## SAT Math 1 & 2 Subject Tests

### Chapter 19

### Level 2 Practice Test Form A Answers and Explanations

**1** **B** Algebraic manipulation is the easiest way to solve this one. Start by adding *s* to each side, producing the inequality *r* > *r* + 2*s*. Then subtract *r* from each side to get 0 > 2*s*. If 2*s* < 0, then *s* < 0 (just divide both sides by 2). This can also be solved by Plugging In, but since only certain values will make the original inequality true, it can take some time to Plug In enough different values to eliminate all the wrong answers. Algebraic manipulation is faster here.

**2** **C** You”ll want to use Process of Elimination here to knock out some answer choices. The best numbers to plug into the function are the ones in the answer choices: 10 and −10. Start by finding out whether the statement *f*(*x*) = *f*(*−x*) is true when *x* = 10. *f*(10) = |10| + 10 = 20. *f*(−10) = |−10| +10 = 10 + 10 = 20. Since *f*(10) and *f*(−10) are equal, *f*(*x*) = *f*(*−x*) is true when *x* = 10, and every answer choice that does not include 10 can be eliminated. That gets rid of (A), (D), and (E). To choose between the remaining answers (B) and (C), Plug In a number included in (C) but not in (B). A simple number like zero works best. *f*(0) = |0| + 10 = 10. *f*(−0) = |−0| + 10 = 0 + 10 = 10. You can see that *f*(0) = *f*(*−*0), so zero must be part of the correct answer. (B) does not include zero, so you can eliminate it. (C) is correct.

**3** **D** Remember how factorials cancel out in fractions. If you expand the factorials in the fraction , then you get

You can see that every factor from 1 to 13 in the numerator is also in the denominator; these factors cancel out, leaving you with , or , which equals 105.

**4** **A** Remember the SOHCAHTOA definitions of the trigonometric functions. In a right triangle, the sine of an angle equals the length of the opposite side over the length of the hypotenuse. You can easily find the lengths of sides *AB* and *BC*, since they are horizontal and vertical, respectively. The hypotenuse”s length can be found with the Pythagorean theorem, but you don”t even need it if you recognize this as a 5-12-13 triangle. To find the sine of ∠*BAC*, then, you just need to find the value of , which equals .

**5** **D** A “system” is another way of describing a set of simultaneous equations. To find the complete solution set of a system of equations, you need to solve the equations simultaneously. Instead of doing the algebra here, however, you should notice that the answer choices all have numbers in them. That means you can PITA. Start with (C), which in this case is the coordinate pair (−4, −3), since it appears in two answer choices. That will allow you to eliminate more efficiently. That makes *x* = −4 and *y* = −3. To PITA, put these numbers into both equations. These numbers work in the first equation, because (−4)^{2} + (−3)^{2} = 16 + 9 = 25. They also work in the second equation, because −3 = −4 + 1. That means that (A), (B), and (E) can be eliminated, because they do not contain the coordinate pair (−4, −3). The only difference between (C) and (D) is the point (3, 4). If you plug it into both equations, it works, so the answer is (D).

**6** **E** When there are variables in the question and the answer choices, Plug In. Remember to select numbers that make your math easy. In this case, it”s important to choose numbers that make the fraction work out conveniently. Making *j* = 4 and *k* = 2 turns out well, because it makes = 2. The expression then works out to = = = 3. To find the correct answer, just go quickly through the answer choices to find the one that also equals 3 when *j* = 4 and *k* = 2. Only (E) works out equal to 3. (E) is correct.

**7** **E** This is a visual perception question, and there”s no hard and fast technique to follow to solve it. The best plan is to experiment with sketches in your test booklet and use common sense. Remember that this is an EXCEPT question, so you”re looking for a shape that *can”t* be made. Any time you find a way to make one of the shapes in the answer choices, that choice can be eliminated.

The intersection of a cube and a plane can be a triangle:

but there”s no way to produce a circle, since none of a cube”s edges or faces is curved. The correct answer is (E).

**8** **B** This is a compound-function question, in which you”ll have to apply two functions in combination. The most important rule to remember is that you have to work from the inside out. Start by finding *g*(2.3). That means putting the number 2.3 in place of the *x* in the definition of *g*(*x*),*g*(2.3) = + 1 = (1.52) + 1 = 0.76 + 1 = 1.76. Once you know that *g*(2.3) = 1.76, you know that *f*(*g*(2.3)) is equal to *f*(1.76), which is easily solved. *f*(1.76) = = 1.21. The correct answer is (B).

**9** **D** Don”t panic because you”ve never seen a term like “*x* mod *y*” before. This isn”t something you slept through in math class. It”s just one of those terms ETS throws at you sometimes. Some will be little-known math terms, and others will be made up. Either way, it doesn”t matter whether you”ve seen it before, because ETS defines it for you. To find the value of any “*x* mod *y*,” just take the number in the *x* position and divide it by the number in the *y* position. The remainder is the value of “*x* mod *y*” for those numbers. The value of 61 mod 7 is 5, because the remainder when 61 is divided by 7 is 5. The value of 5 mod 5 is 0, because the remainder when 5 is divided by 5 is zero. The expression (61 mod 7) − (5 mod 5) can be rewritten as 5 − 0, which equals 5. The correct answer is (D).

**10** **D** Plug In! Trying to solve this question by thinking through the trigonometric theory is a good way to give yourself a brain cramp. Just Plug In a few numbers on your calculator (angles between 0° and 90°) and see what happens. You”ll soon find that Statements I and III always prove true, while it”s easy to find an exception to Statement II. The correct answer is (D).

**11** **B** The statement *f*(*x*) = can be read, “*f* of *x* equals 2 when *x* does not equal 13, and *f* of *x* equals 4 when *x* equals 13.” Since neither of the values you”re given equals 13, the function will always come out to 2. *f*(15) − *f*(14) = 2 − 2 = 0. The correct answer is (B).

**12** **D** You can PITA on this question if you get completely stuck, but algebraic manipulation is simpler and faster. Just isolate *x*. Start by multiplying both sides by 25, so that = 25 becomes *x*^{5} = 625. Then take the fifth root of each side to get *x* alone. On most calculators, you take the fifth root of a value by raising that value to the power of one-fifth, or 0.2. *x* = = 3.62. The correct answer is (D).

**13** **A** On the Math Level 2, most trigonometry questions like this one are solved by using trigonometric identities to change the form of equations. Most often, the most successful strategy is to start by getting everything in terms of sine and cosine. The ratio you”re given can be written in fractional form, like . The secant and cosecant can also be expressed in terms of sine and cosine, = . This fraction simplifies to = , and since , this can be written as tan *x* = . The cotangent is the reciprocal of the tangent, so cot *x* = 4. The ratio of tan *x* to cot *x* is therefore equal to , or . The correct answer is (A).

**14** **D** You can tell by looking at the figure that the *x*-coordinates of the points in region *J* include everything from *x* = 0 to *x* = 6. The *y*-coordinates of the points in the region include everything from *y* = 0 to *y* = 3. A rectangular region containing all points (2*x*, *y* − 1) would therefore stretch from *x* = 0 to *x* = 12 (doubling both values) and from *y* = −1 to *y* = 2 (subtracting 1 from both values). The resulting region would look like this:

It would have a length of 12, a width of 3, and an area of 36. The correct answer is (D).

**15** **D** Draw it! It”s always a good idea to sketch any figure that is described but not shown. The triangle described here would look something like this:

where *h* is the unknown length of the hypotenuse. As you can see, you”re given the measure of an angle and the length of the opposite side in a right triangle, and you”re trying to find the hypotenuse. The easiest way to find it is to use a trigonometric function that relates all these quantities. Since sine = , the sine is the function you want. Just plug the information you have into the formula, and solve for the missing piece, sin 27° = . To solve, start by isolating *h*. *h* = . Then use your calculator to find a numerical value: *h* = = = 19.82.

**16** **B** PITA. Plug each answer choice in for *x* to see which one makes *f*(*x*) = 0. (B) works. Alternately, you could graph *y* = *x*^{2} + 6*x* − 12 on your calculator and see where it crosses the *x*-axis.

**17** **B** There are variables in the answer choices, so it behooves you to Plug In. Pick a nice friendly angle for *x* like 30º. So sin 30º = 0.5 = *m*. tan 30º = 0.577, our target number. Now Plug In 0.5 for *m* to see which answer choice becomes 0.577. It”s (B).

**18** **C** Use the definition of a logarithm to rewrite the equation log* _{y}* 2 = 8 in exponential form,

*y*

^{8}= 2. To find the value of

*y*, take the eighth root of both sides. This can be done on your calculator by raising 2 to the power of one-eighth, or 0.125. You”ll find that

*y*= 1.09. The correct answer is (C). You can also PITA, starting with (C), which works.

**19** **D** An angle of radians is equivalent to an angle of 45°. You”re therefore looking for an angle between 45° and −45° whose sine equals . The easiest way to find this number is to enter the quantity into your calculator and take its inverse sine. This will show you the smallest positive angle whose sine is . This angle, *θ*, is 19.47 degrees, or 0.34 radians. To find cos(2 *θ*), just double this value and take its cosine. You”ll get 0.777…, or . The correct answer is (D).

**20** **A** The best way to solve inverse-function questions on the Math Level 2 is to Plug In numbers. Be sure to pick a number that will work out neatly. For example, for the function *f*(*x*) = − 1, a good value for *x* would be 4. Then, *f*(4) = − 1 = 2 − 1 = 1. You can see that *f*(*x*) turns 4 into 1. So *f*^{−1}(1) = 4. To find the correct answer, plug 1 into all of the answer choices. The correct answer will produce a value of 4. Only (A) comes out to 4, since (1 + 1)^{2} = 4, and so (A) is the correct answer.

**21** **A** Don”t bother doing long division with the polynomial. Just Plug In *x* = 10. The problem now reads, “When 460 + *L* is divided by 11, the remainder is 2.” Well, 460 divided by 11 gives you 41.818. 11 × 41 = 451, and 11 × 42 = 462. To get a remainder of 2, you would divide 464 by 11. So 460 + *L* = 464, and *L* = 4.

**22** **C** If you”ve studied the section of this book dealing with ellipses, then you can tell by looking at the formula that this is an ellipse centered at the origin. You can also tell that the ellipse”s major axis is the vertical axis rather than the horizontal, because the constant under the *y* is larger than the one under the *x*. That means that the ellipse looks something like this:

The major axis is the vertical axis of the ellipse. The endpoints of the major axis are the two points at which the ellipse intersects the *y*-axis. Since every point on the *y*-axis has an *x*-coordinate of 0, then all you have to do to find the coordinates of the endpoints is plug *x* = 0 into the equation of the ellipse, and see what *y*-coordinates that produces.

= 1

The zero causes the entire first term to equal zero, effectively eliminating it:

0 + = 1

= 1

*y*^{2} = 20

*y* = ±

*y* = 4.472, −4.472

The endpoints of the ellipse”s major axis are therefore (0, −4.472) and (0, 4.472). Because the points lie on a vertical line, it”s not even necessary to use the distance formula to find the length of the segment between them. The distance between the points is equal to 4.472 − (−4.472), or 8.944. This is the length of the ellipse”s major axis. The correct answer is (C).

**23** **B** Ah, functions with weird symbols. In this case, [*x*] means the greatest integer less than or equal to *x*. In simple terms, that means that [*x*] equals *x* if *x* is an integer, and if *x* isn”t an integer, then [*x*] is the next smallest integer. The number 2.75 becomes 2.0, −3.54 becomes −4.0, and so on. To find the graph of the function *f*() −1, just Plug In numbers. Start with something very easy, like zero. *f*()− 1 = [0] − 1 = 0 − 1 = −1. The function *f*()− 1 must contain the point (0, −1). Only (B) and (D) contain that point; answer choices (A), (C), and (E) may be eliminated (remember that a function does not include points marked by open circles). To choose between (B) and (D), Plug In another easy number, like 2. *f*() −1 = [1] − 1 = 1 − 1 = 0. The function *f*()− 1 must therefore also contain the point (2, 0). Only (B) contains that point—(D) contains the point (2, 1) instead. The correct answer is (B).

**24** **D** Since cos *θ* and cos^{−1} *θ* are inverses, cos(cos^{−1} *θ*) = *θ*. If you also remember that the secant is the reciprocal of the cosine, then you can rewrite the given equation as = = 2.8353.

**25** **B** The function *f*(*x*) = *x*^{2} + 5*x* + 6 is a quadratic function, which means that its graph will be a parabola. Since the coefficient of the *x*^{2} term is positive, you know the parabola opens upward. The parabola therefore has a minimum value—its vertex. To find the value of *x* at which *f*(*x*) reaches its minimum value, just find the *x*-coordinate of the parabola”s vertex. You can do this by using the vertex formula. For any parabola in the form *y* = *ax*^{2} + *bx* + *c*, the *x*-coordinate of the vertex equals . For the parabola *f*(*x*) = *x*^{2} + 5*x* + 6, that places the vertex at . The correct answer is (B). Of course, you can also PITA. Plug each answer choice into *f*(*x*), to see which one gives you the lowest value.

**26** **B** An arithmetic sequence is one that increases by adding a constant amount again and again. The most important information to have when you”re working with an arithmetic sequence is the size of the interval between any two consecutive terms in the sequence. Since the 20th term in the sequence is 20 and the 50th term is 100, you know that 30 steps in the sequence produce an increase of 80. That makes each step worth or . To find the first term in the sequence, you can use the formula for the *n*th term in an arithmetic sequence, *a _{n}* =

*a*

_{1}+ (

*n*− 1)

*d*, where

*n*is the number of the term and

*d*is the interval between any two consecutive terms. In this case, you know that

*a*

_{20}= 20, and that

*d*= , or . You can then fill those values into the formula, so

20 = *a*_{1} + (20 − 1)()

You can then solve for the value of *a*_{1}.

20 = *a*_{1} + 19 ×

20 = *a*_{1} +

*a*_{1} =

The correct answer is (B).

**27** **A** This question is a lot simpler than it looks. ETS will never require you to do complex calculations with polar coordinates; you need to know only the basics. As you”ve seen in this book, polar coordinates can be converted to rectangular coordinates very simply, with the following two equations: *x* = *r* cos *θ* and *y* = *r* sin *θ*. That means that the equation *r* sin *θ* = 1 simply translates to *y* = 1 in rectangular coordinates. And that”s the equation of a horizontal line. The correct answer is (A).

**28** **B** Remember: A function is a relation in which every element in the domain corresponds to only one element in the range. Simply put, that means that there”s only one *y* for every *x*. If you look at simple tables of values for the functions you”re given, you can see that they obey the rule.

But the inverses of these functions will *reverse* these tables of values, switching the *x* and *f*(*x*) values. After all, the inverse undoes the original function, turning all the *f*(*x*) values back into the *x* values. If you reverse the tables of values for these three functions, you”ll find that only one of them remains a function.

The inverses of functions I and III are not functions themselves. In each case, some *x* values correspond to more than one *f*(*x*) value. That is, there”s more than one *y* for each *x*. The inverse of function II, however, *is* a function, because there”s still only one *y* for every *x*. The correct answer is (B).

**29** **D** This is a classic ETS limit question. Ordinarily, to find the limit of an expression as *x* approaches a certain value, you just plug that value in for *x*, and *voilà*, there”s your answer. This, however, is one of those annoying expressions that becomes undefined (zero in the denominator) when you Plug In the value. That could mean that the limit does not exist; but before you decide that, you”ve got to factor the top and bottom and see whether anything cancels out. In this case, the expression is factorable.

=

=

=

= *x*^{2} − *x*

The whole expression reduces to *x*^{2} − *x*. In order to find the value of *x*^{2} − x, just Plug In −1 for *x*, (−1)^{2} − (−1) = 1 + 1 = 2. The correct answer is (D).

**30** **A** Since there are variables in the answer choices, your best bet is to Plug In an easy number, like *x* = 2, and use your calculator. So *f*(2) = 601,303. *g*(*f*(*x*)) = *x* tells you that *f* and *g* are inverses, so *g*(601,303) = 2, our target number. Plug 601,303 into each answer choice to see which one turns into 2. The nice thing about Plugging In is that you can approximate and eliminate answer choices. You don”t need to compute (B), (C), and (E) to see that they are way too big. Then just calculate the value of (A) and (D). The correct answer is (A).

**31** **D** When a figure is described but not shown, always sketch it.

When a cube is inscribed in a sphere, the long diagonal of the cube is a diameter of the sphere. Since the sphere has a radius of 6 and a diameter of 12, you know that the sphere has a long diagonal of 12. The long diagonal of a cube is related to a side of the cube using the formula *d* =*s* . You can use that formula to find the length of a side: *s* = . In this case *s* = = = 6.93. Once you know the length of one side of the cube, you can easily find the cube”s volume using the formula *V* = *s*^{3}. You”ll find that the cube has a volume of 332.55. Answers (A), (B), and (C) can immediately be eliminated by approximation. To choose between (D) and (E), just calculate the decimal value of one of them. Since 192= 332.55, (D) is the correct answer.

**32** **E** When a figure is described but not shown, always sketch it.

In this problem, you”re actually slicing a small cone off a bigger cone. To find the volume of the remaining solid, all you need to do is find the volumes of the two cones, and subtract the little one from the big one. Because you”re given only variables in the question and answer choices, you”ll be Plugging In your own values. Let”s say the larger cone has a height of *h* = 4 and a radius of *r* = 2. Then you can find its volume using the formula *V* = π*r*^{2}*h*. In this case, you get *V* = π*r*^{2}*h* = π(*2*)^{2}(4) = To find the volume of the smaller cone, you first need to figure out its base and height. The height is easy—since the problem says the larger cone is cut “midway,” you know that the smaller cone”s height equals 2, or half of the larger cone”s height. Because one cone is a piece of the other, they are proportional. The radius of the smaller cone must equal 1, or half of the larger cone”s radius. Now you can find the volume of the smaller cone. *V* = π(1)^{2}(2) = π(2) = Once you know the volumes of both cones, it”s easy to subtract to find the volume of the leftover solid, − = That”s your target number. To find the right answer just plug *h* = 4 and *r* = 2 into the answer choices, and see which answer produces a value of Only (E) does the trick.

**33** **B** The equation *e ^{x}*

^{2}= looks very difficult to solve. Just PITA. Plug each answer choice in for

*x*to see which makes both sides of the equation equal.

**34** **D** The equation *y* = 2*x*^{2} − 6*x* + *c* is in quadratic form, which means that its graph will be a parabola. If it”s tangent to the *x*-axis, that means the vertex of the parabola lies on the *x*-axis (sketch it and you”ll see that”s the only way to get a point of tangency). You know that the *y*-coordinate of the vertex must therefore be zero. Since the quadratic has a “double root,” that is, one distinct zero, then the discriminant in the quadratic formula must equal zero. In this case, *b*^{2} − 4*ac* = 0 becomes (−6)^{2} − 4(2)*c* = 0, so *c* = 4.5. The correct answer is (D). Another way to attack this is to PITA and graph each equation on your calculator.

**35** **C** You”re dealing with imaginary numbers here, so your calculator won”t be much use. To solve this problem, just plug (*i* − 1) into the expression in place of *x* and do the math.

*x*^{2} + 2*x* + 2 =

(*i* − 1)^{2} + 2(*i* − 1) + 2 =

(*i* − 1)(*i* − 1) + 2(*i* − 1) + 2 =

*i*^{2} − 2*i* + 1 + 2*i* − 2 + 2 =

*i*^{2} + 1 =

−1 + 1 = 0

**36** **E** With a graphing calculator, this is easy. Just type in the equations in the answer choices and see what their graphs look like. If you haven”t got a graphing calculator, then you”ve got to proceed by Process of Elimination. The easiest way to eliminate answers is to observe that this curve does not appear to cross the *x*-axis or the *y*-axis. Any answer choice whose graph *does* intercept an axis can therefore be eliminated. That means plugging *x* = 0 and *y* = 0 into the answer choices to see what happens.

You should know roughly what the graph of *y* = *e ^{x}* looks like (it appears earlier in this book). If you remember it, you”ll recall that it crosses the

*y*-axis. Even if you don”t remember the shape of the graph, it”s easy to discover that the curve crosses the axis just by Plugging In

*x*= 0 and seeing that the equations in both (A) and (B) contain the point (0, 1). That”s a

*y*-intercept, so you can eliminate both choices.

Answer choice (C) can be eliminated because it”s the equation of a line—no exponents.

Answer choice (D) is the equation of a parabola, which might fit the graph you”re given. But if you Plug In *x* = 0, you get *y* = 2. The point is clearly not on this graph.

That leaves only (E), which is the equation of a hyperbola. You can easily tell from the equation that *x* ≠ 0 and *y* ≠ 0. That makes (E) the correct answer.

**37** **C** Remember that when you calculate the probability of multiple events, you”ve got to find the probabilities of the individual events and multiply them together. On the first drawing, the odds of getting a dime are , because there are 8 dimes out of a total of 11 coins. On the second drawing, the odds of getting a dime are , because there are 7 dimes remaining out of a total of ten remaining coins. The total probability of drawing two dimes, then, is × = which reduces to .

**38** **C** You can recognize the graph of an even function because it”s symmetrical across the *y*-axis. The graphs in (A), (C), and (D) are symmetrical across the *y*-axis, making them even functions. Graphs (B) and (E) are not even, so you can eliminate them.

You can recognize the graph of an odd function because it looks the same when rotated 180°. The only remaining graph that looks just the same when the page is turned upside-down is (C).

**39** **D** Always draw figures that are described but not shown. The diagram described in this question would look like this:

The circle is just window-dressing here. The important thing in this problem is the triangle. You”ve got an isosceles triangle with two sides of length 5 that meet at a 70° angle. Your mission is to find the length *x* of the mystery side opposite the 70° angle. Estimation is the fastest way to choose the right answer. This is almost an equilateral triangle, but one angle is opened wider than 60°, making the other two angles smaller. Side *x*, opposite the big angle, must be bigger than 5 but much smaller than 10. Only (D) gives you such a number. Of course, it”s also possible to find the exact value of *x*. For this, the Law of Cosines is your best tool, *c*^{2} = *a*^{2} + *b*^{2} − 2 *ab* cos *C*. This formula contains the lengths of all three sides of a triangle (*a*, *b*, and *c*) and the measure of one angle *C*. To solve the problem, just plug the two sides and angle that you know into the formula, and solve for the unknown side. Remember that the unknown side in this case is opposite the 70° angle.

*x*^{2} = (5)^{2} + (5)^{2} − 2(5)(5) cos 70°

*x*^{2} = 25 + 25 − 50 cos 70°

*x*^{2} = 50 − 50(0.342)

*x*^{2} = 50 − 17.10

*x*^{2} = 32.90

*x* = ±5.74

All lengths must be positive, so the length of the missing side must be 5.74. The correct answer is (D).

**40** **D** Plug some points from the graph into the equation. For example, the point (0, 1) is on the graph, so plug 0 in for *x* in the answer choices, to see if *f*(0) = 1 in any of them. Cross off (A), (B), and (C), since they don”t work. Now pick another point on the graph. (, ) looks about right. But when you plug in for *x* in answer choice (E), you get 2.414, which can”t be correct. Doing the same thing in (D) gives you 0.414, which looks right, so the correct answer is (D).

**41** **A** This is a vector addition question with a minor twist—the negative sign. Have no fear—just add the components. Simplifying the expression, you have

*z* = −(*v* + *w*) =

−[(−3, 4) + (12, 5)] =

−[(−3 + 12, 4 + 5)] =

−(9, 9) =

(−9, −9)

So the answer is (A).

**42** **A** This question looks frighteningly complicated, but it”s not actually that bad as long as you PITA. Plug each answer choice in for *x*, and use your calculator to see which one spits out 0.33. The correct answer is (A).

**43** **E** You could certainly try out all five answer choices, and solve five sets of simultaneous equations, but it”s easier to think of the graphs of these simultaneous equations. Three things can happen with simultaneous equations. First, the two given equations may be identical (or one is simply a multiple of the other), which means they describe the same line. That results in infinite solutions, since there are infinite points on a line. Second, the lines are parallel, because one equation has the same slope as the other, but a different *y*-intercept. That results in zero solutions, since the lines never intersect. Third, the lines have different slopes, which means they will always intersect in one point. This results in one solution. In this problem, all the answer choices fit the third situation, except (E), which fits the second situation. If you look at the two equations that result

2*x* + 3*y* = 7

10*x* + 15*y* = 3

you can see that the first equation can be multiplied by 5 to get 10*x* + 15*y* = 35. This line is parallel to the line described by the second equation. Algebraically, you can see that 10*x* + 15*y* could be 3 or 35 for a given point (*x*, *y*), but it can”t be both at the same time! The answer is (E).

**44** **C** The statements *f*(*a*, *b*) = 15, *f*(*b*, *c*) = 20, and *f*(*a*, *c*) = 10 can be rewritten as equations by inserting the definition of the function *f*(*x*, *y*).

= 15, = 20, and = 10

These equations can be simplified easily by getting the 3 out of the denominator. *ab* = 45, *bc* = 60, and *ac* = 30

The easiest way to find the product of *a*, *b*, and *c* is to multiply these three equations together, like

*ab* × *bc* × *ac* = 45 × 60 × 30

*aabbcc* = 81,000

*a*^{2}*b*^{2}*c*^{2} = 81,000

*abc* =

*abc* = 284.60

**45** **B** You”ve got variables in the question and variables in the answer choices. That means it”s time to Plug In. Pick numbers that make the calculation easy. In this case, let”s say that *x* = 3 and *y* = 81. The expression log_{x}^{2} *y* then becomes log_{3}^{2} 81, or log_{9} = 81, and is equal to 2—the exponent that turns 9 into 81. Since log_{3}^{2} 81 = 2, the correct answer will be the one that also equals 2 when *x* = 3 and *y* = 81. Quickly reading through Statements I, II, and III, you”ll find the following:

Statement I isn”t true, because log_{3} 81^{2} ≠ 2 (you don”t need to compute the value of log_{3} 81^{2}; it”s obviously not 2, because 3^{2} ≠ 81^{2}). You can therefore eliminate (A) and (D).

Statement II is true, because log_{3} 9 = 2.

Statement III isn”t true, because log_{3}() ≠ 2 (again, you don”t have to compute the exact value of log_{3}() it”s clearly not 2). You can therefore eliminate answer choices (C) and (E). The correct answer is (B).

**46** **A** Use the formula for volume of a sphere with radius *r*, given in the reference information at the beginning of the test, *V* = π*r*^{3}. Plug In the given volumes, and solve for the radii. When *V* = 4188.79, *r* = 10. When *V* = 14137.167, *r* = 15. Now use the given formula for surface area of a sphere with radius *r*, *S* = 4π*r*^{2}. Plugging In the values of *r* that you found, you discover that *S* increases from 1256.637 to 2827.433, an increase of 1570.796. Now divide this increase in surface area by 12 seconds to find the average rate at which the surface area is changing per second. It”s 130.9, so the answer is (A).

**47** **C** This is a complex number, because it has both real and imaginary components. To find its absolute value, visualize the term 6 − 3*i* on the complex plane, where one axis represents real values and the other represents imaginary values.

The point representing 6 − 3*i* would be six steps along the positive real axis and 3 steps along the imaginary negative axis, as shown. The point”s absolute value is its distance from the origin, which is simply the hypotenuse of a right triangle with legs of lengths 3 and 6, respectively. Just plug those values into the Pythagorean theorem, and you”ve got the absolute value.

*h*^{2} = 3^{2} + 6^{2}

*h*^{2} = 45

*h* = 3

**48** **D** This is a combinations question, since rearranging the order of the dishes doesn”t change the dinner. Since the dinner”s being ordered in two parts (the part from column A and the part from column B), you should calculate the number of combinations in two parts. Start with the five dishes from column A. The number of permutations for five items selected from a group of ten is given by 10 × 9 × 8 × 7 × 6 = 30,240. To find the number of combinations, divide that number by 5! = 252. There are 252 possible combinations of five dishes from column A. You”re also selecting five dishes from column B, which has twenty selections. The number of combinations for column B will therefore be given by = 15,504 So there are 15,504 combinations of 5 dishes from column B. To find the total number of possible combinations, multiply these figures together, 15,504 × 252 = 3,907,008. The correct answer is (D).

**49** **A** If *y* varies directly as the square of *x*, that means that will always have the same value. So set up a proportion. = . Cross-multiply and solve. *x*^{2} = 8, so *x* = ±2. The negative value is answer choice (A).

**50** **D** This probability question starts out with a trick. Since it *tells* you what happens in the first two drawings, they don”t even enter into your calculations. Their outcome is certain. The real question starts after that. Given a container that holds 5 blue marbles and 6 red ones, what is the probability that three drawings will produce *at least* two red marbles?

That “at least” is what makes the question difficult. It”s relatively easy to find the probability of just one outcome—but look at all the ways you can get *at least* two red marbles in three tries—RRB, RBR, BRR, RRR. Each one of these is a separate outcome. To compute the total probability of getting at least two red marbles, you need to find the probability of each outcome that does the trick, and then add them up.

Here”s the probability of drawing RRB.

There are 6 red marbles out of a total of 11 on the first drawing, 5 red marbles out of a total of 10 on the second, and 5 blue marbles out of a total of 9 on the third. The total probability of this outcome is . Here”s the probability of drawing RBR.

And here”s the probability of BRR.

As you can see, RRB, RBR, and BRR are all equally probable. Only one other outcome remains, and that”s RRR.

You now know the probability of every outcome that produces at least two red marbles in three drawings. To find the total overall probability, add up all of the individual probabilities.

The correct answer is (D).