1962 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

The expression

\dfrac{1^{4y-1}}{5^{-1}+3^{-1}}

is equal to:

\dfrac{4y-1}{8}

8

\dfrac{15}{2}

\dfrac{15}{8}

\dfrac18

Concepts:exponent fraction algebraic manipulation

Difficulty rating: 1140

Small hint:

First simplify the power in the numerator

Big hint:

Rewrite each negative first power as a reciprocal

Solution:

The numerator is

1,

while

5^{-1}+3^{-1}=\frac15+\frac13=\frac8{15}.

Therefore the expression equals

\frac{1}{\frac{8}{15}}=\frac{15}{8}.

Thus, the correct answer is D.

2.

The expression

\sqrt{\dfrac43}-\sqrt{\dfrac34}

is equal to:

\dfrac{\sqrt3}{6}

-\dfrac{\sqrt3}{6}

\dfrac{\sqrt{-3}}6

\dfrac{5\sqrt3}{6}

1

Concepts:radical rationalizing denominator fraction

Difficulty rating: 1320

Small hint:

Write the two radicals as

\frac{2}{\sqrt3}

and

\frac{\sqrt3}{2}

Big hint:

Use a common denominator before rationalizing

Solution:

We have

\frac2{\sqrt3}-\frac{\sqrt3}{2} =\frac{4-3}{2\sqrt3} =\frac{\sqrt3}{6}.

Therefore, the correct answer is A.

3.

The first three terms of an arithmetic progression are

x-1,

x+1,

2x+3,

in the order shown. The value of

x

is:

-2

0

2

4

undetermined

Concepts:arithmetic sequence linear equation

Difficulty rating: 1030

Small hint:

Consecutive differences in an arithmetic progression are equal

Big hint:

Set

(x+1)-(x-1)

equal to

(2x+3)-(x+1)

Solution:

Equality of consecutive differences gives

2=(2x+3)-(x+1)=x+2,

so

x=0.

Thus, the correct answer is B.

4.

If

8^x=32,

then

x

equals:

4

\dfrac53

\dfrac32

\dfrac35

\dfrac14

Concepts:exponent algebraic manipulation

Difficulty rating: 1400

Small hint:

Express both sides as powers of

2

Big hint:

Equate the exponents in

2^{3x}=2^5

Solution:

Since

8=2^3

and

32=2^5,

the equation becomes

2^{3x}=2^5.

Hence

3x=5

and

x=\frac{5}{3}.

Therefore, the correct answer is B.

5.

If the radius of a circle is increased by

1

unit, the ratio of the new circumference to the new diameter is:

\pi+2

\dfrac{2\pi+1}{2}

\pi

\dfrac{2\pi-1}{2}

\pi-2

Concepts:circle circumference ratio and proportion

Difficulty rating: 800

Small hint:

Call the new radius

R

Big hint:

Divide the circumference

2\pi R

by the diameter

2R

Solution:

For every circle, regardless of its radius, the circumference divided by the diameter is

\frac{2\pi R}{2R}=\pi.

Thus, the correct answer is C.

6.

A square and an equilateral triangle have equal perimeters. The area of the triangle is

9\sqrt3

square inches. Expressed in inches the diagonal of the square is:

\dfrac92

2\sqrt5

4\sqrt2

\dfrac{9\sqrt2}{2}

none of these

Concepts:equilateral triangle perimeter diagonal area

Difficulty rating: 1410

Small hint:

Use

\frac{\sqrt3}{4}s^2

for the triangle’s area

Big hint:

Equate the two perimeters, then multiply the square’s side by

\sqrt2

Solution:

If

s

is the triangle’s side, then

\frac{\sqrt3}{4}s^2=9\sqrt3,

so

s=6.

Its perimeter is

18,

making the square’s side

\frac{18}{4}=\frac{9}{2}.

The square’s diagonal is therefore

\frac{9\sqrt2}{2}.

Thus, the correct answer is D.

7.

Let the bisectors of the exterior angles at

B

and

C

of triangle

ABC

meet at

D.

Then, if all measurements are in degrees, angle

BDC

equals:

\dfrac12(90-A)

90-A

\dfrac12(180-A)

180-A

180-2A

Concepts:angle bisector angle sum angle chasing

Difficulty rating: 1300

Small hint:

Each exterior-angle bisector makes an angle

90^\circ-\frac{B}{2}

or

90^\circ-\frac{C}{2}

with a side

Big hint:

Apply the angle sum in triangle

BDC

and use

B+C=180^\circ-A

Solution:

The angles of triangle

BDC

at

B

and

C

are

90^\circ-\frac{B}{2}

and

90^\circ-\frac{C}{2}.

Thus

\begin{aligned} \angle BDC &=180^\circ-\left(90^\circ-\frac B2\right) \\ &\quad-\left(90^\circ-\frac C2\right)\\ &=\frac{B+C}{2} \\ &=\frac{180^\circ-A}{2}. \end{aligned}

Therefore, the correct answer is C.

8.

Given the set of

n

numbers,

n\gt1,

of which one is

1-\dfrac1n

and all the others are

1.

The arithmetic mean of the

n

numbers is:

1

n-\dfrac1n

n-\dfrac1{n^2}

1-\dfrac1{n^2}

1-\dfrac1n-\dfrac1{n^2}

Concepts:mean algebraic manipulation

Difficulty rating: 1230

Small hint:

There are

n-1

copies of

1

Big hint:

Add all

n

numbers, then divide the sum by

n

Solution:

The sum is

(n-1)+\left(1-\frac1n\right)=n-\frac1n.

Dividing by

n

gives

1-\frac{1}{n^2}.

Thus, the correct answer is D.

9.

When

x^9-x

is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:

more than

5

5

4

3

2

Concepts:factoring difference of squares polynomial

Difficulty rating: 1280

Small hint:

Begin with

x(x^8-1)

and repeatedly use differences of squares

Big hint:

Over the integers, stop after isolating

x,

x-1,

x+1,

x^2+1,

and

x^4+1

Solution:

Factoring over the integers gives

\begin{aligned} x^9-x &=x(x-1)(x+1)\\ &\quad\cdot(x^2+1)(x^4+1). \end{aligned}

The last two nonconstant factors are irreducible over the integers, so there are

5

factors.

Thus, the correct answer is B.

10.

A man drives

150

miles to the seashore in

3

hours and

20

minutes. He returns from the shore to the starting point in

4

hours and

10

minutes. Let

r

be the average rate for the entire trip. Then the average rate for the trip going exceeds

r,

in miles per hour, by:

5

4\dfrac12

4

2

1

Concepts:distance rate and time unit conversion

Difficulty rating: 1320

Small hint:

Use total distance divided by total time for

r

Big hint:

The outbound time is

\frac{10}{3}

hours and the round-trip time is

\frac{15}{2}

hours

Solution:

The outbound rate is

\frac{150}{\frac{10}{3}}=45

miles per hour. The entire trip covers

300

miles in

3\frac13+4\frac16=7\frac12

hours, so

r=\frac{300}{\frac{15}{2}}=40.

The difference is

45-40=5.

Thus, the correct answer is A.

11.

The difference between the larger root and the smaller root of

x^2-px+\frac{p^2-1}{4}=0

is:

0

1

2

p

p+1

Concepts:quadratic algebraic manipulation

Difficulty rating: 1110

Small hint:

Compute the discriminant of the quadratic

Big hint:

The two roots are

\frac{p\pm\sqrt{\Delta}}{2}

Solution:

The discriminant is

p^2-4\left(\frac{p^2-1}{4}\right)=1.

The roots are

\frac{p+1}{2}

and

\frac{p-1}{2},

whose difference is

1.

Therefore, the correct answer is B.

12.

When

\left(1-\dfrac1a\right)^6

is expanded, the sum of the last three coefficients is:

22

11

10

-10

-11

Concepts:binomial theorem summation

Difficulty rating: 1210

Small hint:

The last three terms use powers

a^{-4},

a^{-5},

and

a^{-6}

Big hint:

Include the alternating signs from

(-\frac{1}{a})^k

Solution:

The last three terms have coefficients

\binom64,\quad-\binom65,\quad\binom66,

so their sum is

15-6+1=10.

Thus, the correct answer is C.

13.

R

varies directly as

S

and inversely as

T.

When

R=\dfrac43

and

T=\dfrac9{14},

S=\dfrac37.

Find

S

when

R=\sqrt{48}

and

T=\sqrt{75}.

28

30

40

42

60

Concepts:ratio and proportion radical algebraic manipulation

Difficulty rating: 1500

Small hint:

Write the variation as

R=\frac{kS}{T}

Big hint:

For fixed

k,

the quantity

\frac{RT}{S}

is constant

Solution:

Because

R=\frac{kS}{T},

we have

S=\frac{RT}{k}.

Hence

\begin{aligned} \frac{S_2}{S_1} &=\frac{R_2T_2}{R_1T_1}\\ &=\frac{\sqrt{48}\sqrt{75}} {(\frac{4}{3})(\frac{9}{14})}\\ &=\frac{60}{\frac{6}{7}}=70. \end{aligned}

Therefore

S_2=(\frac{3}{7})\cdot70=30.

Thus, the correct answer is B.

14.

Let

s

be the limiting sum of the geometric series

4-\dfrac83+\dfrac{16}{9}-\cdots,

as the number of terms increases without bound. Then

s

equals:

a number between

0

and

1

2.4

2.5

3.6

12

Concepts:geometric sequence summation

Difficulty rating: 1110

Small hint:

Find the ratio of the second term to the first

Big hint:

Use

\frac{a}{1-r}

because the ratio has absolute value less than

1

Solution:

The first term is

4

and the common ratio is

-\frac{2}{3}.

Thus

s=\frac4{1-(-\frac{2}{3})}=\frac{12}{5}=2.4.

Therefore, the correct answer is B.

15.

Given triangle

ABC

with base

AB

fixed in length and position. As the vertex

C

moves on a straight line, the intersection point of the three medians moves on:

a circle

a parabola

an ellipse

a straight line

a curve here not listed

Concepts:centroid transformation vector

Difficulty rating: 1280

Small hint:

The centroid lies two-thirds of the way from a vertex to the midpoint of the opposite side

Big hint:

With

AB

fixed, express the centroid as a fixed point plus one-third of the position vector of

C

Solution:

Let

A,

B,

and

C

also denote their position vectors. The centroid is

G=\frac{A+B+C}{3}.

Since

A

and

B

are fixed, this is a translation and scaling by

\frac{1}{3}

of the position of

C.

A straight-line locus therefore maps to a straight-line locus.

Thus, the correct answer is D.

16.

Given rectangle

R_1

with one side

2

inches and area

12

square inches. Rectangle

R_2

with diagonal

15

inches is similar to

R_1.

Expressed in square inches the area of

R_2

is:

\dfrac92

36

\dfrac{135}{2}

9\sqrt{10}

\dfrac{27\sqrt{10}}4

Difficulty rating: 1510

Small hint:

The sides of

R_1

are

2

and

6

Big hint:

Areas scale as the square of the ratio of corresponding diagonals

Solution:

The diagonal of

R_1

is

\sqrt{2^2+6^2}=2\sqrt{10}.

Therefore

\frac{[R_2]}{[R_1]} =\left(\frac{15}{2\sqrt{10}}\right)^2 =\frac{45}{8}.

Thus

[R_2]=12(\frac{45}{8})=\frac{135}{2}.

Therefore, the correct answer is C.

17.

If

a=\log_8 225

and

b=\log_2 15,

then

a,

in terms of

b,

is:

\dfrac b2

\dfrac{2b}{3}

b

\dfrac{3b}{2}

2b

Concepts:logarithm exponent algebraic manipulation

Difficulty rating: 1180

Small hint:

Write

225=15^2

and

8=2^3

Big hint:

Use the change-of-base formula with base

2

Solution:

Changing to base

2

gives

\begin{aligned} a&=\frac{\log_2(15^2)}{\log_2(2^3)}\\ &=\frac{2\log_2 15}{3} =\frac{2b}{3}. \end{aligned}

Thus, the correct answer is B.

18.

A regular dodecagon (

12

sides) is inscribed in a circle with radius

r

inches. The area of the dodecagon, in square inches, is:

3r^2

2r^2

\dfrac{3r^2\sqrt3}{4}

r^2\sqrt3

3r^2\sqrt3

Concepts:regular polygon triangle area trigonometry

Difficulty rating: 1210

Small hint:

Divide the dodecagon into

12

triangles with vertex at the center

Big hint:

Each central angle is

30^\circ

, so use

\frac12r^2\sin30^\circ

Solution:

Each of the

12

central triangles has area

\frac12r^2\sin30^\circ=\frac{r^2}{4}.

Their total area is

12(\frac{r^2}{4})=3r^2.

Thus, the correct answer is A.

19.

If the parabola

y=ax^2+bx+c

passes through the points

(-1,12),

(0,5),

and

(2,-3),

the value of

a+b+c

is:

-4

-2

0

1

2

Concepts:system of equations quadratic substitution

Difficulty rating: 1280

Small hint:

The requested sum is the value of the parabola at

x=1

Big hint:

Use the three given points to solve for

a,

b,

and

c

Solution:

The point

(0,5)

gives

c=5.

The other two points give

a-b=7,\qquad 4a+2b=-8.

Solving yields

a=1

and

b=-6.

Hence

a+b+c=1-6+5=0.

Thus, the correct answer is C.

20.

The angles of a pentagon are in arithmetic progression. One of the angles, in degrees, must be:

108

90

72

54

36

Concepts:angle sum arithmetic sequence

Difficulty rating: 1440

Small hint:

Five terms in arithmetic progression have their middle term equal to their average

Big hint:

The interior angles of a pentagon sum to

540^\circ

Solution:

The average of the five angles is

\frac{540^\circ}{5}=108^\circ.

For five terms in arithmetic progression, the middle term equals the average, so one angle must be

108^\circ.

Thus, the correct answer is A.

21.

It is given that one root of

2x^2+rx+s=0,

with

r

and

s

real numbers, is

3+2i

(i=\sqrt{-1}).

The value of

s

is:

undetermined

5

6

-13

26

Concepts:complex number Vieta’s Formulas

Difficulty rating: 1210

Small hint:

A polynomial with real coefficients also has the conjugate root

Big hint:

Use the product of the roots and Vieta’s formula

Solution:

The other root is

3-2i.

Their product is

(3+2i)(3-2i)=9+4=13.

Vieta’s formula gives

\frac{s}{2}=13,

so

s=26.

Therefore, the correct answer is E.

22.

The number

121_b,

written in the integral base

b,

is the square of an integer, for:

b=10,

only

b=10

and

b=5,

only

2\le b\le10

b\gt2

no value of

b

Concepts:number base place value perfect square

Difficulty rating: 1440

Small hint:

Convert

121_b

to an expression in

b

Big hint:

Remember that the digit

2

requires

b\gt2

Solution:

In ordinary notation,

121_b=b^2+2b+1=(b+1)^2.

This is a square for every allowable base. Since digit

2

occurs, precisely the integral bases

b\gt2

are allowable.

Thus, the correct answer is D.

23.

In triangle

ABC,

CD

is the altitude to

AB

and

AE

is the altitude to

BC.

If the lengths of

AB,

CD,

and

AE

are known, the length of

DB

is:

not determined by the information given

determined only if

A

is an acute angle

determined only if

B

is an acute angle

determined only if

ABC

is an acute triangle

none of these is correct

Concepts:triangle area altitude Pythagorean Theorem

Difficulty rating: 1570

Small hint:

Compute the triangle’s area in two ways using the two known altitudes

Big hint:

After finding

BC,

use right triangle

BCD

Solution:

Let

AB=c,

CD=h,

and

AE=e.

Equating two area formulas gives

\frac12ch=\frac12(BC)e,

so

BC=\frac{ch}{e}.

Since triangle

BCD

is right at

D,

DB=\sqrt{BC^2-CD^2}.

This determines

DB

whether the original triangle is acute, right, or obtuse, so none of the first four choices is correct.

Thus, the correct answer is E.

24.

Three machines

P,

Q,

and

R,

working together, can do a job in

x

hours. When working alone,

P

needs an additional

6

hours to do the job;

Q,

one additional hour; and

R,

x

additional hours. The value of

x

is:

\dfrac23

\dfrac{11}{12}

\dfrac32

2

3

Concepts:rate rational equation

Difficulty rating: 1710

Small hint:

The individual completion times are

x+6,

x+1,

and

2x

Big hint:

Set the sum of the individual hourly rates equal to

\frac{1}{x}

Solution:

The rates satisfy

\frac1{x+6}+\frac1{x+1}+\frac1{2x}=\frac1x.

Thus

\frac1{x+6}+\frac1{x+1}=\frac1{2x},

which simplifies to

3x^2+7x-6=0,

or

(3x-2)(x+3)=0.

A time must be positive, so

x=\frac{2}{3}.

Therefore, the correct answer is A.

25.

Given square

ABCD

with side

8

feet. A circle is drawn through vertices

A

and

D

and tangent to side

BC.

The radius of the circle, in feet, is:

4

4\sqrt2

5

5\sqrt2

6

Concepts:tangent line perpendicular bisector coordinate geometry circle

Difficulty rating: 1570

Small hint:

Place

A=(0,0),

D=(0,8),

and

BC

on the line

x=8

Big hint:

The center lies on the perpendicular bisector of

AD

, and its distance to

BC

equals the radius

Solution:

Place the center at

(h,4).

Since the circle passes through

A,

r^2=h^2+16.

Tangency to

x=8

gives

r=8-h.

Therefore

h^2+16=(8-h)^2,

so

h=3

and

r=5.

Thus, the correct answer is C.

26.

For any real value of

x

the maximum value of

8x-3x^2

is:

0

\dfrac83

4

5

\dfrac{16}{3}

Concepts:quadratic completing the square optimization

Difficulty rating: 1180

Small hint:

Complete the square in

8x-3x^2

Big hint:

A negative square is largest when it equals

0

Solution:

Completing the square,

8x-3x^2 =\frac{16}{3}-3\left(x-\frac43\right)^2.

The square term is nonnegative, so the maximum is

\frac{16}{3}.

Thus, the correct answer is E.

27.

Let

a\mathbin{@}b

represent the operation on two numbers,

a

and

b,

which selects the larger of the two numbers, with

a\mathbin{@}a=a.

Let

a\mathbin{!}b

represent the operation which selects the smaller of the two numbers, with

a\mathbin{!}a=a.

Which of the following three rules is (are) correct?

\begin{aligned} (1)\quad&a\mathbin{@}b=b\mathbin{@}a,\\ (2)\quad&a\mathbin{@}(b\mathbin{@}c)=(a\mathbin{@}b)\mathbin{@}c,\\ (3)\quad&a\mathbin{!}(b\mathbin{@}c)\\ &\quad=(a\mathbin{!}b)\mathbin{@}(a\mathbin{!}c). \end{aligned}

(1)

only

(2)

only

(1)

and

(2)

only

(1)

and

(3)

only

all three

Concepts:custom operation distributive property casework

Difficulty rating: 1500

Small hint:

Translate

\mathbin{@}

as maximum and

\mathbin{!}

as minimum

Big hint:

For rule

(3),

compare both sides separately when

a

is below or above

\max(b,c)

Solution:

Maximum is commutative and associative, so

(1)

and

(2)

hold. Rule

(3)

is the distributive identity

\begin{aligned} &\min(a,\max(b,c))\\ &\quad=\max(\min(a,b),\min(a,c)). \end{aligned}

If

a\ge\max(b,c),

both sides equal

\max(b,c);

if

a\lt\max(b,c),

both sides equal

a.

Thus

(3)

also holds.

Therefore, the correct answer is E.

28.

The set of

x

-values satisfying the equation

x^{\log_{10}x}=\frac{x^3}{100}

consists of:

\dfrac1{10},

only

10,

only

100,

only

10

or

100,

only

more than two real numbers

Concepts:logarithm exponent substitution quadratic

Difficulty rating: 1520

Small hint:

The logarithm requires

x\gt0

; set

y=\log_{10}x

Big hint:

Take base-

10

logarithms to obtain a quadratic in

y

Solution:

Set

y=\log_{10}x,

so

x=10^y.

Taking base-

10

logarithms gives

y^2=3y-2.

Hence

(y-1)(y-2)=0,

so

x=10

or

x=100.

Thus, the correct answer is D.

29.

Which of the following sets of

x

-values satisfy the inequality

2x^2+x\lt6?

-2\lt x\lt\dfrac32

x\gt\dfrac32

or

x\lt-2

x\lt\dfrac32

\dfrac32\lt x\lt2

x\lt-2

Concepts:inequality factoring zero product property

Difficulty rating: 1150

Small hint:

Move all terms to one side and factor the quadratic

Big hint:

A product of two linear factors is negative between its roots

Solution:

The inequality is

2x^2+x-6\lt0.

Factoring gives

(2x-3)(x+2)\lt0.

The product is negative between its roots, so

-2\lt x\lt\frac{3}{2}.

Thus, the correct answer is A.

30.

Consider the statements:

(1)

p

and

q

are both true

(2)

p

is true and

q

is false

(3)

p

is false and

q

is true

(4)

p

is false and

q

is false.

How many of these imply the negation of the statement “

p

and

q

are both true”?

0

1

2

3

4

Concepts:logical deduction

Difficulty rating: 1280

Small hint:

The negation fails only when both statements are true

Big hint:

Check which of the four listed truth assignments are not case

(1)

Solution:

The negation of “

p

and

q

are both true” holds whenever at least one of

p

and

q

is false. This occurs in cases

(2),

(3),

and

(4),

for a total of

3.

Thus, the correct answer is D.

31.

The ratio of the interior angles of two regular polygons with sides of unit length is

3:2.

How many such pairs are there?

1

2

3

4

infinitely many

Concepts:regular polygon angle sum Diophantine Equation

Difficulty rating: 1710

Small hint:

For an

n

-gon, an interior angle is

\frac{180^\circ(n-2)}{n}

Big hint:

If the smaller polygon has

n

sides, solve for the larger side count and test the possible integers

3\le n\lt6

Solution:

Let the smaller and larger polygons have

n

and

N

sides. Then

\frac{\frac{N-2}{N}}{\frac{n-2}{n}}=\frac32,

which gives

N=\frac{4n}{6-n}.

Positivity and

N\gt n

require

n=3,

n=4,

or

n=5.

These give

N=4,

N=8,

and

N=20,

respectively. Thus there are

3

pairs.

Therefore, the correct answer is C.

32.

If

x_{k+1}=x_k+\dfrac12

for

k=1,

2,

\ldots,

n-1

and

x_1=1,

find

x_1+x_2+\cdots+x_n.

\dfrac{n+1}{2}

\dfrac{n+3}{2}

\dfrac{n^2-1}{2}

\dfrac{n^2+n}{4}

\dfrac{n^2+3n}{4}

Concepts:arithmetic sequence summation

Difficulty rating: 1440

Small hint:

The recurrence defines an arithmetic sequence with common difference

\frac{1}{2}

Big hint:

Find

x_n

, then use

\frac{n(x_1+x_n)}{2}

Solution:

We have

x_n=1+\frac{n-1}{2}=\frac{n+1}{2}.

Therefore

\begin{aligned} x_1+\cdots+x_n &=\frac n2\left(1+\frac{n+1}{2}\right)\\ &=\frac{n^2+3n}{4}. \end{aligned}

Thus, the correct answer is E.

33.

The set of

x

-values satisfying the inequality

2\le|x-1|\le5

is:

-4\le x\le-1

or

3\le x\le6

3\le x\le6

or

-6\le x\le-3

x\le-1

or

x\ge3

-1\le x\le3

-4\le x\le6

Concepts:absolute value inequality

Difficulty rating: 1180

Small hint:

Interpret the inequality as distances from

1

between

2

and

5

Big hint:

Solve

2\le x-1\le5

and

-5\le x-1\le-2

Solution:

For

x-1\ge0,

the bounds give

3\le x\le6.

For

x-1\le0,

they give

-4\le x\le-1.

The solution is the union of these two intervals.

Thus, the correct answer is A.

34.

For what real values of

K

does

x=K^2(x-1)(x-2)

have real roots?

none

-2\lt K\lt1

-2\sqrt2\lt K\lt2\sqrt2

K\gt1

or

K\lt-2

all

Concepts:quadratic algebraic manipulation

Difficulty rating: 1550

Small hint:

Expand and collect terms to obtain a quadratic in

x

Big hint:

Its discriminant simplifies to

K^4+6K^2+1

Solution:

Rearranging gives

K^2x^2-(3K^2+1)x+2K^2=0.

Its discriminant is

\begin{aligned} &(3K^2+1)^2-8K^4\\ &\qquad=K^4+6K^2+1\gt0 \end{aligned}

for every real

K.

This also covers

K=0,

when the original equation gives

x=0.

Therefore, the correct answer is E.

35.

A man on his way to dinner shortly after

6{:}00

p.m. observes that the hands of his watch form an angle of

110^\circ.

Returning before

7{:}00

p.m. he notices that again the hands of his watch form an angle of

110^\circ.

The number of minutes that he has been away is:

36\dfrac23

40

42

42.4

45

Concepts:clock relative speed

Difficulty rating: 1570

Small hint:

During the interval, the minute hand gains on the hour hand at

5.5^\circ

per minute

Big hint:

Between the two observations, the signed separation changes from

110^\circ

to

-110^\circ

Solution:

The two observations lie on opposite sides of the instant when the hands coincide. Their signed angular separation changes by

220^\circ.

Since the minute hand gains on the hour hand at

6-0.5=5.5^\circ

per minute, the elapsed time is

\frac{220}{5.5}=40

minutes.

Thus, the correct answer is B.

36.

If both

x

and

y

are integers, how many solutions are there to the equation

(x-8)(x-10)=2^y?

0

1

2

3

more than

3

Concepts:Diophantine Equation power of 2 difference of squares

Difficulty rating: 1900

Small hint:

Rewrite the left side as

(x-9)^2-1

Big hint:

If

n=x-9,

then the consecutive even factors

n-1

and

n+1

must both be powers of

2

Solution:

Put

n=x-9.

Then

2^y=n^2-1=(n-1)(n+1).

Since the product is a power of

2,

both factors must have no odd prime divisor. The only consecutive even integers differing by

2

that are both signed powers of

2

are

(-4,-2)

and

(2,4).

Thus

n=\pm3

and

y=3,

giving

x=6

or

12.

There are

2

ordered pairs

(x,y).

Therefore, the correct answer is C.

37.

ABCD

is a square with side of unit length. Points

E

and

F

are taken respectively on sides

AB

and

AD

so that

AE=AF

and the quadrilateral

CDFE

has maximum area. In square units this maximum area is:

\dfrac12

\dfrac9{16}

\dfrac{19}{32}

\dfrac58

\dfrac23

Concepts:area shoelace formula completing the square optimization

Difficulty rating: 1730

Small hint:

Let

AE=AF=t

and use coordinates or the shoelace formula

Big hint:

The area becomes

\frac12(1+t-t^2)

; complete the square

Solution:

Set

A=(0,0),

B=(1,0),

C=(1,1),

and

D=(0,1).

Then

E=(t,0)

and

F=(0,t).

The shoelace formula gives

\begin{aligned} [CDFE]&=\frac{1+t-t^2}{2}\\ &=\frac58-\frac12 \left(t-\frac12\right)^2. \end{aligned}

Its maximum is

\frac{5}{8}.

Thus, the correct answer is D.

38.

The population of Nosuch Junction at one time was a perfect square. Later, with an increase of

100,

the population was one more than a perfect square. Now, with an additional increase of

100,

the population is again a perfect square.

The original population is a multiple of:

3

7

9

11

17

Concepts:difference of squares perfect square factor systematic listing

Difficulty rating: 1980

Small hint:

Write the populations as

a^2,

b^2+1,

and

c^2

Big hint:

From

b^2-a^2=99,

test the positive factor pairs of

99

, then impose

c^2-a^2=200

Solution:

Let the original population be

a^2.

Then

\begin{aligned} b^2-a^2&=99,\\ c^2-a^2&=200. \end{aligned}

From

(b-a)(b+a)=99,

the positive possibilities for

a

are

49,

15,

and

1.

Checking the second condition, only

a=49

works, since

49^2+200=2601=51^2.

The population is

49^2=7^4,

a multiple of

7.

Thus, the correct answer is B.

39.

The medians

AN

and

BP

of a triangle with unequal sides are, respectively,

3

inches and

6

inches long. Its area is

3\sqrt{15}

square inches. The length of the third median, in inches, is:

4

3\sqrt3

3\sqrt6

6\sqrt3

6\sqrt6

Concepts:median (geometry)law of cosines area

Difficulty rating: 2090

Small hint:

The three medians form the side lengths of a triangle whose area is three-fourths the original area

Big hint:

Use the two known median lengths and that area to find the two possible included angles, then reject the case that makes two medians equal

Solution:

The triangle whose sides are the three medians has area

\frac34\left(3\sqrt{15}\right)=\frac{9\sqrt{15}}4.

If

\theta

is the included angle between its sides

3

and

6,

then

9\sin\theta=\frac{9\sqrt{15}}4,

so

\cos\theta=\pm\frac{1}{4}.

By the law of cosines, the third median

m

satisfies

m^2=3^2+6^2-2(3)(6)\cos\theta,

giving

m^2=36

or

54.

The value

m=6

would make two medians, and hence two sides, equal. Because the triangle has unequal sides,

m=\sqrt{54}=3\sqrt6.

Therefore, the correct answer is C.

40.

The limiting sum of the infinite series

\frac1{10}+\frac2{10^2}+\frac3{10^3}+\cdots,

whose

n

th term is

\frac{n}{10^n},

is:

\dfrac19

\dfrac{10}{81}

\dfrac18

\dfrac{17}{72}

larger than any finite quantity

Concepts:arithmetico-geometric series calculus summation

Difficulty rating: 1570

Small hint:

Start from

\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}

Big hint:

Differentiate and then multiply by

x

Solution:

For

|x|\lt1,

\sum_{n=1}^{\infty}nx^n=\frac{x}{(1-x)^2}.

Taking

x=\frac{1}{10}

gives

\frac{\frac{1}{10}}{(\frac{9}{10})^2}=\frac{10}{81}.

Thus, the correct answer is B.