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David Essner Mathematics Competition XXXI, 2011-2012
Preliminary Exam
1.
Tom, Alice and John took an exam. Alice scored 80. Tom scored 10 more than the average of
the three, while John scored 16 less than the average of the three. The average of the three was
then
(a) 72
(b) 74
(c) 75
2
3
(d) 76
1
3
(e) 78
Soln:
(b)
Let
N
be the average. Then 3
N
= 80 + (
N
+ 10) + (
N
-
16) or 3
N
= 2
N
+ 74
2.
Two sides of an isosceles triangle have length 2 and 5. What is the area of the triangle?
(a) 5
(b) 2
√
6
(c)
√
21
(d) 2
√
5
(e) There is more than one possible value.
Soln:
(b)
The other side is 5. If
h
is the altitude to the side of length 2 then
h
2
= 5
2
-
1
2
= 24.
The area is then (1
/
2)(2)(2
√
6).
3.
What is the sum 1
-
3+5
-
7+9
-
. . .
+81? (The terms increase in magnitude by 2 and alternate
in sign.)
(a)
-
15
(b) 1
(c) 27
(d) 39
(e) 41
Soln:
(e)
(1
-
3) + (5
-
7) + (9
-
11) +
. . .
+ (77
-
79) + 81 = (20)(
-
2) + 81
4.
What is the area of a rectangle if the diagonals have length 1 and 60
◦
is an angle of their
intersection?
(a) 1
/
2
(b)
√
2
/
2
(c)
√
3
/
2
(d)
√
3
/
4
(e) (
√
3 + 1)
/
2
Soln:
(d)
The rectangle is composed of two 30
◦
,
60
◦
right triangles each of which has hy-
potenuse 1 and sides of length 1/2 and
√
3
/
2. The area of each triangle is then
√
3
/
8.
5.
A multiple choice test has 30 questions and 5 choices for each question. If a student answers all
30 questions and the score is [number right - (number wrong/4)] then which of the following is
a possible score?
(a)
-
10
(b) 5
.
25
(c) 7
.
75
(d) 8
.
75
(e) 9
.
25
Soln:
(d)
If
R
is the number right then the score is
R
-
(30
-
R
)
/
4 = 5(
R
-
6)
/
4. Thus the
answer is a multiple of 5/4. Note 8.75 = 7(5/4).
6.
Line
L
1
has slope 1
/
2 and Line
L
2
has slope 1
/
3. If
L
1
and
L
2
have the same
y
–intercept
b
and
the sum of the
x
–intercepts of
L
1
and
L
2
is 10, then
b
equals
(a) 5
/
6
(b)
-
6
/
5
(c)
-
2
(d) 3
/
2
(e) 2
Soln:
(c)
The equations of
L
1
,
L
2
are
y
=
x/
2 +
b
and
y
=
x/
3 +
b
. Setting
y
= 0 gives the
x
intercepts as
-
2
b
and
-
3
b
. From
-
5
b
= 10 it follows that
b
=
-
2.
7.
What is the value of (log
2
3)(log
3
4)?
(a) 3
/
4
(b) 4
/
3
(c) 3
/
2
(d) 2
(e) 8
/
3
Soln:
(d)
Let
x
= log
2
3 and
y
= log
3
4. Then 2
x
= 3 and 3
y
= 4. Hence 2
xy
= 3
y
= 4 and
xy
= 2.
8.
If 0
< x < π/
2 and sin
x
= 2 cos
x
then (sin
x
)(cos
x
) equals
(a) 1
/
3
(b) 2
/
5
(c) 1
/
5
(d) 3
/
8
(e)
√
3
/
4
Soln:
(b)
1
-
cos
2
x
= sin
2
x
= 4 cos
2
x
implies 5 cos
2
x
= 1 and hence cos
2
x
= 1
/
5 and
sin
2
x
= 4
/
5. Thus (sin
2
x
)(cos
2
x
) = 4
/
25 and (sin
x
)(cos
x
) = 2
/
5.
9.
How many positive integer pairs (
m, n
) satisfy the equation 2
m
+ 7
n
= 835?
(a) 44
(b) 51
(c) 60
(d) 71
(e) 119
Soln:
(c)
n
can be any odd integer from 1 to 119 inclusive. There are (1 + 119)
/
2 = 60 such
integers.
10.
From a point
P
two tangent lines are drawn to a circle
C
. If
A, B
are the tangent points,
O
is
the center of
C
, and
negationslash
APB
= 30
◦
then
negationslash
AOB
equals
(a) 60
◦
(b) 90
◦
(c) 120
◦
(d) 150
◦
(e) 180
◦
Soln:
(d)
negationslash
AOB
= 360
◦
-
negationslash
APB
-
negationslash
PAO
-
negationslash
PBO
= 360
◦
-
30
◦
-
90
◦
-
90
◦
= 150
◦
.
11.
Let the function
f
satisfy
f
(
xy
) =
f
(
x
)
/y
for all positive numbers
x, y
. If
f
(5) = 10 then what
is the value of
f
(8)?
(a) 4
(b) 16
/
5
(c) 32
/
5
(d) 25
/
4
(e) There are many possible answers.
Soln:
(d)
f
(8) =
f
(5(8
/
5)) = 10
/
(8
/
5) = 25
/
4
12.
A full glass contains a mixture of 4
/
5 water and 1
/
5 alcohol. First 1
/
4 of the content is removed
and replaced with alcohol.
Then 1
/
3 of the resulting mixture is removed and replaced with
alcohol. What fraction of the final mixture is alcohol?
(a) 13
/
20
(b) 11
/
20
(c) 2
/
5
(d) 3
/
5
(e) 47
/
60
Soln:
(d)
After the first step, the fraction of water is (4/5)(3/4) = 3/5. After the second step,
the fraction of water is (3/5)(2/3) = 2/5. The final fraction of alcohol is then 1 - 2/5 = 3/5.
13.
If
a, b
are real numbers,
a
+
b
= 3 and
a
2
+
b
2
= 45 then the value of
a
3
+
b
3
is
(a) 27
(b) 87
(c) 189
(d) 135
(e) not uniquely determined.
Soln:
(c)
a
3
+
b
3
= (
a
+
b
)(
a
2
-
ab
+
b
2
). From 9
-
45 = (
a
+
b
)
2
-
(
a
2
+
b
2
) = 2
ab
it follows
that
ab
=
-
18. Thus
a
3
+
b
3
= 3(45 + 18) = 189.
14.
Which of the following sets of three numbers form the lengths of the sides of an obtuse triangle?
(a) 5
,
6
,
12
(b) 4
,
6
,
7
(c) 4
,
5
,
6
(d) 3
,
4
,
5
(e) 2
,
3
,
4
Soln:
(e)
An obtuse angle is opposite the longest side. From the law of cosines if the sides
are
a, b, c
and
c
is the longest side then
c
2
-
a
2
-
b
2
is positive if and only if the angle opposite
c
is obtuse. Note 4
2
>
2
2
+ 3
2
. Also note the numbers in (a) are not the sides of a triangle.
15.
In the Cartesian plane points
P, Q
are respectively in the first and fourth quadrants and each is
a distance of 10 from the origin
O
. If segments
OP
,
OQ
respectively make an angle of 40
◦
, 20
◦
with the positive
x
–axis and
M
is the midpoint of
PQ
, then the distance from
O
to
M
is
(a) 4
√
2
(b) 5
√
3
(c) 5
(d) 5
√
2
(e) 4
√
3
Soln:
(b)
The triangle
OPQ
is equilateral and triangles
OPM
and
OQM
are congruent
30
◦
,
60
◦
right triangles. The length of
OM
is 10 sin 60
◦
.
16.
What is the real number value of
x
such that 2
x
+ 4
x
= 12?
(a) log
2
3
(b) log
3
2
(c) log
3
4
(d) log
4
3
(e) log
2
12
Soln:
(a)
Let
y
= 2
x
, then
y
2
= 4
x
.
Hence
y
2
+
y
-
12 = (
y
+ 4)(
y
-
3) = 0 from which
y
=
-
4
,
3. Only
y
= 3 gives a real number value for
x
, and 2
x
= 3 implies
x
= log
2
3.
17.
The length of a chord of a circle is 14 and the shortest distance from the midpoint of the chord
to the circle is 5. Then the radius of the circle equals
(a) 28
/
5
(b) 37
/
5
(c) 28
/
3
(d) 19
/
2
(e) 2
√
6
Soln:
(b)
Let
O
be the center of the circle,
A
be the center of the chord and
B
an endpoint of
the chord. Then applying the Pythagorean Theorem to the triangle
AOB
gives (
r
-
5)
2
+7
2
=
r
2
from which
r
= 37
/
5.
18.
In a race
A
,
B
, and
C
run at a constant speed. When
A
finishes,
B
is 20 feet behind and
C
is
29 feet behind. When
B
finishes,
C
is 10 feet behind. What was the distance of the race?
(a) 100 feet
(b) 150 feet
(c) 200 feet
(d) 250 feet
(e) 300 feet
Soln:
(c)
Let
d
be the distance and
r
A
,
r
B
and
r
C
respectively be the speeds of
A
,
B
and
C
. Then
d/
(
d
-
29) =
r
A
/r
C
= (
r
A
/r
B
)(
r
B
/r
C
) = [
d/
(
d
-
20)][
d/
(
d
-
10)] which simplifies to
d
2
-
29
d
=
d
2
-
30
d
+ 200 and has solution
d
= 200.
19.
Three standard six-sided dice are rolled. What is the probability that the largest number that
occurs is a five?
(a) 16
/
216
(b) 25
/
216
(c) 48
/
216
(d) 61
/
216
(e) 75
/
216
Soln:
(d)
There are 6
3
= 216 possible outcomes. The numbers of outcomes with one, two and
three 5s, and other numbers less than 5, are respectively 3(4
2
), 3(4), 1 giving a total of 48 + 12
+ 1 = 61. Thus the answer is 61/216.
20.
How many non-congruent rectangles are there which satisfy both (i) the lengths of the sides are
integers and (ii) the perimeter equals the area?
(a) 0
(b) 1
(c) 2
(d) 3
(e) more than 3
Soln:
(c)
Let
L
and
W
be the integer valued lengths of the sides of a rectangle which has the
same area and perimeter. Then 2
L
+ 2
W
=
LW
. Therefore
W
= 2 + 4
/
(
L
-
2) and the only
solutions with integer values are
L
= 3
, W
= 6 and
L
= 4
, W
= 4 and
L
= 6
, W
= 3. The cases
L
= 3
, W
= 6 and
L
= 6
, W
= 3 give congruent rectangles.
21.
If the polynomial
P
(
x
) =
x
3
-
4
x
2
+
ax
+ 30, where
a
is a real number, has roots 2
, r, s
, then
what is
|
r
-
s
|
?
(a) 2
(b) 4
(c) 6
(d) 8
(e) There is more than one possible value.
Soln:
(d)
Since
P
(
x
) = (
x
-
2)(
x
-
r
)(
x
-
s
), we have
r
+
s
+ 2 =
-
(
-
4) and 2
rs
=
-
30.
Solving gives -3, 5 as values of
r, s
.
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22.
Mr. Jones invested the amount $
P
. During the first year the value of the investment increased
by
x
% and during the second year it decreased by
y
% of the amount at the end of the first year.
If the value of the investment was exactly $
P
at the end of the second year then
x
equals
(a)
y
(b) 101
y/
100
(c) 100
y/
101
(d) 100
y/
(100
-
y
)
(e) 101
y/
(101
-
y
)
Soln:
(d)
Solve
P
(1 +
x/
100)(1
-
y/
100) =
P
for
x
in terms of
y
.
23.
How many numbers between 1 and 1000 are integer multiples of either 6 or 15?
(a) 265
(b) 232
(c) 221
(d) 199
(e) 33
Soln:
(d)
Every sixth number is a multiple of 6. Thus from 1 to 1000 there are 166 multiples
of 6. Similarly, from 1 to 1000 there are 66 multiples of 15, but half of these are also multiples
of 6. Therefore the answer is 166 + 33 = 199.
24.
Three men and three women are assigned different numbers selected at random from the integers
1 through 9. What is the probability the three numbers assigned to the men are all greater than
the three numbers assigned to the women?
(a) 1
/
2
(b) 1
/
6
(c) 1
/
9
(d) 1
/
20
(e) 1
/
120
Soln:
(d)
. There are 6! = 720 orderings of the six assigned numbers and for 3!
×
3! = 36 of
these orderings the numbers assigned to the men are all greater than the numbers assigned to
the women. Thus the answer is 36
/
720 = 1
/
20.
25.
If
y
=
x
2
-
2
x
, what is the sum of all distinct real numbers
x
such that (
y
+ 2)
y
= 1?
(a)
-
1
(b) 1
(c) 2
(d) 3
(e) 3
/
2
Soln:
(d)
(
y
+ 2)
y
= 1 if
y
= 0 or if
y
+ 2 = 1. Solving
x
2
-
2
x
= 0 gives
x
= 0
,
2 and solving
x
2
-
2
x
=
-
1 gives
x
= 1. (Note:
y
+ 2 =
-
1 does not give a solution for
x
).
26.
The difference
√
10
-
3 is nearest to which of the following fractions?
(a) 1
/
3
(b) 3
/
10
(c) 2
/
9
(d) 1
/
4
(e) 1
/
6
Soln:
(e)
Method 1:
Use the binomial expansion (9 + 1)
1
/
2
= 9
1
/
2
+ (1
/
2)9
−
1
/
2
(1) +
[(1
/
2)(
-
1
/
2)9
−
3
/
2
]
/
2!+
. . .
= 3+1
/
6
-
1
/
216+
· · ·
. The sum of the terms after 1/6 is very small
compared to 1/6.
Method 2:
√
10
-
3 = 1
/
(
√
10 + 3) and 6
<
√
10 + 3
<
3
.
2 + 3 = 6
.
2
27.
In the Cartesian plane if a line through the point (10
,
0) is tangent to the circle
x
2
+
y
2
= 25 at
the point (
a, b
) then
a
equals
(a) 5
/
2
(b) 5
/
√
2
(c) 5
/
√
3
(d) 5
√
3
/
2
(e) 12
/
5
Soln:
(a)
The slope of the line from (0
,
0) to (
a, b
) is the negative reciprocal of the slope of
the tangent line. Solve simultaneously
b/a
= (10
-
a
)
/b
and
a
2
+
b
2
= 25 to get 10
a
= 25.
28.
If 50! =
N
(10
k
), where
N
is not a multiple of 10, then
k
equals
(a) 5
(b) 7
(c) 12
(d) 15
(e) 20
Soln:
(c)
The number 5 appears 12 times in the prime factorization of 50!. Since 5 is a prime
factor of 10, all multiples of 10 must have a factor of 5.
29.
At a high school 2/5 of the students are boys and 1/3 of the seniors are boys. If 1/5 of the boys
are seniors then what fraction of the girls are seniors?
(a) 7
/
25
(b) 3
/
10
(c) 6
/
25
(d) 4
/
15
(e) 1
/
4
Soln:
(d)
Let
B, G, B
S
, G
S
respectively denote the number of boys, girls, senior boys and
senior girls. Then (2
/
5)(
B
+
G
) =
B
, (1
/
3)(
B
S
+
G
S
) =
B
S
and
B/
5 =
B
S
. Thus
G
= (3
/
2)
B
and
G
S
= 2
B
S
. Hence
G
S
/G
= 2
B
S
/
(3
/
2)
B
= (4
/
3)(
B
S
/B
) = 4
/
15.
30.
The ratio
(1 +
i
)
10
(1
-
i
)
7
, where
i
2
=
-
1, equals
(a) 4
-
i
(b) 2 + 2
i
(c) 1
-
2
i
(d) 3
-
3
i
(e) 4
i
-
1
Soln:
(b)
(1 +
i
)
/
(1
-
i
) = (1 +
i
)(1 +
i
)
/
(1
-
i
)(1 +
i
) = 2
i/
2 =
i
. Also (1 +
i
)
2
= 2
i
and
(1 +
i
)
3
=
-
2 + 2
i
. Thus (1 +
i
)
10
/
(1
-
i
)
7
= (1 +
i
)
3
[(1 +
i
)
/
(1
-
i
)]
7
= (
-
2 + 2
i
)(
-
i
) = 2 + 2
i
.