Sat - MC Grawhill part 28 pot

When you solve an algebraic equation, you use logical laws of equality such as the addition law of equality to prove the equation you want.. To prove that a state-ment doesn’t have to be

b – 1, is 6 The only positive integer pairs with a

prod-uct of 6 are 2 × 3 and 1 × 6, so one possibility is that

a = 2 and b = 4 This gives a(b – 1) = 2(4 – 1) = 2(3) = 6,

and it satisfies the condition that a < b Now check the

statements Statement I is true here because 4/2 = 2, which is an integer Statement II is also true here because 4 is an even number Statement III is also true because 2 × 4 = 8, which is 6 greater than 2 So the answer is (E) I, II and III, right?

Wrong Remember that the question asks what

must be true, not just what can be true We’ve only

shown that the statements can be true We can prove that statement I must be true by testing all the

possi-ble cases Since there is only one other possipossi-ble

solu-tion that satisfies the condisolu-tions: a = 1 and b = 7, and

since 7/1 = 7 is an integer, we can say with confidence

that statement I must be true But statement II

does-n’t have to be true because b can equal 7, which is not

even We have found a counterexample Next, we can prove that statement III must be true by checking both cases: 2 × 4 is 6 greater than 2, and 1 × 7 is 6 greater

than 1 (We can prove it algebraically too! If we add a

to both sides of the original equation, we get ab = a + 6, which proves that ab is 6 greater than a.)

Process of Elimination (POE)

On multiple-choice questions (and especially

“must be true” questions), it helps to cross off wrong answers right away Sometimes POE simplifies the problem dramatically

What if, in the preceding question, the first

solu-tion we found was a = 1 and b = 7 For this solusolu-tion,

statements I and III are true, but statement II is not Therefore, we could eliminate those choices contain-ing II—(B), (D), and (E) Since the two remaincontain-ing choices contain statement I, it must be true—we don’t even need to prove it!

Numerical and Algebraic Proof

Logical proofs aren’t just for geometry class

They apply to arithmetic and algebra, too In

arithmetic, you often need to apply the laws of

arithmetic (such as odd × even = even, negative

÷ positive = negative—see Chapter 9, Lesson 3)

to prove what you’re looking for When you

solve an algebraic equation, you use logical

laws of equality (such as the addition law of

equality) to prove the equation you want

“Must Be True” Questions

Logic is especially useful in solving SAT “must

be true” questions You know them and hate

them—they usually have those roman

numer-als I, II, and III To prove that a statement

“must be true,” apply the laws of equality or

the laws of arithmetic To prove that a

state-ment doesn’t have to be true, just find one

counterexample, a valid example for which the

statement is false

If a and b are positive integers such that a < b and

ab – a = 6, which of the following must be true?

I is an integer

II b is an even number.

III ab is 6 greater than a.

(A) I only

(B) I and II only

(C) I and III only

(D) II and III only

(E) I, II, and III

This requires both numerical and algebraic logic

First, let’s see how far we can get trying to solve the

equation for a and b.

ab – a = 6

Factor out the a: a(b – 1) = 6

Okay, we’ve got a problem We have two

un-knowns but only one equation, which means we can’t

solve it uniquely Fortunately, we know that a and b

must be positive integers, so the equation basically

says that the product of two positive integers, a and

b

a

Concept Review 7: Thinking Logically

1 What is a proof, and why is understanding proofs helpful on the SAT?

2 How can POE help on the SAT?

3 What is the difference between geometric, algebraic, and numerical proofs?

4 Name two geometric theorems that are useful on the SAT.

5 Name two algebraic theorems that are useful on the SAT.

6 Name two numerical theorems that are useful on the SAT.

4. If (m + 1)(n + 1) = 1, which of the following can

be true?

I m and n are both positive.

II m and n are both negative.

III m is positive and n is negative.

(A) II only (B) III only (C) I and II only (D) I and III only (E) II and III only

5. If p is a prime number greater than 5 and q is

an odd number greater than 5, which of the fol-lowing must be true?

I p + q is not a prime number.

II pq has at least three positive integer

factors greater than 1

III is not an integer.

(A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, and III

q p

SAT Practice 7: Thinking Logically

Use logical methods to solve each of the following

SAT questions

1. If a > b and b(b – a) > 0, which of the following

must be true?

I b < 0

II a < 0

III ab < 0

(A) I only

(B) II only

(C) I and II only

(D) I and III only

(E) I, II, and III

A, B, C, and D are the consecutive vertices of a

quadrilateral

∠ABC and ∠BCD are right angles.

2. If the two statements above are true, then which

of the following also must be true?

(A) ABCD is a rectangle.

(B) is parallel to .

(C) is parallel to .

(D) Triangle ACD is a right triangle.

(E) Triangle ABD is a right triangle.

3. The statement a ⇔ b is defined to be true if and

only if Which of the following is true?

(A) 3 ⇔ 5

(B) 5 ⇔ 3

(C) 4 ⇔ 2

(D) 6 ⇔ 4

(E) 7 ⇔ 5

a b

5>3

AD BC

DC AB

SAT Practice 7

1 A Since a is greater than b, b – a must be a negative

number Since b(b – a) must be positive, but b – a is

negative, b also must be negative because negative×

negative = positive, but positive × negative = negative.

This proves that statement I must be true However,

statement II does not have to be true because a

coun-terexample is a = 1 and b = −1 Notice that this

satis-fies the conditions that a > b and that b(b – a) > 0.

Statement III also isn’t necessarily true because a

counterexample is a = −1 and b = −2 Notice that this

also satisfies the conditions that a > b and b(b – a) > 0

but contradicts the statement that ab < 0.

2 B First draw a diagram that illustrates the given

conditions, such as the one above This diagram

shows that the only true statement among the

choices is (B) This fact follows from the fact that

“if a line (BC), crosses two other lines (AB and DC)

in a plane so that same-side interior angles are

supplementary, then the two lines are parallel.”

3 C First, translate each choice according to the

definition of the bizarre new symbol This gives us

(A) 3/5 > 5/3, (B) 5/5 > 3/3, (C) 4/5 > 2/3, (D) 6/5 > 4/3,

and (E) 7/5 > 5/3 The only true statement among

these is (C)

4 E The question asks whether the statements can

be true, not whether they must be true The

equa-tion says that two numbers have a product of 1

You might remember that such numbers are

reci-procals, so we want to find values such that m + 1

and n + 1 are reciprocals of each other One pair

of reciprocals is 2 and 1/2, which we can get if m = 1 and n = −1/2 Therefore, statement III can be true, and we can eliminate choices (A) and (C) Next, think of negative reciprocals, such as −2 and −1/2,

which we can get if m = −3 and n = −1/2 Therefore, statement II can be true, and we can eliminate choices (B) and (D), leaving only (E), the correct

answer Statement I can’t be true because if m and

n are both positive, then both m + 1 and n + 1 are

greater than 1 But, if a number is greater than 1,

its reciprocal must be less than 1.

5 A You might start by just choosing values for

p and q that satisfy the conditions, such as p = 7

and q = 9 When you plug these values in, all three

statements are true Bummer, because this nei-ther proves any statement true nor proves any

statement false Are there any interesting possible values for p and q that might disprove one or more

of the statements? Notice that nothing says that

p and q must be different, so choose p = 7 and q = 7.

Now pq = 49, which only has 1, 7, and 49 as fac-tors Therefore, it does not have at least three

pos-itive integer factors greater than 1, and statement

II is not necessarily true Also, q/p = 1, which is an

integer, so statement III is not necessarily true So

we can eliminate any choices with II or III, leav-ing only choice (A)

Concept Review 7

1 A proof is a sequence of logical statements that

begins with a set of assumptions and proceeds to

a desired conclusion You construct a logical proof

every time you solve an equation or determine a

geometric or arithmetic fact

2 The process of elimination (POE) is the process of

eliminating wrong answers Sometimes it is easier

to show that one choice is wrong than it is to show

that another is right, so POE may provide a

quicker path to the right answer

3 Geometric proofs depend on geometric facts such as

“angles in a triangle have a sum of 180°,” algebraic

proofs use laws of equality such as “any number can

be added to both sides of an equation,” and

numeri-cal proofs use facts such as “an odd number plus an

odd number always equals an even number.”

4 The most important geometric theorems for the SAT are given in Chapter 10 They include parallel lines theorems such as “if two parallel lines are cut by

a transversal, then alternate interior angles are congruent” and triangle theorems such as “if two sides of a triangle are congruent, then the angles opposite those sides are also congruent.”

5 The most important algebraic theorems are the laws of equality, such as “you can subtract any number from both sides of an equation.”

6 The most important numerical theorems are dis-cussed in Chapter 9, Lesson 3, and Chapter 7,

Les-son 7 They include “odd × odd = odd” and “positive

× negative = negative.”

Answer Key 7: Thinking Logically

A

D

Check the Question

Always quickly reread the question before

marking your answer to make sure that you’ve

answered the right question and to make sure

that your solution makes sense in the context

of the question

A bin contains 20 basketballs and soccer balls If there

are 8 more basketballs than soccer balls in the bin,

how many soccer balls are in the bin?

(A) 4

(B) 6

(C) 8

(D) 10

(E) 12

Many students think that since there are 8 more

basketballs than soccer balls, they should just subtract

8 from the total to get the number of soccer balls,

get-ting 20 – 8 = 12 soccer balls The answer is (E), right?

Wrong If there are 12 soccer balls, then there

must be 8 basketballs, for a total of 20 But the

ques-tion says that there are 8 more basketballs than

soc-cer balls, and 8 sure isn’t more than 12! So now what?

Eliminate choice (E) first of all Since there are fewer

basketballs than soccer balls, soccer balls must make

up fewer than half the balls, so there must be fewer

than 10 soccer balls, eliminating choice (D) Checking

the remaining choices shows (B) 6 works because if

there are 6 soccer balls, there are 14 basketballs, and

14 is 8 greater than 6!

The “direct” method for solving is to subtract 8

from 20 and then divide the result by 2 to get the

num-ber of soccer balls

Check Your Algebra

When solving an equation, check two things:

first your answer, and then your steps If the

answer works when you plug it back into the

equation, there’s no need to check the steps If

it doesn’t, then check your steps When solving

equations, write out every step, and make sure

that each one is logical You’re likely to make

mistakes if you skip steps or do them too

quickly in your head

If , then what is the value of x?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4 You might notice that since both fractions have the same numerator, the equation can be simplified and solved without much trouble

Multiply by x – 2: 2x = x2

Divide by x: 2 = x

Piece of cake The answer is (C), right? Wrong Notice

that substituting x = 2 into the original equation gives you

4/0 = 4/0 Although this seems true at first, it’s not, be-cause 4/0 isn’t a number! (Just ask your calculator.) What now? The check suggests a solution: you can just test the

choices Plugging in the other choices for x shows that

only (A) 0 produces a true equation: 0/–2 = 0/–2

What went wrong the first time? Check the steps

Our second step was to divide by x We’re allowed to divide both sides of an equation by any number except

0 because division by 0 is undefined That’s where we

went wrong: We didn’t check division by 0 Notice that division by 0 also explains why x = 2 doesn’t work in

the original equation

Check by Estimating

Estimation is one of the simplest and most ef-fective checking strategies you can use on the

SAT Getting an approximate answer can help

you to narrow down the choices quickly

If Carla drives at 40 miles per hour for n miles and then drives at 60 miles per hour for another n miles, what is

her average speed, in miles per hour, for the entire trip? (A) 42

(B) 48 (C) 50 (D) 52 (E) 54 Many students average 40 and 60 and get 50 But

this is wrong because Carla is not spending equal times

at 40 and 60 miles an hour Since 40 mph is slower than

60 mph, she spends more time at that speed So her

av-erage speed is closer to 40 than 60 This eliminates choices (C), (D), and (E) The correct answer is (B) (For more on rate problems, see Chapter 9, Lesson 4.)

2

2

x x

x x

2

2

x x

x x

Concept Review 8: Checking Your Work

1 What are the best strategies for avoiding mistakes when solving an equation?

2 When should you estimate on an SAT math question?

3 Why should you estimate on certain SAT math questions?

4 What is the last thing to check before marking an answer to an SAT math question?

5 What steps “aren’t allowed” when solving equations?

4. If the 30 students in Ms Harkin’s class scored an average of 80% on their final exams, and if the 20 students in Ms Johnson’s class scored an aver-age of 70% on their final exams, what was the av-erage score for the two classes combined? (A) 74%

(B) 75%

(C) 76%

(D) 77%

(E) 78%

5. What is the least positive integer m such that 168m is the square of an integer?

SAT Practice 8: Checking Your Work

Check your work carefully before choosing your

an-swer to the following questions

1. If 3x – 5 = 20, what is the value of 3x + 5?

2. If s2– 1 = 2s – 1 , which of the following gives all

possible values of s2?

(A) 2 only

(B) 4 only

(C) 0 and 2 only

(D) 0 and 4 only

(E) All even integers

3. Last year, Tom was twice as old as Julio This

year, the sum of their ages is 65 How old is Tom

now?

(A) 22

(B) 32

(C) 41

(D) 42

(E) 43

1

2 3 4 5 7 8 9 6

1 0 2 3 4 5 7 8 9 6

1 0 2 3 4 5 7 8 9 6

1 0 2 3 4 5 7 8 9 6

Concept Review 8

1 Check your answer by plugging it back into the

original equation and checking your steps.

Write out each step, one beneath the other, so

that checking your logic and arithmetic is

easier

2 Estimate only when it is easy to do If the answer

choices are numerically far apart, estimation can

help you to eliminate obviously wrong answers

3 If you can quickly “ballpark” a numerical answer

and rule out those choices that are “out of the

ball-park,” you can often avoid doing complicated cal-culations or algebra

4 Reread the question and make sure that you’ve

answered the right question, and make sure that

your answer makes sense in the context of the question

5 Dividing by 0 and taking the square root of an ex-pression that can be negative are not allowed be-cause they are undefined

Answer Key 8: Checking Your Work

SAT Review 8

1 30 The simplest way to solve this problem is to

add 10 to both sides of the equation, which gives

3x + 5 = 30 However, many students do the

“knee-jerk” of solving for x and become prone to

silly arithmetic mistakes If you did solve for x,

you should have checked your answer by

plug-ging it back in to the original equation

2 D Did you say (C)? Then you misread the question

Always reread before marking your answer It

asks for the value of s2, not s Although s can be 0

or 2, s2is either 0 or 4 Did you say (A) or (B)?

Then you may have made this common mistake:

s2– 1 = 2s – 1

Add 1: s2= 2s

Divide by s: s = 2

What went wrong? In the second step, we divided

by s without checking whether it could equal 0.

(Remember that division by 0 is undefined and

usually causes trouble.) Indeed, plugging in

shows that s = 0 is in fact a solution The correct

method is

s2– 1 = 2s – 1

Subtract 2s: s2– 2s = 0

Factor: s(s – 2) = 0

Use 0 product property: s = 0 or 2

3 E You might start by approximating Since the sum of their ages is about 60, and since Tom is about twice as old as Julio, Tom is about

40 and Julio is about 20 This rules out (A) and (B) From here, you may just want to “test” the remaining choices until you find what works If

you prefer algebra, you may want to let t equal Tom’s age now and j equal Julio’s age now You are told that t + j = 65 and that 2(j – 1) = t – 1 Since you only need the value of t, solve the first equation for j, getting j = 65 – t, and

sub-stitute this into the second equation This gives

2(65 – t – 1) = t – 1

Simplify: 2(64 – t) = t – 1

Distribute: 128 – 2t = t – 1 Add 2t and 1: 129 = 3t

Divide by 3: 43 = t

Therefore, Tom is now 43 and Julio is now 65 – 43 =

22 Notice that last year they were 42 and 21, respec-tively, and 42 is twice as old as 21

4 E Since more students averaged 80% than 70%, the overall average should be closer to 80% This rules out

(A) and (B) To get the precise answer, let x be the

overall average There are two ways to calculate the

sum of all the scores: (30)(80) + (20)(70) and

(50)(x) Since these must be equal,

2,400 + 1,400 = 50x

Simplify: 3,800 = 50x

Divide by 50: 76 = x

5 42 Do the prime factorization of 168: 23× 3 × 7 Since, in a perfect square, all prime factors “pair up,” we need to multiply at least one more factor

of 2, one more factor of 3, and one more factor of

7 to make a perfect square: 24× 32× 72= 7,056 = 842.

(Notice that now every factor appears an even

number of times.) Therefore, k = 2 × 3 × 7 = 42

ESSENTIAL

PRE-ALGEBRA SKILLS CHAPTER 7

✓

1 Numbers and Operations

2 Laws of Arithmetic

3 Fractions

4 Ratios and Proportions

5 Percents

6 Negatives

7 Divisibility

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