SAT Math Diagnostic Quiz

Calculator is allowed on all questions. Figures are not necessarily drawn to scale.

Question 1

If \(5x + 8 = 33\), what is the value of \(x\)?

A) \(4\)
B) \(5\)
C) \(8\)
D) \(25\)

Show solution

\[5x + 8 = 33 \Rightarrow 5x = 25 \Rightarrow x = 5\]

• 1. \(4\): subtraction error when isolating the \(x\)-term
• 3. \(8\): reported the constant subtracted from both sides rather than the final value of \(x\)
• 4. \(25\): reported the value of \(5x\) instead of dividing by 5

Answer: B

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Question 2

The table below shows the quiz scores of 6 students.

Student	Score
A	72
B	85
C	90
D	68
E	91
F	74

What is the mean score of these 6 students?

A) \(78\)
B) \(80\)
C) \(83\)
D) \(85\)

Show solution

Sum: \(72 + 85 + 90 + 68 + 91 + 74 = 480\).

Mean \(= \dfrac{480}{6} = 80\).

• 1. \(78\): addition error — likely mis-summed one of the values
• 3. \(83\): divided the sum by 5 instead of 6
• 4. \(85\): reported the second-highest score rather than the mean

Answer: B

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Question 3

Which of the following is equivalent to \(x^2 + 7x + 12\)?

A) \((x + 3)(x + 4)\)
B) \((x + 2)(x + 6)\)
C) \((x + 1)(x + 12)\)
D) \((x + 3)(x - 4)\)

Show solution

Find two numbers that multiply to \(12\) and add to \(7\): \(3 \times 4 = 12\) and \(3 + 4 = 7\) ✓

\[x^2 + 7x + 12 = (x + 3)(x + 4)\]

• 2. \((x+2)(x+6)\): product \(= 12\) ✓ but sum \(= 8 \neq 7\)
• 3. \((x+1)(x+12)\): product \(= 12\) ✓ but sum \(= 13 \neq 7\)
• 4. \((x+3)(x-4)\): product \(= -12 \neq 12\); wrong sign on constant

Answer: A

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Question 4

A car travels at a constant speed of 60 miles per hour. How many miles does the car travel in 2 hours and 30 minutes?

Enter your answer:

Show solution

Convert 2 hours 30 minutes to hours: \(2.5\) hours.

Distance \(= 60 \times 2.5 = 150\) miles.

Answer: 150

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Question 5

In the \(xy\)-plane, the graph of \(y = -2x + 7\) is a line. Which of the following correctly identifies the slope and \(y\)-intercept?

A) Slope \(2\), \(y\)-intercept \(-7\)
B) Slope \(-2\), \(y\)-intercept \(7\)
C) Slope \(7\), \(y\)-intercept \(-2\)
D) Slope \(-7\), \(y\)-intercept \(2\)

Show solution

In slope-intercept form \(y = mx + b\), \(m\) is the slope and \(b\) is the \(y\)-intercept.

For \(y = -2x + 7\): slope \(= -2\), \(y\)-intercept \(= 7\).

• 1. Dropped the negative sign on the slope and negated the intercept
• 3. Swapped the roles of \(-2\) and \(7\)
• 4. Used the \(y\)-intercept value as slope (negated) and the slope’s magnitude as intercept

Answer: B

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Question 6

A right triangle has a hypotenuse of length \(15\) and one leg of length \(9\). What is the length of the other leg?

A) \(6\)
B) \(\sqrt{306}\)
C) \(12\)
D) \(\sqrt{144}\)

Show solution

By the Pythagorean theorem:

\[b^2 = c^2 - a^2 = 15^2 - 9^2 = 225 - 81 = 144 \Rightarrow b = 12\]

This is the 3-4-5 triple scaled by 3: \(9 = 3(3)\), \(12 = 3(4)\), \(15 = 3(5)\).

Note: choices C and D are both equal to 12, so D is a valid numerical answer but in an unsimplified form — the cleaner answer is C.

• 1. \(6\): subtracted the legs (\(15 - 9 = 6\)) instead of applying the theorem
• 2. \(\sqrt{306}\): added the squares instead of subtracting (\(\sqrt{225 + 81} = \sqrt{306}\)); treats the known leg as a second hypotenuse
• 4. \(\sqrt{144}\): correctly equals 12 but is an unsimplified form

Answer: C

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Question 7

The system of equations below has solution \((x, y)\). What is the value of \(x\)?

\[x + y = 10\] \[x - y = 4\]

A) \(3\)
B) \(4\)
C) \(7\)
D) \(14\)

Show solution

Add the two equations to eliminate \(y\):

\[(x + y) + (x - y) = 10 + 4\] \[2x = 14 \Rightarrow x = 7\]

• 1. \(3\): solved correctly for \(x = 7\) then found \(y = 3\), but reported \(y\) instead of \(x\)
• 2. \(4\): used the right-hand side of the second equation as the answer
• 4. \(14\): found \(2x = 14\) but forgot to divide by 2

Answer: C

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Question 8

The function \(g\) is defined by \(g(x) = 3x^2 - 12\). For what positive value of \(x\) does \(g(x) = 0\)?

Enter your answer:

Show solution

Set \(g(x) = 0\):

\[3x^2 - 12 = 0 \Rightarrow 3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\]

The positive value is \(x = 2\).

Answer: 2

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Question 9

A store is offering a \(15\%\) discount on all items. After the discount, a jacket costs \(\$68\). What was the original price of the jacket, in dollars?

A) \(\$57.80\)
B) \(\$78.20\)
C) \(\$80.00\)
D) \(\$83.00\)

Show solution

After a \(15\%\) discount, the customer pays \(85\%\) of the original price \(p\):

\[0.85p = 68 \Rightarrow p = \frac{68}{0.85} = 80\]

• 1. \(\$57.80\): applied an additional 15% discount to the sale price (\(68 \times 0.85\)) — double-discounted
• 2. \(\$78.20\): added \(15\%\) of \(68\) to \(68\) (\(68 + 10.20\)) — added the discount back additively rather than reversing it multiplicatively
• 4. \(\$83.00\): divided by \(0.82\) instead of \(0.85\)

Answer: C

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Question 10

In the \(xy\)-plane, the parabola \(y = (x - 2)^2 - 9\) intersects the \(x\)-axis at two points. What are the \(x\)-coordinates of those two points?

A) \(x = 2\) and \(x = 9\)
B) \(x = -1\) and \(x = 5\)
C) \(x = -5\) and \(x = 1\)
D) \(x = -7\) and \(x = 11\)

Show solution

Set \(y = 0\):

\[(x - 2)^2 - 9 = 0 \Rightarrow (x - 2)^2 = 9 \Rightarrow x - 2 = \pm 3\]

\[x = 2 + 3 = 5 \quad \text{or} \quad x = 2 - 3 = -1\]

• 1. \(x = 2\) and \(x = 9\): used the vertex coordinates \((h, -k)\) instead of solving for zeros
• 3. \(x = -5\) and \(x = 1\): used \(-h = -2\) as the center instead of \(h = 2\), giving \(-2 \pm 3\)
• 4. \(x = -7\) and \(x = 11\): used \(k = 9\) instead of \(\sqrt{k} = 3\) when taking the square root

Answer: B

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Question 11

A circle in the \(xy\)-plane has equation \(x^2 + y^2 - 6x + 4y - 3 = 0\). What is the radius of the circle?

A) \(3\)
B) \(4\)
C) \(16\)
D) \(\sqrt{22}\)

Show solution

Complete the square for \(x\) and \(y\):

\[x^2 - 6x + y^2 + 4y = 3\]

\[(x^2 - 6x + 9) + (y^2 + 4y + 4) = 3 + 9 + 4\]

\[(x - 3)^2 + (y + 2)^2 = 16\]

This is standard form \((x-h)^2 + (y-k)^2 = r^2\), so \(r^2 = 16\) and \(r = 4\).

• 1. \(3\): used only \(\sqrt{9}\) from the \(x\) completing-the-square term, ignoring the \(y\) term
• 3. \(16\): reported \(r^2\) instead of \(r = \sqrt{16} = 4\)
• 4. \(\sqrt{22}\): added \(3 + 9 + 4 + 6 = 22\) — included the original constant incorrectly on the right side

Answer: B

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Question 12

A line passes through the points \((1, 4)\) and \((4, 13)\). What is the \(y\)-intercept of the line?

Enter your answer:

Show solution

Find the slope: \(m = \dfrac{13 - 4}{4 - 1} = \dfrac{9}{3} = 3\).

Use point-slope form with \((1, 4)\):

\[y - 4 = 3(x - 1) \Rightarrow y = 3x - 3 + 4 \Rightarrow y = 3x + 1\]

The \(y\)-intercept is \(1\).

Answer: 1

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Question 13

For which of the following values of \(k\) does the equation \(kx^2 - 6x + 3 = 0\) have no real solutions?

A) \(k = -3\)
B) \(k = 0\)
C) \(k = 3\)
D) \(k = 5\)

Show solution

For no real solutions, the discriminant must be negative: \(\Delta = b^2 - 4ac < 0\).

With \(a = k\), \(b = -6\), \(c = 3\):

\[\Delta = 36 - 12k < 0 \Rightarrow k > 3\]

Only \(k = 5 > 3\) among the choices.

Checking all options:

• 1. \(k = -3\): \(\Delta = 36 + 36 = 72 > 0\) → two real solutions
• 2. \(k = 0\): equation becomes \(-6x + 3 = 0\) (linear) → one solution
• 3. \(k = 3\): \(\Delta = 36 - 36 = 0\) → exactly one repeated solution
• 4. \(k = 5\): \(\Delta = 36 - 60 = -24 < 0\) → no real solutions ✓

Answer: D

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Question 14

For the polynomial \(p(x) = 2x^3 + x^2 - 13x + 6\), it is given that \(p(2) = 0\). Which of the following must be true?

A) \((x + 2)\) is a factor of \(p(x)\).
B) \((x - 2)\) is a factor of \(p(x)\).
C) \(p(x)\) has no other real roots.
D) The remainder when \(p(x)\) is divided by \((x - 2)\) is \(2\).

Show solution

By the Factor Theorem, if \(p(c) = 0\) then \((x - c)\) is a factor of \(p(x)\).

Since \(p(2) = 0\), it follows that \((x - 2)\) is a factor of \(p(x)\).

Verify: \(p(2) = 2(8) + 4 - 26 + 6 = 16 + 4 - 26 + 6 = 0\) ✓

• 1. \((x + 2)\) is a factor only if \(p(-2) = 0\). Check: \(p(-2) = 2(-8) + 4 + 26 + 6 = 20 \neq 0\). Not a factor.
• 3. A cubic has up to 3 real roots; knowing one root doesn’t eliminate the others. In fact \(p(x) = (x-2)(2x^2+5x-3) = (x-2)(2x-1)(x+3)\), which has two additional real roots.
• 4. By the Remainder Theorem, the remainder when \(p(x)\) is divided by \((x - 2)\) equals \(p(2) = 0\), not \(2\).

Answer: B

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Question 15

In the \(xy\)-plane, line \(\ell\) has slope \(1\) and passes through the point \((1, 4)\). Line \(m\) is perpendicular to \(\ell\) and passes through the point \((5, 4)\). At what \(y\)-value do lines \(\ell\) and \(m\) intersect?

Enter your answer:

Show solution

Line \(\ell\): slope \(= 1\), passes through \((1, 4)\):

\[y - 4 = 1(x - 1) \Rightarrow y = x + 3\]

Line \(m\): perpendicular to \(\ell\), so slope \(= -1\) (negative reciprocal of \(1\)). Passes through \((5, 4)\):

\[y - 4 = -1(x - 5) \Rightarrow y = -x + 9\]

Find intersection — set equal:

\[x + 3 = -x + 9 \Rightarrow 2x = 6 \Rightarrow x = 3\]

\[y = 3 + 3 = 6\]

The lines intersect at \((3, 6)\), so the \(y\)-value is \(6\).

Answer: 6

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