# SAT Data Perfect Score

To get a perfect score on SAT Math, you need to:

- Get every single problem correct.
- Have complete mastery of all of the SAT skills
- Remember the Tips and use them
- Figure out your common mistakes and avoid them

## SAT Hardest Problems

If the median and mode of seven positive integers are \(5\) and \(6,\) respectively, what is the smallest possible mean of the seven numbers?

(A)\(\ \ \frac{26}{7}\)

(B)\(\ \ \frac{27}{7}\)

(C)\(\ \ 4\)

(D)\(\ \ \frac{29}{7}\)

(E)\(\ \ \frac{30}{7}\)

Correct Answer: B

Solution:Let \(a_1, a_2, a_3, a_4, a_5, a_6, a_7\) be the seven integers arranged in ascending order. Then since the number of data is seven, which is odd, it must be true that the median is \(a_4=5.\) Then it must also be true that the mode is \(a_5=a_6=6.\)

If \(a_7\) is also equal to \(6,\) the way to minimize the sum of the first three numbers is to have two \(1\)'s and a \(2:\) \(a_1=1, a_2=1, a_3=2.\) In this case, the mean of the seven numbers is \[\frac{1+1+2+5+6+6+6}{7}=\frac{27}{7}. \qquad (1) \]

If \(a_7=7,\) the first three numbers must all be distinct. Then the way to minimize the sum of the first three numbers is to have \(a_1=1, a_2=2, a_3=3.\) In this case, the mean of the seven numbers is \[\frac{1+2+3+5+6+6+7}{7}=\frac{30}{7}. \qquad (2) \] The same logic would apply for any number larger than \(7\) for \(a_7.\)

Thus, from \((1)\) and \((2)\) we can deduce that \((1)\) is the smallest possible mean of the seven numbers, and the correct answer is (B).

Incorrect Choices:

(A)

If you forgot about the mode and got \(a_1=a_2=a_3=1,\) you would get this wrong answer.

(C)

If you got \(a_1=a_2=1, a_3=2, a_7=7,\) you would get this wrong answer.

(D)

If you thought \(a_1, a_2, a_3\) must all be distinct and got \(a_1=1, a_2=2, a_3=3,\) you would get this wrong answer.

(E)

If you followed the steps in \((2)\) of the above solution, you would get this wrong answer.

A class of \(20\) students took a math quiz and the results are as follows:

What is the sum of the mean, median and mode of their scores?(A)\(\ \ 7.0\)

(B)\(\ \ 7.1\)

(C)\(\ \ 7.2\)

(D)\(\ \ 7.3\)

(E)\(\ \ 7.4\)

Correct Answer: B

Solution:Since the total number of students is \(20,\) we have \[3+A+6+3+1=20 \Rightarrow A=7.\] Then since the score \(2\) has the most frequency of \(7,\) the mode is \(2. \qquad (1)\)

Let \(a_1, a_2, a_3, \ldots, a_{19}, a_{20}\) be the twenty scores arranged in ascending order. Then since the number of data is twenty, an even number, the median is the average of \(a_{10}=2\) and \(a_{11}=3:\) \[\text{Median}=\frac{2+3}{2}=2.5. \qquad (2)\]

Now, the the mean is \[\frac{3\times 1+7\times 2+6\times 3+3\times 4+1\times 5}{20}=\frac{52}{20}=2.6. \qquad (3)\]

Therefore, from \((1), (2)\) and \((3),\) the sum of the mean, median and mode of the twenty scores is \(2+2.5+2.6=7.1\) and and the correct answer is (B).

Incorrect Choices:

(A),(C),(D), and(E)

The solution explains why these choices are wrong.

The above graph shows the relative frequency distributions of the numbers of questions correctly answered by boys and girls in a school on a math test with \(8\) questions. If the same number of boys and girls got six questions right and \(50\) girls got at most three questions right, which of the following statements is true?

\(\begin{array}{r r l} & \text{I.} & \text{The total number of students in the school is } 500.\\

& \text{II.} & \text{Three times more boys got four questions right than the girls.}\\

& \text{III.} & 45 \text{ more girls got seven questions right than the boys.} \\

\end{array}\)(A)\(\ \ \) I only

(B)\(\ \ \) II only

(C)\(\ \ \) I and III only

(D)\(\ \ \) II and III only

(E)\(\ \ \) I, II and III

Correct Answer: E

Solution:I. Let \(B\) and \(G\) be the numbers of boys and girls in the school, respectively. Then since the same number of boys and girls got six questions right, \[0.1B=0.15G \Rightarrow B=1.5G.\] Now, since \(50\) girls got at most three questions right, \[(0.05+0.05+0.15)G=50 \Rightarrow G=200 \Rightarrow B=1.5G=300.\] Hence, there are \(300\) boys and \(200\) girls in the school, making the total number of student \(500.\) Thus statement \(\text{ I }\) is true.

II. The number of boys who got four questions right is \(300 \times 0.4=120.\) The number of girls who got four questions right is \(200 \times 0.2=40.\) Hence, \(120\div 40=3\) times more boys got four questions right than the girls, and thus statement II is true.

III. The number of girls who got seven questions right is \(200 \times 0.3=60,\) and the number of boys who got seven questions right is \(300 \times 0.05=15.\) Hence, \(60-15=45\) more girls got seven questions right than the boys, and thus statement III is true.

Therefore, all three of the statements are true and the correct answer is (E).

Incorrect Choices:

(A),(B),(C), and(D)

The solution explains why these choices are wrong.

## SAT Tips for Data Analysis, Statistics and Probability

## Mean, Median, and Mode

- The average of \(n\) numbers is the sum of the numbers divided by \(n.\)
- If the average of a set of numbers is \(A\) and a new number \(x=A\) is introduced to the set, the new average will also equal \(A.\)
- If \(n\) numbers are arranged in increasing order, the median is the middle value if \(n\) is odd, and it is the average of the two middle values if \(n\) is even.
- In a set of numbers, the mode is the number that appears most frequently.
- To find the weighted mean of a some numbers, find the product of each number and its weight, then divide the sum of these products by the sum of the weights.
## Data-Tables

## Data-Graphs and Charts

## Sets and Venn Diagrams

- The union of two sets, \(A\) and \(B,\) is that collection of elements that are in \(A,\) or in \(B,\) or in both \(A\) and \(B.\)
- The intersection of two sets, \(A\) and \(B,\) is that collection of elements that are only in both \(A\) and \(B.\)
- If every element in set \(A\) is an element in set \(B,\) then \(A\) is a subset of \(B.\)
## Counting and Probability

- If \(a<b\) are two integers, the number of integers between \(a\) and \(b\) when one endpoint is included is \(b-a.\)
- If \(a<b\) are two integers, the number of integers between \(a\) and \(b\) when both endpoints are included is \(b-a+1.\)
- If \(a<b\) are two integers, the number of integers between \(a\) and \(b,\) endpoints NOT included is \(b-a-1.\)
- If there are \(n\) ways for an event to happen and \(m\) ways for another event to happen, then the number of ways for both events to happen is \(m\cdot n.\)
- If \(P(A)\) is the probability that event \(A\) will occur, then \(0 \leq P(A) \leq 1.\)
- If \(P(B)\) is the probability that event \(A\) does NOT occur, then \(P(A) = 1- P(B).\)
- Assuming that all the possible outcomes of an event \(A\) are equally likely, the probability that \(A\) will occur is \(P(A) = \frac{\text{ # of favorable outcomes}}{\text{total # of outcomes}}.\)
- Two events are independent if the outcome of one does not affect the outcome of the other.
- If events \(A\) and \(B\) are independent, then \(P(A\ \text{and}\ B) = P(A) \cdot P(B).\)
- If events \(A\) and \(B\) are mutually exclusive, then \(P(A\ \text{or}\ B) = P(A) + P(B).\)
- If a point is chosen at random in a geometric figure, the probability that the point lies in a particular region is: \(\frac{\text{area of region}}{\text{area of whole figure}}.\)

**Cite as:**SAT Data Perfect Score.

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