SAT Math Hard Practice Quiz Answers
Numbers and Operations
1.
B
(Estimated Difficulty Level: 5)
The number of green and red tomatoes are 4
n
and 3
n
,
respectively, for some integer
n
. In this way, we can be
sure that the green-to-red ratio is 4
n/
3
n
= 4
/
3. We
need to solve the equation:
4
n
−
5
3
n
−
5
=
3
2
.
Cross-multiplying, 8
n
−
10 = 9
n
−
15 so that
n
= 5.
There were 3
n
, or 15, red tomatoes in the bag.
Working with the answers may be easier. If answer A
is correct, then there were 16 green tomatoes and 12
red tomatoes, in order to have the 4 to 3 ratio. But
removing five of each gives 11 green and 7 red, which is
not in the ratio of 3 to 2. If answer B is correct, then
there were 20 green tomatoes and 15 red tomatoes, since
20
/
15 = 4
/
3. Removing five of each gives 15 green and
10 red, and 15
/
10 = 3
/
2, so answer B is correct.
2.
C
(Estimated Difficulty Level: 4)
From 10 to 19, 12 and up (eight numbers) are mono-
tonic. Among the numbers from 20 to 29, seven (23 and
up) are monotonic. If you can see a pattern in counting
problems like this, you can save a lot of time. Here, the
30s will have 6 monotonic numbers, the 40s will have 5,
and so forth. You should find 8 + 7 + 6 + 5 + 4 + 3 +
2 + 1 + 0 = 36 total monotonic numbers.
3.
113
(Estimated Difficulty Level: 5)
Since the second term is 7 greater than the first term,
(2
a
−
1)
−
a
= 7 so that
a
= 8. The sequence is 8,
15, 22,
. . .
You can either continue to write out the
sequence until the 16
th
term, or realize that the 16
th
term is 16
a
−
15 = 16(8)
−
15 = 128
−
15 = 113.
4.
D
(Estimated Difficulty Level: 4)
The answer must be true for any value of
p
, so plug in
an easy (prime) number for
p
, such as 2. The factors of
2
3
= 8 are 1, 2, 4, and 8, so answer D is correct.
In general, since
p
is prime, the only numbers that go
into
p
3
without a remainder are 1,
p
,
p
2
, and
p
3
.
5.
11
(Estimated Difficulty Level: 4)
For the two-digit numbers, only 33 begins and ends in
3. For three-digit numbers, the only possibilities are:
303, 313,
. . .
, 383, and 393. We found ten three-digit
numbers, and one two-digit number, for a total of 11
numbers that begin and end in 3.
Yes, this was a counting problem soon after another
counting problem. But this one wasn’t so bad, was it?
6.
C
(Estimated Difficulty Level: 4)
This type of SAT math question contains three separate
mini-problems. (This kind of question is also known as
“one of those annoying, long, SAT math questions with
roman numerals”). Let’s do each mini-problem in order.
First, recall that a prime number is only divisible by
itself and 1, and that 1 is not a prime number. So,
statement I must be true, since a number that can be
divided by two prime numbers can’t itself be prime.
Next, recall that every number can be written as a prod-
uct of a particular bunch of prime numbers. Let’s say
that
N
is divisible by 3 and 5. Then,
N
is equal to
3
·
5
·
p
1
·
p
2
· · ·
, where
p
1
,
p
2
, etc. are some other primes.
So,
N
is divisible by 3
·
5 = 15. Statement II must be
true.
Finally, remember that 2 is a prime number. So,
N
could be 6, since 6 = 2
·
3. Statement III isn’t always
true, making C the correct answer.
erikthered.com/tutor
pg. 10