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}).

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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)

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