To help answer this question, we can construct a two-way table and fill in our known quantities from the question.

The columns of the table will represent the students who have a curfew or do not have a curfew and the rows will contain the students who are on the honor roll or are not on the honor roll. The first bit of information that we were given from the question was that students had a curfew; therefore, needs to go in the "curfew" column as the row total. Next, we were told that of those students, were on the honor roll; therefore, we need to put in the "curfew" column and in the "honor roll" row. Then, we were told that students do not have a curfew but were on the honor roll, so we need to put in the "no curfew" column and the "honor roll" row. Finally, we were told that students do not have a curfew or were on the honor roll, so needs to go in the "no curfew" column and "no honor roll" row. If done correctly, you should create a table similar to the following:

Our question asked how many students were not on the honor roll. We add up the numbers in the "no honor roll" row to get the total, but first we need to fill in a gap in our table, students who have a curfew but were not on the honor roll. We can take the total number of students that have a curfew, , and subtract the number of students who are on the honor roll,

This means that students who have a curfew, aren't on the honor roll.

Now, we add up the numbers in the "no honor roll" row to get the total:

This means that students were not on the honor roll.

According to the graph, 8 of her peers eat 1 fruit daily while 3 of her peers eat 3 fruits. 8 - 3 = 5.

To help answer this question, we can construct a two-way table and fill in our known quantities from the question.

The columns of the table will represent the students who have a curfew or do not have a curfew and the rows will contain the students who are on the honor roll or are not on the honor roll. The first bit of information that we were given from the question was that students had a curfew; therefore, needs to go in the "curfew" column as the row total. Next, we were told that of those students, were on the honor roll; therefore, we need to put in the "curfew" column and in the "honor roll" row. Then, we were told that students do not have a curfew but were on the honor roll, so we need to put in the "no curfew" column and the "honor roll" row. Finally, we were told that students do not have a curfew or were on the honor roll, so needs to go in the "no curfew" column and "no honor roll" row. If done correctly, you should create a table similar to the following:

Our question asked how many students were not on the honor roll. We add up the numbers in the "no honor roll" row to get the total, but first we need to fill in a gap in our table, students who have a curfew but were not on the honor roll. We can take the total number of students that have a curfew, , and subtract the number of students who are on the honor roll,

This means that students who have a curfew, aren't on the honor roll.

Now, we add up the numbers in the "no honor roll" row to get the total:

This means that students were not on the honor roll.

According to the graph, 8 of her peers eat 1 fruit daily while 3 of her peers eat 3 fruits. 8 - 3 = 5.

This question is a great example of a case where the SAT might try to trap you into doing more work than is necessary. Since the problem asks you whether that mean is exactly 3.0, it is possible to determine this without tedious work. If the average were 3.0, the values on either side of the twelve B (3.0) grades would all even out. And they ALMOST do: there are 6 As (4.0 each) and 6 Cs (2.0 each, so those grades all average out to 3.0); there are 8 A-s and 8 C+s, so those grades (3.7 and 2.3) all average out to 3.0; and there are 11 B+s and 11 B-s, so those all average out to 3. On the list from A on down to C, all those grades average out to 3.0…but then there are the 3 C− grades, and those do not get “averaged out” by anything above an A. So without fully calculating, you can tell that the mean is lower than 3.0. So, we can eliminate “The mean math grade of the 65 students was 3.0.”

Since two of our options deal with the median, we can focus there next. The middle of the 65 terms will be the 33rd term (32 terms above it and 32 terms below it). So when you go from the top down and see that there are 25 grades from A through B+ and then 12 grades of B, you should see that the 33rd term will fall in that group of Bs. This means that the median is 3.0.

This knowledge allows us to both eliminate “The median math grade of the 65 students is less than a 3.0” and select “The median math grade of the 65 students is higher than the mean math grade of the 65 students.​” Since the mean is somewhat less than 3.0, and the median *is* 3.0, the mean is less than the median.

“If three new students were added to the class and each scored an A, the new class average would be greater than or equal to 3.0” can be eliminated on the logic of the first paragraph: The 62 grades from A down to C all average out to a 3.0, leaving the three C− grades of 1.7 to weigh down that average. In order to balance three 1.7s you’d need three 4.3s; three 4.0s, as this option presents, aren’t quite enough, so the average would still be below 3.0.

From the initial information presented to us in the stem, we can create a two way table using the categories left vs. right-handed, and take the bus vs. do not take the bus. We can fill in the table as follows: (*note - be very cautious mixing percent/fraction information with actual value information. It may be easier to leave the “30” out of the table and solve for right-handed seniors who take the bus in terms of a fraction.)

From here, if we know that half of the students who are left-handed take the bus, we can split up the of our total that is left-handed into and .

If of the students take the bus, and of the students are left-handed students who take the bus, the remaining are right-handed students who take the bus. Since we know the value of right-handed students who take the bus is 30, and of our total students, the total number of students must be

*Note, we did not need to complete the entire table to answer this question, but we could! The rest of the table is as follows:

The graph shows an increase (a positive slope) from Monday to Tuesday and Friday to Saturday. However, there is a sharp decline from Thursday to Friday.

This question is a great example of a case where the SAT might try to trap you into doing more work than is necessary. Since the problem asks you whether that mean is exactly 3.0, it is possible to determine this without tedious work. If the average were 3.0, the values on either side of the twelve B (3.0) grades would all even out. And they ALMOST do: there are 6 As (4.0 each) and 6 Cs (2.0 each, so those grades all average out to 3.0); there are 8 A-s and 8 C+s, so those grades (3.7 and 2.3) all average out to 3.0; and there are 11 B+s and 11 B-s, so those all average out to 3. On the list from A on down to C, all those grades average out to 3.0…but then there are the 3 C− grades, and those do not get “averaged out” by anything above an A. So without fully calculating, you can tell that the mean is lower than 3.0. So, we can eliminate “The mean math grade of the 65 students was 3.0.”

Since two of our options deal with the median, we can focus there next. The middle of the 65 terms will be the 33rd term (32 terms above it and 32 terms below it). So when you go from the top down and see that there are 25 grades from A through B+ and then 12 grades of B, you should see that the 33rd term will fall in that group of Bs. This means that the median is 3.0.

This knowledge allows us to both eliminate “The median math grade of the 65 students is less than a 3.0” and select “The median math grade of the 65 students is higher than the mean math grade of the 65 students.​” Since the mean is somewhat less than 3.0, and the median *is* 3.0, the mean is less than the median.

“If three new students were added to the class and each scored an A, the new class average would be greater than or equal to 3.0” can be eliminated on the logic of the first paragraph: The 62 grades from A down to C all average out to a 3.0, leaving the three C− grades of 1.7 to weigh down that average. In order to balance three 1.7s you’d need three 4.3s; three 4.0s, as this option presents, aren’t quite enough, so the average would still be below 3.0.

From the initial information presented to us in the stem, we can create a two way table using the categories left vs. right-handed, and take the bus vs. do not take the bus. We can fill in the table as follows: (*note - be very cautious mixing percent/fraction information with actual value information. It may be easier to leave the “30” out of the table and solve for right-handed seniors who take the bus in terms of a fraction.)

From here, if we know that half of the students who are left-handed take the bus, we can split up the of our total that is left-handed into and .

If of the students take the bus, and of the students are left-handed students who take the bus, the remaining are right-handed students who take the bus. Since we know the value of right-handed students who take the bus is 30, and of our total students, the total number of students must be

*Note, we did not need to complete the entire table to answer this question, but we could! The rest of the table is as follows:

The graph shows an increase (a positive slope) from Monday to Tuesday and Friday to Saturday. However, there is a sharp decline from Thursday to Friday.