Sat - MC Grawhill part 26 ppt

What You Can!If you can’t ﬁnd what you want right away, just look at the parts of the problem one at a time, and ﬁnd what you can.. Look for Simple Relationships Once you see the parts o

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What You Can!

If you can’t ﬁnd what you want right away, just look at the parts of the problem one at a time,

and ﬁnd what you can Often, going step by

step and noticing the relationships among the parts will lead you eventually to the answer you need

Simple Ones

Analyzing is key to solving many SAT math

problems Analyzing a problem means simply

looking at its parts and seeing how they relate

Often, a complicated problem can be greatly

simpliﬁed by looking at its individual parts If

you’re given a geometry diagram, mark up the

angles and the sides when you can ﬁnd them

If you’re given algebraic expressions, notice

how they relate to one another

For a certain fence, vertical posts must be placed

6 feet apart with supports in between, as shown

above How many vertical posts are needed for a

fence 120 feet in length?

You may want to divide 120 by 6 and get 20, which

seems reasonable But how can you check this

with-out drawing a fence with 20 posts? Just change the

question to a much simpler one to check the

relation-ship between length and posts How many posts are

needed for a 12-foot fence? The ﬁgure above provides

the answer Obviously, it’s 3 But 12 ÷ 6 isn’t 3; it’s 2

What gives? If you think about it, you will see that

di-viding only gives the number of spaces between the

posts, but there is always one more post than spaces

So a 120-foot fence requires 20 + 1 = 21 vertical posts

Look for Simple Relationships

Once you see the parts of a problem, look for

simple relationships between them Simple

relationships usually lead to simple solutions

If 2x2+ 5y = 15, then what is the value of 12x2+ 30y?

Don’t worry about solving for x and y You only

need to see the simple relationship between the

ex-pressions The expression you’re looking for, 12x2+

30y, is 6 times the expression you’re given, 2x2+ 5y So,

by substitution, 12x2+ 30y must equal 6 times 15, or 90.

C D

G

In the ﬁgure above, ABCD is a rectangle with area

60, and AB = 10 If E, F, and G are the midpoints of

their respective sides, what is the area of the shaded region?

This looks complicated at ﬁrst, but it becomes much simpler when you analyze the diagram You probably know that the formula for the area of a

tangle is a = bh, but the shaded region is not a

rec-tangle So how do you ﬁnd its area? Analyze the diagram using the given information First, write the

fact that AB = 10 into the diagram Since the area of

the rectangle is 60 and its base is 10, its height must

be 6 Then, knowing that E, F, and G are midpoints,

you can mark up the diagram like this:

C D

G

10

3

3 3

5 5

Notice that the dotted lines divide the shaded re-gion into three right triangles, which are easy to work with The two bottom triangles have base 5 and height 3 (flip them up if it helps you to see), and the top triangle has base 10 and height 3 Since the

formula for the area of a triangle is a = bh, the

areas of the triangles are 7.5, 7.5, and 15, for a total area of 30

1 2

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Concept Review 2: Analyzing Problems

1 What does it mean to analyze a problem?

2 Analyze the diagram above by indicating the measures of as many angles as possible

3 If $20,000 is divided among three people in the ratio of 2:3:5, how much does each person get?

4 If (x)(x − 1)(x − 2) is negative and x is greater than 0, then what can be concluded about x − 1 and x − 2?

95°

l3

l2

Given: l1|| l2||l3

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3. In the sophomore class at Hillside High School, the ratio of boys to girls is 2 to 3 The junior class contains as many boys as the sophomore class does, and the ratio of boys to girls in the junior class is 5 to 4 If there are 200 students in the sophomore class, how many students are there in the junior class?

Note: Figure not drawn to scale

4. In the ﬁgure above, ᐉ1 ᐉ2, AD = 8, EF = 4,

GF = 3, and BG = 9 What is the total area of the

shaded regions?

(A) 32 (B) 36 (C) 40 (D) 42 (E) 44

1. How many odd integers are there between 1 and

99, not including 1 and 99?

(A) 46

(B) 47

(C) 48

(D) 49

(E) 50

2. In the ﬁgure above, equilateral triangle ABC has

an area of 20, and points D, E, and F are the

mid-points of their respective sides What is the area

of the shaded region?

A

D

E F

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

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

G

l1

l2

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Concept Review 2

1 To analyze a problem means to look at its parts

and ﬁnd how they relate to each other

2 Your diagram should look like this:

3 If the total is divided in the ratio of 2:3:5, then it is divided into 2 + 3 + 5 = 10 parts The individual parts, then, are 2/10, 3/10, and 5/10 of the total Multiplying these fractions by $20,000 gives parts

of $4,000, $6,000, and $10,000

4 If (x)(x − 1)(x − 2) is negative, and x is greater than 0, then (x − 1)(x − 2) must be negative, which

means that one of the factors is positive and the

other negative Since x − 2 is less than x − 1, x − 2 must be negative and x− 1 must be positive

SAT Practice 2

1 C You might start by noticing that every other

number is odd, so that if we have an even number

of consecutive integers, half of them will always be

odd But this one is a little trickier Start by solving

a simpler problem: How many odd numbers are

be-tween 1 and 100, inclusive? Simple: there are 100

consecutive integers, so 50 of them must be odd

Now all we have to do is remove 1, 99, and 100 That

removes 2 odd numbers, so there must be 48 left

2 5 Don’t worry about ﬁnding the base and height

of the triangle and using the formula area =

(base × height)/2 This is needlessly complicated.

Just notice that the four smaller triangles are all

equal in size, so the shaded region is just 1/4 of the

big triangle Its area, then, is 20/4 = 5

3 144 If the ratio of boys to girls in the sophomore

class is 2 to 3, then 2/5 are boys and 3/5 are girls

If the class has 200 students, then 80 are boys and

120 are girls If the junior class has as many boys

as the sophomore class, then it has 80 boys, too

If the ratio of boys to girls in the junior class is 5

to 4, then there must be 5n boys and 4n girls.

Since 5n = 80, n must be 16 Therefore, there are

80 boys and 4(16) = 64 girls, for a total of 144 stu-dents in the junior class

4 C Write what you know into the diagram Because the lines are parallel, ∠GEF is congruent to ∠GCB,

and the two triangles are similar (To review simi-larity, see Lesson 6 in Chapter 10.) This means that

the corresponding sides are proportional Since GF and BG are corresponding sides, the ratio of corre-sponding sides is 3/9, or 1/3 Therefore, EF is 1/3 of

BC, so BC= 12 To ﬁnd the areas of the triangles, you need the heights of the triangles The sum of the two heights must be 8, and they must be in a ratio of 1:3 You can guess and check that they are

2 and 6, or you can ﬁnd them algebraically: if the

height of the smaller triangle is h, then the height

of the larger is 8 − h.

Cross-multiply: 3h = 8 − h

So the shaded area is (4)(2)/2 + (12)(6)/2 = 4 + 36 = 40

h h

8

1 3

95 °

20 °

95 °

160 °

85 °

75 °

A

D

E F

G

l1

l2 8

4 3 9

2 6 12

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Finding patterns means looking for simple

rules that relate the parts of a problem One

key to simplifying many SAT math problems is

exploiting repetition If something repeats, you

usually can cancel or substitute to simplify.

If 5x2+ 7x + 12 = 4x2+ 7x + 12, then what is the value

of x?

This question is much simpler than it looks at

first because of the repetition in the equation If

you subtract the repetitive terms from both sides of

the equation, it reduces to 5x2= 4x2 Subtracting 4x2

from both sides then gives x2= 0, so x = 0.

Patterns in Geometric Figures

Sometimes you need to play around with the

parts of a problem until you ﬁnd the patterns or

relationships For instance, it often helps to

treat geometric ﬁgures like jigsaw puzzle pieces

The ﬁgure above shows a circle with radius 3 in

which an equilateral triangle has been inscribed

Three diameters have been drawn, each of which

in-tersects a vertex of the triangle What is the sum of

the areas of the shaded regions?

This ﬁgure looks very complicated at ﬁrst But

look closer and notice the symmetry in the ﬁgure

No-tice that the three diameters divide the circle into six

congruent parts Since a circle has 360°, each of the

central angles in the circle is 360° ÷ 6 = 60° Then

no-tice that the two shaded triangles ﬁt perfectly with the

other two shaded regions to form a sector such as this:

Moving the regions is okay because it doesn’t change their areas Notice that this sector is 1/3 of the entire circle Now ﬁnding the shaded area is easy The

total area of the circle is a = r2= (3)2= 9 So the area of 1/3 of the circle is 9/3 = 3

Patterns in Sequences

Some SAT questions will ask you to analyze a sequence When given a sequence question, write out the terms of the sequence until you notice the pattern Then use whole-number di-vision with remainders to ﬁnd what the ques-tion asks for

1, 0, −1, 1, 0, −1, 1, 0, −1,

If the sequence above continues according to the pattern shown, what will be the 200th term of the sequence?

Well, at least you know it’s either 1, 0, or −1, right?

Of course, you want a better than a one-in-three guess, so you need to analyze the sequence more deeply The sequence repeats every 3 terms In 200 terms, then the pattern repeats itself 200 ÷ 3 = 66 times with a remainder of 2 This means that the 200th term is the same as the second term, which is 0 What is the units digit of 2740?

The units digit is the “ones” digit or the last digit You can’t ﬁnd it with your calculator because when

2740is expressed as a decimal, it has 58 digits, and your calculator can only show the ﬁrst 12 or so To ﬁnd the units digit, you need to think of 2740as a term in the sequence 271, 272, 273, 274, If you look at these terms in decimal form, you will notice that the units digits follow a pattern: 7, 9, 3, 1, 7, 9, 3, 1, The sequence has a repeating pattern of four terms Every fourth term is 1, so the 40th term is also 1 Therefore, the units digit of 2740is 1

3

60° 60°

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Concept Review 3: Finding Patterns

Solve the following problems by taking advantage of repetition

1 If 5 less than 28% of x2is 10, then what is 15 less than 28% of x2?

2 If m is the sum of all multiples of 3 between 1 and 100, and n is the sum of all multiples of 3 between 5 and

95, what is m − n?

3 How much greater is the combined surface area of two cylinders each with a height of 4 cm and a radius

of 2 cm than the surface area of a single cylinder with a height of 8 cm and a radius of 2 cm?

Solve each of the following problems by analyzing a sequence

4 What is the units digit of 4134?

5 The ﬁrst two terms of a sequence are 1 and 2 If every term after the second term is the sum of the previous two, then how many of the ﬁrst 100 terms are odd?

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(B) 15

(C) 20

(D) 40

(E) 60

2. Every term of a sequence, except the ﬁrst, is 6 less

than the square of the previous term If the ﬁrst

term is 3, what is the ﬁfth term of this sequence?

(B) 15

(C) 19

(D) 30

(E) 43

−4, 0, 4, −4, 0, 4, −4, 0, 4,

3. If the sequence above continues according to the

pattern shown, what is the sum of the ﬁrst 200

terms of the sequence?

(A) −800

(B) −268

(C) −4

(D) 0

(E) 268

y

x

y x

4. In the figure above, ABCD is a square and BC =

10 What is the total area of the shaded regions?

5. What is the units digit of 340? (A) 1

(B) 3 (C) 6 (D) 7 (E) 9

C D

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

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4 50 Move the shaded regions around, as shown above, to see that they are really half of the square Since the area of the square is (10)(10) = 100, the area of the shaded region must be half of that, or 50

5 A The number 340is so big that your calculator is useless for telling you what the last digit is Instead, think of 340as being an element in the sequence

31, 32, 33, 34, If you write out the ﬁrst six terms

or so, you will see that there is a clear pattern to the units digits: 3, 9, 27, 81, 243, 729, So the pattern

in the units digits is 3, 9, 7, 1, 3, 9, 7, 1, The sequence repeats every four terms Since 40 is a mul-tiple of 4, the 40th term is the same as the 4th and the 8th and the 12th terms, so the 40th term is 1

Concept Review 3

1 Don’t worry about the percent or about ﬁnding x.

Translate: 5 less than 28% of x2is 10 means

.28(x2) − 5 = 10

Subtract 10: .28(x2) − 15 = 0

So 15 less than 28% of x2is 0

2 m= 3 + 6 + 9 + + 93 + 96 + 99

n= 6 + 9 + + 93

When you subtract n from m, all the terms cancel

except 3 + 96 + 99 = 198

extra bases Each base has an area of π(2)2= 4π, so the surface area of the smaller cylinders is 2(4π) =

8π greater than that of the larger cylinder

4 Your calculator is no help on this one because 4134

is so huge Instead, think of 4134as a term in the sequence 41, 42, 43, 44, What is the units digit

of 4134? If you write out the first few terms, you will see a clear pattern to the units digits: 4, 16,

64, 256, Clearly, every odd term ends in a 4 and every even term ends in a 6 So 4134must end

in a 6

5 The ﬁrst few terms are 1, 2, 3, 5, 8, 13, 21, Since

we are concerned only about the “evenness” and

“oddness” of the numbers, think of the sequence as odd, even, odd, odd, even, odd, odd, even, No-tice that the sequence repeats every three terms: (odd, even, odd), (odd, even, odd), (odd, even, odd), In the ﬁrst 100 terms, this pattern re-peats 100/3 = 331⁄3times Since each pattern con-tains 2 odd numbers, the first 33 repetitions contain 66 odd numbers and account for the ﬁrst

99 terms The last term must also be odd because each pattern starts with an odd number There-fore, the total number of odds is 66 + 1 = 67

Answer Key 3: Finding Patterns

SAT Practice 3

1

2 A If every term is 6 less than the square of the

previous term, then the second term must be

(3)2− 6 = 9 − 6 = 3 The third term, then, is also

(3)2− 6 = 3, and so on Every term, then, must be

3, including the ﬁfth

3 C The sequence repeats every three terms:

(−4, 0, 4), (−4, 0, 4), (−4, 0, 4), Each one of the

groups has a sum of 0 Since 200/3 = 672⁄3, the ﬁrst

200 terms contain 67 repetitions of this pattern,

plus two extra terms The 67 repetitions will have a

sum of 67(0) = 0, but the last two terms must be −4

and 0, giving a total sum of −4

C 6

x

⎝⎜

⎞

3 Don’t calculate the total surface area Instead, just

notice that the two small cylinders, stacked

to-gether, are the same size as the large cylinder But

remember that you are comparing surface areas,

not volumes The surface areas are almost the

same, except that the smaller cylinders have two

8

4

C D

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When a problem seems overwhelming, try one

of these four simpliﬁcation strategies: beelining,

substituting, combining, and canceling.

Look for the Beeline—The Direct Route

Many SAT problems have “beelines”—direct

paths from the given information to the answer

We sometimes miss the “beeline” because we

get trapped in a knee-jerk response—for

in-stance, automatically solving every equation

or using the Pythagorean theorem on every

right triangle Avoid the knee-jerk response

Instead, step back and look for the “beeline.”

If and what is the value of ?

This problem looks tough because of all the

un-knowns You might do the knee-jerk thing and try to

solve for a, b, and c Whoa, there! Step back The

ques-tion doesn’t ask for a, b, and c It asks for a fracques-tion that

you can get much more directly Notice that just

mul-tiplying the two given fractions gets you almost there:

This is close to what you want—all

you have to do is divide by 3 to get Substituting

the given values of the fractions gives you

which is the value of

Simplify by Substituting

The simplest rule in algebra is also the most

powerful: Anything can be substituted for its

equal When you notice a complicated

expres-sion on the SAT, just notice if it equals

some-thing simpler, and substitute!

If 3x2+ 5x + y = 8 and x≠ 0, then what is the value

of ?

Again, take a deep breath Both the equation and

the fraction look complicated, but you can simplify by

16 2

−

+

y

a c

10

1

1 4

× ÷ = ,

a c

10

3

10

a

b

c

a

c

a c

10

b c

3

2

1

4

a

b=

equal Notice that 3x2+ 5x appears in both the equation and the fraction What does it equal? Subtract y from both sides of the equation to get 3x2+ 5x = 8 − y If

you substitute 8 − y for 3x2+ 5x in the fraction, you get

Nice!

Simplify by Combining or Canceling

Many algebraic expressions can be simpliﬁed

by combining or canceling terms Always keep

your eye out for like terms that can be com-bined or canceled and for common factors in

fractions that can be canceled.

If m and n are positive integers such that m > n and

what is the value of m + n?

To simplify this one, it helps to know a basic

factor-ing formula from Chapter 8, Lesson 5: m2– n2= (m – n) (m + n) If you factor the numerator and denominator

of the fraction, a common factor reveals itself, and it can be canceled:

If f(x) = 2x2 – 5x + 3 and g(x) = 2x2 + 5x + 3, then for how many values of x does f(x) = g(x)?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

Remember the simple rule that anything can be

substituted for its equal, and then cancel to simplify.

Since f(x) = g(x), you can say that 2x2 – 5x + 3 = 2x2 + 5x + 3.

Subtract 2x2and 3: −5x = 5x

So the answer is (A) 0, right? Wrong! Remember that

the question asks for how many values of x are the

function values equal Since we only got one solution

for x, the answer is (B) 1.

m+ n =

2

9

2,

m n m n

m n

2 2

−

2 2

9 2

−

16 2 8

2 8

−

y y

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Concept Review 4: Simplifying Problems

Simplify the following expressions

1 −2n − (6 − 5n) − n

2

3

4

5

Solve the following problems with substitution

6 If y = 1 − x and 2y + x = 5, then what is the value of x?

7 If 3 + m + n = n2+ m2, what is the value of ?

8 For all real numbers x, let <x> = (1 − x)2 What is the value of <<4>>?

n2 n m2 m

6

−

x x6

2

3 2

x

2

x+

2

x

−

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