Digital SAT Math: Hard Grid-In Practice (Set 1)

The key to this problem is the formal definition of the principal square root: \(\sqrt{u^2} = |u|\).

Step 1: Simplify the radical.

\[ \sqrt{(-x+3)^2} = |-x+3| \]

Because \(|-x+3|\) is equivalent to \(|x-3|\), the equation simplifies to:

\[ |x-3| = 2x - 3 \]

Step 2: Solve the two cases.

• Case 1 (Positive): \(x - 3 = 2x - 3 \Rightarrow x = 0\)
• Case 2 (Negative): \(-(x - 3) = 2x - 3 \Rightarrow -x + 3 = 2x - 3 \Rightarrow 3x = 6 \Rightarrow x = 2\)

Step 3: Check for extraneous solutions.

• Checking \(x = 0\): \(\sqrt{(-(0)+3)^2} = 2(0) - 3 \Rightarrow \sqrt{9} = -3 \Rightarrow 3 \neq -3\) (Extraneous)
• Checking \(x = 2\): \(\sqrt{(-(2)+3)^2} = 2(2) - 3 \Rightarrow \sqrt{1} = 1 \Rightarrow 1 = 1\) (Valid)

Answer: 2

The equation \(f(x+2) = f(x+4)\) indicates that the function repeats its output every 2 units along the x-axis. It has a period of 2.

To find \(f(2)\) using the known value of \(f(4)\), we need to pick an \(x\) value that makes the expressions inside the parentheses equal to 2 and 4. Let \(x = 0\).

\[ f(0+2) = f(0+4) \]

\[ f(2) = f(4) \]

Since we are given that \(f(4) = 3\), substitute that value in:

\[ f(2) = 3 \]

Answer: 3

This function accumulates value. Every time the \(x\)-value increases by 3, the output increases by 2. We must step forward from our known value.

• Step 1 (Find f(5)): Let \(x = 2\).
  \(f(2+3) = f(2) + 2 \Rightarrow f(5) = 5 + 2 = 7\)
• Step 2 (Find f(8)): Let \(x = 5\).
  \(f(5+3) = f(5) + 2 \Rightarrow f(8) = 7 + 2 = 9\)
• Step 3 (Find f(11)): Let \(x = 8\).
  \(f(8+3) = f(8) + 2 \Rightarrow f(11) = 9 + 2 = 11\)

Answer: 11

The rule \(f(x) = f(x+4)\) means the function has a period of 4. To evaluate \(f(19)\), we must "walk backward" by intervals of 4 until we find an equivalent \(x\)-value that falls within the defined base domain of \(0 \le x < 4\).

Divide 19 by 4, which gives 4 with a remainder of 3. Therefore:

\[ f(19) = f(15) = f(11) = f(7) = f(3) \]

Now, evaluate the base quadratic function at \(x = 3\):

\[ f(3) = 2(3)^2 - (3) \]

\[ f(3) = 2(9) - 3 = 18 - 3 = 15 \]

Answer: 15

To determine how circles interact, compare the distance between their centers to the sum of their radii.

Circle 1: Center \(C_1 = (3, 1)\), Radius \(r_1 = \sqrt{4} = 2\)

Circle 2: Center \(C_2 = (-1, -1)\), Radius \(r_2 = \sqrt{1} = 1\)

Calculate Distance (d):

\[ d = \sqrt{(-1 - 3)^2 + (-1 - 1)^2} = \sqrt{(-4)^2 + (-2)^2} = \sqrt{20} \approx 4.47 \]

Compare to Radii: The sum of the radii is \(r_1 + r_2 = 2 + 1 = 3\).

Because the distance between the centers (\(\approx 4.47\)) is strictly greater than the sum of their radii (\(3\)), the circles are completely separated and do not intersect.

Answer: 0

Because the sun's rays strike the ground at the same angle, the poles and their shadows form two similar right triangles. Set up a proportion of height to shadow length:

\[ \dfrac{\text{Height}_A}{\text{Shadow}_A} = \dfrac{\text{Height}_B}{\text{Shadow}_B} \]

\[ \dfrac{1}{1.25} = \dfrac{\text{Height}_B}{12.5} \]

Solve for the unknown height:

\[ \text{Height}_B = \dfrac{12.5}{1.25} = 10 \]

Answer: 10

This is a nested similar triangles problem. Let \(d\) be the distance from the lamppost to the person.

• The small triangle (person and shadow) has height 2 and base 3.
• The large triangle (lamppost and total ground distance) has height 6 and a base of \((d + 3)\).

Set up the proportion:

\[ \dfrac{6}{d + 3} = \dfrac{2}{3} \]

Cross-multiply to solve for \(d\):

\[ 18 = 2(d + 3) \]

\[ 18 = 2d + 6 \]

\[ 12 = 2d \Rightarrow d = 6 \]

Answer: 6

A fundamental circle theorem states that a tangent line is always perpendicular to the radius at the point of tangency. Therefore, angle \(OTP\) is \(90^\circ\), making \(\triangle OTP\) a right triangle.

We know the hypotenuse (\(OP = 13\)) and one leg (\(OT = 5\), the radius). Use the Pythagorean theorem to find the tangent segment \(PT\):

\[ 5^2 + PT^2 = 13^2 \Rightarrow 25 + PT^2 = 169 \Rightarrow PT^2 = 144 \Rightarrow PT = 12 \]

The base and height of the right triangle are the two legs, \(OT\) and \(PT\):

\[ \text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 5 \times 12 = 30 \]

Answer: 30

Because \(MA\) and \(MB\) are tangent lines, they form right angles with radii \(OA\) and \(OB\). This splits the quadrilateral \(MAOB\) into two identical right triangles: \(\triangle OAM\) and \(\triangle OBM\).

In right triangle \(\triangle OAM\), the leg \(OA = 5\) and the hypotenuse \(OM = 13\). Using the Pythagorean theorem (or recognizing the 5-12-13 triple), the tangent leg \(AM = 12\).

Calculate the area of one triangle:

\[ \text{Area of } \triangle OAM = \dfrac{1}{2} \times 5 \times 12 = 30 \]

Since the quadrilateral is made of two identical triangles, multiply by 2:

\[ \text{Total Area} = 2 \times 30 = 60 \]

Answer: 60

There are \(6 \times 6 = 36\) total possible outcomes in the sample space.

Outcomes where the sum is less than 4:

• Sum of 2: (1,1) → 1 outcome
• Sum of 3: (1,2), (2,1) → 2 outcomes
• Total: 3 outcomes

Outcomes where the sum is \(\ge 8\):

• Sum of 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes
• Sum of 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes
• Sum of 10: (4,6), (5,5), (6,4) → 3 outcomes
• Sum of 11: (5,6), (6,5) → 2 outcomes
• Sum of 12: (6,6) → 1 outcome
• Total: 15 outcomes

Because these are mutually exclusive events, we add the successful outcomes together: \(3 + 15 = 18\).

The probability is \(18 / 36\), which reduces to \(1/2\).

Answer: 1/2 (or .5)

Initially, there are 13 balls in total. Since one red ball is drawn and not replaced, the bag's new state is:

• 2 Red balls
• 4 Blue balls
• 6 Green balls
• New Total = 12 balls

The probability of drawing a blue ball from this new state is the number of blue balls divided by the new total:

\[ \text{Probability} = \dfrac{4}{12} = \dfrac{1}{3} \]

Answer: 1/3

This is a reverse conditional probability problem. We want to find \(P(\text{1st is Red} \mid \text{2nd is Blue})\).

Step 1: Calculate the probability of the specific path (Red then Blue).

\[ P(\text{Red then Blue}) = \dfrac{3}{13} \times \dfrac{4}{12} = \dfrac{12}{156} \]

Step 2: Calculate the total probability of getting Blue second.

There are three ways this could happen:

• Red then Blue: \(\dfrac{3}{13} \times \dfrac{4}{12} = \dfrac{12}{156}\)
• Blue then Blue: \(\dfrac{4}{13} \times \dfrac{3}{12} = \dfrac{12}{156}\)
• Green then Blue: \(\dfrac{6}{13} \times \dfrac{4}{12} = \dfrac{24}{156}\)

Total \(P(\text{2nd is Blue}) = \dfrac{12}{156} + \dfrac{12}{156} + \dfrac{24}{156} = \dfrac{48}{156}\)

Step 3: Divide the specific path by the total paths.

\[ P(\text{1st Red} \mid \text{2nd Blue}) = \dfrac{12/156}{48/156} = \dfrac{12}{48} = \dfrac{1}{4} \]

Answer: 1/4 (or .25)