Full-Length SAT Math Practice Test-Answers and Explanations

44- Choice D is correct
The equation of a circle with center \((h, k)\) and radius \(r\) is \((x-h)^2+(y-k)^2=r^2\). To put the equation \(x^2+y^2+10x+4y=3\) in this form, complete the square as follows:
\(x^2+y^2+10x+4y=3, (x^2+10x)+(y^2+4y)=3\),
\((x^2+10x+25)-25+(y^2+4y+4)-4=3, (x+5)^2+(y+2)^2=32\)
\((x+5)^2+(y+2)^2=(\sqrt{32})^ 2\), Therefore, the radius of the circle is \(\sqrt{32}\).

Best SAT Math Prep Resource for 2026

SAT Math in 10 Days The Most Effective SAT Math Crash Course

Download

$29.99 Original price was: $29.99.$16.99Current price is: $16.99.

45- Choice D is correct
By definition, the sine of an acute angle is equal to the cosine of its complement.
Since, angle A and B are complementary angles, therefore: sin A = cos B

46- Choice D is correct
Substituting 10 for \(y\) in \(y=cx^2+d\) gives \(10=cx^2+d\) which can be rewritten as \(10-d=cx^2 \). Since \(y = 10\) is one of the equations in the given system, any solution \(x\) of \(10-d=cx^2\) corresponds to the solution \((x,10)\) of the given system. Since the square of a real number is always nonnegative, and a positive number has two square roots, the equation \(10-d=cx^2\) will have two solutions for \(x\) if and only if (1) \(c > 0\) and \(d <10\) or (2) \(c < 0\) and \(d >10\). Of the values for \(c\) and d given in the choices, only \(c=5\), \(d=5\) satisfy one of these pairs of conditions. Alternatively, if \(c=5\) and \(d=5\), then the second equation would be, \(y=5x^2+5\), The equation above has two real answer.

47- Choice B is correct
\(x\) and \(2z\) are colinear. \(y\) and \(4x\) are colinear. Therefore,
\(x+2z=y+4x\),subtract \(x\) from both sides,then, \(y=2z-3x\)

48- Choice A is correct
It is given that \(g(2)=6\). Therefore, to find the value of \(f(g(6))\), substitute 6 for \(g(2)\).
\(f(g(2))=f(6)=12\).

49- Choice D is correct
Let’s review the options: I. \(|a|<\frac{3}{4}→-\frac{3}{4}<a<\frac{3}{4}\)
Multiply all sides by b. Since, \(b < 0→-b \frac{3}{4}>ab>b \frac{3}{4}\)
II. Since, \(-\frac{3}{4}<a<\frac{3}{4}\), and \(a < 0\)→ \(-\frac{3}{4}a^2>\frac{3}{4} a\) (plug in \(-\frac{1}{2}\), and check!)
III. \(-\frac{3}{4}<a<\frac{3}{4}\), multiply all sides by 4,then: \(-3<4a<3\), add 3 from all sides,then:
\(3-3<4a+3<3+3→0<4a+3<6\), II and III are correct.

50- Choice B is correct
The equation can be rewritten as
\(c-d=2ac\)→(divide both sides by \(c\)) \(1-\frac{d}{c}=2a\), since \(c> 0\) and \(d< 0\), the value of \(-\frac{d}{c}\) is positive. Therefore, 1 plus a positive number is positive. a must be greater than \(\frac{1}{2}\). \(a >\frac{1}{2}\)

51- The answer is 136.5
This is a simple matter of substituting values for variables.
We are given that the 30 cars were washed today, therefore we can substitute that for \(a\).
Giving us the expression \(\frac{15(30)-45}{30}+b\), We are also given that the profit was $150, which we can substitute for \(f(a)\). Which gives us the equation \(150=\frac{15(30)-45}{30}+b\)
Simplifying the fraction gives us the equation \(150=13.5+b\)
And subtracting both sides of the equation by 13.5 gives us \(b=136.5\), which is the answer.

52- The answer is 6.3
The equation of a circle with center \((h, k)\) and radius r is \((x-h)^2+(y-k)^2=r^2\). To put the equation \(x^2+y^2-8x+4y+18=0\) in this form, complete the square as follows:
\(x^2+y^2-8x+4y+18=0, (x^2-8x)+(y^2+4y)+18=0\)
\((x^2-8x+16)-16+(y^2+4y+4)-4+18=0\)
\((x-4)^2+(y+2)^2-20+18=0, (x-4)^2+(y+2)^2-2=0\)
\((x-4)^2+(y+2)^2=2, r^2\) equals 2. Then, \(r=\sqrt{2}\)
The radius of the circle is \(\sqrt{2}\). Thus, the area of the circle is:
\(A=πr^2=3.14(\sqrt{2})^2=3.14×2=6.28\) Round the answer to one decimal place to get 6.3.

53- The answer is 4
Since we are dealing with an absolute value, \(f(a)=10\) means that either \(6-a^2=10\) or
\(6-a^2=-10\), Let’s start with the negative value \((-10)\) and see what we get. If \(6-a^2=-10\), then \(a^2=16\), hence, we get \(a=+4\) or \(-4\), On the other hand, if \(6-a^2=10\), then
\(a=±2i\), Notice that the question states that a is a positive integer, therefore the answer is 4.

54- The answer is 6.
Let \(x\) represent the number of liters of the \(40\%\) solution. The amount of salt in the \(40\%\) solution \((0.40x)\) plus the amount of salt in the \(60\%\) solution \((0.6) × (4)\) must be equal to the amount of salt in the \(48\%\) mixture \((0.48 × (x + 4))\). Write the equation and solve for \(x\).
\(0.40x+0.60(4)=0.48(x+4)→0.40x+2.4=0.48x+1.92→
0.40x-0.48x=1.92-2.4→0.08x=0.48→x=\frac{0.48}{0.08}=6\)

55- The answer is 3.
The function \(f(x)\) is undefined when the denominator of \(\frac{1}{(x-8)^2+25(x-8)+25}\) is equal to zero. The expression \((x-8)^2+10(x-8)+25\) is a perfect square.
\((x-8)^2+10(x-8)+25=((x-8)+5)^2\) which can be rewritten as \((x-3)^2\). The expression \((x-3)^2\) is equal to zero if and only if \(x=3\). Therefore, the value of \(x\) for which \(f(x)\) is undefined is 3.

56- The answer is 336
One of the four numbers is \(x\); let the other three numbers be \(y\), \(z\) and \(w\). Since the sum of four numbers is 560, the equation \(x + y + z+w = 560\) is true. The statement that \(x\) is \(150\%\) more than the sum of the other three numbers can be represented as
\(x = 1.5(y + z+w)\) or \(\frac{x}{1.5}=y+z+w→\frac{2x}{3}=y+z+w\)
Substituting the value \(y+z+w\) in the equation \(x+y+z+w=560\)
gives \(x+\frac{2x}{3}=560→\frac{5x}{3}=560→5x=1,680→x=\frac{1,680}{5}=336\)

57- The answer is 1
Squaring both sides of the equation gives \(12-8a=4a^2\)
Subtracting both sides by \(12-8a\) gives us the equation \(4a^2+8a-12=0\)
Here you can solve the quadratic by factoring to get \((2a+6)(2a-2)=0\)
For the expression \((2a+6)(2a-2)\) to equal zero, \(a=-3\) or \(a=1\)
Since a is a positive integer, \(1\) is the answer.

58- The answer is 52
The area of ∆BED is 30, then: \(\frac{6×AB}{2} =30→6×AB=60→AB=10\)
The area of ∆BDF is 32, then: \(\frac{4×BC}{2}=32→4×BC=64→BC=16\)
The perimeter of the rectangle is = \(2×(10+16)=52\)

College Entrance Tests

The Best Books to Ace the SAT Math Test

SAT Math in 10 Days The Most Effective SAT Math Crash Course

Download

$29.99 Original price was: $29.99.$16.99Current price is: $16.99.