Average Problems - SSAT Middle Level Quantitative

  1. During a road trip, the car's speeds are 60 mph, 55 mph, 70 mph, and 65 mph. A student adds these four numbers to get a total. Next, the student divides by 4 because four speeds are recorded. The result is the mean speed for the listed parts. What is the average speed over the entire trip?
    Answer: The average speed is 62.5 mph This question tests middle school math skills in solving average problems using sums and counts. To find the average, add all the numbers together and divide the sum by the number of items. In this problem, you are given the numbers 60 mph, 55 mph, 70 mph, and 65 mph and need to calculate their average. The correct answer is choice A because it accurately reflects the sum of 250 divided by 4 which equals 62.5. Choice B is incorrect because it involves a common error of rounding to the nearest whole number. To help students master this skill, encourage them to carefully count the number of items, double-check their sums, and ensure they divide by the correct count. Practice with varied scenarios to build confidence in identifying and calculating averages.
  2. In a class of 24 students, the average score on a quiz was 82. If 8 students scored 90 and the remaining students all scored the same, what did each of the remaining students score?
    Answer: 78 When you encounter average problems involving groups with different scores, you need to use the total points to work backwards and find unknown values. Start by finding the total points for all students: 24 students×82 average=1,968 total points24 \text{ students} \times 82 \text{ average} = 1,968 \text{ total points} Next, calculate the points from the high-scoring group: 8 students×90 points each=720 points8 \text{ students} \times 90 \text{ points each} = 720 \text{ points} The remaining students contributed: 1,968−720=1,248 points1,968 - 720 = 1,248 \text{ points} Since there are 24−8=1624 - 8 = 16 remaining students who all scored the same, each scored: 1,248÷16=781,248 ÷ 16 = 78 This confirms answer A is correct. Looking at the wrong answers: B) 76 would give a total of 720+(16×76)=1,936720 + (16 \times 76) = 1,936 points, making the class average 1,936÷24=80.671,936 ÷ 24 = 80.67, which is too low. C) 74 would yield 720+(16×74)=1,904720 + (16 \times 74) = 1,904 total points and an average of 79.3379.33, also too low. D) 80 would produce 720+(16×80)=2,000720 + (16 \times 80) = 2,000 total points and an average of 83.3383.33, which exceeds the given average of 82. Study tip: In weighted average problems, always verify your answer by checking that it produces the original average when you calculate backwards. This catch errors and builds confidence in your solution method.
  3. For a full week, the recorded temperatures are 70°F, 72°F, 68°F, 75°F, 73°F, 71°F, and 69°F. A student adds all seven temperatures to get one total. Then the student divides by 7 to find the average for the week. The student reports the result to the nearest tenth. Determine the average temperature for the week.
    Answer: The average temperature is 71.1°F This question tests middle school math skills in solving average problems using sums and counts. To find the average, add all the numbers together and divide the sum by the number of items. In this problem, you are given the numbers 70°F, 72°F, 68°F, 75°F, 73°F, 71°F, and 69°F and need to calculate their average. The correct answer is choice A because it accurately reflects the sum of 498 divided by 7 which equals 71.1. Choice B is incorrect because it involves a common error of not rounding to the nearest tenth. To help students master this skill, encourage them to carefully count the number of items, double-check their sums, and ensure they divide by the correct count. Practice with varied scenarios to build confidence in identifying and calculating averages.
  4. Based on the table shown, what is the average number of books sold per day during the 5-day period?
    Answer: 46.4 books Total books sold: 52 + 38 + 45 + 51 + 46 = 232 books. Average per day: 232 ÷ 5 = 46.4 books. Choice (A) results from addition errors. Choice (B) uses incorrect totals. Choice (D) overestimates the average. Choice (E) underestimates due to calculation mistakes.
  5. A family records four driving speeds on a trip, driving for the same amount of time at each speed: 60 mph, 55 mph, 70 mph, and 65 mph. They calculate the sum of the speeds to get 250 mph. Then they divide 250 by 4 because there are four recorded speeds. This quotient is the average speed of the recorded values. What is the average speed over the entire trip?
    Answer: The average speed is 62.5 mph Average speed for a whole trip is total distance divided by total time. Because the family drove for the same amount of time at each speed, each speed counts equally, so the trip's average speed equals the mean of the four speeds. Add them: 60+55+70+65=25060 + 55 + 70 + 65 = 250. Divide by the number of speeds: 250÷4=62.5250 \div 4 = 62.5 mph. An answer of 250 mph is the sum, before dividing by 4. An answer of 50 mph divides 250 by 5 instead of by the 4 recorded speeds. An answer of 64.5 mph comes from an addition slip, such as a sum of 258 divided by 4.
  6. On four parts of a trip, a car travels at 60 mph, 55 mph, 70 mph, and 65 mph. The driver adds the speeds to find the total. Then the driver divides by 4 since four speeds are listed. This gives one average speed for the trip parts. What is the average speed over the entire trip?
    Answer: The average speed is 62.5 mph This question tests middle school math skills in solving average problems using sums and counts. To find the average, add all the numbers together and divide the sum by the number of items. In this problem, you are given the numbers 60 mph, 55 mph, 70 mph, and 65 mph and need to calculate their average. The correct answer is choice A because it accurately reflects the sum of 250 divided by 4 which equals 62.5. Choice B is incorrect because it involves a common error of miscalculating the sum or dividing incorrectly. To help students master this skill, encourage them to carefully count the number of items, double-check their sums, and ensure they divide by the correct count. Practice with varied scenarios to build confidence in identifying and calculating averages.
  7. A weather app records daily temperatures for seven days: 70°F, 72°F, 68°F, 75°F, 73°F, 71°F, and 69°F. A student adds all seven temperatures to find the total. Then the student divides by 7 because there are seven days. This gives the average temperature for the week. Determine the average temperature for the week.
    Answer: The average temperature is 71.1°F This question tests middle school math skills in solving average problems using sums and counts. To find the average, add all the numbers together and divide the sum by the number of items. In this problem, you are given the numbers 70°F, 72°F, 68°F, 75°F, 73°F, 71°F, and 69°F and need to calculate their average. The correct answer is choice C because it accurately reflects the sum of 498 divided by 7 which equals 71.1. Choice D is incorrect because it involves a common error of using the sum instead of dividing by the count. To help students master this skill, encourage them to carefully count the number of items, double-check their sums, and ensure they divide by the correct count. Practice with varied scenarios to build confidence in identifying and calculating averages.
  8. A student receives allowance for four weeks: $10, $15, $12, and $13. The student finds the sum of these four amounts. Then the student divides by 4 to get one typical weekly amount. That number is the average weekly allowance. Find the average allowance received.
    Answer: The average allowance is $12.5 This question tests middle school math skills in solving average problems using sums and counts. To find the average, add all the numbers together and divide the sum by the number of items. In this problem, you are given the numbers $10, $15, $12, and $13 and need to calculate their average. The correct answer is choice A because it accurately reflects the sum of 50 divided by 4 which equals 12.5. Choice C is incorrect because it involves a common error of miscalculating the sum or rounding incorrectly. To help students master this skill, encourage them to carefully count the number of items, double-check their sums, and ensure they divide by the correct count. Practice with varied scenarios to build confidence in identifying and calculating averages.
  9. The average age of 7 family members is 28 years. If the youngest member is 8 years old and the oldest is 52 years old, what is the average age of the remaining 5 family members?
    Answer: 27.2 years When you encounter average problems involving subgroups, the key strategy is to work with totals first, then find the average of the remaining group. Start by finding the total age of all 7 family members. Since their average age is 28 years, the total is 7×28=1967 \times 28 = 196 years. Next, subtract the ages of the youngest and oldest members from this total: 196−8−52=136196 - 8 - 52 = 136 years. This gives you the combined age of the remaining 5 family members. Finally, find the average age of these 5 members: 136÷5=27.2136 \div 5 = 27.2 years. Let's examine why the other answers are incorrect. Choice A (26.0 years) would result if you miscalculated the total as 130 years instead of 136. Choice B (24.8 years) comes from the common error of subtracting the youngest and oldest ages from the original average (28 - 8 - 52 = -32, then somehow arriving at 24.8). Choice D (25.6 years) results from incorrectly calculating the remaining total as 128 years instead of 136. The correct answer is C) 27.2 years. Study tip: For average problems involving subgroups, always convert to totals first. The formula "Average = Total ÷ Count" works both ways, so use "Total = Average × Count" to find sums, then work backwards to find new averages. This systematic approach prevents calculation errors that create the trap answers.
  10. Over four weeks, a student earns allowance amounts of $10, $15, $12, and $13. The student adds all four numbers to get the total allowance. Then the student divides the total by 4 to find the average per week. The student writes the average in dollars and cents. Find the average allowance received.
    Answer: The average allowance is $12.5 This question tests middle school math skills in solving average problems using sums and counts. To find the average, add all the numbers together and divide the sum by the number of items. In this problem, you are given the numbers $10, $15, $12, and $13 and need to calculate their average. The correct answer is choice A because it accurately reflects the sum of 50 divided by 4 which equals 12.5. Choice D is incorrect because it involves a common error of using the sum instead of dividing by the count. To help students master this skill, encourage them to carefully count the number of items, double-check their sums, and ensure they divide by the correct count. Practice with varied scenarios to build confidence in identifying and calculating averages.
  11. Five students earn these scores on a science test: 78, 85, 92, 88, and 81. Their teacher adds the scores to get a total of points. The teacher divides by 5 to share the total equally across students. This gives the class average for the test. Based on the information provided, calculate the average score.
    Answer: The average score is 84.8 This question tests middle school math skills in solving average problems using sums and counts. To find the average, add all the numbers together and divide the sum by the number of items. In this problem, you are given the numbers 78, 85, 92, 88, and 81 and need to calculate their average. The correct answer is choice B because it accurately reflects the sum of 424 divided by 5 which equals 84.8. Choice D is incorrect because it involves a common error of using the sum instead of dividing by the count. To help students master this skill, encourage them to carefully count the number of items, double-check their sums, and ensure they divide by the correct count. Practice with varied scenarios to build confidence in identifying and calculating averages.
  12. Jordan gets weekly allowance amounts of $10, $15, $12, and $13 over four weeks. Jordan adds the four amounts to find the total money received. Next, Jordan divides the total by 4 because there are four weeks. This gives the average allowance per week. Find the average allowance received.
    Answer: The average allowance is $12.5 This question tests middle school math skills in solving average problems using sums and counts. To find the average, add all the numbers together and divide the sum by the number of items. In this problem, you are given the numbers $10, $15, $12, and $13 and need to calculate their average. The correct answer is choice A because it accurately reflects the sum of 50 divided by 4 which equals 12.5. Choice B is incorrect because it involves a common error of using the sum instead of dividing by the count. To help students master this skill, encourage them to carefully count the number of items, double-check their sums, and ensure they divide by the correct count. Practice with varied scenarios to build confidence in identifying and calculating averages.
  13. Jake's average score on his first 4 tests is 88. His average score on his first 6 tests is 85. What is his average score on just the 5th and 6th tests?
    Answer: 79 When you encounter average problems involving different groups of tests or scores, think systematically about what information you have and what you need to find. The key is working with totals rather than just averages. Start by finding the total points from the given averages. Jake's average on his first 4 tests is 88, so his total points for those tests is 4×88=3524 \times 88 = 352. His average on all 6 tests is 85, so his total for all 6 tests is 6×85=5106 \times 85 = 510. To find the combined score on just the 5th and 6th tests, subtract the first four tests' total from the six tests' total: 510−352=158510 - 352 = 158. Since this represents the sum of two test scores, the average is 158÷2=79158 \div 2 = 79. Looking at the wrong answers: Choice A (80) is close but represents a common calculation error where students might round incorrectly or make an arithmetic mistake. Choice B (82) could result from incorrectly calculating the difference between the two given averages (88 - 85 = 3, then perhaps adding this to 79). Choice D (81) might come from averaging the two given averages incorrectly: (88+85)÷2=86.5(88 + 85) \div 2 = 86.5, then making further errors. The correct answer is C (79). Strategy tip: Always convert averages to totals first in multi-group problems. This makes the arithmetic clearer and helps you avoid the trap of trying to work directly with averages, which often leads to incorrect shortcuts.
  14. Sarah bowled 4 games with scores of 120, 135, 110, and 115. In her 5th game, she wants her overall average to be at least 125. What is the minimum score she needs in the 5th game?
    Answer: 145 When you encounter average problems like this, you're working with the relationship between sum, count, and average. The key insight is that if you know the desired average and the number of values, you can find the required total sum. To find the minimum score Sarah needs, start by calculating what her total points must be across all 5 games. If she wants an average of at least 125 over 5 games, her total points must be at least 125×5=625125 \times 5 = 625 points. Next, find her current total from the first 4 games: 120+135+110+115=480120 + 135 + 110 + 115 = 480 points. Therefore, her 5th game score must be at least 625−480=145625 - 480 = 145 points to achieve her goal. Looking at the wrong answers: Choice B (140) falls short because 480+140=620480 + 140 = 620 total points, giving an average of 620÷5=124620 ÷ 5 = 124, which doesn't meet her minimum requirement of 125. Choices C (150) and D (155) would certainly work since they exceed the minimum, but the question asks for the minimum score needed, making these unnecessarily high. The answer is A) 145. Strategy tip: In minimum/maximum average problems, always set up the equation using the target average times the number of values to find the required total, then subtract what you already have. This systematic approach prevents calculation errors and ensures you find the exact minimum rather than just any score that works.
  15. A student lists four travel speeds from a trip that spent an equal amount of time at each speed: 60 mph, 55 mph, 70 mph, and 65 mph. The student adds them to get a total of 250 mph. Then the student divides by 4 to find the mean of the listed speeds. This mean is the average speed for those parts. What is the average speed over the entire trip?
    Answer: The average speed is 62.5 mph Average speed for a whole trip is total distance divided by total time. Because the trip spent an equal amount of time at each speed, each speed counts equally, so the trip's average speed equals the mean of the four speeds. Add them: 60+55+70+65=25060 + 55 + 70 + 65 = 250. Divide by the number of speeds: 250÷4=62.5250 \div 4 = 62.5 mph. An answer of 62 mph rounds 62.5 down, but the exact value is 62.5. An answer of 55 mph is just the slowest speed, and 65 mph is one of the listed speeds rather than their mean. The equal-time condition matters: if the trip had covered equal distances instead, the average speed would be a little less than 62.5 mph.
  16. During basketball practice, Alex made 12 free throws out of 20 attempts on Monday, 8 out of 15 attempts on Tuesday, and 15 out of 18 attempts on Wednesday. What was his overall free throw percentage for the three days?
    Answer: 66.0% When you encounter percentage problems involving multiple events, remember that you need to find the overall rate, not the average of individual rates. This means combining all successes and all attempts first. To find Alex's overall free throw percentage, add up all his successful shots and all his attempts across the three days. On Monday: 12 out of 20. On Tuesday: 8 out of 15. On Wednesday: 15 out of 18. Total successful free throws: 12+8+15=3512 + 8 + 15 = 35 Total attempts: 20+15+18=5320 + 15 + 18 = 53 Overall percentage: 3553=0.660...=66.0%\frac{35}{53} = 0.660... = 66.0\% This confirms that A) 66.0% is correct. Now let's see why the other answers are wrong. B) 65.1% likely comes from a calculation error, perhaps misadding the totals. C) 68.2% and D) 67.5% represent common traps where students incorrectly average the daily percentages instead of using the total shots and attempts. If you calculated the daily percentages (60%, 53.3%, 83.3%) and averaged them, you'd get about 65.5%, which is close to some wrong answers but not the correct approach. This method fails because it doesn't account for the different number of attempts each day. Strategy tip: For combined percentage problems, always use total successes divided by total attempts. Never average individual percentages unless each group has the same size—the varying number of attempts on each day makes simple averaging incorrect.
  17. A coach tracks five practice scores: 78, 85, 92, 88, and 81. The coach adds all five scores to find the total points. Then the coach divides by 5 because there are five scores. The team uses this number as the average performance score. What is the average of the given numbers?
    Answer: The average is 84.8 This question tests middle school math skills in solving average problems using sums and counts. To find the average, add all the numbers together and divide the sum by the number of items. In this problem, you are given the numbers 78, 85, 92, 88, and 81 and need to calculate their average. The correct answer is choice A because it accurately reflects the sum of 424 divided by 5 which equals 84.8. Choice B is incorrect because it involves a common error of misadding the numbers or dividing by the wrong count. To help students master this skill, encourage them to carefully count the number of items, double-check their sums, and ensure they divide by the correct count. Practice with varied scenarios to build confidence in identifying and calculating averages.
  18. In a math class, 15 students scored an average of 85 on a test, and 10 students scored an average of 92 on the same test. What is the overall average score for all 25 students?
    Answer: 87.8 When you encounter weighted average problems, you're dealing with groups of different sizes that each have their own average. The key insight is that larger groups have more influence on the overall average than smaller groups. To find the overall average, you need to calculate the total points earned by all students, then divide by the total number of students. Start by finding the total points for each group: the 15 students averaging 85 earned 15×85=1,27515 \times 85 = 1,275 total points, while the 10 students averaging 92 earned 10×92=92010 \times 92 = 920 total points. The combined total is 1,275+920=2,1951,275 + 920 = 2,195 points earned by all 25 students. Therefore, the overall average is 2,19525=87.8\frac{2,195}{25} = 87.8. Choice A (88.5) represents the trap of simply averaging the two given averages: 85+922=88.5\frac{85 + 92}{2} = 88.5. This ignores the fact that the groups have different sizes. Choice C (88.0) might result from rounding errors or similar miscalculations. Choice D (87.4) could come from computational mistakes when handling the weighted calculation. Strategy tip: On weighted average problems, never just average the averages unless the groups are the same size. Always calculate total points (or total whatever is being measured) for each group first, then find the overall average. Remember that the final answer should be closer to the average of the larger group—here, 87.8 is closer to 85 than to 92, which makes sense since the group of 15 students is larger.
  19. In a group of 12 people, 4 people are 25 years old, 6 people are 30 years old, and 2 people are 40 years old. If 2 more people join the group, both aged 35, what is the new average age?
    Answer: 30.7 years When you encounter average problems where new data points are added to a group, you need to recalculate using all the data, not adjust the existing average. First, find the total age of the original 12 people. You have 4 people at 25 years old (4 × 25 = 100), 6 people at 30 years old (6 × 30 = 180), and 2 people at 40 years old (2 × 40 = 80). The total age is 100 + 180 + 80 = 360 years. When 2 more people join, both aged 35, you add 2 × 35 = 70 years to your total. The new total age becomes 360 + 70 = 430 years, and you now have 12 + 2 = 14 people total. The new average age is 43014=30.714...\frac{430}{14} = 30.714... years, which rounds to 30.7 years. Choice (A) 30.0 years is the original average age of the initial group (360 ÷ 12 = 30), but this ignores the two new people entirely. Choice (B) 31.4 years might result from incorrectly calculating the weighted contributions or making an arithmetic error. Choice (D) 29.6 years is too low and could come from miscalculating the original total or the final division. Remember that when new data points are added to find a new average, always recalculate from scratch using the total of all values divided by the new count. Don't try to "adjust" the existing average—this often leads to errors on SSAT problems.
  20. Jenny's science test scores are 88, 92, 85, 90, and 95. If her teacher drops the lowest score and counts the remaining four scores equally, what is Jenny's new average, rounded to the nearest tenth?
    Answer: 91.3 Put the scores in order: 85, 88, 90, 92, 95. The lowest score, 85, is dropped. Add the four remaining scores: 88+90+92+95=36588 + 90 + 92 + 95 = 365. Divide by 4: 3654=91.25\frac{365}{4} = 91.25. Rounded to the nearest tenth, 91.25 becomes 91.3, because the hundredths digit is 5. An answer of 90.5 comes from dropping the wrong score, 88 instead of 85 (3624=90.5\frac{362}{4} = 90.5). An answer of 91.0 comes from cutting off the decimal or rounding down instead of rounding to the nearest tenth. An answer of 91.8 comes from an addition slip that gives a sum of 367 (3674=91.75\frac{367}{4} = 91.75). Keeping all five scores gives 4505=90\frac{450}{5} = 90, which is lower still. When a score is dropped, divide by the number of scores that remain, not the original count.

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