SAT Math Test 10 Section 3 solved questions

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SAT Practice Test 10 Section 3 

Answer Explanations 

Question 1 Answer B

Step 1: Isolate variable on one side of the equation and solve
Move 𝑧 to the left side and 1 to the right side of the equation.
2𝑧 + 1 =  𝑧   →   2𝑧 − 𝑧 =  −1      𝑧  =  −1
There are several categories that demonstrate how to solve for a variable in an equation.

Question 2 Answer C
Step 1: Identify the components of the equation
Initial payment = $60.
Weekly payment = $30. Number of weekly payments = 𝑤. Hence, payments in 𝑤 weeks = 30 × number of weeks = 30𝑤. 
Total price = 300. 
Step 2: Determine the equation
total price = initial payment + payments in 𝑤 weeks
300 = 60 + 30𝑤
Maps to Category 14 – Word problems on linear equations with one variable.

Question 3 Answer B 
Step 1: Determine the slope
The question mentions that the relationship between weight and shipping charge is linear. Remember that linear = line. Hence, each row in the table is a point (𝑥, 𝑦) on the line of linear function 𝑓(𝑥). Since it is given that 𝑥 is the weight, 𝑦 must be the shipping charge.
Below slope equation is set up using point (5, 16.94) of row 1 and point (10, 21.89) of row 2.
$${slope=\frac {y_2-y_1} {x_2-x_1}=\frac {21.89-16.94}{10-5}=\frac {4.95}{5}=0.99}$$
Only answer choices A and B have slope = 0.99. This eliminates answer choices C and D.
Step 2: Determine the 𝑦-intercept 
In the slope intercept equation, the value of 𝑏 is the 𝑦-intercept of the line. Plug in the slope and any given point in the slope intercept equation, as shown below.
𝑦 = 𝑚𝑥 + 𝑏   →   16.94 = (0.99 × 5) + 𝑏   →   16.94 = 4.95 + 𝑏   →   𝑏 = 11.99
This eliminates answer choice A.
Alternatively, if you are getting good with slope questions, then see the following logic that can eliminate Step 2 calculation. 
At the 𝑦-intercept, 𝑥 = 0. It can be observed from the given table that for every 5 pounds increase in weight the shipping charge increases by 4.95. Hence, for 5  5 = 0 pounds (𝑥), the shipping charge (𝑦) is 16.94  4.95 = 11.99. This is the 𝑦-intercept = 𝑏 in the slope intercept equation. 
Maps to Category 1 – Line equation in slope intercept form and Category 6 – Linear functions. 

Question 4 Answer C 
Step 1: Read the 𝑦-axis value from the graph for the given 𝑥-axis value 
𝑥-axis = base diameter. 
𝑦-axis = height. 
For diameter = 𝑥 = 2, the height = 𝑦 =14. 
For diameter = 𝑥 = 5, the height = 𝑦 = 35. 
Step 2: Determine the difference 
35 − 14 = 21 
There are several categories that demonstrate how to read points on a graph.

Question 5 Answer A 
Step 1: Determine the square root
Both terms under the square root are perfect squares. 
Maps to Category 49 – Exponents. 

Question 6 Answer A 
Step 1: Determine the approach
The numerator of the fraction can be factorized as (𝑥 − 1)(𝑥 + 1). (𝑥 − 1) from the numerator and denominator can cancel out, simplifying the equation.
Step 2: Solve
$${\frac {(x^2-1)}{(x-1)}=-2}  →  {\frac {\sout{(x-1)}(x+1)}{\sout{(x-1)}}=-2}$$
$${x+1=-2}   →   {x=-3}$$
An alternative approach to solve this question is to plug in each answer choice in the given equation as the value of 𝑥 and determine which value of 𝑥 gives 2.
Maps to Category 40 – Fractions with expressions in the denominator. 

Question 7 Answer D
Step 1: Read the value of 𝑦 from the graph for 𝑥 = 0 
In a function, 𝑓(𝑥) is the 𝑦 value for an input value of 𝑥. Hence, 𝑓(0) refers to the value of 𝑦 when 𝑥 = 0. 
From the graph, it is seen that for 𝑥 = 0, the value of 𝑦 is 4. 
Maps to Category 22 – Polynomial functions as tables and graphs.

Question 8 Answer C 
Step 1: Determine the value of 𝑥 
The angles of any point on a straight line add up to 180 degrees. The figure shows that angle 𝐴𝐵𝐶 is 90 degrees. Hence, angle 𝐶𝐵𝐷 is 180 − 90 = 90 degrees. 
The figure shows that angle 𝐶𝐵𝐷 is comprised of three angles: 𝑥, 2𝑥, and 2𝑥. Hence,
𝐶𝐵𝐷 =  𝑥 + 2𝑥 + 2𝑥 = 90  →    5𝑥 = 90    →    𝑥 = 18 
Step 2: Determine the value of 3𝑥
3𝑥  = 3 × 18 = 54   
Maps to Category 69 – Parallel and intersecting lines.

Question 9 Answer C
Step 1: Determine the slope of the line
If you are good at slope, it can be seen from the graph that slope = rise/run = 1. 
Alternatively, select any two points on the line to determine the slope using the slope equation. Points (0, 4) and (4, 0) are selected for the calculation below.
$$\frac {0-(-4)}{-4-0}=\frac {4}{-4}=-1$$
Step 2: Determine the 𝑦-intercept of the line 
From the graph it is seen that the 𝑦-intercept = 4 (𝑦-intercept of a line is the value of 𝑦 when 𝑥 = 0). 
The answer choices are given in the standard form equation, 𝑎𝑥 + 𝑏𝑦 = 𝑐. If you are familiar with matching the slope and the 𝑦-intercept in a standard form equation, then you know the correct answer is C. If not, proceed to step 3.
Step 3: Determine the standard form equation 
Plug in slope = 1 and 𝑏 = 𝑦-intercept = 4 in 𝑦 = 𝑚𝑥 + 𝑏. 
𝑦 = 𝑚𝑥 + 𝑏   →   𝑦 = (−1 × 𝑥) − 4   →   𝑦 = −𝑥 − 4
Rearrange the above equation to the standard form. 
𝑦 = −𝑥 − 4    →    𝑥 + 𝑦 = −4 
Maps to Category 1 – Line equation in slope intercept form and Category 2 – Line equation in standard form.

Question 10 Answer D
Step 1: Determine the 𝑦-coordinate of the 𝑦-intercept of the parabola 
The point where the parabola crosses or intersects the 𝑦-axis is the 𝑦-intercept of the parabola. The value of 𝑥 at this point is 0. Hence, the given point (0, 𝑘) is the 𝑦-intercept of the parabola and 𝑘 is the 𝑦-coordinate of the 𝑦-intercept of the parabola. 
In this question, the equation of the parabola is given in the standard form. The value of 𝑐 in the standard form equation is the 𝑦-coordinate of the 𝑦-intercept of the parabola. Hence,  
𝑘 = 𝑐 = 12 
Maps to Category 30 – Standard form equation of a parabola. 

Question 11 Answer A
Step 1: Determine the equation
The equation a circle in the standard form is 
(𝘩, 𝑘) are the (𝑥, 𝑦) coordinates of the center of the circle and 𝑟 is the radius. In the question, the coordinates of the center are given as (5, 7) and the radius is given as 2. Plug these values in the equation,
Maps to Category 68 – Equation of a circle. 

Question 12 Answer B
Step 1: Identify corresponding angles
In two similar triangles, (i) the corresponding sides are in proportion, (ii) the corresponding angles are congruent, and (iii) the values of the sine, the cos, and the tan for the corresponding angles are same. Based on the given figure, angle 𝐵 and angle 𝐸 are corresponding angles. 
Hence, cos 𝐵 = cos 𝐸. 
Step 2: Determine cos 𝐵 = cos 𝐸
Since the side lengths of triangle 𝐴𝐵𝐶 are given in the figure, cos 𝐵 can be determined.
In triangle 𝐴𝐵𝐶, the side adjacent to angle 𝐵 is 𝐴𝐵 = 12 and the hypothenuse 𝐵𝐶 = 13. Hence,
$$cosE=cosB=\frac {adjacent}{hypotenuse}=\frac{AB}{AC}=\frac{12}{13}$$
Maps to Category 80 – Right triangles and trigonometry.

Question 13 Answer C
Step 1: Factorize the equation 
The factors of a quadratic function are the 𝑥-intercepts of the graph of the function. At the 𝑥-intercept, 𝑦 = 0. Hence, the given equation can be factorized as 
$$x^2+5x+4=0   →   (x+1)(x+4)=0$$
The two values of 𝑥 are 1 and 4. 
Step 2: Determine the distance 
The difference between the two values is
(−1) − (−4) = 3 
Maps to Category 27 – Quadratic equations and factors.

Question 14 Answer B
Step 1: Square both sides and remove the square root
When an equation contains a square root expression, the square root can be removed by squaring both sides of the equation. It is important to square the entire left and the right expression, as shown below.
$${(\sqrt{4x}})^2=(x-3)^2   →    4x=(x-3)^2$$
Foil the right expression and form a quadratic equation. 
$$4x=x^2-6x+9   →   x^2-10x+9=0$$
Step 2: Factorize the equation and determine the solutions
The above equation can be factorized as (𝑥  1)(𝑥  9) = 0.
The two values of 𝑥 are 1 and 9. 
Step 3: Check for extraneous solutions 
In equations with square root expressions, it is possible that the solutions may not solve the equation. These are known as extraneous solutions. This is more likely when the equations with square root expression have more than one solution. To check for them, plug in the solutions in the given equation and check if they solve the equation. 
Plug in 𝑥 = 1: 
$${\sqrt{4\times 1}}=1-3   →   {\sqrt{4}}=-2   →   2=-2$$This is incorrect. Hence, 1 is an extraneous solution. 
Plug in 𝑥 = 9:
$${\sqrt{4\times 9}}=9-3   →   {\sqrt{36}}=6   →   6=6$$
This solves the equation. Hence, 9 is a solution to the given equation. 
Maps to Category 42 – Expressions with square root.

Question 15 Answer A
Step 1: Set up ratios of 𝑎 and 𝑏 and equate them
For two linear equations in a system of equations in the 𝑎𝑥 + 𝑏𝑦 = 𝑐 form, the system has no solution when the ratios for 𝑎 and 𝑏 are same but different than the ratio of 𝑐. See the ratios below for two equations.
$$a_1x+b_1y=c_1\ \ and\ \ a_2x+b_2y+c_2$$
$${\frac {a_1}{a_2}=}\ {\frac {b_1}{b_2}\ne}\ {\frac {c_1}{c_2}}$$
In equation 3𝑥 + 𝑦 = 6: 𝑎 = 3, 𝑏 = 1.
In equation 𝑎𝑥 + 2𝑦 = 4: 𝑎 = 𝑎, 𝑏 = 2.
$${\frac {a_1}{a_2}=}\ {\frac {b_1}{b_2}}  →  {\frac {-3}{a}=}\ {\frac {1}{2}}$$
Step 2: Solve for 𝑎
The value of 𝑎 must result in the same ratios of 𝑎 and 𝑏. Cross multiply the above equation to solve for 𝑎.
𝑎 = −3 × 2 = −6
Maps to Category 9 – System of linear equations with no solution.

Question 16 Answer 2200
Step 1: Plug in the given values in the given equation and solve
𝑇 = 47,000.  𝑓 = 3000. 
𝑇 = 5𝑐 + 12𝑓  →  47000 = 5𝑐 + (12 × 3000)  →   
47000 = 5𝑐 + 36000    5𝑐 = 47000 − 36000 = 11000  →  𝑐 = 2200 
Maps to Category 15 – Word problems on linear equations with two variables.

Question 17 Answer 5
Step 1: Remove the absolute value bars 
An expression within the absolute value bars is solved by removing the absolute bars and giving the equation two values: one positive and one negative, as shown below.
|2𝑥 + 1| = 5    2𝑥 + 1 = 5 and 2𝑥 + 1 = 5 
Step 2: Solve for the positive value 
2𝑥 + 1 = 5    2𝑥 = 5 − 1 = 4    𝑥 = 2 
Step 3: Solve for the negative value 
2𝑥 + 1 = −5    2𝑥 = −5 − 1 = −6    𝑥 = −3 
Step 4: Determine |𝑎  𝑏
The questions labels the two solutions of 𝑥 as 𝑎 and 𝑏. Since the question is asking for the absolute value of 𝑎  𝑏, it does not matter which value of 𝑥 is 𝑎 or 𝑏
|𝑎  𝑏|  =  |−3 −2| =  |−5|  =  5 or
 |𝑎  𝑏|  =  |2 − (−3)|  = |5|  =  5
Maps to Category 37 – Absolute value and linear equations.

Question 18 Answer 1.21
Step 1: Determine the components of exponential growth equation 
Since the value of the antique increased each year over its previous year value, the question is on exponential growth. The equation for exponential growth is 
𝑎 = initial value = 200.
𝑥 = rate of change as percent decimal = 1 + 0.1 = 1.1. 
𝑡 = time = 2. 
𝑦 = end value in 𝑡 = 2 years = 200𝑎. 
Step 2: Determine the equation 
Plug in the above values in the equation.
$$y=a(x)^t  →  200a=200(1.1)^2  →  a=1.21$$
Note that 200 from both sides cancels out.
Maps to Category 51 – Exponential growth and decay.

Question 19 Answer 2500
Step 1: Determine 5𝑥 + 5𝑦
Note that if the 𝑥 and 𝑦 variables from both the equations are added, the result will be 5𝑥 + 5𝑦. Hence, there is no need to solve for the individual values of 𝑥 and 𝑦. Add the two equations. 
2𝑥 + 3𝑦 = 1200
3𝑥 + 2𝑦 = 1300
5𝑥 + 5𝑦 = 2500 
Maps to Category 11 – System of linear equations with one solution.

Question 20 Answer 20
Step 1: Determine the approach 
The values of (𝑢  𝑡) and (𝑢 + 𝑡) are given.
Step 2: Solve by plugging in the given values 
$$(u-t)(u-t)(u+t)=2\times 2\times 5=20$$
Maps to Category 40 – Fractions with expressions in the denominator.