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Contests
National and Regional Contests
USA Contests
MAA AMC
AMC 12/AHSME
1965 AMC 12/AHSME
1965 AMC 12/AHSME
Part of
AMC 12/AHSME
Subcontests
(40)
40
1
Hide problems
Integer Values in Polynomial
Let
n
n
n
be the number of integer values of
x
x
x
such that P \equal{} x^4 \plus{} 6x^3 \plus{} 11x^2 \plus{} 3x \plus{} 31 is the square of an integer. Then
n
n
n
is:
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<span class='latex-bold'>(A)</span>\ 4 \qquad <span class='latex-bold'>(B)</span>\ 3 \qquad <span class='latex-bold'>(C)</span>\ 2 \qquad <span class='latex-bold'>(D)</span>\ 1 \qquad <span class='latex-bold'>(E)</span>\ 0
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39
1
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Inserting Plugs
A foreman noticed an inspector checking a
3
"
3"
3"
-hole with a
2
"
2"
2"
-plug and a
1
"
1"
1"
-plug and suggested that two more gauges be inserted to be sure that the fit was snug. If the new gauges are alike, then the diameter,
d
d
d
, of each, to the nearest hundredth of an inch, is:
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87
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86
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83
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75
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71
<span class='latex-bold'>(A)</span>\ .87 \qquad <span class='latex-bold'>(B)</span>\ .86 \qquad <span class='latex-bold'>(C)</span>\ .83 \qquad <span class='latex-bold'>(D)</span>\ .75 \qquad <span class='latex-bold'>(E)</span>\ .71
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38
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Doing Work
A
A
A
takes
m
m
m
times as long to do a piece of work as
B
B
B
and
C
C
C
together;
B
B
B
takes
n
n
n
times as long as
C
C
C
and
A
A
A
together; and
C
C
C
takes
x
x
x
times as long as
A
A
A
and
B
B
B
together. Then
x
x
x
, in terms of
m
m
m
and
n
n
n
, is:
(A)
\ \frac {2mn}{m \plus{} n} \qquad
(B)
\ \frac {1}{2(m \plus{} n)} \qquad
(C)
\ \frac {1}{m \plus{} n \minus{} mn} \qquad
(D)
\ \frac {1 \minus{} mn}{m \plus{} n \plus{} 2mn} \qquad
(E)
\ \frac {m \plus{} n \plus{} 2}{mn \minus{} 1}
37
1
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Ratios of Various Lengths on Triangle
Point
E
E
E
is selected on side
A
B
AB
A
B
of triangle
A
B
C
ABC
A
BC
in such a way that AE: EB \equal{} 1: 3 and point
D
D
D
is selected on side
B
C
BC
BC
such that CD: DB \equal{} 1: 2. The point of intersection of
A
D
AD
A
D
and
C
E
CE
CE
is
F
F
F
. Then \frac {EF}{FC} \plus{} \frac {AF}{FD} is:
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2
<span class='latex-bold'>(A)</span>\ \frac {4}{5} \qquad <span class='latex-bold'>(B)</span>\ \frac {5}{4} \qquad <span class='latex-bold'>(C)</span>\ \frac {3}{2} \qquad <span class='latex-bold'>(D)</span>\ 2 \qquad <span class='latex-bold'>(E)</span>\ \frac {5}{2}
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36
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Limit of sum of lengths
Given distinct straight lines
O
A
OA
O
A
and
O
B
OB
OB
. From a point in
O
A
OA
O
A
a perpendicular is drawn to
O
B
OB
OB
; from the foot of this perpendicular a line is drawn perpendicular to
O
A
OA
O
A
. From the foot of this second perpendicular a line is drawn perpendicular to
O
B
OB
OB
; and so on indefinitely. The lengths of the first and second perpendiculars are
a
a
a
and
b
b
b
, respectively. Then the sum of the lengths of the perpendiculars approaches a limit as the number of perpendiculars grows beyond all bounds. This limit is:
(A)
\ \frac {b}{a \minus{} b} \qquad
(B)
\ \frac {a}{a \minus{} b} \qquad
(C)
\ \frac {ab}{a \minus{} b} \qquad
(D)
\ \frac {b^2}{a \minus{} b} \qquad
(E)
\ \frac {a^2}{a \minus{} b}
35
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Rectangle Length and Width
The length of a rectangle is
5
5
5
inches and its width is less than
4
4
4
inches. The rectangle is folded so that two diagonally opposite vertices coincide. If the length of the crease is
6
\sqrt {6}
6
, then the width is:
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2
<span class='latex-bold'>(A)</span>\ \sqrt {2} \qquad <span class='latex-bold'>(B)</span>\ \sqrt {3} \qquad <span class='latex-bold'>(C)</span>\ 2 \qquad <span class='latex-bold'>(D)</span>\ \sqrt {5} \qquad <span class='latex-bold'>(E)</span>\ \sqrt {\frac {11}{2}}
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Minimum Value
For
x
≥
0
x \ge 0
x
≥
0
the smallest value of \frac {4x^2 \plus{} 8x \plus{} 13}{6(1 \plus{} x)} is:
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12
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<span class='latex-bold'>(A)</span>\ 1 \qquad <span class='latex-bold'>(B)</span>\ 2 \qquad <span class='latex-bold'>(C)</span>\ \frac {25}{12} \qquad <span class='latex-bold'>(D)</span>\ \frac {13}{6} \qquad <span class='latex-bold'>(E)</span>\ \frac {34}{5}
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25
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33
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Factorial Problem
If the number
15
!
15!
15
!
, that is,
15
⋅
14
⋅
13
…
1
15 \cdot 14 \cdot 13 \dots 1
15
⋅
14
⋅
13
…
1
, ends with
k
k
k
zeros when given to the base
12
12
12
and ends with
h
h
h
zeros when given to the base
10
10
10
, then k \plus{} h equals:
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<span class='latex-bold'>(A)</span>\ 5 \qquad <span class='latex-bold'>(B)</span>\ 6 \qquad <span class='latex-bold'>(C)</span>\ 7 \qquad <span class='latex-bold'>(D)</span>\ 8 \qquad <span class='latex-bold'>(E)</span>\ 9
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x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
9
32
1
Hide problems
Selling and Reselling
An article costing
C
C
C
dollars is sold for
$
100
\$100
$100
at a lostt of
x
x
x
percent of the selling price. It is then resold at a profit of
x
x
x
percent of the new selling price
S
′
S'
S
′
. If the difference between
S
′
S'
S
′
and
C
C
C
is
1
1
9
1\frac {1}{9}
1
9
1
dollars, then
x
x
x
is:
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
undetermined
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
80
9
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
10
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
95
9
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
100
9
<span class='latex-bold'>(A)</span>\ \text{undetermined} \qquad <span class='latex-bold'>(B)</span>\ \frac {80}{9} \qquad <span class='latex-bold'>(C)</span>\ 10 \qquad <span class='latex-bold'>(D)</span>\ \frac {95}{9} \qquad <span class='latex-bold'>(E)</span>\ \frac {100}{9}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
undetermined
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
9
80
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
10
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
9
95
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
9
100
31
1
Hide problems
Logarithm Equality
The number of real values of
x
x
x
satisfying the equality (\log_2x)(\log_bx) \equal{} \log_ab, where
a
>
0
a > 0
a
>
0
,
b
>
0
b > 0
b
>
0
,
a
≠
1
a \neq 1
a
=
1
,
b
≠
1
b \neq 1
b
=
1
, is:
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
a finite integer greater than 2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
not finite
<span class='latex-bold'>(A)</span>\ 0 \qquad <span class='latex-bold'>(B)</span>\ 1 \qquad <span class='latex-bold'>(C)</span>\ 2 \qquad <span class='latex-bold'>(D)</span>\ \text{a finite integer greater than 2} \qquad <span class='latex-bold'>(E)</span>\ \text{not finite}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
a finite integer greater than 2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
not finite
30
1
Hide problems
Proving in Geometry
Let
B
C
BC
BC
of right triangle
A
B
C
ABC
A
BC
be the diameter of a circle intersecting hypotenuse
A
B
AB
A
B
in
D
D
D
. At
D
D
D
a tangent is drawn cutting leg
C
A
CA
C
A
in
F
F
F
. This information is not sufficient to prove that
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
D
F
bisects
C
A
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
D
F
bisects
∠
C
D
A
<span class='latex-bold'>(A)</span>\ DF \text{ bisects }CA \qquad <span class='latex-bold'>(B)</span>\ DF \text{ bisects }\angle CDA
<
s
p
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l
a
ss
=
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l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
D
F
bisects
C
A
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
D
F
bisects
∠
C
D
A
(C)
\ DF \equal{} FA \qquad
(D)
\ \angle A \equal{} \angle BCD \qquad
(E)
\ \angle CFD \equal{} 2\angle A
29
1
Hide problems
Students Taking Classes
Of
28
28
28
students taking at least one subject the number taking Mathematics and English only equals the number taking Mathematics only. No student takes English only or History only, and six students take Mathematics and History, but not English. The number taking English and History only is five times the number taking all three subjects. If the number taking all three subjects is even and non-zero, the number taking English and Mathematics only is:
<
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c
l
a
s
s
=
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l
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−
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o
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>
(
A
)
<
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>
5
<
s
p
a
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c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
B
)
<
/
s
p
a
n
>
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
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a
n
>
7
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
8
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
9
<span class='latex-bold'>(A)</span>\ 5 \qquad <span class='latex-bold'>(B)</span>\ 6 \qquad <span class='latex-bold'>(C)</span>\ 7 \qquad <span class='latex-bold'>(D)</span>\ 8 \qquad <span class='latex-bold'>(E)</span>\ 9
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
7
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
8
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
9
28
1
Hide problems
Escalator
An escalator (moving staircase) of
n
n
n
uniform steps visible at all times descends at constant speed. Two boys,
A
A
A
and
Z
Z
Z
, walk down the escalator steadily as it moves,
A
A
A
negotiating twice as many escalator steps per minute as
Z
Z
Z
.
A
A
A
reaches the bottom after taking
27
27
27
steps while
Z
Z
Z
reaches the bottom after taking
18
18
18
steps. Then
n
n
n
is:
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
63
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
54
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
45
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
36
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
30
<span class='latex-bold'>(A)</span>\ 63 \qquad <span class='latex-bold'>(B)</span>\ 54 \qquad <span class='latex-bold'>(C)</span>\ 45 \qquad <span class='latex-bold'>(D)</span>\ 36 \qquad <span class='latex-bold'>(E)</span>\ 30
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
63
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
54
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
45
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
36
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
30
27
1
Hide problems
Remainders of Polynomial Division
When y^2 \plus{} my \plus{} 2 is divided by y \minus{} 1 the quotient is
f
(
y
)
f(y)
f
(
y
)
and the remainder is
R
1
R_1
R
1
. When y^2 \plus{} my \plus{} 2 is divided by y \plus{} 1 the quotient is
g
(
y
)
g(y)
g
(
y
)
and the remainder is
R
2
R_2
R
2
. If R_1 \equal{} R_2 then
m
m
m
is:
(A)
\ 0 \qquad
(B)
\ 1 \qquad
(C)
\ 2 \qquad
(D)
\ \minus{} 1 \qquad
(E)
\ \text{an undetermined constant}
26
1
Hide problems
Arithmetic Means
For the numbers
a
a
a
,
b
b
b
,
c
c
c
,
d
d
d
,
e
e
e
define
m
m
m
to be the arithmetic mean of all five numbers;
k
k
k
to be the arithmetic mean of
a
a
a
and
b
b
b
;
l
l
l
to be the arithmetic mean of
c
c
c
,
d
d
d
, and
e
e
e
; and
p
p
p
to be the arithmetic mean of
k
k
k
and
l
l
l
. Then, no matter how
a
a
a
,
b
b
b
,
c
c
c
,
d
d
d
, and
e
e
e
are chosen, we shall always have:
(A)
\ m \equal{} p \qquad
(B)
\ m \ge p \qquad
(C)
\ m > p \qquad
(D)
\ m < p \qquad
(E)
\ \text{none of these}
25
1
Hide problems
Quadrilateral
Let
A
B
C
D
ABCD
A
BC
D
be a quadrilateral with
A
B
AB
A
B
extended to
E
E
E
so that \overline{AB} \equal{} \overline{BE}. Lines
A
C
AC
A
C
and
C
E
CE
CE
are drawn to form angle
A
C
E
ACE
A
CE
. For this angle to be a right angle it is necessary that quadrilateral
A
B
C
D
ABCD
A
BC
D
have:
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
all angles equal
<span class='latex-bold'>(A)</span>\ \text{all angles equal}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
all angles equal
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
all sides equal
<span class='latex-bold'>(B)</span>\ \text{all sides equal}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
all sides equal
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
two pairs of equal sides
<span class='latex-bold'>(C)</span>\ \text{two pairs of equal sides}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
two pairs of equal sides
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
one pair of equal sides
<span class='latex-bold'>(D)</span>\ \text{one pair of equal sides}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
one pair of equal sides
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
one pair of equal angles
<span class='latex-bold'>(E)</span>\ \text{one pair of equal angles}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
one pair of equal angles
24
1
Hide problems
Sequence
Given the sequence
1
0
1
11
,
1
0
2
11
,
1
0
3
11
,
…
,
1
0
n
11
10^{\frac {1}{11}},10^{\frac {2}{11}},10^{\frac {3}{11}},\ldots,10^{\frac {n}{11}}
1
0
11
1
,
1
0
11
2
,
1
0
11
3
,
…
,
1
0
11
n
, the smallest value of
n
n
n
such that the product of the first
n
n
n
members of this sequence exceeds
100000
100000
100000
is:
<
s
p
a
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c
l
a
s
s
=
′
l
a
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x
−
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7
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8
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11
<span class='latex-bold'>(A)</span>\ 7 \qquad <span class='latex-bold'>(B)</span>\ 8 \qquad <span class='latex-bold'>(C)</span>\ 9 \qquad <span class='latex-bold'>(D)</span>\ 10 \qquad <span class='latex-bold'>(E)</span>\ 11
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9
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11
23
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Absolute Value Inequality
If we write |x^2 \minus{} 4| < N for all
x
x
x
such that |x \minus{} 2| < 0.01, the smallest value we can use for
N
N
N
is:
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.
0301
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.
0349
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.
0399
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.
0401
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.
0499
<span class='latex-bold'>(A)</span>\ .0301 \qquad <span class='latex-bold'>(B)</span>\ .0349 \qquad <span class='latex-bold'>(C)</span>\ .0399 \qquad <span class='latex-bold'>(D)</span>\ .0401 \qquad <span class='latex-bold'>(E)</span>\ .0499 \qquad
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.0301
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.0349
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.0399
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.0401
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.0499
22
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Equality holds when...
If
a
2
≠
0
a_2 \neq 0
a
2
=
0
and
r
r
r
and
s
s
s
are the roots of a_0 \plus{} a_1x \plus{} a_2x^2 \equal{} 0, then the equality a_0 \plus{} a_1x \plus{} a_2x^2 \equal{} a_0\left (1 \minus{} \frac {x}{r} \right ) \left (1 \minus{} \frac {x}{s} \right ) holds:
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)
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for all values of
x
,
a
0
≠
0
<span class='latex-bold'>(A)</span>\ \text{for all values of }x, a_0\neq 0
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for all values of
x
,
a
0
=
0
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for all values of
x
<span class='latex-bold'>(B)</span>\ \text{for all values of }x
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for all values of
x
(C)
\ \text{only when }x \equal{} 0
(D)
\ \text{only when }x \equal{} r \text{ or }x \equal{} s
(E)
\ \text{only when }x \equal{} r \text{ or }x \equal{} s, a_0 \neq 0
21
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Choosing x
It is possible to choose
x
>
2
3
x > \frac {2}{3}
x
>
3
2
in such a way that the value of \log_{10}(x^2 \plus{} 3) \minus{} 2 \log_{10}x is
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negative
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zero
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one
<span class='latex-bold'>(A)</span>\ \text{negative} \qquad <span class='latex-bold'>(B)</span>\ \text{zero} \qquad <span class='latex-bold'>(C)</span>\ \text{one}
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negative
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zero
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(
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one
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(
D
)
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smaller than any positive number that might be specified
<span class='latex-bold'>(D)</span>\ \text{smaller than any positive number that might be specified}
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smaller than any positive number that might be specified
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greater than any positive number that might be specified
<span class='latex-bold'>(E)</span>\ \text{greater than any positive number that might be specified}
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greater than any positive number that might be specified
20
1
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Arithmetic Progression
For every
n
n
n
the sum of
n
n
n
terms of an arithmetic progression is 2n \plus{} 3n^2. The
r
r
r
th term is:
(A)
\ 3r^2 \qquad
(B)
\ 3r^2 \plus{} 2r \qquad
(C)
\ 6r \minus{} 1 \qquad
(D)
\ 5r \plus{} 5 \qquad
(E)
\ 6r \plus{} 2 \qquad
19
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Polynomial Divisibility
If x^4 \plus{} 4x^3 \plus{} 6px^2 \plus{} 4qx \plus{} r is exactly divisible by x^3 \plus{} 3x^2 \plus{} 9x \plus{} 3, the value of (p \plus{} q)r is:
(A)
\ \minus{} 18 \qquad
(B)
\ 12 \qquad
(C)
\ 15 \qquad
(D)
\ 27 \qquad
(E)
\ 45 \qquad
18
1
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Approximation
If 1 \minus{} y is used as an approximation to the value of \frac {1}{1 \plus{} y},
∣
y
∣
<
1
|y| < 1
∣
y
∣
<
1
, the ratio of the error made to the correct value is:
(A)
\ y \qquad
(B)
\ y^2 \qquad
(C)
\ \frac {1}{1 \plus{} y} \qquad
(D)
\ \frac {y}{1 \plus{} y} \qquad
(E)
\ \frac {y^2}{1 \plus{} y}\qquad
17
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Biconditional Statements
Given the true statement: The picnic on Sunday will not be held only if the weather is not fair. We can then conclude that:
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If the picnic is held, Sunday’s weather is undoubtedly fair.
<span class='latex-bold'>(A)</span>\ \text{If the picnic is held, Sunday's weather is undoubtedly fair.}
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If the picnic is held, Sunday’s weather is undoubtedly fair.
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)
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>
If the picnic is not held, Sunday’s weather is possibly unfair.
<span class='latex-bold'>(B)</span>\ \text{If the picnic is not held, Sunday's weather is possibly unfair.}
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(
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)
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>
If the picnic is not held, Sunday’s weather is possibly unfair.
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>
(
C
)
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>
If it is not fair Sunday, the picnic will not be held.
<span class='latex-bold'>(C)</span>\ \text{If it is not fair Sunday, the picnic will not be held.}
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(
C
)
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If it is not fair Sunday, the picnic will not be held.
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−
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>
(
D
)
<
/
s
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>
If it is fair Sunday, the picnic may be held.
<span class='latex-bold'>(D)</span>\ \text{If it is fair Sunday, the picnic may be held.}
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(
D
)
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/
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>
If it is fair Sunday, the picnic may be held.
<
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s
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(
E
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<
/
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p
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>
If it is fair Sunday, the picnic must be held.
<span class='latex-bold'>(E)</span>\ \text{If it is fair Sunday, the picnic must be held.}
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If it is fair Sunday, the picnic must be held.
16
1
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Area of Triangle
Let line
A
C
AC
A
C
be perpendicular to line
C
E
CE
CE
. Connect
A
A
A
to
D
D
D
, the midpoint of
C
E
CE
CE
, and connect
E
E
E
to
B
B
B
, the midpoint of
A
C
AC
A
C
. If
A
D
AD
A
D
and
E
B
EB
EB
intersect in point
F
F
F
, and \overline{BC} \equal{} \overline{CD} \equal{} 15 inches, then the area of triangle
D
F
E
DFE
D
FE
, in square inches, is:
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2
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75
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(
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)
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>
15
2
105
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100
<span class='latex-bold'>(A)</span>\ 50 \qquad <span class='latex-bold'>(B)</span>\ 50\sqrt {2} \qquad <span class='latex-bold'>(C)</span>\ 75 \qquad <span class='latex-bold'>(D)</span>\ \frac {15}{2}\sqrt {105} \qquad <span class='latex-bold'>(E)</span>\ 100
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2
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75
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2
15
105
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100
15
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Base numbers
The symbol
2
5
b
25_b
2
5
b
represents a two-digit number in the base
b
b
b
. If the number
5
2
b
52_b
5
2
b
is double the number
2
5
b
25_b
2
5
b
, then
b
b
b
is:
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7
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8
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9
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11
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>
12
<span class='latex-bold'>(A)</span>\ 7 \qquad <span class='latex-bold'>(B)</span>\ 8 \qquad <span class='latex-bold'>(C)</span>\ 9 \qquad <span class='latex-bold'>(D)</span>\ 11 \qquad <span class='latex-bold'>(E)</span>\ 12
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Hide problems
Sum of coefficients
The sum of the numerical coefficients in the complete expansion of (x^2 \minus{} 2xy \plus{} y^2)^7 is:
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128
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2
<span class='latex-bold'>(A)</span>\ 0 \qquad <span class='latex-bold'>(B)</span>\ 7 \qquad <span class='latex-bold'>(C)</span>\ 14 \qquad <span class='latex-bold'>(D)</span>\ 128 \qquad <span class='latex-bold'>(E)</span>\ 128^2
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8
2
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System with Equation, Inequality
Let
n
n
n
be the number of number-pairs
(
x
,
y
)
(x,y)
(
x
,
y
)
which satisfy 5y \minus{} 3x \equal{} 15 and x^2 \plus{} y^2 \le 16. Then
n
n
n
is:
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more than two, but finite
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greater than any finite number
<span class='latex-bold'>(A)</span>\ 0 \qquad <span class='latex-bold'>(B)</span>\ 1 \qquad <span class='latex-bold'>(C)</span>\ 2 \qquad <span class='latex-bold'>(D)</span>\ \text{more than two, but finite} \qquad <span class='latex-bold'>(E)</span>\ \text{greater than any finite number}
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more than two, but finite
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greater than any finite number
12
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Side length of Rhombus
A rhombus is inscribed in triangle
A
B
C
ABC
A
BC
in such a way that one of its vertices is
A
A
A
and two of its sides lie along
A
B
AB
A
B
and
A
C
AC
A
C
. If \overline{AC} \equal{} 6 inches, \overline{AB} \equal{} 12 inches, and \overline{BC} \equal{} 8 inches, the side of the rhombus, in inches, is:
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1
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5
<span class='latex-bold'>(A)</span>\ 2 \qquad <span class='latex-bold'>(B)</span>\ 3 \qquad <span class='latex-bold'>(C)</span>\ 3 \frac {1}{2} \qquad <span class='latex-bold'>(D)</span>\ 4 \qquad <span class='latex-bold'>(E)</span>\ 5
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Statement Truth
Consider the statements: I: (\sqrt { \minus{} 4})(\sqrt { \minus{} 16}) \equal{} \sqrt {( \minus{} 4)( \minus{} 16)}, II: \sqrt {( \minus{} 4)( \minus{} 16)} \equal{} \sqrt {64}, and \sqrt {64} \equal{} 8. Of these the following are incorrect.
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none
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I only
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II only
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III only
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I and III only
<span class='latex-bold'>(A)</span>\ \text{none} \qquad <span class='latex-bold'>(B)</span>\ \text{I only} \qquad <span class='latex-bold'>(C)</span>\ \text{II only} \qquad <span class='latex-bold'>(D)</span>\ \text{III only} \qquad <span class='latex-bold'>(E)</span>\ \text{I and III only}
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I and III only
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Quadratic Inequality
The statement x^2 \minus{} x \minus{} 6 < 0 is equivalent to the statement:
(A)
\ \minus{} 2 < x < 3 \qquad
(B)
\ x > \minus{} 2 \qquad
(C)
\ x < 3
(D)
\ x > 3 \text{ and }x < \minus{} 2 \qquad
(E)
\ x > 3 \text{ and }x < \minus{} 2
9
1
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Vertex of Parabola
The vertex of the parabola y \equal{} x^2 \minus{} 8x \plus{} c will be a point on the
x
x
x
-axis if the value of
c
c
c
is:
(A)
\ \minus{} 16 \qquad
(B)
\ \minus{} 4 \qquad
(C)
\ 4 \qquad
(D)
\ 8 \qquad
(E)
\ 16
8
1
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Length of segment inside Triangle
One side of a given triangle is
18
18
18
inches. Inside the triangle a line segment is drawn parallel to this side forming a trapezoid whose area is one-third of that of the triangle. The length of this segment, in inches, is:
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2
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9
<span class='latex-bold'>(A)</span>\ 6\sqrt {6} \qquad <span class='latex-bold'>(B)</span>\ 9\sqrt {2} \qquad <span class='latex-bold'>(C)</span>\ 12 \qquad <span class='latex-bold'>(D)</span>\ 6\sqrt {3} \qquad <span class='latex-bold'>(E)</span>\ 9
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2
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3
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Reciprocals of Roots
The sum of the reciprocals of the roots of the equation ax^2 \plus{} bx \plus{} c \equal{} 0 is:
(A)
\ \frac {1}{a} \plus{} \frac {1}{b} \qquad
(B)
\ \minus{} \frac {c}{b} \qquad
(C)
\ \frac {b}{c} \qquad
(D)
\ \minus{} \frac {a}{b} \qquad
(E)
\ \minus{} \frac {b}{c}
6
1
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Logarithm as Exponent
If 10^{\log_{10}9} \equal{} 8x \plus{} 5 then
x
x
x
equals:
(A)
\ 0 \qquad
(B)
\ \frac {1}{2} \qquad
(C)
\ \frac {5}{8} \qquad
(D)
\ \frac {9}{8} \qquad
(E)
\ \frac {2\log_{10}3 \minus{} 5}{8}
5
1
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Repeating Decimal as Fraction
When the repeating decimal
0.363636
…
0.363636\ldots
0.363636
…
is written in simplest fractional form, the sum of the numerator and denominator is:
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150
<span class='latex-bold'>(A)</span>\ 15 \qquad <span class='latex-bold'>(B)</span>\ 45 \qquad <span class='latex-bold'>(C)</span>\ 114 \qquad <span class='latex-bold'>(D)</span>\ 135 \qquad <span class='latex-bold'>(E)</span>\ 150
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114
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135
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150
4
1
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Number of Points Equidistant from Lines
Line
l
2
l_2
l
2
intersects line
l
1
l_1
l
1
and line
l
3
l_3
l
3
is parallel to
l
1
l_1
l
1
. The three lines are distinct and lie in a plane. The number of points equidistant from all three lines is:
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1
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2
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4
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8
<span class='latex-bold'>(A)</span>\ 0 \qquad <span class='latex-bold'>(B)</span>\ 1 \qquad <span class='latex-bold'>(C)</span>\ 2 \qquad <span class='latex-bold'>(D)</span>\ 4 \qquad <span class='latex-bold'>(E)</span>\ 8
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1
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2
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4
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8
3
1
Hide problems
Negative Exponent
The expression (81)^{ \minus{} 2^{ \minus{} 2}} has the same value as:
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1
81
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3
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81
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8
1
4
<span class='latex-bold'>(A)</span>\ \frac {1}{81} \qquad <span class='latex-bold'>(B)</span>\ \frac {1}{3} \qquad <span class='latex-bold'>(C)</span>\ 3 \qquad <span class='latex-bold'>(D)</span>\ 81 \qquad <span class='latex-bold'>(E)</span>\ 81^4
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81
1
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3
1
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3
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81
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8
1
4
2
1
Hide problems
Ratio of lengths of sides
A regular hexagon is inscribed in a circle. The ratio of the length of a side of the hexagon to the length of the shorter of the arcs intercepted by the side, is:
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6
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:
π
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:
π
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:
π
<span class='latex-bold'>(A)</span>\ 1: 1 \qquad <span class='latex-bold'>(B)</span>\ 1: 6 \qquad <span class='latex-bold'>(C)</span>\ 1: \pi \qquad <span class='latex-bold'>(D)</span>\ 3: \pi \qquad <span class='latex-bold'>(E)</span>\ 6: \pi
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:
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:
6
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:
π
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:
π
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6
:
π
1
1
Hide problems
Number of real values of x
The number of real values of
x
x
x
satisfying the equation 2^{2x^2 \minus{} 7x \plus{} 5} \equal{} 1 is:
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more than 4
<span class='latex-bold'>(A)</span>\ 0 \qquad <span class='latex-bold'>(B)</span>\ 1 \qquad <span class='latex-bold'>(C)</span>\ 2 \qquad <span class='latex-bold'>(D)</span>\ 3 \qquad <span class='latex-bold'>(E)</span>\ \text{more than 4}
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more than 4