is 21. What is the value of the largest of these integers?
Answer: 25 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.
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?
Answer: 8 The sum of the three numbers is $3 \times 16 = 48$ , so $a + b + c = 48$ . We are given relationships between the numbers: $b = 2a$ and $c = b + 8$ . We can express $c$ in terms of $a$ : $c = (2a) + 8$ . Now substitute these expressions into the sum equation: $a + (2a) + (2a + 8) = 48$ . Combine like terms: $5a + 8 = 48$ . Subtract 8 from both sides: $5a = 40$ . Divide by 5: $a = 8$ .
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?
Answer: 23 First, find the sum of the original nine numbers: $9 \times 20 = 180$ . Next, add the new number to the sum: $180 + 50 = 230$ . This is the sum of the new set of ten numbers. To find the new average, divide the new sum by the new count: $230 / 10 = 23$ .
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?
Answer: 94 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 $90 \times 5 = 450$ . Next, find the sum of Jordan's first four scores: $88 + 92 + 85 + 91 = 356$ . The score needed on the fifth test is the difference between the required total and the current total: $450 - 356 = 94$ .
What is the average of the first five positive even integers that are multiples of 4?
Answer: 12 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 $4+8+12+16+20=60$ and divide by the count, 5. The average is $60/5 = 12$ .
A set contains five distinct positive integers. The average of the set is 20. What is the greatest possible value of the median of this set?
Answer: 31 The sum of the five integers is $5 \times 20 = 100$ . Let the integers in increasing order be $a, b, c, d, e$ . The median is $c$ . To maximize $c$ , we must minimize $a$ and $b$ , and keep $d$ and $e$ as close to $c$ as possible. The smallest distinct positive integers for $a$ and $b$ are 1 and 2. Since the integers must be distinct, the smallest possible values for $d$ and $e$ are $c+1$ and $c+2$ . The sum is $1 + 2 + c + (c+1) + (c+2) = 100$ . This simplifies to $3c + 6 = 100$ , so $3c = 94$ , and $c = 31.33...$ . Since $c$ must be an integer, the greatest possible value for $c$ is 31.

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