1970 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

The fourth power of

\sqrt{1+\sqrt{1+\sqrt1}}

is:

\sqrt2+\sqrt3

\tfrac12(7+3\sqrt5)

1+2\sqrt3

3

3+2\sqrt2

Concepts:radical exponent algebraic manipulation

Difficulty rating: 1740

Small hint:

Simplify from the innermost radical outward

Big hint:

After the innermost radical becomes

1,

square

\sqrt{1+\sqrt2}

twice

Solution:

The innermost radical is

\sqrt1=1,

so the given number is

\sqrt{1+\sqrt2}.

Its fourth power is

\begin{aligned} (1+\sqrt2)^2&=1+2\sqrt2+2\\ &=3+2\sqrt2. \end{aligned}

Therefore, the correct answer is E.

2.

A square and a circle have equal perimeters. The ratio of the area of the circle to the area of the square is:

\frac{4}{\pi}

\frac{\pi}{\sqrt2}

\frac{4}{1}

\frac{\sqrt2}{\pi}

\frac{\pi}{4}

Concepts:circle area circumference area ratio square (geometry)

Difficulty rating: 1560

Small hint:

Let the common perimeter be

p

Big hint:

Express the circle radius and square side in terms of

p

, then divide their areas

Solution:

If the common perimeter is

p,

then the circle has radius

\frac{p}{2\pi}

and the square has side

\frac{p}{4}.

Therefore

\frac{\pi\left(\frac{p}{2\pi}\right)^2}{\left(\frac p4\right)^2} =\frac{\frac{p^2}{4\pi}}{\frac{p^2}{16}}=\frac4\pi.

Therefore, the correct answer is A.

3.

If

x=1+2^p

and

y=1+2^{-p},

then

y

in terms of

x

is:

\dfrac{x+1}{x-1}

\dfrac{x+2}{x-1}

\dfrac{x}{x-1}

2-x

\dfrac{x-1}{x}

Concepts:exponent substitution rational equation

Difficulty rating: 1360

Small hint:

Solve the first equation for

2^p

Big hint:

Use

2^{-p}=\frac{1}{2^p}

and simplify

1+\frac{1}{x-1}

Solution:

From

x=1+2^p

we get

2^p=x-1,

so

\begin{aligned} y&=1+2^{-p}\\ &=1+\frac1{x-1}=\frac{x}{x-1}. \end{aligned}

Therefore, the correct answer is C.

4.

Let

S

be the set of all numbers which are the sum of the squares of three consecutive integers. Then we can say that:

No member of

S

is divisible by

2

No member of

S

is divisible by

3

but some member is divisible by

11

No member of

S

is divisible by

3

or by

5

No member of

S

is divisible by

3

or by

7

None of these

Concepts:modular arithmetic divisibility algebraic manipulation

Difficulty rating: 1760

Small hint:

Write the integers as

n-1,n,n+1

Big hint:

Their squared sum is

3n^2+2;

examine it modulo

3

and test a residue modulo

11

Solution:

The sum is

\begin{gathered} (n-1)^2+n^2+(n+1)^2\\ =3n^2+2. \end{gathered}

It is always congruent to

2\pmod3,

so no member of

S

is divisible by

3.

On the other hand, taking

n=5

gives

3(25)+2=77,

which is divisible by

11.

Therefore, the correct answer is B.

5.

If

f(x)=\dfrac{x^4+x^2}{x+1},

then

f(i),

where

i=\sqrt{-1},

is equal to:

1+i

1

-1

0

-1-i

Concepts:complex number exponent function

Difficulty rating: 1560

Small hint:

Use

i^2=-1

and

i^4=1

Big hint:

The numerator vanishes while the denominator does not

Solution:

Because

i^2=-1

and

i^4=1,

f(i)=\frac{i^4+i^2}{i+1}=\frac{1-1}{i+1}=0.

Therefore, the correct answer is D.

6.

The smallest value of

x^2+8x

for real values of

x

is:

-16.25

-16

-15

-8

None of these

Concepts:completing the square quadratic inequality

Difficulty rating: 1360

Small hint:

Complete the square

Big hint:

Rewrite the expression as

(x+4)^2-16

Solution:

Completing the square gives

x^2+8x=(x+4)^2-16.

Since

(x+4)^2\ge0,

the minimum occurs at

x=-4

and equals

-16.

Therefore, the correct answer is B.

7.

Inside square

ABCD

with side

s,

quarter-circle arcs with radii

s

and centers at

A

and

B

are drawn. These arcs intersect at a point

X

inside the square. How far is

X

from side

CD?

\tfrac12s(\sqrt3+4)

\tfrac12s\sqrt3

\tfrac12s(1+\sqrt3)

\tfrac12s(\sqrt3-1)

\tfrac12s(2-\sqrt3)

Concepts:equilateral triangle altitude square (geometry)circle

Difficulty rating: 1980

Small hint:

The two radii to

X

and side

AB

form an equilateral triangle

Big hint:

The altitude from

X

to

AB

is

\frac{s\sqrt3}{2}

; subtract it from the square’s height

Solution:

Since

AX=BX=AB=s,

triangle

ABX

is equilateral. Thus the perpendicular distance from

X

to

AB

is

\frac{s\sqrt3}{2}.

The distance between the parallel sides

AB

and

CD

is

s,

so the requested distance is

s-\frac{s\sqrt3}{2}=\frac s2(2-\sqrt3).

Therefore, the correct answer is E.

8.

If

a=\log_8 225

and

b=\log_2 15,

then:

a=\frac{b}{2}

a=\frac{2b}{3}

a=b

b=\frac{a}{2}

a=\frac{3b}{2}

Concepts:logarithm exponent algebraic manipulation

Difficulty rating: 1710

Small hint:

Rewrite

225

as

15^2

and

8

as

2^3

Big hint:

Convert both logarithms to base

2

Solution:

Using

225=15^2

and

8=2^3,

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

Therefore, the correct answer is B.

9.

Points

P

and

Q

are on line segment

AB,

and both points are on the same side of the midpoint of

AB.

Point

P

divides

AB

in the ratio

2:3,

and

Q

divides

AB

in the ratio

3:4.

If

PQ=2,

then the length of segment

AB

is:

12

28

70

75

105

Concepts:ratio and proportion linear equation

Difficulty rating: 1610

Small hint:

Express

AP

and

AQ

as fractions of

AB

Big hint:

Use

AP=\frac{2AB}{5}

and

AQ=\frac{3AB}{7}

, then subtract

Solution:

The division ratios give

AP=\frac25AB,\qquad AQ=\frac37AB.

Because the two points are on the same side of the midpoint,

\begin{aligned} PQ&=AQ-AP\\ &=\left(\frac37-\frac25\right)AB\\ &=\frac1{35}AB. \end{aligned}

Since

PQ=2,

we obtain

AB=70.

Therefore, the correct answer is C.

10.

Let

F=0.4818181\ldots

be an infinite repeating decimal with the digits

8

and

1

repeating. When

F

is written as a fraction in lowest terms, the denominator exceeds the numerator by:

13

14

29

57

126

Concepts:repeating decimal geometric sequence fraction

Difficulty rating: 1590

Small hint:

Separate the initial

0.4

from the repeating tail

Big hint:

The tail is

0.081(1+0.01+0.01^2+\cdots)

Solution:

We have

\begin{aligned} F&=0.4+0.081(1+0.01+\cdots)\\ &=\frac25+\frac{0.081}{0.99}\\ &=\frac25+\frac9{110}\\ &=\frac{53}{110}. \end{aligned}

The denominator exceeds the numerator by

110-53=57.

Therefore, the correct answer is D.

11.

If two factors of

2x^3-hx+k

are

x+2

and

x-1,

the value of

|2h-3k|

is:

4

3

2

1

0

Concepts:polynomial substitution system of equations

Difficulty rating: 1590

Small hint:

Apply the factor theorem at

x=-2

and

x=1

Big hint:

Solve

-16+2h+k=0

and

2-h+k=0

Solution:

The factor theorem gives

\begin{aligned} -16+2h+k&=0,\\ 2-h+k&=0. \end{aligned}

Subtracting the second equation from the first gives

3h=18,

so

h=6

and then

k=4.

Hence

|2h-3k|=|12-12|=0.

Therefore, the correct answer is E.

12.

A circle with radius

r

is tangent to sides

AB,

AD,

and

CD

of rectangle

ABCD

and passes through the midpoint of diagonal

AC.

The area of the rectangle, in terms of

r,

is:

4r^2

6r^2

8r^2

12r^2

20r^2

Concepts:rectangle tangent circles midpoint area

Difficulty rating: 2090

Small hint:

Tangency to the two opposite sides makes one side of the rectangle equal to

2r

Big hint:

The line from the third tangency point through the center bisects the diagonal, making the corresponding chord a diameter

Solution:

Let the circle be tangent to the parallel sides

AB

and

CD.

Their separation is its diameter, so

AD=BC=2r.

Let

R

be the tangency point on

AD,

let

Q

be the center, and let

M

be the midpoint of

AC.

The line

RQ

is parallel to

AB

and halfway between

AB

and

CD,

so

R

is the midpoint of

AD

and the line contains

M.

Since

M

lies on the circle and

R,Q,M

are collinear,

RM=2r.

Thus

RM

is a midsegment of triangle

ADC,

so

DC=2RM=4r.

Therefore the area is

(2r)(4r)=8r^2.

Therefore, the correct answer is C.

13.

Given the binary operation

*

defined by

a*b=a^b

for all positive numbers

a

and

b.

Then for all positive

a,

b,

c,

n,

we have:

a*b=b*a

a*(b*c)=(a*b)*c

a*(b^n)=(a*n)*b

(a*b)^n=a*(bn)

None of these

Concepts:custom operation exponent algebraic manipulation

Difficulty rating: 1710

Small hint:

Replace every

*

using the definition before comparing the two sides

Big hint:

For choice D, both sides simplify to a power with exponent

bn

Solution:

For choice D,

\begin{aligned} (a*b)^n&=(a^b)^n=a^{bn},\\ a*(bn)&=a^{bn}. \end{aligned}

Thus that identity always holds. The other choices would assert false general identities such as

a^b=b^a

or

a^{b^c}=a^{bc}.

Therefore, the correct answer is D.

14.

Consider

x^2+px+q=0,

where

p

and

q

are positive numbers. If the roots of this equation differ by

1,

then

p

equals:

\sqrt{4q+1}

q-1

-\sqrt{4q+1}

q+1

\sqrt{4q-1}

Concepts:quadratic Vieta’s Formulas

Difficulty rating: 1670

Small hint:

Use the quadratic formula to express the difference of the roots

Big hint:

The difference has magnitude

\sqrt{p^2-4q}

Solution:

The two roots differ in absolute value by

\sqrt{p^2-4q}.

Hence

p^2-4q=1,

so

p^2=4q+1.

Since

p

is positive,

p=\sqrt{4q+1}.

Therefore, the correct answer is A.

15.

Lines in the

xy

-plane are drawn through the point

(3,4)

and the trisection points of the line segment joining the points

(-4,5)

and

(5,-1).

One of these lines has the equation:

3x-2y-1=0

4x-5y+8=0

5x+2y-23=0

x+7y-31=0

x-4y+13=0

Concepts:coordinate geometry slope linear equation

Difficulty rating: 1620

Small hint:

Find the two trisection points by moving one-third and two-thirds of the displacement vector

Big hint:

The trisection points are

(-1,3)

and

(2,1)

; find the line from either one to

(3,4)

Solution:

The displacement from

(-4,5)

to

(5,-1)

is

(9,-6).

The trisection points are therefore

(-4,5)+\tfrac13(9,-6)=(-1,3)

and

(2,1).

The line through

(-1,3)

and

(3,4)

has slope

\frac{1}{4},

so

y-4=\frac14(x-3),

or

x-4y+13=0.

Therefore, the correct answer is E.

16.

If

F(n)

is a function such that

F(1)=F(2)=F(3)=1,

and such that

F(n+1)=\frac{F(n)F(n-1)+1}{F(n-2)}

for

n\ge3,

then

F(6)

is equal to:

2

3

7

11

26

Concepts:recursion function substitution

Difficulty rating: 1450

Small hint:

Compute the terms one at a time beginning with

F(4)

Big hint:

Use the recurrence successively for

n=3,4,5

Solution:

Applying the recurrence successively,

\begin{aligned} F(4)&=\frac{1\cdot1+1}{1}=2,\\ F(5)&=\frac{2\cdot1+1}{1}=3, \end{aligned}

and

F(6)=\frac{3\cdot2+1}{1}=7.

Therefore, the correct answer is C.

17.

If

r\gt0,

then for all

p

and

q

such that

pq\ne0

and

pr\gt qr,

we have:

-p\gt-q

-p\gt q

1\gt-\frac{q}{p}

1\lt \frac{q}{p}

None of these

Concepts:inequality counterexample ratio and proportion

Difficulty rating: 1850

Small hint:

Because

r\gt0,

the hypothesis gives only

p\gt q

Big hint:

Test the remaining claims with both positive and negative choices of

p,q

Solution:

Since

r\gt0,

the hypothesis is equivalent to

p\gt q.

Then

-p\lt-q,

so A is false. Taking

(p,q)=(2,1)

makes B and D false. Taking

(p,q)=(1,-2)

makes C false because

-\frac{q}{p}=2.

Thus none of A-D follows in every case.

Therefore, the correct answer is E.

18.

\sqrt{3+2\sqrt2}-\sqrt{3-2\sqrt2}

is equal to:

2

2\sqrt3

4\sqrt2

\sqrt6

2\sqrt2

Concepts:radical perfect square algebraic manipulation

Difficulty rating: 1740

Small hint:

Recognize each radicand as a squared binomial

Big hint:

Use

3\pm2\sqrt2=(\sqrt2\pm1)^2

Solution:

Since

\begin{gathered} 3+2\sqrt2=(\sqrt2+1)^2,\\ 3-2\sqrt2=(\sqrt2-1)^2, \end{gathered}

and both

\sqrt2+1

and

\sqrt2-1

are positive, the difference is

(\sqrt2+1)-(\sqrt2-1)=2.

Therefore, the correct answer is A.

19.

The sum of an infinite geometric series with common ratio

r

such that

|r|\lt1

is

15,

and the sum of the squares of the terms of this series is

45.

The first term of the series is:

12

10

5

3

2

Concepts:geometric sequence system of equations algebraic manipulation

Difficulty rating: 1990

Small hint:

If the first term is

a,

write one equation for each infinite sum

Big hint:

Use

\frac{a}{1-r}=15

and

\frac{a^2}{1-r^2}=45

, then cancel a factor of

1-r

Solution:

Let the first term be

a.

Then

\frac{a}{1-r}=15,\qquad \frac{a^2}{1-r^2}=45.

Since

1-r^2=(1-r)(1+r),

dividing the second equation by the first gives

\frac{a}{1+r}=3.

Thus

a=15(1-r)=3(1+r).

Solving gives

r=\frac{2}{3}

and

a=5.

Therefore, the correct answer is C.

20.

Lines

HK

and

BC

lie in a plane.

M

is the midpoint of line segment

BC,

and

BH

and

CK

are perpendicular to

HK.

Then we:

always have

MH=MK

always have

MH\gt BK

sometimes have

MH=MK

but not always

always have

MH\gt MB

always have

BH\lt BC

Concepts:coordinate geometry midpoint perpendicular bisector

Difficulty rating: 1790

Small hint:

Put

HK

on the

x

-axis

Big hint:

Use

H=(h,0),

B=(h,u),

K=(k,0),

and

C=(k,v)

; compare squared distances from their midpoint

Solution:

Choose coordinates

\begin{gathered} H=(h,0),\quad B=(h,u),\\ K=(k,0),\quad C=(k,v). \end{gathered}

Then

M=\left(\frac{h+k}{2},\frac{u+v}{2}\right).

The horizontal displacements from

M

to

H

and

K

are opposites, while the vertical displacements are equal. Therefore

MH^2=MK^2,

so

MH=MK

always.

Therefore, the correct answer is A.

21.

On an auto trip, the distance read from the instrument panel was

450

miles. With snow tires on for the return trip over the same route, the reading was

440

miles. Find, to the nearest hundredth of an inch, the increase in radius of the wheels if the original radius was

15

inches.

0.33

0.34

0.35

0.38

0.66

Concepts:ratio and proportion circumference distance rate and time

Difficulty rating: 1650

Small hint:

For a fixed true distance, the odometer reading is inversely proportional to wheel radius

Big hint:

If the radii are

r_1,r_2,

use

450r_1=440r_2

Solution:

For a fixed true distance, the odometer reading is proportional to the number of wheel revolutions. Thus

450r_1=440r_2, \qquad \frac{r_2}{r_1}=\frac{45}{44}.

With

r_1=15,

the increase is

\begin{aligned} r_2-r_1 &=15\left(\frac{45}{44}-1\right)\\ &=\frac{15}{44}\\ &\approx0.3409. \end{aligned}

To the nearest hundredth this is

0.34

inch.

Therefore, the correct answer is B.

22.

If the sum of the first

3n

positive integers is

150

more than the sum of the first

n

positive integers, then the sum of the first

4n

positive integers is:

300

350

400

450

600

Concepts:triangular number quadratic summation

Difficulty rating: 1590

Small hint:

Use

1+2+\cdots+m=\frac{m(m+1)}{2}

Big hint:

The given difference simplifies to

4n^2+n

Solution:

Let

S_m=\frac{m(m+1)}{2}.

Then

\begin{aligned} 2(S_{3n}-S_n) &=3n(3n+1)\\ &\quad{}-n(n+1)\\ &=8n^2+2n. \end{aligned}

Since

S_{3n}-S_n=150,

this gives

4n^2+n=150.

Hence

(n-6)(4n+25)=0,

and the positive solution is

n=6.

Therefore

S_{4n}=S_{24}=\frac{24\cdot25}{2}=300.

Therefore, the correct answer is A.

23.

The number

10!

(

10

is written in base

10

), when written in the base

12

system, ends in exactly

k

zeros. The value of

k

is:

1

2

3

4

5

Concepts:number base trailing zeros prime factorization factorial

Difficulty rating: 1830

Small hint:

A trailing base-

12

zero contributes one factor of

12=2^2\cdot3

Big hint:

Compare the exponents of

2

and

3

in

10!

Solution:

The prime exponents in

10!

are

\begin{gathered} v_2(10!)=5+2+1=8,\\ v_3(10!)=3+1=4. \end{gathered}

Each factor of

12=2^2\cdot3

uses two factors of

2

and one of

3.

Hence the number of trailing base-

12

zeros is

\min\left(\left\lfloor\frac82\right\rfloor,4\right)=4.

Therefore, the correct answer is D.

24.

An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is

2,

then the area of the hexagon is:

2

3

4

6

12

Concepts:equilateral triangle regular polygon area ratio perimeter

Difficulty rating: 1400

Small hint:

Let the hexagon side be

s

; equal perimeters make the triangle side

2s

Big hint:

The triangle splits into four small equilateral triangles of side

s

, while the hexagon splits into six

Solution:

Let the hexagon side be

s.

Its perimeter is

6s,

so the equilateral triangle has side

2s.

The large triangle consists of

4

equilateral triangles of side

s,

while the hexagon consists of

6.

Their area ratio is therefore

\frac{6}{4}=\frac{3}{2}.

Since the triangle’s area is

2,

the hexagon’s area is

3.

Therefore, the correct answer is B.

25.

For every real number

x,

let

\lfloor x\rfloor

be the greatest integer which is less than or equal to

x.

If the postal rate for first class mail is six cents for every ounce or portion thereof, then the cost in cents of first-class postage on a letter weighing

W

ounces is always:

6W

6\lfloor W\rfloor

6(\lfloor W\rfloor-1)

6(\lfloor W\rfloor+1)

-6\lfloor-W\rfloor

Concepts:floor and ceiling functions money function

Difficulty rating: 1850

Small hint:

The number of charged ounces is the least integer at least

W

Big hint:

Use the identity

\lceil W\rceil=-\lfloor-W\rfloor

Solution:

Charging for every ounce or portion thereof means the number of charged ounces is

\lceil W\rceil.

The floor and ceiling functions satisfy

\lceil W\rceil=-\lfloor-W\rfloor.

Thus the cost is

6\lceil W\rceil=-6\lfloor-W\rfloor.

Therefore, the correct answer is E.

26.

The number of distinct points in the

xy

-plane common to the graphs of

(x+y-5)(2x-3y+5)=0

and

(x-y+1)(3x+2y-12)=0

is:

0

1

2

3

4

infinite

Concepts:system of equations zero product property counting intersections

Difficulty rating: 1640

Small hint:

Each product equation represents a union of two lines

Big hint:

Check the intersection of the two lines in each pair before considering all four pairings

Solution:

The first graph is the union of

x+y=5,\qquad 2x-3y=-5,

and those two lines meet at

(2,3).

The second graph is the union of

x-y=-1,\qquad 3x+2y=12,

and these also meet at

(2,3).

In fact, substituting

(2,3)

satisfies all four line equations. Since the four lines have distinct slopes, no other point can belong to a line from each graph. There is exactly one common point.

Therefore, the correct answer is B.

27.

In a triangle, the area is numerically equal to the perimeter. What is the radius of the inscribed circle?

2

3

4

5

6

Concepts:incircle, incenter, and inradius triangle area perimeter

Difficulty rating: 1830

Small hint:

Express the area in terms of the inradius and semiperimeter

Big hint:

Use

K=rs

and perimeter

=2s

Solution:

If

r

is the inradius and

s

the semiperimeter, the triangle’s area is

K=rs.

Its perimeter is

2s.

The given equality yields

rs=2s.

Since

s\gt0,

we obtain

r=2.

Therefore, the correct answer is A.

28.

In triangle

ABC,

the median from vertex

A

is perpendicular to the median from vertex

B.

If the lengths of sides

AC

and

BC

are

6

and

7,

respectively, then the length of side

AB

is:

\sqrt{17}

4

4\tfrac12

2\sqrt5

4\tfrac14

Concepts:centroid median (geometry)vector Pythagorean Theorem

Difficulty rating: 2300

Small hint:

Place the centroid at the origin and let the position vectors of

A,B

be perpendicular vectors

u,v

Big hint:

Then

C=-u-v,

so

AC^2=|2u+v|^2

and

BC^2=|u+2v|^2

Solution:

Place the centroid at the origin. Let the position vectors of

A

and

B

be

u

and

v.

The two medians lie along

u

and

v,

so

u\cdot v=0.

Also the centroid condition gives

C=-u-v.

Let

U=|u|^2

and

V=|v|^2.

Then

AC^2=|2u+v|^2=4U+V=36

and

\begin{aligned} BC^2&=|u+2v|^2\\ &=U+4V=49. \end{aligned}

Adding appropriate multiples gives

U+V=17.

Finally,

AB^2=|u-v|^2=U+V=17,

so

AB=\sqrt{17}.

Therefore, the correct answer is A.

29.

It is now between

10{:}00

and

11{:}00

o’clock, and six minutes from now, the minute hand of a watch will be exactly opposite the place where the hour hand was three minutes ago. What is the exact time now?

10{:}05\tfrac5{11}

10{:}07\tfrac12

10{:}10

10{:}15

10{:}17\tfrac12

Concepts:clock linear equation

Difficulty rating: 1890

Small hint:

Let

x

be the number of minutes after

10{:}00

now

Big hint:

In minute-space units, the future minute hand is at

x+6

, while the point opposite the past hour hand is at

20+\frac{x-3}{12}

Solution:

Let

x

be the number of minutes after

10{:}00

now. Six minutes from now, the minute hand is

x+6

minute spaces clockwise from

12.

Three minutes ago, the hour hand was

\frac{x-3}{12}

minute spaces past the

10,

so the point opposite it was

20+\frac{x-3}{12}

minute spaces past

12.

Hence

x+6=20+\frac{x-3}{12}.

Solving gives

x=15,

so the time is

10{:}15.

Therefore, the correct answer is D.

30.

In the accompanying figure, segments

AB

and

CD

are parallel, the measure of angle

D

is twice that of angle

B,

and the measures of segments

AD

and

CD

are

a

and

b,

respectively. Then the measure of

AB

is equal to:

\tfrac12a+2b

\tfrac32b+\tfrac34a

2a-b

4b-\tfrac12a

a+b

Concepts:angle bisector parallel lines isosceles triangle parallelogram

Difficulty rating: 2190

Small hint:

Bisect angle

D

and let the bisector meet

AB

at

P

Big hint:

Angle chasing makes

\triangle APD

isosceles, while

PBCD

is a parallelogram

Solution:

Let the bisector of

\angle D

meet

AB

at

P.

Because

\angle D=2\angle B,

each half has measure

\angle B.

Since

AB\parallel CD,

the alternate interior angle

\angle APD

also equals

\angle PDC.

Thus triangle

APD

is isosceles and

AP=AD=a.

Also

PB\parallel CD

and

BC\parallel PD,

so

PBCD

is a parallelogram. Hence

PB=CD=b.

Therefore

AB=AP+PB=a+b.

Therefore, the correct answer is E.

31.

If a number is selected at random from the set of all five-digit numbers in which the sum of the digits is equal to

43,

what is the probability that this number will be divisible by

11?

\frac{2}{5}

\frac{1}{5}

\frac{1}{6}

\frac{1}{11}

\frac{1}{15}

Concepts:digits divisibility basic counting basic probability

Difficulty rating: 2300

Small hint:

Start from

99999

; a digit sum of

43

means a total deficit of

2

Big hint:

The possibilities are one

7

or two

8

s; apply the divisibility-by-

11

alternating-sum test

Solution:

The maximum five-digit digit sum is

45,

so a sum of

43

has total deficit

2

from

99999.

Either one digit is

7

and the others are

9

, giving

5

numbers, or two digits are

8

and the others are

9

, giving

\binom52=10

numbers. Thus there are

15

numbers total.

Using the divisibility-by-

11

test, exactly

97999,\qquad 99979,\qquad 98989

are divisible by

11.

The probability is therefore

\frac{3}{15}=\frac{1}{5}.

Therefore, the correct answer is B.

32.

A

and

B

travel around a circular track at uniform speeds in opposite directions, starting from diametrically opposite points. If they start at the same time, meet first after

B

has travelled

100

yards, and meet a second time

60

yards before

A

completes one lap, then the circumference of the track in yards is:

400

440

480

560

880

Concepts:relative speed circumference ratio and proportion system of equations

Difficulty rating: 2190

Small hint:

Let the circumference be

2C

and compare the two runners’ traveled-distance ratios at each meeting

Big hint:

At the first meeting their distances are

C-100

and

100

; at the second they are

2C-60

and

C+60

Solution:

Let the circumference be

2C.

At the first meeting,

B

has traveled

100

yards and

A

has traveled

C-100,

so their speed ratio is

\frac{v_A}{v_B}=\frac{C-100}{100}.

At the second meeting,

A

is

60

yards short of one lap, so it has traveled

2C-60.

By then their combined distance is

3C,

so

B

has traveled

C+60.

Hence

\frac{C-100}{100}=\frac{2C-60}{C+60}.

Cross-multiplying gives

C=240,

so the circumference is

2C=480

yards.

Therefore, the correct answer is C.

33.

Find the sum of the digits of all the numerals in the sequence

1,

2,

3,

4,

\ldots,

10000.

180{,}001

154{,}756

45{,}001

154{,}755

270{,}001

Concepts:digits place value symmetry (algebra)summation

Difficulty rating: 1890

Small hint:

First sum the digits from

0000

through

9999

Big hint:

In each of the four positions, every digit appears exactly

1000

times

Solution:

Write the integers from

0

through

9999

using four digits with leading zeros. In each of the four positions, every digit

0,

1,

\ldots,

9

appears

1000

times. Thus their total digit sum is

\begin{gathered} 4\cdot1000(0+1+\cdots+9)\\ =4000\cdot45\\ =180000. \end{gathered}

Replacing

0

by

10000

adds

1,

so the requested sum is

180001.

Therefore, the correct answer is A.

34.

The greatest integer that will divide

13{,}511,

13{,}903,

and

14{,}589

and leave the same remainder is:

28

49

98

an odd multiple of

7

greater than

49

an even multiple of

7

greater than

98

Concepts:greatest common divisor divisibility algebraic manipulation

Difficulty rating: 1650

Small hint:

A common remainder means the divisor divides every pairwise difference

Big hint:

Find the GCD of

13903-13511

and

14589-13903

Solution:

A divisor leaves the same remainder on all three numbers exactly when it divides their differences. The two successive differences are

\begin{gathered} 13903-13511=392,\\ 14589-13903=686. \end{gathered}

Therefore the greatest possible divisor is

\begin{gathered} \gcd(392,686)\\ =\gcd(392,294)\\ =\gcd(294,98)=98. \end{gathered}

Therefore, the correct answer is C.

35.

A retiring employee receives an annual pension proportional to the square root of the number of years of his service. Had he served

a

years more, his pension would have been

p

dollars greater, whereas, had he served

b

years more (

b\ne a

), his pension would have been

q

dollars greater than the original annual pension. Find his annual pension in terms of

a,

b,

p,

and

q.

\dfrac{p^2-q^2}{2(a-b)}

\dfrac{(p-q)^2}{2\sqrt{ab}}

\dfrac{ap^2-bq^2}{2(ap-bq)}

\dfrac{aq^2-bp^2}{2(bp-aq)}

\sqrt{(a-b)(p-q)}

Concepts:ratio and proportion radical system of equations algebraic manipulation

Difficulty rating: 2440

Small hint:

Let the current pension be

X=k\sqrt n

and write the two hypothetical pensions

Big hint:

Square

X+p=k\sqrt{n+a}

and

X+q=k\sqrt{n+b}

, then eliminate both

n

and

k^2

Solution:

Let the current pension be

X=k\sqrt n.

Since

X+p=k\sqrt{n+a}

and

X+q=k\sqrt{n+b},

squaring and using

X^2=k^2n

gives the first line below. Multiplying its equations by

b

and

a,

respectively, and subtracting gives the second:

\begin{gathered} 2pX+p^2=k^2a,\\ 2qX+q^2=k^2b,\\ 2X(bp-aq)=aq^2-bp^2. \end{gathered}

Hence

X=\dfrac{aq^2-bp^2}{2(bp-aq)}.

Therefore, the correct answer is D.