Sat Math Hard Practice Quiz Worksheet With Answer Key Page 10
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18SAT Math Hard Practice Quiz Answers
Numbers and Operations
4. D
(Estimated Difficulty Level: 4)
The answer must be true for any value of p, so plug in
1. B
(Estimated Difficulty Level: 5)
an easy (prime) number for p, such as 2. The factors of
3
2
= 8 are 1, 2, 4, and 8, so answer D is correct.
The number of green and red tomatoes are 4n and 3n,
In general, since p is prime, the only numbers that go
respectively, for some integer n. In this way, we can be
3
2
3
sure that the green-to-red ratio is 4n/3n = 4/3. We
into p
without a remainder are 1, p, p
, and p
.
need to solve the equation:
4n
5
3
=
.
3n
5
2
5. 11
(Estimated Difficulty Level: 4)
Cross-multiplying, 8n
10 = 9n
15 so that n = 5.
There were 3n, or 15, red tomatoes in the bag.
For the two-digit numbers, only 33 begins and ends in
3. For three-digit numbers, the only possibilities are:
Working with the answers may be easier. If answer A
303, 313, . . ., 383, and 393. We found ten three-digit
is correct, then there were 16 green tomatoes and 12
numbers, and one two-digit number, for a total of 11
red tomatoes, in order to have the 4 to 3 ratio. But
numbers that begin and end in 3.
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
Yes, this was a counting problem soon after another
there were 20 green tomatoes and 15 red tomatoes, since
counting problem. But this one wasn’t so bad, was it?
20/15 = 4/3. Removing five of each gives 15 green and
10 red, and 15/10 = 3/2, so answer B is correct.
6. C
(Estimated Difficulty Level: 4)
This type of SAT math question contains three separate
2. C
(Estimated Difficulty Level: 4)
mini-problems. (This kind of question is also known as
“one of those annoying, long, SAT math questions with
From 10 to 19, 12 and up (eight numbers) are mono-
roman numerals”). Let’s do each mini-problem in order.
tonic. Among the numbers from 20 to 29, seven (23 and
up) are monotonic. If you can see a pattern in counting
First, recall that a prime number is only divisible by
problems like this, you can save a lot of time. Here, the
itself and 1, and that 1 is not a prime number. So,
30s will have 6 monotonic numbers, the 40s will have 5,
statement I must be true, since a number that can be
and so forth. You should find 8 + 7 + 6 + 5 + 4 + 3 +
divided by two prime numbers can’t itself be prime.
2 + 1 + 0 = 36 total monotonic numbers.
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
· p
· · ·, where p
, p
, etc. are some other primes.
1
2
1
2
3. 113
(Estimated Difficulty Level: 5)
So, N is divisible by 3 · 5 = 15. Statement II must be
true.
Since the second term is 7 greater than the first term,
(2a
1)
a = 7 so that a = 8. The sequence is 8,
Finally, remember that 2 is a prime number. So, N
15, 22, . . . You can either continue to write out the
could be 6, since 6 = 2 · 3. Statement III isn’t always
th
th
sequence until the 16
term, or realize that the 16
true, making C the correct answer.
term is 16a
15 = 16(8)
15 = 128
15 = 113.
pg. 10
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