SAT Math Hard Practice Quiz Answers
13.
A
(Estimated Difficulty Level: 4)
A good “skip-the-algebra” way to do this problem is to
use the answers by plugging them into
x
until the two
given equations work. Using answer A, you should find
that (
−
3+1)
2
= (
−
2)
2
= 4 and (
−
3
−
1)
2
= (
−
4)
2
= 16,
so answer A is correct.
You must have the algebraic solution, you say? Try
taking the square root of both sides of the equations,
but don’t forget that there are two possible solutions
when you do this. The first equation gives:
x
+ 1 =
±
2
so that
x
= 1 or
x
=
−
3. The second equation gives
x
−
1 =
±
4 so that
x
= 5 or
x
=
−
3. The only solution
that works for both equations is
x
=
−
3.
14.
96
(Estimated Difficulty Level: 5)
Since the tick marks correspond to consecutive integers,
and it takes four “steps” to go from
x/
12 to
x/
8, we
know that
x/
8 is four greater than
x/
12. (Or, think of
the spaces between the tick marks: there are four spaces
and each space is length 1, so the distance from
x/
12
to
x/
8 is 4.) In equation form:
x
8
=
x
12
+ 4
.
Multiplying both sides by 24 gives: 3
x
= 2
x
+ 4
·
24 so
that
x
= 96.
15.
B
(Estimated Difficulty Level: 5)
First, recall that if
y
is proportional to
x
, then
y
=
kx
for some constant
k
. So, “
y
increased by 12 is directly
proportional to
x
decreased by 6” translates into the
math equation:
y
+ 12 =
k
(
x
−
6). Plugging in
y
= 2
and
x
= 8 gives 14 =
k
·
2 so that
k
= 7. Our equation
is now:
y
+ 12 = 7(
x
−
6). Plugging in 16 for
y
gives
28 = 7(
x
−
6) so that
x
−
6 = 4, or
x
= 10.
16.
75
/
2 or 37
.
5
(Estimated Difficulty Level: 5)
To make this problem more concrete, make up a number
for the circumference of the racetrack. It doesn’t really
matter what number you use; I’ll use 75 feet. Since
speed is distance divided by time, the speed of car A
is 75
/
15 = 5 feet per second, and the speed of car B is
75
/
25 = 3 feet per second. (I picked 75 mostly because
it is divided evenly by 15 and 25.) Every second, car A
gains 2 feet on car B. To pass car B, car A must gain
75 feet on car B. This will require 75
/
2 = 37
.
5 seconds.
You may be thinking, “Whoa, tricky solution!” Here
is the mostly straightforward but somewhat tedious al-
gebraic solution. Once again, I’ll use 75 feet for the
circumference of the track. Suppose that you count
time from when car A first passes car B. Then, car A
travels a distance (75
/
15)
t
= 5
t
feet after
t
seconds.
(Remember that distance = speed
×
time.) For ex-
ample, after 15 seconds, car A has traveled a distance
5
·
15 = 75 feet, and after 30 seconds, car A has traveled
a distance 5
·
30 = 150 feet. Similarly, car B travels a
distance (75
/
25)
t
= 3
t
feet after
t
seconds. When the
two cars pass again, car A has traveled 75 feet more
than car B: 5
t
= 3
t
+ 75. Solving for
t
gives: 2
t
= 75,
or
t
= 75
/
2 = 37
.
5 seconds.
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