SAT Practice 5
- D Use the FCP: 3 × 2 × 8 =48.
- B “Build” the number: If it’s between 30 and 70,
it must be a two-digit number that begins with 4
or 6. That’s two choices. The second digit can be
anything in the list, so that’s 5 choices. 2 × 5 = 10 - B Since there are four spaces, there are four deci-
sions to make, so four numbers to multiply. You can
choose from four paintings for the first spot, then three
paintings for the second spot, etc. 4 × 3 × 2 × 1 = 24 - D If there are 31 students in calculus but 10 of
these are also taking physics, then 31 − 10 =21 stu-
dents are taking only calculus. If there are 36 stu-
dents taking either physics or calculus, but only 31
are taking calculus, then 36 − 31 =5 students are
taking only physics. Therefore, the Venn diagram
should look like this:
As you can see, 5 + 10 = 15
students are taking physics.- A “Build” the lineup. You have six spots to fill,
and thus six decisions to make and six numbers to
multiply. You only have two choices for the first
spot (Zack or Paul) and one choice for the second
spot (Dave), then you have four players left to fill
the other slots, so you have four choices for the
third spot, then three for the fourth spot, etc.
2 × 1 × 4 × 3 × 2 × 1 = 48- B David gets xcards. Tina gets two more cards
than David, which is x+2. Samuel gets five fewer
cards than Tina, which is x+ 2 − 5 =x−3. So all
together, x+x+ 2 +x− 3 = 3 x−1. - C You have six choices for the first spot, then five
for the second, then four for the third and three for
the fourth. 6 × 5 × 4 × 3 = 360 - E Set up the Venn diagram: Since
the ratio of striped marbles
to nonstriped marbles is 3:1, x+ 6 =
3 y,and since the ratio of dotted
marbles to nondottedmarbles is 2:3, y+ 6 =2/3xand
therefore y=2/3x−6. Substituting, we get x+
6 =3(2/3x−6) or x+ 6 = 2 x−18, so x=24. Plug
this back in to get y=2/3(24) − 6 =10. Total =
24 + 6 + 10 =40. - 36 Draw the cube. To get from any vertex to its
opposite vertex, the ant has 3 × 2 ×1 possible paths.
To see why, trace a path and notice it has three
choices for the first edge, then two for the second,
then only one option for the third. Since it must
return to the opposite vertex, it has 3 × 2 ×1 differ-
ent paths it can take back. 3 × 2 × 1 × 3 × 2 × 1 = 36
CHAPTER 9 / SPECIAL MATH PROBLEMS 353
Concept Review 5
- The number of ways an event can happen is equal
to the product of the choices that must be made
to “build” the event. - Try listing all the “words” that start with L, then
all that start with M, and so on:
LMNO MLNO NLMO OLMN
LMON MLON NLOM OLNM
LNMO MNLO NMLO OMLN
LNOM MNOL NMOL OMNL
LOMN MOLN NOLM ONML
LONM MONL NOML ONLM
Whew! 24 in total. Annoying, but not impossible.
Using the FCP makes it a lot easier: 4 × 3 × 2 × 1 =
24. That’s it! - There are too many possibilities to list, but the
FCP makes it easy: We have 9 choices for the first
digit, 2 choices for the second digit, and 10
choices for the last digit, and 9 × 2 × 10 =180.
4. “Build” the outfield from left to right. You have
6 players to choose from for left field, but then just
5 for center field and 4 for right field. 6 × 5 × 4 =120.
5. Since 20 + 27 − 40 =7, there must be 7 students
who take both. This Venn diagram shows how it
works out:
6. This one’s tough. Say there are ggreen buttons. If
there are four times as many blue buttons as
green buttons, then there are 4gblue buttons, so
g+ 4 g= 5 gbuttons in total. So the total number of
buttons must be a multiple of 5. Similarly, if there
are ntwo-holed buttons, there must be 6nfour-
holed buttons, so the total number of buttons is n- 6 n= 7 n,so the total number of buttons is also a
multiple of 7. The least common multiple of 5 and
7 is 35, so there are 35 buttons: 5 two-holed and
30 four-holed, and 7 green and 28 blue.
- 6 n= 7 n,so the total number of buttons is also a
713 20FSAnswer Key 5: Counting Problems
52110PCxy 6DS