SAT Examples v6 - Collier High School

```SAT
1.
The figure shown is a square divided into two non-overlapping regions. What is the
greatest number of non-overlapping regions that can be obtained by drawing any
(A) 4
(B) 5
(C) 6
(D) 7
(E)
8
(D) 7
This is a good example of “thinking outside the box.” The
obvious choice is 6, which is wrong.
The NOT so obvious but correct choice is D (7).
SAT
2.
A 50 foot wire runs from the roof of a building to the top of a 10 foot pole 14 feet
from across the street. How much taller would the pole have to be if the street
were 16 feet wider and the wire remained the same length ?
(A)
2 feet
(B) 8 feet
50
(C) 14 feet
(D) 16 feet
(E)
18 feet
(E) 8 feet
a
a
50
x
14
a2 + 142 = 502
a2 + 196 = 2500
a2 = 2304
a = 48
10
x
14
10
16
(a – x) 2 + 302 = 502
(48 – x) 2 + 900 = 2500
(48 – x) 2 = 1600
48 – x = 40
8=x
SAT
3.
A dress is selling for \$100 after a 20 percent discount. What was the original selling
price?
(A) \$ 200
(B) \$ 125
(C) \$ 120
(D) \$ 80
(E)
\$ 75
(B) \$ 125
What do the given values really represent?
This question really says that \$100 is 80% of
the original selling price.
So,
\$ 100 = 80 % of x
100 = .8 x
125 = x
SAT
4.
If 3 parallel lines are cut by 3 non-parallel lines, what is the maximum number of
intersections possible ?
1
(A) 9
2
(B) 10
3
4
5
(C) 11
(D) 12
6
(E)
9
13
7
10
8
12
11
(D) 12
The trick here is to make sure each non-parallel line
intersects all 3 parallel lines and each other where
there are NO COMMON intersections.
SAT
5.
If the average of x, y, and 80 is 6 more than the average of y, z, and 80, what is the
value of x – z ?
(A) 2
(B) 3
(C) 6
(D) 18
(E)
It cannot be determined from
the information given.
x + y + 80 = y + z + 80 + 6
3
3
x + y + 80 = y + z + 80 + 18
x = z + 18
x – z = 18
(D) 18
SAT
If x > 0, which of the following is equivalent to x3 ?
I.
6.
x+x½
II. (x½ ) 3
III. (x2) (x –½ )
(A) None
(B) I and II only
(C) I and III only
(D) II and III only
(E)
I, II, and III
(D) II and III only
Choice option I. x + x ½ has no exponent rule for
adding exponent expressions with like bases, which
means choices B,C and E are NOT viable. Choice II
works because x½ is the same as square root of x,
which means A is not valid, leaving Choice D as the only
possible choice.
SAT
7.
If
(A)
5
(B)
7
5 = x – 2 , what is the value of (x – 2)2 ?
(C) 5
(D) 9
(E)
25
If
(C) 5
5 = x – 2 , then ( 55 )2 = (x – 2)2
5 = (x – 2)2
SAT
8.
In the expression x2 + kx + 12, k is an integer and k < 0. Which of the following is a
possible value of k ?
(A) – 13
(B) – 12
(C) – 6
(D) 7
(E)
It cannot be determined from the information given
x2 + kx + 12 = (x – 1 ) ( x – 12 ) = x2 – 13x + 12
(A) – 13
SAT
9.
If 2x – 3y = 5, what is the value of 4x2 – 12 xy + 9y2 ?
(A)
5
(B) 5
(C) 12
(D) 25
(E)
100
D) 25
Since 2x – 3y = 5 and 4x2 – 12 xy + 9y2 = (2x – 3y)2
= 52 = 25
SAT
10. x
(3 – 3 ) = x (x + 7 )
x
(A) – 7
(B) – 1
(C) 1
(D) 3
(E)
7
(C) – 1
Since 3x – 3 = x (x + 7)
Then, 0 = x2 + 4 x + 3
and 0 = (x + 3) (x + 1)
Thus, x = – 3 or – 1
```