This is an estimate based on your percent correct on our practice tests. The official GED score conversion is not public, so your real score may differ. Scores of 145+ pass; 165+ is College Ready.
Question 1 of 10 · Mean, median, mode, range, weighted average
Fill in
Keisha bowled four games with scores of 142, 165, 138, and 151. What score does she need in her fifth game so that her mean score for all five games is exactly 150?
Type your answer in the box. Fractions like 3/4 are fine.
Hint
A mean of 150 over 5 games means the five scores must add up to a certain total. What is that total?
Step-by-step solution
Total needed for a mean of 150 over 5 games:
5 \times 150 = 750
Total of the four games so far:
142 + 165 + 138 + 151 = 596
The fifth score must make up the difference:
750 - 596 = 154
Tip: For any "what score do I need" question: (target mean × total number of values) − (sum so far) = missing value. Don't just average the four scores and compare to 150.
Question 2 of 10 · Mean, median, mode, range, weighted average
In Ms. Patel's class, the final grade is a weighted average:
Category
Weight
Ana's score
Tests
40%
85
Homework
35%
92
Final exam
25%
70
What is Ana's final grade?
Hint
Multiply each score by its weight (as a decimal), then add. The weights already add to 100%, so you do not divide by 3.
Step-by-step solution
Change each weight to a decimal: 40\% = 0.4, 35\% = 0.35, 25\% = 0.25.
Multiply each score by its weight:
0.4 \times 85 = 34 \qquad 0.35 \times 92 = 32.2 \qquad 0.25 \times 70 = 17.5
Add the products:
34 + 32.2 + 17.5 = 83.7
Ana's final grade is 83.7.
Check: The weights add to 0.4 + 0.35 + 0.25 = 1 ✓, and 83.7 lies between the lowest score (70) and the highest (92) ✓. It is a bit above the simple mean (82.3) because the low final-exam score counts the least.
Tip: In a weighted average, values with larger weights pull the result toward them. If the weights add to 100%, add the weighted products and stop; if they are counts (like 40 shirts), divide by the total count.
Question 3 of 10 · Mean, median, mode, range, weighted average
Fill in
A community center surveyed 30 families about how many children live in their home.
Number of children
0
1
2
3
4
Number of families
6
9
12
2
1
What is the mean number of children per family? Round to the nearest tenth.
children
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Each row of the table stands for many families. Multiply to find the total number of children, then divide by the total number of families.
Step-by-step solution
Multiply each number of children by how many families have it:
0 \times 6 = 0 \qquad 1 \times 9 = 9 \qquad 2 \times 12 = 24 \qquad 3 \times 2 = 6 \qquad 4 \times 1 = 4
Add to get the total number of children:
0 + 9 + 24 + 6 + 4 = 43
Add the frequencies to get the number of families:
6 + 9 + 12 + 2 + 1 = 30
Divide:
\frac{43}{30} \approx 1.433 \approx 1.4
The mean is about 1.4 children per family.
Check: The answer must be between 0 and 4, and most families have 1 or 2 children, so about 1.4 makes sense ✓
Tip: In a frequency table, do not average the top row (\frac{0+1+2+3+4}{5} = 2). Each value counts as many times as its frequency, so multiply first, then divide by the total frequency (30), not by the number of columns (5).
Question 4 of 10 · Mean, median, mode, range, weighted average
Fill in
The dot plot shows how many hours each member of a neighborhood cleanup club volunteered last month. Each dot is one member.
What is the median number of hours volunteered?
hours
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Count the dots to find how many members there are. Then count from the left to the middle dot.
Step-by-step solution
Count the dots above each number of hours:
Hours
1
2
3
4
5
6
7
Members
2
3
4
3
1
0
2
Total members: 2 + 3 + 4 + 3 + 1 + 0 + 2 = 15.
With 15 values, the median is the \frac{15 + 1}{2} = 8th value in order.
Count from the left: 1 hour covers values 1–2, 2 hours covers values 3–5, and 3 hours covers values 6–9. The 8th value is 3.
The median is 3 hours.
Check: 5 members volunteered less than 3 hours and 6 members volunteered more than 3 hours; with the 4 members at 3 hours, the middle member (8th) is at 3 ✓
Tip: A dot plot is already sorted from left to right, so you just count dots to the middle position. Don't pick the middle of the axis (4 hours) or the tallest stack unless it happens to contain the middle dot.
Question 5 of 10 · Mean, median, mode, range, weighted average
No calculator
A youth soccer team scored these numbers of goals in its first 10 games:
2, 0, 3, 1, 2, 4, 1, 0, 2, 1
Which statement about the mode of the data is true?
Hint
Make a tally: how many times does each number of goals appear? Can more than one value be the mode?
Step-by-step solution
Tally how often each value appears:
Goals
0
1
2
3
4
Games
2
3
3
1
1
The mode is the value (or values) that appears most often. The highest count is 3.
Both 1 and 2 appear 3 times. When two values tie for most frequent, the data set has two modes (it is bimodal).
The data set has two modes: 1 goal and 2 goals.
Check: Recount from the original list: 1 appears in games 4, 7, 10 (3 times) and 2 appears in games 1, 5, 9 (3 times). No other value appears 3 times ✓
Tip: The mode is a data value, not the number of times it appears (3 here). A data set can have one mode, more than one mode, or no mode (if every value appears the same number of times).
Question 6 of 10 · Mean, median, mode, range, weighted average
Fill in
A store sold T-shirts at three prices last month:
Price per shirt
Number sold
$11
42
$16
23
$24
15
What was the mean price of all the shirts sold? Round to the nearest cent.
$
Type your answer in the box. Fractions like 3/4 are fine.
Hint
This is a weighted average: each price counts as many times as the number of shirts sold at that price.
Step-by-step solution
Find the money taken in at each price:
11 \times 42 = 462 \qquad 16 \times 23 = 368 \qquad 24 \times 15 = 360
Total money:
462 + 368 + 360 = 1{,}190
Total number of shirts:
42 + 23 + 15 = 80
Mean price:
\frac{1{,}190}{80} = 14.875 \approx 14.88
The mean price was $14.88.
Check: The answer lies between $11 and $24 ✓, and is closer to $11 because most shirts (42) sold at $11. The simple mean of the three prices, \tfrac{11 + 16 + 24}{3} = 17, is too high because it ignores how many sold at each price.
Tip: Weighted average with counts = \dfrac{\sum(\text{value} \times \text{count})}{\sum \text{count}}. Divide by the number of items (80), not the number of prices (3).
Question 7 of 10 · Mean, median, mode, range, weighted average
Drop-down
Seven students in a GED class took a practice test. Their scores were:
88, 91, 85, 90, 42, 87, 89
The mean score is about 81.7.
Choose the correct answer for each blank.
The outlier in this data set is . The outlier makes the mean the median, so the better describes a typical score.
Choose an answer from each menu.
Hint
Find the value far from the others. Then sort the data to find the median and compare it with the mean of 81.7.
Step-by-step solution
Outlier: six scores are between 85 and 91. The score 42 is far below them, so it is the outlier.
Median: sort the scores: 42, 85, 87, \mathbf{88}, 89, 90, 91. The middle (4th) value is 88.
Mean vs median: the mean is \frac{88 + 91 + 85 + 90 + 42 + 87 + 89}{7} = \frac{572}{7} \approx 81.7 That is less than the median of 88, because the low outlier drags the mean down.
Typical score: six of the seven students scored 85 or higher. The median (88) sits in that group; the mean (81.7) is lower than all six. The median better describes a typical score.
Outlier: 42. The mean is less than the median. The median is the better typical value.
Check: Without the 42, the mean would be 530 \div 6 \approx 88.3, very close to the median. The one low score moved the mean down by more than 6 points ✓
Tip: A low outlier pulls the mean down; a high outlier pulls it up. The median barely moves either way, so it is the safer "typical" value when an outlier is present.
Question 8 of 10 · Mean, median, mode, range, weighted average
A hiking club has 40 members with a mean age of 31 years. The 25 women in the club have a mean age of 34 years. What is the mean age of the men in the club?
Hint
Work with totals: find the total age of all members and the total age of the women. How many men are there?
Step-by-step solution
Total age of all 40 members:
40 \times 31 = 1{,}240
Tip: Group-mean problems are solved with totals, not by averaging means. Mean \times count = total; subtract totals to find the missing group, then divide by that group's own count.
Question 9 of 10 · Mean, median, mode, range, weighted average
In a training course, the final grade is 60% test average and 40% homework average. Carlos has a homework average of 95. What is the lowest test average he can have and still earn a final grade of at least 80?
Hint
Write the weighted-average equation with the test average as the unknown, set it equal to 80, and solve.
Step-by-step solution
Let T be the test average. The final grade is:
0.6T + 0.4(95)
Set it equal to 80 (the lowest grade allowed):
0.6T + 38 = 80
Subtract 38 from both sides:
0.6T = 42
Divide by 0.6:
T = \frac{42}{0.6} = 70
The lowest test average is 70.
Check:0.6(70) + 0.4(95) = 42 + 38 = 80 ✓. Any test average below 70 would make the final grade less than 80.
Tip: "Work backwards" with a weighted average: write weight × value for every part, put the unknown in, and solve the equation. Because homework (95) is above the target, Carlos can afford a test average below 80.
Question 10 of 10 · Mean, median, mode, range, weighted average
Drag and drop
At a community college, a grade of A is worth 4.0 points and a grade of B is worth 3.0 points. A student's GPA is the mean of the grade points, weighted by credits.
Course
Credits
Grade
Biology
4
A
English
3
B
Math
4
B
Art
1
A
Drag the correct value into each box. Not every value is used.
Total grade points
Total credits
GPA (to the nearest hundredth)
Tap a tile, then tap a box to place it.
Hint
Each course's grade points are credits × grade value. The GPA divides the total grade points by the total credits, not by the number of courses.
Step-by-step solution
Grade points for each course (credits \times grade value):
Biology: 4 \times 4.0 = 16
English: 3 \times 3.0 = 9
Math: 4 \times 3.0 = 12
Art: 1 \times 4.0 = 4
Total grade points:16 + 9 + 12 + 4 = 41
Total credits:4 + 3 + 4 + 1 = 12
GPA:\frac{41}{12} = 3.41\overline{6} \approx 3.42
Total grade points 41; total credits 12; GPA 3.42.
Check: The student has 5 credits of A and 7 credits of B, so the GPA must be between 3.0 and 4.0 and closer to 3.0. 3.42 fits ✓
Tip: A GPA is a weighted average: courses with more credits count more. Averaging the four letter grades as if they were equal (\frac{4 + 3 + 3 + 4}{4} = 3.50) gives too much weight to the 1-credit Art class.