CHAPTER 6 / WHAT THE SAT MATH IS REALLYTESTING 243
Concept Review 2
- To analyze a problem means to look at its parts
and find how they relate to each other. - 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 xis 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.
Answer Key 2:Analyzing Problems
SAT Practice 2
- 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. - 5 Don’t worry about finding the base andheight
of the triangle and using the formulaarea=
(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. - 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 5nboys and 4ngirls.
Since 5n=80, nmust be 16. Therefore, there are
80 boys and 4(16) =64 girls, for a total of 144 stu-
dents in the junior class.- C Write what you know into the diagram. Because
the lines are parallel, ∠GEFis 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 BGare corresponding sides, the ratio of corre-
sponding sides is 3/9, or 1/3. Therefore, EFis 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: 3 h= 8 −h
Add h: 4 h= 8
Divide by 4: h= 2
So the shaded area is (4)(2)/2 +(12)(6)/2 = 4 + 36 =40.h
8 h1
− 3
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B C
D
E
F
A B C
D EF
G
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