## SAT Test Prep

**CHAPTER 6**

WHAT THE SAT MATH IS REALLY TESTING

WHAT THE SAT MATH IS REALLY TESTING

**Lesson 2: Analyzing Problems**

**Break Complicated Problems into 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 without drawing a fence with 20 posts? Just change the question to a much simpler one to check the relationship 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 isn’t 3; it’s 2. What gives? If you think about it, you will see that dividing only gives the number of *spaces* between the posts, but there is always one more post than spaces. So a 120-foot fence requires 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 , then what is the value of ?

Don’t worry about solving for *x* and *y*. You only need to see the simple relationship between the expressions. The expression you’re looking for, , is 6 times the expression you’re given, . So, by substitution, must equal 6 times 15, or 90.

**If You Can’t Find What You Want, Find 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.

In the figure above, *ABCD* is a rectangle with area 60, and . 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 rectangle is , but the shaded region is not a rectangle. So how do you find its area? Analyze the diagram using the given information. First, write the fact that 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:

Notice that the dotted lines divide the shaded region 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 , the areas of the triangles are 7.5, 7.5, and 15, for a total area of 30.

**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 is negative and *x* is greater than 0, then what can be concluded about and ?

**SAT Practice 2: Analyzing Problems**

**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 midpoints of their respective sides. What is the area of the shaded region?

**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?

**4**__.__ In the figure above, , , , , and . What is the total area of the shaded regions?

(A) 32

(B) 36

(C) 40

(D) 42

(E) 44

**Answer Key 2: Analyzing Problems**

**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 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 is negative, and *x* is greater than 0, then must be negative, which means that one of the factors is positive and the other negative. Since is less than , must be negative and 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 between 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 . 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 .

__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 5*n* boys and 4*n* girls. Since , *n* must be 16. Therefore, there are 80 boys and girls, for a total of 144 students 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 similarity, see Lesson 6 in __Chapter 10__.) This means that the corresponding sides are proportional. Since *GF* and *BG* are corresponding sides, the ratio of corresponding sides is 3/9, or 1/3. Therefore, *EF* is 1/3 of *BC*, so . 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 .

So the shaded area is .