Sat - MC Grawhill part 35 potx

330 MCGRAW-HILL’S SAT
7. When the Apex Pet Store first opened, the ratio of
cats to dogs was 4 to 5. Since then, the number of
cats has doubled, while the number of dogs has
increased by 12. If the ratio of dogs to cats is now
1 to 1, how many cats did the store have when it
opened?
8. Hillside High School has 504 students. One-quarter
of the students are sophomores, and 3/7 of the
sophomores are boys. If one-third of the sopho-
more girls take French, how many sophomore
girls do not take French?
(A) 24
(B) 36
(C) 48
(D) 72
(E) 126
9. A jar contains only red, green, and blue marbles.
If it is three times as likely that you randomly
pick a red marble as a green marble, and five
times as likely that you pick a green one as a blue
one, which of the following could be the number
of marbles in the jar?
(A) 38
(B) 39
(C) 40
(D) 41
(E) 42

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
CHAPTER 8 / ESSENTIAL ALGEBRA I SKILLS 331
Concept Review 7
1. m = Mike’s current age, d = Dave’s current age;
m = 2(d − 5)
2. a = the population of town A, b = the population
of town B; a = 1.4b
3. n = number of marbles in the jar;
n − (2/3)n = 5 + (1/6)n
4. b = number of blue marbles, r = number of red
marbles; b = 4 + 2r
5. c = cost of one candy bar, l = cost of one lollipop;
3c + 2l = 2.20, and 4c + 2l = 2.80.
Subtract: 4c + 2l = 2.80
– (3c + 2l = 2.20)
c = .60
Plug in to find l: 3(.60) + 2l = 2.20
Simplify: 1.80 + 2l = 2.20
Subtract 1.80:
2l = .40
Divide by 2: l = .20
6. n = number of seats in the stadium;
(2/3)n − 1,000 = (3/7)n
Subtract (2/3)n: −1,000 =−(5/21)n

Multiply by −(21/5): 4,200 = n
7.
Simplify:
Multiply by 4: 2m + 2n = s + t
Subtract t: 2m + 2n − t = s
8. b = value of blue chip, r = value of red chip, g = value
of green chip; b = 2 + r, r = 2 + g, and 5g = m, so
Cost of 10 blue and 5 red chips: 10b + 5r
Substitute b = 2 + r: 10(2 + r) + 5r
Simplify: 20 + 15r
Substitute r = 2 + g: 20 + 15(2 + g)
Simplify:
50 + 15g
Substitute g = m/5: 50 + 3m
mn st+
=
+
⎛
⎝
⎜
⎞
⎠
⎟
24
mn st+
=
+
⎛
⎝
⎜

⎞
⎠
⎟
2
1
22
Answer Key 7: Word Problems
SAT Practice 7
1. C You could test the choices here, or do the
algebra:
Multiply by x: 24 − x = 3x
Add x: 24 = 4x
Divide by 4: 6 = x
2. n = Nora’s current age, m = Mary’s current age.
Interpret first sentence: n − 3 = (1/2)m
Interpret second sentence: m = n + 4
Subtract 4: m − 4 = n
Substitute n = m − 4: m − 4 − 3 = (1/2)m
Simplify: m − 7 = (1/2)m
Subtract m: −7 =−(1/2)m
Multiply by −2: 14 = m
3. E p/q = 9/7, q/r = 14/3.
Multiply:
4. 40 J = number of books Joan had originally.
E = number of books Emily had originally. J = 2
E.
After the exchange, Emily has E + 5 and Joan has
J − 5 books, so J − 5 = 10 + (E + 5).
Simplify: J − 5 = E + 15
Subtract 15: J − 20 = E

Substitute into J = 2(J − 20)
first equation:
Solve for J:
J = 40
(Reread and check)
5. C Let xbe the cost of living in 1960. In 1970, the cost
of living was 1.2x, and in 1980 it was 1.5x. Use the per-
cent change formula: (1.5x − 1.2x)/1.2x = .25 = 25%.
p
q
q
r
p
r
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟
==
⎛
⎝
⎜

⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟
=
9
7
14
3
6
1
24
3
−
=
x
x
6. 8 Let w = the number of games won and l = the
number of games lost. w/l = 7/5 and w + l = 48.
Multiply by l: w = (7/5)l
Substitute into 2nd eq.: (7/5)l + l = 48
Simplify: (12/5)l = 48
Multiply by 5/12: l = 20
Plug in to find w: w + 20 = 48
Subtract 20: w = 28

How many more games won than lost?
w − l = 28 − 20 = 8
7. 16 Let c = number of cats originally, d = number
of dogs originally. c/d = 4/5. Now the number of
cats is 2c and the number of dogs is d + 12. If the
ratio of dogs to cats is now 1 to 1, 2c = d + 12.
Cross-multiply: 5c = 4d
Divide by 4: (5/4)c = d
Substitute: 2c = (5/4)c + 12
Subtract (5/4)c: (3/4)
c
= 12
Multiply by 4/3: c = 16 (Reread
and check)
8. C Number of sophomores = (1/4)(504) = 126.
If 3/7 of the sophomores are boys, 4/7 are girls:
(4/7)(126) = 72. If 1/3 of the sophomore girls take
French, 2/3 do not: (2/3)(72) = 48.
9. E r, g, and b are the numbers of red, green, and
blue marbles. r = 3g and g = 5b. Total marbles =
r + g + b.
Substitute r = 3g:3g + g + b = 4g + b
Substitute g = 5b:
4(5b) + b = 21b
So the total must be a multiple of 21, and 42 = 2(21).
SPECIAL MATH PROBLEMS
CHAPTER 9
332
1. New Symbol or Term Problems
2. Mean/Median/Mode Problems

3. Numerical Reasoning Problems
4. Rate Problems
5. Counting Problems
6. Probability Problems
✓
Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.
CHAPTER 9 / SPECIAL MATH PROBLEMS 333
New Symbol or Term Problems
Example:
For all real numbers a and b, let the expression
a ¿ b be defined by the equation a ¿ b = 10a + b.
Question 3: What is 5 ¿ 10?
Just substitute 5 for a and 10 for b in the given equa-
tion: 5 ¿ 10 = 10(5) + 10 = 60.
Question 4: If 2.5 ¿ x = 50, what is the value of x?
Just translate the left side of the equation:
2.5 ¿ x = 10(2.5) + x = 50
Then solve for x: 25 + x = 50
x = 25
Question 5: What is 1.5 ¿ (1.5 ¿ 1.5)?
According to the order of operations, evaluate what is
in parentheses first:
1.5 ¿ (1.5 ¿ 1.5)
Substitute: 1.5 ¿ (10(1.5) + 1.5)
Simplify: 1.5 ¿ (16.5)
Substitute again: 10(1.5) + 16.5
Simplify: 15 + 16.5 = 31.5
Lesson 1: New Symbol or Term Problems
Don’t be intimidated by SAT questions with
strange symbols, like Δ, φ, or ¥, or new terms

that you haven’t seen before. These crazy sym-
bols or terms are just made up on the spot, and
the problems will always explain what they
mean. Just read the definition of the new sym-
bol or term carefully and use it to “translate” the
expressions with the new symbol or term.
Example:
Let the “kernel” of a number be defined as the
square of its greatest prime factor. For instance, the
kernel of 18 is 9, because the greatest prime factor
of 18 is 3 (prime factorization: 18 = 2 × 3 × 3), and
3
2
equals 9.
Question 1: What is the kernel of 39?
Don’t worry about the fact that you haven’t heard of
a “kernel” before. Just read the definition carefully. By
the definition, the kernel of 39 is the square of its
greatest prime factor. So just find the greatest prime
factor and square it. First, factor 39 into 3 × 13, so its
greatest prime factor is 13, and 13
2
= 169.
Question 2: What is the greatest integer less than 20
that has a kernel of 4?
This requires a bit more thinking. If a number has a
kernel of 4, then 4 must be the square of its greatest
prime factor, so its greatest prime factor must be 2.
The only numbers that have a greatest prime factor of
2 are the powers of 2. The greatest power of 2 that is

less than 20 is 2
4
= 16.
There’s a lot of detail to learn and understand to do well on the SAT. For more tools and resources that will help, visit our Online
Practice Plus at www.MHPracticePlus.com/SATmath.
334 MCGRAW-HILL’S SAT
Concept Review 1:
New Symbol or Term Problems
For questions 1–6, translate each expression into its simplest terms, using the definition of the new symbol.
The following definition pertains to questions 1–3:
For any real number x, let § x be defined as the greatest integer less than or equal to x.
1. §−4.5 = __________
2. §−1.5 + §1.5 = __________
3. = __________
The following definition pertains to questions 4–6:
If q is any positive real number and n is an integer, let q @ n be defined by the equation
.
4. 8 @ 3 = __________
5. 9 @ (k − 1) = __________
6. x
2
@ 0 = __________
7. If q is any positive real number and n is an integer, let q @ n be defined by the equation .
If y @ 2 = 64, what is the value of y?
8. For any integer n and real number x, let x ^ n be defined by the equation x ^ n = nx
n−1
. If y ^ 4 =−32, what is
the value of y?
9. For any integer n, let Ωn be defined as the sum of the distinct prime factors of n. For instance, Ω36 = 5, because
2 and 3 are the only prime factors of 36 and 2 + 3 = 5. What is the smallest value of w for which Ωw = 12?

qn q
n
@ =
+1
qn q
n
@ =
+1
§§15 17+
CHAPTER 9 / SPECIAL MATH PROBLEMS 335
1. For all real numbers d, e, and f, let
d * e * f = de + ef + df. If 2 * 3 * x = 12, then x =
(A)
(B)
(C)
(D) 2
(E) 6
2. If b ≠ 0, let . If x # y = 1, then which of
the following statements must be true?
(A) x = y
(B) x =|y|
(C) x =−y
(D) x
2
− y
2
= 0
(E) x and y are both positive
3. On a digital clock, a time like 6:06 is called a
“double” time because the number representing

the hour is the same as the number represent-
ing the minute. Other such “doubles” are 8:08
and 9:09. What is the smallest time period
between any two such doubles?
(A) 11 mins. (B) 49 mins.
(C) 60 mins. (D) 61 mins.
(E) 101 mins.
4. Two numbers are “complementary” if their
reciprocals have a sum of 1. For instance, 5 and
are complementary because .
If x and y are complementary, and if ,
what is y?
(A) −2 (B) (C)
(D) (E) 3
1
3
−
1
3
−
1
2
x =
2
3
1
5
4
5
1+=

5
4
ab
a
b
# =
2
2
8
5
6
5
5
6
5. For x ≠ 0, let . What is the value of $$5?
6. For all nonnegative real numbers x, let ◊x be
defined by the equation . For what
value of x does ◊x = 1.5?
(A) 0.3 (B) 6 (C) 12
(D) 14 (E) 36
7. For any integer n, let [n] be defined as the sum of
the digits of n. For instance, [341] = 3 + 4 + 1 = 8.
If a is an integer greater than 0 but less than
1,000, which of the following must be true?
I. [10a] < [a]+1
II. [[a]] < 20
III. If a is even, then [a] is even
(A) none
(B) II only
(C) I and II only

(D) II and III only
(E) I, II, and III
8. For all integers, n, let
What is the value of 13&&?
(A) 10 (B) 13 (C) 20
(D) 23 (E) 26
n
nn
nn
& =
−
⎧
⎨
⎪
⎩
⎪
2
3
if is even
if is odd
◊=x
x
4
$x
x
=
1
SAT Practice 1: New Symbol or Term Problems

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
336 MCGRAW-HILL’S SAT
Answer Key 1: New Symbol or Term Problems
8. y^4 =−32
Translate: 4y
4–1
=−32
Simplify and divide by 4: y
3
=−8
Take the cube root: y =−2
9. If Ωw = 12, then w must be a number whose dis-
tinct prime factors add up to 12. The prime num-
bers less than 12 are 2, 3, 5, 7, and 11. Which of
these have a sum of 12? (Remember you can’t re-
peat any, because it says the numbers have to be
distinct.) A little trial and error shows that the
only possibilities are 5 and 7, or 2, 3, and 7. The
smallest numbers with these factors are 5 × 7 = 35
and 2 × 3 × 7 = 42. Since the question asks for the
least such number, the answer is 35.
SAT Practice 1
1. B 2 * 3 * x = 12

Translate: (2)(3) + (3)(x) + (2)(x) = 12
Simplify: 6 + 5x = 12
Subtract 6: 5x = 6
Divide by 5: x = 6/5
2. D If x # y = 1, then (x
2
/y
2
) = 1, which means x
2
= y
2
.
Notice that x =−1 and y = 1 is one possible solu-
tion, which means that
(A) x = y
(B) x =⏐y⏐
(E) x and y are both positive
is not necessarily true. Another simple
solution is x = 1 and y = 1, which means that
(C) x =−y
is not necessarily true, leaving only
(D) as an answer.
3. B All of the consecutive “double times” are
1 hour and 1 minute apart except for 12:12 and
1:01, which are only 49 minutes apart.
4. A If
2
⁄3 and y are complementary, then the sum
of their reciprocals is 1:

3
⁄2 + 1/y = 1
Subtract
3
⁄2:1/y =−1/2
Take the reciprocal of both sides: y =−2
5. 5 The “double” symbol means you simply per-
form the operation twice. Start with 5, then $5 =
1/5. Therefore, $$5 = $(1/5) = 1/(1/5) = 5.
6. E
Multiply by 4:
Square both sides:
Plug in x = 36 to the original and see that it works.
7. C If a is 12, which is even, then [12] = 1 + 2 = 3 is
odd, which means that statement III is not
necessarily true. (Notice that this eliminates choices
(D) and (E).) Statement I is true because [10a] will
always equal [a] because 10a has the same digits
as a, but with an extra 0 at the end, which con-
tributes nothing to the sum of digits. Therefore,
[10a] < [a] + 1 is always true. Notice that this leaves
only choice (C) as a possibility. To check statement
II, though (just to be sure!), notice that the biggest
sum of digits that you can get if a is less than 1,000
is from 999. [999] = 9 + 9 + 9 = 27; therefore,
[[999]] = [27] = 2 + 7 = 9. It’s possible to get a slightly
bigger value for [[a]] if a is, say, 991: [[991]] =
[19] = 10, but you can see that [[a]] will never ap-
proach 20.
8. C Since 13 is odd, 13& = 13 − 3 = 10. Therefore,

13&& = 10&. Since 10 is even, 10& = 2(10) = 20.
◊= =
=
=
x
x
x
x
4
15
6
36
.
Concept Review 1
1. §−4.5 =−5
2. §−1.5 + §1.5 =−2 + 1 =−1
3.
4.
5.
6.
7.
Simplify:
Take the cube root:
Square:
yy
y
y
y
@2 64
64

4
16
21
3
=
()
=
()
=
=
=
+
xxx
22
01
0@ =
()
=
+
919 3
11
@ k
k
k
−
(
)
=
()
=

−+
83 8 64
4
@ =
()
=
§15 17 3 4 7+=+=
CHAPTER 9 / SPECIAL MATH PROBLEMS 337
Average (Arithmetic Mean) Problems
You probably know the procedure for finding an
average of a set of numbers: add them up and divide
by how many numbers you have. For instance, the av-
erage of 3, 7, and 8 is (3 + 7 + 8)/3 = 6. You can de-
scribe this procedure with the “average formula”:
Since this is an algebraic equation, you can manipu-
late it just like any other equation, and get two more
formulas:
Sum = average × how many numbers
This is a great tool for setting up tough problems. To find
any one of the three quantities, you simply need to find
the other two, and then perform the operation between
them. For instance, if the problem says, “The average
(arithmetic mean) of five numbers is 30,” just write 30 in
the “average” place and 5 in the “how many” place. No-
tice that there is a multiplication sign between them, so
multiply 30 ×5 =150 to find the third quantity: their sum.
Medians
How many numbers =
sum
average

Average =
sum
how many numbers
Lesson 2: Mean/Median/Mode Problems
Just about every SAT will include at least one
question about averages, otherwise known as
arithmetic means. These won’t be simplistic
questions like “What is the average of this set
of numbers?” You will have to really under-
stand the concept of averages beyond the basic
formula.
Occasionally the SAT may ask you about the
mode of a set of numbers. A mode is the num-
ber that appears the most frequently in a set. (Just
remember: MOde = MOst.) It’s easy to see that
not every set of numbers has a mode. For
instance, the mode of [−3, 4, 4, 1, 12] is 4, but
[4, 9, 14, 19, 24] doesn’t have a mode.
The average (arithmetic mean) and the me-
dian are not always equal, but they are equal
whenever the numbers are spaced symmetri-
cally around a single number.
it splits the highway exactly in half. The median
of a set of numbers, then, is the middle number
when they are listed in increasing order. For in-
stance, the median of {−3, 7, 65} is 7, because
the set has just as many numbers bigger than
7 as less than 7. If you have an even number of
numbers, like {2, 4, 7, 9}, then the set doesn’t
have one “middle” number, so the median is

the average of the two middle numbers. (So
the median of {2, 4, 7, 9} is (4+7)/2 = 5.5.)
All three of these formulas can be summarized
in one handy little “pyramid”:
When you take standardized tests like the SAT, your
score report often gives your score as a percentile,
which shows the percentage of students whose scores
were lower than yours. If your percentile score is
50%, this means that you scored at the median of all
the scores: just as many (50%) of the students scored
below your score as above your score.
Example:
Consider any set of numbers that is evenly spaced,
like 4, 9, 14, 19, and 24:
Notice that these numbers are spaced symmetrically
about the number 14. This implies that the mean
and the median both equal 14. This can be helpful to
know, because finding the median of a set is often
much easier than calculating the mean.
Modes
41492419
A median is something that splits a set into two
equal parts. Just think of the median of a
highway:
average
how
many
sum
×
÷

÷
1. Draw the “average pyramid.”
2. Explain how to use the average pyramid to solve a problem involving averages.
3. Define a median.
4. Define a mode.
5. In what situations is the mean of a set of numbers the same as its median?
6. The average (arithmetic mean) of four numbers is 15. If one of the numbers is 18, what is the average of the
remaining three numbers?
7. The average (arithmetic mean) of five different positive integers is 25. If none of the numbers is less than
10, then what is the greatest possible value of one of these numbers?
8. Ms. Appel’s class, which has twenty students, scored an average of 90% on a test. Mr. Bandera’s class, which
has 30 students, scored an average of 80% on the same test. What was the combined average score for the
two classes?
338 MCGRAW-HILL’S SAT
Concept Review 2:
Mean/Median/Mode Problems
1. If y = 2x + 1, what is the average (arithmetic
mean) of 2x, 2x, y, and 3y, in terms of x?
(A) 2x (B) 2x + 1 (C) 3x
(D) 3x + 1 (E) 3x + 2
2. The average (arithmetic mean) of seven inte-
gers is 11. If each of these integers is less than
20, then what is the least possible value of any
one of these integers?
(A) −113 (B) −77 (C) −37
(D) −22 (E) 0
3. The median of 8, 6, 1, and k is 5. What is k?
4. The average (arithmetic mean) of two numbers is
z. If one of the two numbers is x, what is the value
of the other number in terms of x and z?

(A) z − x (B) x − z (C) 2z − x
(D) x − 2z (E)
5. A set of n numbers has an average (arithmetic
mean) of 3k and a sum of 12m, where k and m
are positive. What is the value of n in terms of k
and m?
(A) (B) (C)
(D) (E) 36km
6. The average (arithmetic mean) of 5, 8, 2, and k
is 0. What is the median of this set?
(A) 0 (B) 3.5 (C) 3.75
(D) 5 (E) 5.5
m
k4
k
m4
4k
m
4m
k
xz
+
2
7. A die is rolled 20 times, and the outcomes are as
tabulated above. If the average (arithmetic
mean) of all the rolls is a, the median of all the
rolls is b, and the mode of all the rolls is c, then
which of the following must be true?
I. a = b II. b > c III. c = 5
(A) I only (B) II only

(C) I and II only (D) II and III only
(E) I, II, and III
8. If a 30% salt solution is added to a 50% salt so-
lution, which of the following could be the con-
centration of the resulting mixture?
I. 40%
II. 45%
III. 50%
(A) I only (B) I and II only
(C) I and III only (D) II and III only
(E) I, II, and III
9. Set A consists of five numbers with a median of
m. If Set B consists of the five numbers that are
two greater than each of the numbers in Set A,
which of the following must be true?
I. The median of Set B is greater
than m.
II. The average (arithmetic mean) of
Set B is greater than m.
III. The greatest possible difference
between two numbers in Set B is
greater than the greatest possible
difference between two numbers
in Set A.
(A) I only (B) I and II only
(C) I and III only (D) II and III only
(E) I, II, and III
SAT Practice 2: Mean/Median/Mode Problems
CHAPTER 9 / SPECIAL MATH PROBLEMS 339
Roll Frequency

15
23
33
43
53
63

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