Sat Math Medium Practice Quiz With Answer Key Page 16
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16SAT Math Medium Practice Quiz Answers
Data, Statistics, and Probability
3. 3/5 or 0.6
(Estimated Difficulty Level: 3)
There are 15 possible values of xy:
1. D
(Estimated Difficulty Level: 3)
2 × 11 = 22
2 × 13 = 26
One way to do this is to find all the numbers in set N ,
2 × 15 = 30
and then calculate the average of those numbers. (Your
calculator may be useful here.) Set N is:
3 × 11 = 33
. . .
0.5, 1, 1.5, 2, 2.5, 3, 3.5
and so forth. You should find that 9 of these are even, so
so that the average of the numbers in set N is 14/7 = 2.
that the probability that xy is even is 9/15 = 3/5 = 0.6.
It may be easier and less error-prone to notice that if
You may also notice that an even product will result
you divide a set of numbers by 2, the average of the set
only when you multiply by one of 2, 4, or 6, so that
will get divided by 2 as well. Since the average of set
there will be 3 × 3 = 9 even numbers out of 3 × 5 = 15
M is 4, the average of set N is 2.
numbers total.
4. D
(Estimated Difficulty Level: 3+)
For choice I, suppose that the set is a group of ten 2’s.
2. 1/2 or 0.5
(Estimated Difficulty Level: 3)
The median of this set is 2, which is an integer, so choice
I could be true.
Often, a good way to start probability problems on the
For choice II, note that for a set of ten numbers, there
SAT is to list all the possibilities, and then count the
is no single “middle” number, so the median is the av-
number that match the given criteria. Using “c” for cat,
th
erage of the two middle numbers (the 5
largest and
“d” for dog, and so forth, all the possible combinations
th
6
largest). But the average of any two consecutive
of two pets are:
integers is never an integer, so choice II can never be
true. (As an example, suppose that the set is the inte-
cd
gers from 1 to 10. The two middle numbers are 5 and
cb
6 and the median is (5 + 6)/2 = 5.5.)
cl
db
This reasoning helps with choice III: suppose that the
dl
th
th
middle two numbers (the 5
largest and 6
largest) are
bl
7 and 11. Then the median is (7 + 11)/2 = 9, which is
an integer, so choice III could be true, making answer
Since three of the six possible combinations include the
D correct.
lizard, the probability that the two pets chosen by the
family include the lizard is 3/6 = 1/2 = 0.5.
pg. 16
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