Master gmat 2010 part 27 docx

Except for square roots, the radical sign will indicate the root to be taken.. =8a35=~4!~2!a35 2a=2a 8 and a 3 are both perfect cubes, which contain perfect-square factors; remove the pe

ROOTS AND RADICALS

On the flip side of exponents and powers are roots and radicals The square root of a number

n is a number that you “square” (multiply by itself, or raise to the power of 2) to obtain n.

2 5=4 (the square root of 4) because 2 3 2 (or 22) 5 4

The cube root of a number n is a number that you raise to the power of 3 (multiply by itself

twice) to obtain n You determine greater roots (for example, the “fourth root”) in the same

way Except for square roots, the radical sign will indicate the root to be taken

2 5=3

8 (the cube root of 8) because 2 3 2 3 2 (or 23) 5 8

2 5=4

16 (the fourth root of 16) because 2 3 2 3 2 3 2 (or 24) 5 16

For the GMAT, you should know the rules for simplifying and for combining radical

expressions

Simplifying Radicals

On the GMAT, always look for the possibility of simplifying radicals by moving what’s under

the radical sign to the outside of the sign Check inside your square-root radicals for perfect

squares: factors that are squares of nice tidy numbers or other terms The same advice applies

to perfect cubes, and so on

from under the radical sign, and find each one’s square root

=8a35=~4!~2!a35 2a=2a 8 and a

3 are both perfect cubes, which contain perfect-square factors; remove the perfect squares from under the radical sign, and find each one’s square root

You can simplify radical expressions containing fractions in the same way Just be sure that

what’s in the denominator under the radical sign stays in the denominator when you remove

it from under the radical sign

Î20x

x3 5Î~4!~5!

x2 5

2=5

x

Î3 3

85Î3 3

235

1

2=3

3

TIP

Whenever you see a non-prime number under a square-root radical sign, factor it to see whether it contains perfect-square factors you can move outside the radical sign More than likely, you need to do so to solve the problem

at hand.

10. Î28a b

36a4b65

(A) a

bÎa

2b

2bÎa b

(C) |a|

3|b|=7

2

3b2=2

(E) 2a

3b

The correct answer is (C) Divide a4 and b4 from the numerator and denominator of the fraction Also, factor out 4 from 28 and 36 Then, remove perfect squares from under the radical sign:

Î28a6b4

36a4b65Î7a2

9b25

|a|=7

3|b| , or

|a|

3|b|=7

In GMAT questions involving radical terms, you might want to remove a radical term from a fraction’s denominator to match the correct answer To accomplish this, multiply both numerator and denominator by the radical value (This process is called “rationalizing the denominator.”) Here’s an example of how to do it:

3

3=15

3=15

15 or

1

5=15

Combining Radical Terms

The rules for combining terms that include radicals are quite similar to those for exponents Keep the following two rules in mind; one applies to addition and subtraction, while the other applies to multiplication and division

Addition and subtraction: If a term under a radical is being added to or subtracted from a

term under a different radical, you cannot combine the two terms under the same radical

=x 1=y Þ=x 1 y

=x 2=y Þ=x 2 y

=x 1=x 5 2=x, not=2x

On the GMAT, if you’re asked to combine radical terms by adding or subtracting, chances are

11. =24 2=16 2=6 5

(A) =6 2 4

(B) 4 2 2=2

(C) 2

(D) =6

E. 2=2

The correct answer is (A) Although the numbers under the three radicals combine to equal

2, you cannot combine terms this way Instead, simplify the first two terms, then combine the

first and third terms:

=24 2=16 2=6 5 2=6 2 4 2=6 5=6 2 4

Multiplication and Division: Terms under different radicals can be combined under a common

radical if one term is multiplied or divided by the other, but only if the radical is the same

=x=x 5~=x!2, or x

y

x=x 5 ?

(you cannot easily combine =3

x=x 5 x

1

3x

1

25 x

1

31 1

25 x

5 6

12 ~2=2a!2

5

(A) 4a

(B) 4a2

(C) 8a

(D) 8a2

(E) 6a

The correct answer is (C) Square each of the two terms, 2 and =2a, separately Then

combine their squares by multiplication: ~2=2a!25 223 ~=2a!25 4 3 2a 5 8a.

Roots You Should Know

For the GMAT, 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 on the following page, notice that the cube root of a negative number is negative

and the cube root of a positive number is positive

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

~2!8 5 ~2!2

~2!27 5 ~2!3

~2!64 5 ~2!4

~2!125 5 ~2!5

~2!216 5 ~2!6

~2!343 5 ~2!7

~2!512 5 ~2!8

~2!729 5 ~2!9

~2!1000 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 (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,Î1

64,Î3 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

2 27 5 23 (23)(23)(23) 5 227

232 5 22 (22)(22)(22)(22)(22) 5 232

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

13 Which of the following inequalities, if true, is sufficient alone to show that

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 GMAT are linear equations, in which the

variables don’t come with 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

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.

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.

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

14 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:

2x 2 6 2 x 5 x 2 9 2 x

x 2 6 5 29

Next, isolate x by adding 6 to both sides:

x 2 6 1 6 5 29 1 6

x 5 23

Multiply or divide both sides of the equation by the same non-zero term

To solve for x, you may need to either multiply or divide a term from both sides of

an equation Or, you may need to multiply and divide

15 If 12 511

x 2

3

x , then x 5

11

(B) 1

2

(C) 2

3

(D) 11

12

(E) 11

3

The correct answer is (C) First, combine the x-terms:

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.)

16 If,7a

a 1 1

3 , then a 5

13

(B) 7

8

(C) 2

(D) 7

3

(E) 15

The correct answer is (A) First, cross-multiply as we’ve described:

(3)(7a) 5 (8)(a 1 1)

Next, combine terms (distribute 8 to both a and 1):

21a 5 8a 1 8

Next, isolate a-terms on one side by subtracting 8a from both sides; then 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

8

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 equa-tion (Use a similar technique for cube roots and other roots.)

17 If 3=2x 5 2, then x 5

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:

~32!~=2x!25 22

~9!~2x! 5 4 18x 5 4 Next, isolate x by dividing both sides by 18:

x 5 4

18, or

2 9

LINEAR EQUATIONS WITH TWO VARIABLES

What we’ve covered up to this point is pretty basic stuff 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

Solve for x in terms of y:

4x 53

2y 1 9

x 53

8y 1

9 4

Solve for y in terms of x:

4x 2 9

3

2

5 y

2

3~4x 2 9! 5 y

8

3x 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, there are two different methods for finding the values

of the two variables:

The substitution method

The addition-subtraction method

Next, we’ll apply each method to determine the values of two variables in a two-equation

system

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, plug it into either equation to find the value of x.

ALERT!

You can’t solve

an 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.

18 If2

5p 1 q 5 3 q 2 10, and if q 5 10 2 p, then

p

q5

(A) 5

7

(B) 3

2

(C) 5

3

(D) 25

6

(E) 36

6

The correct answer is (A) Don’t let the fact that the question asks for p

q (rather than

simply p or q) throw you Because you’re given two linear equations with two unknowns, you know that you can first solve for p and q, then divide p by q First things first: Combine the

q-terms in the first equation:

2

5p 5 2 q 2 10

Next, substitute (10 2 p) for q (from the second equation) in the first equation:

2

5p 5 2~10 2 p! 2 10 2

5p 5 20 2 2 p 2 10 2

5p 5 10 2 2 p

Move the p-terms to the same side, then isolate p:

2

5p 1 2 p 5 10 12

5 p 5 10

p 5S5

12D~10!

25