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Ratio & Proportion 301

26. Since the circle with center O and the circle with center A touch each other internally, the distance between their centers equals

(Radius of larger circle) – (Radius of circle with center A) = Column A

Similarly, since the circle with center O and the circle with center B touch each other internally, the distance between their centers equals

(Radius of larger circle) – (Radius of circle with center B) = Column B

Since the ratio of the radii of the circles with center A and center B is given to be 7 : 9 and since 7 < 9, (Radius of circle with center A) < (Radius of circle with center B). Multiplying the inequality by –1 and flipping direction yields

–(Radius of circle with center A) > –(Radius of circle with center B)

Adding “Radius of larger circle” to both sides yields

(Radius of larger circle) – (Radius of circle with center A) > (Radius of larger circle) – (Radius of circle with center B)

Column A > Column B.
The answer is (A).
27. Forming the ratio yields		a + 6	=	5	. Multiplying both sides of the equation by 6(b + 6) yields
27. Forming the ratio yields		b + 6	=	6	. Multiplying both sides of the equation by 6(b + 6) yields
		b + 6		6
6(a + 6) = 5(b + 6)
6a + 36 = 5b + 30
6a = 5b – 6
a = 5b/6 – 1
0	< 5b/6 – 1				since a is positive
1	< 5b/6
6/5 < b
1.2 < b
1	< 1.2 < b				since 1 < 1.2

Column B < 1.2 < Column A

Hence, the answer is (A).

28. Let the income of A and B be 3s and 2s, respectively, and let their expenditures be 4t and 3t. Then since savings is defined as income minus expenditure, Column A = the saving of A = 3s – 4t, and Column B = saving of B = 2s – 3t.

Column A is greater than Column B when 3s – 4t > 2s – 3t, or s > t.

Since money is positive, s and t are positive. Since income is greater than expenditure (given), the Income of B = 2s > Expenditure of B = 3t. Hence, s > 3 t/2. Clearly, s > t. Since we know that Column A > Column B when s > t, the answer is (A).

basic element.

Then the total compositions of the two basic elements in the 7x ounces of alloy X would contain 5x/2 ounces (from A) + x ounces (from B) = 7x/2 ounces of first basic element, and 3/2 x (from A) + 2x (from B) = 7x/2 ounces of the second basic element. Hence, the composition of the two basic elements in alloy X is 7x/2 : 7x/2 = 1 : 1. The answer is (A).

30. Let 1 : k be the ratio in which Joseph mixed the two types of rice. Then a sample of (1 + k) ounces of the mixture should equal 1 ounce of rice of the first type, and k ounces of rice of the second type. The rice of the first type costs 5 cents an ounce and that of the second type costs 6 cents an ounce. Hence, it cost him

(1 ounce 5 cents per ounce) + (k ounces 6 cents per ounce) = 5 + 6k

Since he sold the mixture at 7 cents per ounce, he must have sold the net 1 + k ounces of the mixture at 7(1 + k).

Since he earned 20% profit doing this, 7(1 + k) must be 20% more than 5 + 6k. Hence, we have the equation

7(1 + k) = (1 + 20/100)(5 + 6k) 7 + 7k = (120/100)(5 + 6k)

7 + 7k = (6/5)(5 + 6k)

7 + 7k = 6/5 5 + 6/5 6k 7 + 7k = 6 + 36k/5

1 = k/5 k = 5

Hence, the required ratio is 1 : k = 1 : 5. The answer is (B).

31. We have that the incomes of A, B, and C are in the ratio 1 : 2 : 3. Let their incomes be i, 2i, and 3i, respectively. Also, their expenses ratio is 3 : 2 : 1. Hence, let their expenses be 3e, 2e, and e. Since the Saving = Income – Expenditure, the savings of the three employees A, B, and C is i – 3e, 2i – 2e, and 3i – e, respectively.

Now, the saving of C is greater the saving of B when 3i – e > 2i – 2e, or i + e > 0 which surely is correct, since the income and expenditure, i and e, are both money and therefore positive.

Now, the saving of B is greater the saving of A when 2i – 2e > i – 3e, or i + e > 0 which is surely correct, since the income and the expenditure, i and e, are both money and therefore positive.

Hence, the employees A, B, and C in the order of their savings is C > B > A. The answer is (E).

302 GRE Math Bible

Very Hard

29. The new alloy X is formed from the two alloys A and B in the ratio 4 : 3. Hence, 7 parts of the alloy contains 4 parts of alloy A and 3 parts of alloy B. Let 7x ounces of alloy X contain 4x ounces of alloy A and 3x ounces of alloy B.

Now, alloy A is formed of the two basic elements mentioned in the ratio 5 : 3. Hence, 4x ounces of alloy A

contains	5	4 x =	5x	ounces of first basic element and	3	4 x =	3x	ounces of the second basic
	5+ 3		2		5+ 3		2

element.

Also, alloy B is formed of the two basic elements mentioned in the ratio 1 : 2. Hence, let the 3x ounces of

alloy A contain	1	3x = x ounces of the first basic element and	2	3x = 2x ounces of the second
	1+ 2		1+ 2

Ratio & Proportion 303

32. We have that the volume of the cask is 15 liters.

Emptying + Filling up (first time):

The volume of alcohol initially available is 15 liters. When 5 liters are removed, the cask has 10 liters of alcohol. When filled back to the brim (by 5 liters of water), the composition of the solution now becomes 10 liters of alcohol and 5 liters of water.

* The cask now has 10 liters of alcohol and 5 liters of water.

Now, the ratio of alcohol to water is 10 : 5 = 2 : 1.
Emptying (second time):	2		10
The next time 5 liters are removed, the removed solution is		5 =		liters of alcohol and 5/3 liters of
	2 +1		3

water. Hence, the remaining solution is 10 – 10/3 = 20/3 liters of alcohol and 5 – 5/3 = 10/3 liters of water.

* The cask now has 20/3 liters of alcohol and 10/3 liters of water.

Filling up (second time):

When the cask is filled the second time (with water), the solution is now 20/3 liters of alcohol (already existing) and 10/3 + 5 = 25/3 liters of water.

* The cask now has 20/3 liters of alcohol and 25/3 liters of water.

Hence, the solution now has alcohol and water in the ratio 203 : 253 = 4 : 5. The answer is (D).

Exponents & Roots

EXPONENTS

Exponents afford a convenient way of expressing long products of the same number. The expression bn is called a power and it stands for b b b L b , where there are n factors of b. b is called the base, and n

is called the exponent. By definition, b0 = 1.*

There are six rules that govern the behavior of exponents:

Rule 1:

Rule 2:

Rule 3:

Rule 4:

Rule 5:

Rule 6:

xa xb = xa +b

(xa )b					= xab
(xy )a				= x a ya
x a				=		xa

						ya
		y
	xa		= xa b , if a > b.
	xb

	xa		=	1			, if b > a.
	xb			xb a
	x a =				1
					xa

Example, 23 22 = 23+2 = 25 = 32 . Caution, xa + xb x a+ b

Example, (23 )2				= 23 2 = 26 = 64
Example, (2y)3 = 23 y3 = 8y3
Example,		x 2				x2				x2
				=				=
		3				32			9

Example,	2	6	= 26 3				= 23 = 8
	2	3

Example,	2	3	=	1			=		1		=	1
	2	6		26 3					23			8
Example,	z 3 =			1			Caution, a negative exponent does not make
					z 3

the number negative; it merely indicates that the base should be reciprocated. For example, 3 2 312 or 19 .

Problems involving these six rules are common on the GRE, and they are often listed as hard problems. However, the process of solving these problems is quite mechanical: simply apply the six rules until they can no longer be applied.

Example 1: If x 0,

(x5 )2

x 4

(A) x 5

(B) x 6

(D) x 8

(E) x 9

First, apply the rule (x a )b

= x ab to the expression

x(x5 )2

x x5 2

x x10

Next, apply the rule xa xb = xa +b :

x 4

x + 5

* Any term raised to the zero power equals 1, no matter how complex the term is. For example,

= 1.

304

= x a ya :

Column A

Exponents & Roots 305

				x x10				x11
							=
					x4		=	x4
Finally, apply the rule	xa	= xa b :
Finally, apply the rule	xb	= xa b :
	xb
			x11		= x	11 4		= x7
			x4		= x	11 4		= x7
			x4

The answer is (C).

Note: Typically, there are many ways of solving these types of problems. For this example, we could have

xa 1 begun with Rule 5, xb = xb a :

x(x5 )2

(x5 )2

x4 1

Then apply Rule 2, (x a )b

= x ab :

(x 5)2

x10

x 3

Finally, apply the other version of Rule 5,

= xa b

x10

= x7

x 3

Example 2:

Column A

Column B

3 3 3 3

1 4

9 9 9 9

1 1 1 1

Canceling the common factor 3 in Column A yields

, or

3 . Now, by the definition of a

3 3 3 3

power,

Hence, the columns are equal and the answer is (C).

Example 3:

First, factor Column A:

Next, apply the rule (xy )a

Finally, apply the rule xa = xa b : xb

Hence, the columns are equal and the answer is (C).

Column B

2 4 3 2

(2 3)4

24 34

24 32

306GRE Math Bible

ROOTS

The symbol

is read the nth root of b, where n is called the index, b is called the base, and

is called

the radical.

denotes that number which raised to the nth power yields b. In other words,

a is the nth

root of b if an = b . For example,

= 3* because 32 = 9 , and 3

= 2 because

(2)3 = 8 . Even

roots occur in pairs: both a positive root and a negative root. For example, 4

= 2 since

24 = 16 , and

= 2 since (2)4 = 16 . Odd roots occur alone and have the same sign as the base:

= 3 since

(3)3 = 27 . If given an even root, you are to assume it is the positive root. However, if you introduce even roots by solving an equation, then you must consider both the positive and negative roots:

x2 = 9

x 2 = ±9 x = ±3

Square roots and cube roots can be simplified by removing perfect squares and perfect cubes, respectively. For example,

8 = 4 2 = 42 = 22 354 = 327 2 = 327 32 = 332

Radicals are often written with fractional exponents. The expression nb can be written as b1 n . This can be generalized as follows:

bmn = (nb)m = nbm

Usually, the form (nb)m is better when calculating because the part under the radical is smaller in this

		)2
case. For example, 272 3 = (3			= 32 = 9 . Using the form n	bm	would be much harder in this case:
	27

272 3 = 3272 = 3729 = 9 . Most students know the value of 327 , but few know the value of 3729 . If n is even, then

nx n = x

For example, 4( 2)4 = 2 = 2. With odd roots, the absolute value symbol is not needed. For example, 3( 2)3 = 38 = 2.

To solve radical equations, just apply the rules of exponents to undo the radicals. For example, to solve the radical equation x2 3 = 4 , we cube both sides to eliminate the cube root:

(x 2 3)3 = 43

x2 = 64 x 2 = 64 x = 8

x = ±8

Even roots of negative numbers do not appear on the GRE. For example, you will not see expressions of the form 4 ; expressions of this type are called complex numbers.

* With square roots, the index is not written, 29 = 9 .

Exponents & Roots 307

The following rules are useful for manipulating roots:

= n

For example,

3x =

For example,

= n y

8 =

3 8

2 .

Caution: n

+ n

. For example,

. Also,

x 2 + y 2 x + y . This common

x + y

x + 5

mistake occurs because it is similar to the following valid property: (x + y)2 = x + y (If x + y can be negative, then it must be written with the absolute value symbol: x + y ). Note, in the valid formula, it’s the whole term, x + y, that is squared, not the individual x and y.

To add two roots, both the index and the base must be the same. For example, 32 + 42 cannot be added because the indices are different, nor can 2 +3 be added because the bases are different. However,

3 2 + 3 2 = 23 2 . In this case, the roots can be added because both the indices and bases are the same. Sometimes radicals with different bases can actually be added once they have been simplified to look alike.

For example, 28 + 7 = 4 7 + 7 = 4	7	+ 7 = 2 7 +	7 = 3 7 .

You need to know the approximations of the following roots:			2 1.4
You need to know the approximations of the following roots:			2 1.4	3 1.7	5 2.2

x 2 = 4

Example 4: Given the system y3 = 8 , which of the following is NOT necessarily true?

(A) y < 0	(B) x < 5	(C)	y is an integer	(D) x > y	(E)	x	is an integer
						y

y3 = 8 yields one cube root,	y = –2. However,		x2 = 4 yields two square roots,		x = ±2. Now, if x = 2,

then x > y; but if x = –2, then x = y. Hence, choice (D) is not necessarily true. The answer is (D).

Example 5: If x < 0 and y is 5 more than the square of x, which one of the following expresses x in terms of y?

(A) x = y 5 (B) x = y 5 (C) x = y + 5 (D) x = y 2 5 (E) x = y 2 5 Translating the expression “y is 5 more than the square of x” into an equation yields:

y = x 2 + 5

y 5 = x2

±y 5 = x

Since we are given that x < 0, we take the negative root, y 5 = x . The answer is (B).

308GRE Math Bible

RATIONALIZING

A fraction is not considered simplified until all the radicals have been removed from the denominator. If a denominator contains a single term with a square root, it can be rationalized by multiplying both the numerator and denominator by that square root. If the denominator contains square roots separated by a plus or minus sign, then multiply both the numerator and denominator by the conjugate, which is formed by merely changing the sign between the roots.

Example: Rationalize the fraction 325 .

																				2
Multiply top and bottom of the fraction by														:													5			2	5				2 5	= 2 5
												5															5		=	2	5			=	2 5
												5																	=					=	3 5
																				3			5			5				3	25				3 5	15
Example:									2
	Rationalize the fraction												.
										3
										3	5
Multiply top and bottom of the fraction by the conjugate 3+																						:
Multiply top and bottom of the fraction by the conjugate 3+																					5	:
									2(3+							)				2(3+					)			2(3+				)
															5									5							5
	2			3+ 5			=								5				=					5		=					5		=		3+ 5

																		2		9			5							4					2

	3 5 3+				5	3		2	+ 3 5 3 5 (								5)
	3 5 3+				5			2

Problem Set R:

Easy
1.		Column A																																	Column B
		x(x2)4																																			(x3)3
Medium
2.		Column A															0 < x < y																		Column B
	1		1																																		7x – y

7x y

3.If x = 101.4, y = 100.7, and xz = y3, then what is the value of z ?

(A)0.5

(B)0.66

(D)2

(E)3

4.A perfect square is a positive integer which when square rooted results in an integer. If N = 34 53 7, then what is the biggest perfect square that is a factor of N ?

(A)32

(B)52

(D)(9 5)2

(E)(3 5 7)2


5.	If p =	3 2	, then which one of the following equals p – 4?

		2 +1
		2 +1

(A)3 – 2

(B)3 + 2

(D)22 + 6 – 3 – 2

(E)22 + 6 – 3 + 2

Exponents & Roots 309

6.		Column A	Column B

	12.5 + 12.5		25
Hard

7. If p = 216–1/3 + 243–2/5 + 256–1/4, then which one of the following is an integer?

(A) p/19

(B) p/36

(D) 19/p

(E) 36/p

4(	6	+		2	)
4(	6	+		2	)	2 +	3
8.								=
8.								=
6			2			2	3

(A)1

(B)6 2

(D)8

(E)12

9.If m27 = 33m and 4m > 1, then what is the value of m ?

(A)–1

(B)–1/4

(D)1/4

(E)1

10.	Column A					Column B
					x 2			7x
					x 2			7x
	(	7)
	(	7)					11
						7
	(			)11
	(		7	)11
11.	Column A					Column B

	3 +				2	5 +			2


	3				2	5			2

Very Hard

12.In which one of the following choices must p be greater than q?

(A)0.9p = 0.9q

(B)0.9p = 0.92q

(D)9p < 9q

(E)9p > 9q

310 GRE Math Bible

Answers and Solutions to Problem Set R

Easy

1. Column A: x(x2)4 = x x2 4 = x x8 = x1 + 8 = x9.

Column B: (x3)3 = x3 3 = x9.

The answer is (C).

Medium

2. From the inequality 0 < x < y, x and y are positive and x < y. Hence, the sides of the inequality x < y can be inverted and the direction safely flipped. Hence, 1/x > 1/y.

Subtracting y from both sides of the inequality x < y yields x – y < 0.

Finally, subtracting 1/y from both sides of the inequality 1/x > 1/y yields 1/x – 1/y > 0.

Thus, Column A has 7 raised to a positive number, while Column B has 7 raised to a negative number. Hence, Column A is greater than 1, and Column B is less than 1. The answer is (A).

3. We are given that x = 101.4 and y = 100.7. Substituting these values in the given equation xz = y3 yields

(101.4)z = (100.7)3

101.4z = 100.7 • 3

101.4z = 102.1

1.4z = 2.1 since the bases are the same, the exponents must be equal z = 2.1/1.4 = 3/2

The answer is (C).

4. We are given that N = 34 53 7. Writing N as a product of perfect squares yields (34 52) 5 7 = (32 5)2 5 7 = (9 5)2 5 7. The product of the remaining numbers 5 and 7 cannot be written as a perfect square. Hence, (9 5)2 is the biggest perfect square factor of N. The answer is (D).

5. Since none of the answers are fractions, let’s rationalize the given fraction by multiplying top and bottom by the conjugate of the bottom of the fraction: