Tài liệu Master the Gre 2010 - Part 26 pdf

MULTIPLICATION AND DIVISION Terms under different radicals can be combined under a common radical if one term is multiplied or divided by theother. But if the radicals are different (for example, square root and cube root), combining terms is a bit tricky—as in the last example below. = x = x 5~ = x! 2 or x = x = y 5 = xy = x = y 5 Î x y = 3 x = x 5 x 1/3 x 1/2 5 x 1/3 1 1/2 5 x 5/6 14. ~2 = 2a! 2 5 (A) 4a (B) 4a 2 (C) 8a (D) 8a 2 (E) 6a The correct answer is (C). Square each of the two terms, 2 and = 2a, sepa- rately. Then combine their squares by multiplication: ~2 = 2a! 2 5 2 2 3~ = 2a! 2 5 4 3 2a 5 8a. Roots You Should Know For the GRE, memorize the roots in the following table. If you encounter one of these radical terms on the exam, chances are you’ll need to know its equivalent integer to answer the question. In the table, notice that the cube root of a negative number is negative and the cube root of a positive number is positive. Chapter 10: Math Review: Number Theory and Algebra 233 www.petersons.com Square Roots of “Perfect Square” Integers Cube Roots of “Perfect Cube” Integers (Positive and Negative) = 121 5 11 = 144 5 12 = 169 5 13 = 196 5 14 = 225 5 15 = 625 5 25 = 3 ~2!8 5~2!2 = 3 ~2!27 5~2!3 = 3 ~2!64 5~2!4 = 3 ~2!125 5~2!5 = 3 ~2!216 5~2!6 = 3 ~2!343 5~2!7 = 3 ~2!512 5~2!8 = 3 ~2!729 5~2!9 = 3 ~2!1,000 5~2!10 Roots and the Real Number Line As with exponents, the root of a number can bear a surprising relationship to the magnitude and/or sign (negative vs. positive) of the number. This is another of the test makers’ favorite areas. Here are three rules you should remember: If n . 1, then 1 , = 3 n , = n , n (the greater the root, the lesser the value). However, if n lies between 0 and 1, then n , = n , = 3 n , 1 (the greater the root, the greater the value). n 5 64 1 , = 3 64 , = 64 ,64 1 , 4 , 8 , 64 n 5 1 64 1 64 , Î 1 64 , Î 3 1 64 , 1 1 64 , 1 8 , 1 4 , 1 Every negative number has exactly one cube root, and that root is a negative number. The same holds true for all other odd-numbered roots of negative numbers. = 3 2 27 523 (23)(23)(23) 5227 = 5 232 522 (22)(22)(22)(22)(22) 5232 PART IV: Quantitative Reasoning234 NOTE The square root (or other even-number root) of any negative number is an imaginary number, not a real number. That’s why the preceding rules don’t cover these roots. www.petersons.com Every positive number has only one cube root, and that root is always a positive number. The same holds true for all other odd-numbered roots of positive numbers. 15. Which of the following inequalities, if true, is sufficient alone to show that = 3 x , = 5 x ? (A) 21 , x , 0 (B) x . 1 (C) |x| ,21 (D) |x| . 1 (E) x ,21 The correct answer is (E). If x ,21, then applying a greater root yields a lesser negative value—further to the left on the real number line. LINEAR EQUATIONS WITH ONE VARIABLE Algebraic expressions are usually used to form equations, which set two expressions equal to each other. Most equations you’ll see on the GRE are linear equations, in which the variables don’t have exponents. To solve any linear equation containing one variable, your goal is always the same: Isolate the unknown (variable) on one side of the equation. To accomplish this, you may need to perform one or more of the following four operations on both sides, depending on the equation: Add or subtract the same term from both sides. Multiply or divide by the same term on both sides. Clear fractions by cross-multiplication. Clear radicals by raising both sides to the same power (exponent). Performing any of these operations on both sides does not change the equality; it merely restates the equation in a different form. Let’s take a look at each of these four operations to see when and how to use each one. Add or subtract the same term from both sides of the equation. To solve for x, you may need to either add or subtract a term from both sides of an equation—or do both. 16. If 2x 2 6 5 x 2 9, then x 5 (A) 29 (B) 26 (C) 23 (D) 2 (E) 6 The correct answer is (C). First, put both x-terms on the left side of the equation by subtracting x from both sides. Then combine x-terms: Chapter 10: Math Review: Number Theory and Algebra 235 ALERT! The operation you perform on one side of an equation must also be performed on the other side; otherwise, the two sides won’t be equal. www.petersons.com 2x 2 6 2 x 5 x 2 9 2 x x 2 6 529 Next, isolate x by adding 6 to both sides: x 2 6 1 6 529 1 6 x 523 Multiply or divide both sides of the equation by the same non-zero term. To solve for x, you may need either to multiply or divide a term from both sides of an equation. Or you may need to multiply and divide. 17. If 12 5 11 x 2 3 x , then x 5 (A) 3 11 (B) 1 2 (C) 2 3 (D) 11 12 (E) 11 3 The correct answer is (C). First combine the x-terms: 12 5 11 2 3 x Next, clear the fraction by multiplying both sides by x: 12x 5 11 2 3 12x 5 8 Finally, isolate x by dividing both sides by 12: x 5 8 12 or 2 3 If each side of the equation is a fraction, your best bet is to cross-multiply. Where the original equation equates two fractions, use cross-multiplication to eliminate the fractions. Multiply the numerator from one side of the equation by the denominator from the other side. Set the product equal to the product of the other numerator and denominator. (In effect, cross- multiplication is a shortcut method of multiplying both sides of the equation by both denominators.) PART IV: Quantitative Reasoning236 www.petersons.com 18. If 7a 8 5 a 1 1 3 , then a 5 (A) 8 13 (B) 7 8 (C) 2 (D) 7 3 (E) 15 The correct answer is (A). First, cross-multiply as described above: (3)(7a) 5 (8)(a 1 1) Next, combine terms (distribute 8 to both a and 1): 21a 5 8a 1 8 Then isolate a-terms on one side by subtracting 8a from both sides and combine the a-terms: 21a 2 8a 5 8a 1 8 2 8a 13a 5 8 Finally, isolate a by dividing both sides by its coefficient 13: 13a 13 5 8 13 a 5 8 13 Square both sides of the equation to eliminate radical signs. Where the variable is under a square-root radical sign, remove the radical sign by squaring both sides of the equation. (Use a similar technique for cube roots and other roots.) 19. If 3 = 2x 5 2, then x 5 (A) 1 18 (B) 2 9 (C) 1 3 (D) 5 4 (E) 3 The correct answer is (B). First, clear the radical sign by squaring all terms: ~3 2 !~ = 2x! 2 5 2 2 ~9!~2x!54 18x 5 4 Next, isolate x by dividing both sides by 18: x 5 4 18 or 2 9 Chapter 10: Math Review: Number Theory and Algebra 237 www.petersons.com LINEAR EQUATIONS WITH TWO VARIABLES What we’ve covered up to this point is pretty basic. If you haven’t quite caught on, you should probably stop here and consult a basic algebra workbook for more practice. On the other hand, if you’re with us so far, let’s forge ahead and add another variable. Here’s a simple example: x 1 3 5 y 1 1 Quick . . . what’s the value of x? It depends on the value of y, doesn’t it? Similarly, the value of y depends on the value of x. Without more information about either x or y, you’re stuck—but not completely. You can express x in terms of y, and you can express y in terms of x: x 5 y 2 2 y 5 x 1 2 Let’s look at one more: 4x 2 9 5 3 2 y Solving for x in terms of y: 4x 5 3 2 y 1 9 x 5 3 8 y 1 9 4 Solving for y in terms of x: 4x 2 9 3 2 5 y 2 3 ~4x 2 9!5y 8 3 x 2 6 5 y To determine numerical values of x and y, you need a system of two linear equations with the same two variables. Given this system, you can find the values of x and y by using either the substitution method or the addition-subtraction method. Next, we’ll apply each method to determine the values of two variables in a two- equation system. PART IV: Quantitative Reasoning238 ALERT! You can’t solve one equation if it contains two unknowns (variables). You either need to know the value of one of the variables or you need a second equation. www.petersons.com The Substitution Method To solve a system of two equations using the substitution method, follow these four steps (we’ll use x and y here): In either equation, isolate one variable (x) on one side. Substitute the expression that equals x in place of x in the other equation. Solve that equation for y. Now that you know the value of y,plugitintoeither equation to find the value of x. 20. If 2 5 p 1 q 5 3q 2 10, and if q 5 10 2 p, then p q 5 (A) 5 7 (B) 3 2 (C) 5 3 (D) 25 6 (E) 36 6 The correct answer is (D). First, combine the q-terms in the first equation: 2 5 p 5 2q 2 10 Next, substitute (10 2 p)forq (from the second equation) in the first equation: 2 5 p 5 2~10 2 p!210 2 5 p 5 20 2 2p 2 10 2 5 p 5 10 2 2p Move the p-terms to the same side, then isolate p: 2 5 p 1 2p 5 10 12 5 p 5 10 p 5 S 5 12 D ~10! p 5 25 6 Chapter 10: Math Review: Number Theory and Algebra 239 www.petersons.com The Addition-Subtraction Method Another way to solve for two unknowns in a system of two equations is with the addition-subtraction method. Here are the five steps: Make the coefficient of either variable the same in both equations (you can disregard the sign) by multiplying every term in one of the equations. Make sure the equations list the same variables in the same order. Place one equation above the other. Add the two equations (work down to a sum for each term), or subtract one equation from the other, to eliminate one variable. You can repeat steps 1 to 3 to solve for the other variable. 21. If 3x 1 4y 528, and if x 2 2y 5 1 2 , then x 5 (A) 212 (B) 2 7 5 (C) 1 3 (D) 14 5 (E) 9 The correct answer is (B). To solve for x, you want to eliminate y.You can multiply each term in the second equation by 2, then add the equations: 3x 1 4y 528 2x 2 4y 5 1 5x 1 0y 527 x 5 2 7 5 Since the question asked only for the value of x, stop here. If the question had asked for both x and y (or for y only), you could have multiplied both sides of the second equation by 3, then subtracted the second equation from the first: 3x 1 4y 528 3x 2 6y 5 3 2 0x 1 10y 5 29 1 2 10y 5 2 19 2 y 5 2 19 20 PART IV: Quantitative Reasoning240 NOTE If a question requires you to find values of both unknowns, combine the two methods. For example, after using addition- subtraction to solve for x, substitute the value of x into either equation to find y. www.petersons.com Which Method Should You Use? Which method—substitution or addition-subtraction—you should use depends on what the equations look like to begin with. To see what we mean, look again at this system: 2 5 p 1 q 5 3q 2 10 q 5 10 2 p Notice that the second equation is already set up nicely for the substitution method. You could use addition-subtraction instead; however, you’d just have to rearrange the terms in both the equations first: 2 5 p 2 2q 5210 p 1 q 5 10 Now, look again at the following system: 3x 1 4y 528 x 2 2y 5 1 2 Notice that the x-term and y-term already line up nicely here. Also, notice that it’s easy to match the coefficients of either x or y: multiply both sides of the second equation by either 3 or 2. This system is an ideal candidate for addition-subtraction. To appreciate this point, try using substitution instead. You’ll discover that it takes far more number crunching. SIMPLE ALGEBRAIC FUNCTIONS In a function (or functional relationship), the value of one variable depends upon the value of—or is “a function of”—another variable. In mathematics, the relationship is expressed in the form y = f(x), where y is a function of x. To find the value of the function for any value x, simply substitute the x-value for x wherever it appears in the function. For instance, suppose that f(x)=x 2 +3x + 4. Here’s how you’d find f(2 + a): f~2 1 a!5~2+a! 2 +3~2+a!24 5 4+4a + a 2 +6+3a 2 4 5 a 2 +7a +6 A relatively simple GRE function problem amounts to a single plug-in exercise like the above example. But a more complex function problem might require that you apply the same function twice. Chapter 10: Math Review: Number Theory and Algebra 241 TIP To solve a system of two linear equations with two variables, use addition-subtraction if you can quickly and easily eliminate one of the variables. Otherwise, use substitution. www.petersons.com 22. If f(x)=x + 1, then 1 f~x! 3 f S 1 x D 5 (A) x(x +1) (B) x+1 x (C) x (D) x+1 2 (E) 1 x The correct answer is (E). To determine 1 f~x! , substitute x +1forf(x): 1 f~x! = 1 x +1 To determine f S 1 x D , substitute 1 x for x in the function: f S 1 x D = 1 x +1= 1+x x To solve the problem, multiply as follows: 1 x +1 3 1+x x = 1 x LINEAR EQUATIONS THAT CANNOT BE SOLVED Never assume that one linear equation with one variable is solvable. If you can reduce the equation to 0 5 0, then you can’t solve it. In other words, the value of the variable could be any real number. 23. 3x 2 3 2 4x = x 2 7 2 2x +4 Column A Column B x 0 (A) The quantity in Column A is greater. (B) The quantity in Column B is greater. (C) The quantities are equal. (D) The relationship cannot be determined from the information given. The correct answer is (D). All terms on both sides cancel out: 3x 2 3 2 4x = x 2 7 2 2x +4 2x 2 3=2x 2 3 0=0 ∴x = any real number In some cases, what appears to be a system of two equations with two variables is actually one equation expressed in two different ways. PART IV: Quantitative Reasoning242 www.petersons.com . (the greater the root, the lesser the value). However, if n lies between 0 and 1, then n , = n , = 3 n , 1 (the greater the root, the greater the value). n. eliminate the fractions. Multiply the numerator from one side of the equation by the denominator from the other side. Set the product equal to the product of the