SAT Math Hard Practice Quiz Answers
Data, Statistics, and Probability
1.
A
(Estimated Difficulty Level: 4)
The first nine terms of the sequence are:
n
,
n
+ 5,
n
+ 10,
n
+ 15,
. . .
,
n
+ 40
.
(You should probably write all nine terms out to avoid
mistakes.) Adding these terms up gives: 9
n
+ 180. The
average is the sum (9
n
+ 180) divided by the number of
terms (9). The average is then: (9
n
+ 180)
/
9 =
n
+ 20.
2.
B
(Estimated Difficulty Level: 5)
The average of a set of numbers is the sum of the num-
bers divided by the number of numbers:
average =
sum
N
.
We can solve this equation for the sum:
sum = average
×
N.
Here, since there are 7 numbers and the average is 12,
the sum of the numbers is 7
×
12 = 84. The sum of the
new set of numbers is 7
×
15 = 105. Now, suppose that
the seven numbers are
a
,
b
,
c
,
d
,
e
,
f
, and
g
, and that
g
gets replaced with the number 6. Then, we have:
a
+
b
+
c
+
d
+
e
+
f
+
g
= 84
,
and
a
+
b
+
c
+
d
+
e
+
f
+ 6 = 105
.
The second equation says that
a
+
b
+
c
+
d
+
e
+
f
= 99.
Substituting into the first equation gives 99 +
g
= 84 so
that
g
=
−
15.
3.
E
(Estimated Difficulty Level: 4)
Using the definition of average gives:
a
+
b
+
c
3
= 100
so that
a
+
b
+
c
= 300. Since
a
,
b
, and
c
are all positive,
the smallest possible value for any of the numbers is 1.
The largest possible value of one of the three numbers
then occurs when the other two numbers are both 1. In
this case, the numbers are 1, 1, and 298, so that the
largest possible value is 298. Answer E can not be a
possible value, so it is the correct answer.
4.
E
(Estimated Difficulty Level: 5)
Plug in real numbers for set
M
to make this problem
concrete. For example, if
M
is the set of consecutive
integers from 1 to 5, then the median and average are
both 3. If
M
is the set of consecutive integers from 1
to 4, then the median and average are both 2
.
5. From
these examples, we can see that the number of numbers
in set
M
needs to odd, otherwise the median is not an
integer. Choice II must be true.
Also, if the number of numbers in a set of consecutive
integers is odd, then when the first number is odd, the
last number is odd. Or, when the first number is even,
the last number is even. This is because the difference
of the largest number and the smallest number will be
even when the number of numbers is odd. Choice III
must then be true, since the sum of two odd numbers
or two even numbers is an even number.
At this point, the only answer with choices II and III is
answer E, so that must be the correct answer. Why is
choice I also correct? The average of a set of consecutive
integers is equal to the average of the first and the last
integers in the set. The average of two integers that are
both odd or both even is the integer halfway between
the two, which is also the median of the set. Whew!
erikthered.com/tutor
pg. 18
