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AMC 12/AHSME
1990 AMC 12/AHSME
1
1990 AHSME #1: Equation in X
1990 AHSME #1: Equation in X
Source:
January 1, 2013
AMC
Problem Statement
If
x
/
4
2
=
4
x
/
2
\dfrac{x/4}{2}=\dfrac{4}{x/2}
2
x
/4
=
x
/2
4
then
x
=
x=
x
=
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
±
1
/
2
<
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
>
±
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
±
8
<span class='latex-bold'>(A) </span>\pm 1/2\qquad <span class='latex-bold'>(B) </span>\pm 1\qquad <span class='latex-bold'>(C) </span>\pm 2\qquad <span class='latex-bold'>(D) </span>\pm 4\qquad <span class='latex-bold'>(E) </span>\pm 8
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
±
1/2
<
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
>
±
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
±
8
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