AB is a fixed diameter of a circle whose center is O. From C, any point on the circle, a chord CD is drawn perpendicular to AB. Then, as C moves over a semicircle, the bisector of angle OCD cuts the circle in a point that always:

(D) Is as far from AB as from D

(E) Is equidistant from B and C

1951 AMC 12 — Problems & Online Timed Exam

1.

The percent that

M

is greater than

N

is:

\dfrac{100(M-N)}{M}

\dfrac{100(M-N)}{N}

\dfrac{M-N}{N}

\dfrac{M-N}{M}

\dfrac{100(M+N)}{N}

Answer: B

Concepts:percentage algebraic manipulation

Difficulty rating: 1020

Small hint:

Percent increase is measured relative to the original quantity

Big hint:

Here

N

is the original quantity, and the increase is

M-N

Solution:

The increase from

N

to

M

is

M-N.

Dividing by the original quantity and multiplying by

100

gives

\frac{100(M-N)}{N}.

Thus, the correct answer is B.

2.

A rectangular field is half as wide as it is long and is completely enclosed by

x

yards of fencing. The area in terms of

x

is:

\dfrac{x^2}{2}

2x^2

\dfrac{2x^2}{9}

\dfrac{x^2}{18}

\dfrac{x^2}{72}

Answer: D

Concepts:rectangle perimeter area

Difficulty rating: 1180

Small hint:

Let the width be

w

, so the length is

2w

Big hint:

Use the perimeter equation

2(w+2w)=x

before finding the area

Solution:

Let the width be

w

, so the length is

2w.

The fencing gives

2(w+2w)=6w=x,

hence

w=\frac{x}{6}.

Therefore

A=w(2w)=2\left(\frac{x}{6}\right)^2=\frac{x^2}{18}.

Thus, the correct answer is D.

3.

If the length of a diagonal of a square is

a+b,

then the area of the square is:

(a+b)^2

\dfrac12(a+b)^2

a^2+b^2

\dfrac12(a^2+b^2)

None of these

Answer: B

Concepts:square (geometry)diagonal area

Difficulty rating: 1270

Small hint:

Relate the diagonal

d

of a square to its side

s

Big hint:

Since

d=s\sqrt2

, the area

s^2

equals

\frac{d^2}{2}

Solution:

If

s

is the side length and

d=a+b

is the diagonal, then

d=s\sqrt2.

Thus

s^2=\frac{d^2}{2}=\frac12(a+b)^2.

Thus, the correct answer is B.

4.

A barn with a flat roof is rectangular in shape,

10

yd. wide,

13

yd. long, and

5

yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of square yards to be painted is:

360

460

490

590

720

Answer: D

Concepts:rectangular prism surface area

Difficulty rating: 1220

Small hint:

Count both the inside and outside faces of each of the four walls

Big hint:

The ceiling contributes one

10

-by-

13

rectangle

Solution:

The four walls have one-sided area

2(13\cdot5)+2(10\cdot5)=230.

Painting them inside and outside contributes

2(230)=460.

The ceiling adds

10\cdot13=130,

so the total is

460+130=590.

Thus, the correct answer is D.

5.

Mr.

A

owns a home worth

\$10{,}000.

He sells it to Mr.

B

at a

10\%

profit based on the worth of the house. Mr.

B

sells the house back to Mr.

A

at a

10\%

loss. Then:

A

comes out even

A

makes

\$1100

on the deal

A

makes

\$1000

on the deal

A

loses

\$900

on the deal

A

loses

\$1000

on the deal

Answer: B

Concepts:percentage money

Difficulty rating: 1180

Small hint:

Compute the two sale prices separately

Big hint:

The first sale is for

10{,}000(1.10)

, and the return sale is

10\%

below that amount

Solution:

The first sale price is

10{,}000(1.10)=11{,}000.

The return sale is

11{,}000(0.90)=9{,}900.

Mr.

A

receives

11{,}000

and pays back

9{,}900,

so he gains

1{,}100.

Thus, the correct answer is B.

6.

The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:

The volume of the box

The square root of the volume

Twice the volume

The square of the volume

The cube of the volume

Answer: D

Concepts:rectangular prism volume algebraic manipulation

Difficulty rating: 1270

Small hint:

Call the three edge lengths

l,w,h

Big hint:

Multiply the three face areas

lw,

wh,

and

hl

Solution:

If the edge lengths are

l,w,h,

the three face areas are

lw,

wh,

and

hl.

Their product is

(lw)(wh)(hl)=l^2w^2h^2=(lwh)^2,

the square of the volume.

Thus, the correct answer is D.

7.

An error of

0.02''

is made in the measurement of a line

10''

long, while an error of only

0.2''

is made in a measurement of a line

100''

long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:

Greater by

0.18

The same

Less

10

times as great

Correctly described by both

(A)

and

(D)

Answer: B

Concepts:ratio and proportion decimal

Difficulty rating: 1240

Small hint:

Relative error is the absolute error divided by the measured length

Big hint:

Compare

\frac{0.02}{10}

with

\frac{0.2}{100}

Solution:

The two relative errors are

\frac{0.02}{10}=0.002

and

\frac{0.2}{100}=0.002.

They are equal.

Thus, the correct answer is B.

8.

The price of an article is cut

10\%.

To restore it to its former value, the new price must be increased by:

10\%

9\%

11\dfrac19\%

11\%

None of these answers

Answer: C

Concepts:percentage

Difficulty rating: 1320

Small hint:

After the cut, the price is

90\%

of the original

Big hint:

If the required increase is

p

, solve

0.9(1+p)=1

Solution:

Let the original price be

1.

The reduced price is

0.9.

The required fractional increase is

\frac{1-0.9}{0.9}=\frac19,

which is

11\dfrac19\%.

Thus, the correct answer is C.

9.

An equilateral triangle is drawn with a side of length

a.

A new equilateral triangle is formed by joining the midpoints of the sides of the first one. Then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. The limit of the sum of the perimeters of all the triangles thus drawn is:

Infinite

\left(5\dfrac14\right)a

2a

6a

\left(4\dfrac12\right)a

Answer: D

Concepts:geometric sequence perimeter equilateral triangle

Difficulty rating: 1400

Small hint:

Each midpoint triangle has half the side length of the preceding triangle

Big hint:

The perimeters form

3a+\frac{3a}{2}+\frac{3a}{4}+\cdots

Solution:

The first perimeter is

3a,

and every later perimeter is half the preceding one. Hence the total is

\begin{aligned} &3a\left(1+\frac12+\frac14+\cdots\right)\\ &\quad=3a\cdot\frac{1}{1-\frac12}\\ &\quad=6a. \end{aligned}

Thus, the correct answer is D.

10.

Of the following statements, the one that is incorrect is:

Doubling the base of a given rectangle doubles the area.

Doubling the altitude of a triangle doubles the area.

Doubling the radius of a given circle doubles the area.

Doubling the divisor of a fraction and dividing its numerator by

2

changes the quotient.

Doubling a given quantity may make it less than it originally was.

Answer: C

Concepts:power scaling of length, area, and volume counterexample

Difficulty rating: 1180

Small hint:

Check how each area formula changes when one length is doubled

Big hint:

A circle’s area depends on the square of its radius

Solution:

A circle has area

\pi r^2.

Replacing

r

by

2r

gives

\pi(2r)^2=4\pi r^2,

so doubling the radius quadruples, rather than doubles, the area. The other statements can hold as written.

Thus, the correct answer is C.

11.

The limit of the sum of an infinite number of terms in a geometric progression is

\dfrac{a}{1-r},

where

a

denotes the first term and

-1\lt r\lt1

denotes the common ratio. The limit of the sum of their squares is:

\dfrac{a^2}{(1-r)^2}

\dfrac{a^2}{1+r^2}

\dfrac{a^2}{1-r^2}

\dfrac{4a^2}{1+r^2}

None of these

Answer: C

Concepts:geometric sequence

Difficulty rating: 1470

Small hint:

Squaring every term also squares the common ratio

Big hint:

The squared series begins

a^2+a^2r^2+a^2r^4+\cdots

Solution:

The squared terms form a geometric series with first term

a^2

and ratio

r^2.

Since

|r|\lt1,

its sum is

\frac{a^2}{1-r^2}.

Thus, the correct answer is C.

12.

At

2{:}15

o’clock, the hour and minute hands of a clock form an angle of:

30^\circ

5^\circ

22\dfrac12^\circ

7\dfrac12^\circ

28^\circ

Answer: C

Concepts:clock angle chasing

Difficulty rating: 1200

Small hint:

At

2{:}15

the minute hand is at

90^\circ

from

12

Big hint:

The hour hand moves

0.5^\circ

per minute in addition to its position at

2{:}00

Solution:

At

2{:}15

the minute hand is

90^\circ

from

12.

The hour hand is

2\cdot30^\circ+15\cdot0.5^\circ=67.5^\circ

from

12.

Their angle is

90^\circ-67.5^\circ=22.5^\circ=22\dfrac12^\circ.

Thus, the correct answer is C.

13.

A

can do a piece of work in

9

days.

B

is

50\%

more efficient than

A.

The number of days it takes

B

to do the same piece of work is:

13\dfrac12

4\dfrac12

6

3

None of these answers

Answer: C

Concepts:rate ratio and proportion

Difficulty rating: 1150

Small hint:

Translate days per job into jobs per day

Big hint:

Multiply

A

’s rate

\frac{1}{9}

by

1.5

Solution:

A

’s rate is

\frac{1}{9}

job per day. Thus

B

’s rate is

\frac32\cdot\frac19=\frac16

job per day, so

B

needs

6

days.

Thus, the correct answer is C.

14.

In connection with proof in geometry, indicate which one of the following statements is incorrect:

Some statements are accepted without being proved.

In some instances there is more than one correct order in proving certain propositions.

Every term used in a proof must have been defined previously.

It is not possible to arrive by correct reasoning at a true conclusion if, in the given, there is an untrue proposition.

Indirect proof can be used whenever there are two or more contrary propositions.

Answer: C

Concepts:logical deduction

Difficulty rating: 1260

Small hint:

Distinguish undefined primitive terms from unproved axioms

Big hint:

An axiomatic system cannot define every term without an infinite regress

Solution:

An axiomatic development begins with primitive terms that are deliberately left undefined, as well as statements accepted without proof. Therefore it is not true that every term in a proof must previously have been defined.

Thus, the intended correct answer is C.

15.

The largest number by which the expression

n^3-n

is divisible for all possible integral values of

n

is:

2

3

4

5

6

Answer: E

Concepts:divisibility factoring

Difficulty rating: 1390

Small hint:

Factor

n^3-n

Big hint:

The factors

n-1,n,n+1

are three consecutive integers

Solution:

We have

n^3-n=n(n-1)(n+1),

a product of three consecutive integers. One is divisible by

3

and at least one is even, so the product is always divisible by

6.

Taking

n=2

gives exactly

6,

so no larger integer always divides it.

Thus, the correct answer is E.

16.

If in applying the quadratic formula to a quadratic equation

f(x)=ax^2+bx+c=0,

it happens that

c=\dfrac{b^2}{4a},

then the graph of

y=f(x)

will certainly:

Have a maximum

Have a minimum

Be tangent to the

x

-axis

Be tangent to the

y

-axis

Lie in one quadrant only

Answer: C

Concepts:quadratic tangent line

Difficulty rating: 1400

Small hint:

Substitute the condition into the discriminant

b^2-4ac

Big hint:

A quadratic with one repeated real root touches the horizontal axis once

Solution:

The condition gives

b^2-4ac=b^2-4a\left(\frac{b^2}{4a}\right)=0.

Hence the quadratic has a repeated real root, so its parabola is tangent to the

x

-axis.

Thus, the correct answer is C.

17.

Indicate in which one of the following equations

y

is neither directly nor inversely proportional to

x

:

x+y=0

3xy=10

x=5y

3x+y=10

\dfrac{x}{y}=\sqrt3

Answer: D

Concepts:ratio and proportion linear equation

Difficulty rating: 1150

Small hint:

Direct proportion has the form

y=kx

, while inverse proportion has the form

xy=k

Big hint:

Rewrite each equation by solving for

y

Solution:

Choices A, C, and E rearrange to

y=kx,

and choice B rearranges to

xy=k.

But choice D gives

y=10-3x,

which is neither form.

Thus, the correct answer is D.

18.

The expression

21x^2+ax+21

is to be factored into two linear prime binomial factors with integer coefficients. This can be done if

a

is:

Any odd number

Some odd number

Any even number

Some even number

Zero

Answer: D

Concepts:factoring parity

Difficulty rating: 1450

Small hint:

Write the factors as

(Ax+B)(Cx+D)

Big hint:

Because

AC=BD=21,

all four integer factors are odd

Solution:

Suppose

\begin{aligned} &21x^2+ax+21\\ &\quad=(Ax+B)(Cx+D). \end{aligned}

Then

AC=BD=21,

so

A,B,C,D

are odd. Therefore

a=AD+BC

is even. Some even values work; for example,

\begin{aligned} &(3x+7)(7x+3)\\ &\quad=21x^2+58x+21. \end{aligned}

Not every even value works.

Thus, the correct answer is D.

19.

A six-place number is formed by repeating a three-place number; for example,

256{,}256,

or

678{,}678,

etc. Any number of this form is always exactly divisible by:

7

only

11

only

13

only

101

1001

Answer: E

Concepts:place value divisibility

Difficulty rating: 1440

Small hint:

Represent the repeated three-digit block by

N

Big hint:

The six-digit number is

1000N+N

Solution:

If the repeated block is

N,

then the six-place number is

1000N+N=1001N.

It is therefore always divisible by

1001.

Thus, the correct answer is E.

20.

When simplified and expressed with negative exponents, the expression

(x+y)^{-1}(x^{-1}+y^{-1})

is equal to:

x^{-2}+2x^{-1}y^{-1}+y^{-2}

x^{-2}+2^{-1}x^{-1}y^{-1}+y^{-2}

x^{-1}y^{-1}

x^{-2}+y^{-2}

\dfrac{1}{x^{-1}y^{-1}}

Answer: C

Concepts:algebraic manipulation exponent

Difficulty rating: 1320

Small hint:

Rewrite

x^{-1}+y^{-1}

as one fraction

Big hint:

The factor

x+y

cancels after combining the reciprocals

Solution:

For nonzero

x,y

with

x+y\ne0,

\begin{aligned} &(x+y)^{-1}(x^{-1}+y^{-1})\\ &\quad=\frac{1}{x+y}\left(\frac{x+y}{xy}\right)\\ &\quad=\frac1{xy}=x^{-1}y^{-1}. \end{aligned}

Thus, the correct answer is C.

21.

Given

x\gt0,

y\gt0,

x\gt y,

and

z\ne0.

The inequality which is not always correct is:

x+z\gt y+z

x-z\gt y-z

xz\gt yz

\dfrac{x}{z^2}\gt\dfrac{y}{z^2}

xz^2\gt yz^2

Answer: C

Concepts:inequality counterexample

Difficulty rating: 1400

Small hint:

Adding the same number preserves order, but multiplying does not always do so

Big hint:

Test the statements when

z\lt0

Solution:

Because

z

may be negative, multiplying

x\gt y

by

z

reverses the inequality and gives

xz\lt yz.

The operations in the other choices preserve the inequality because

z^2\gt0.

Thus, the correct answer is C.

22.

The values of

a

in the equation

\log_{10}(a^2-15a)=2

are:

\dfrac{15\pm\sqrt{233}}{2}

20,

-5

\dfrac{15\pm\sqrt{305}}{2}

\pm20

None of these

Answer: B

Concepts:logarithm quadratic

Difficulty rating: 1400

Small hint:

Convert the logarithmic equation to exponential form

Big hint:

Solve

a^2-15a=100

Solution:

The equation is equivalent to

a^2-15a=10^2=100,

so

\begin{aligned} &a^2-15a-100\\ &\quad=(a-20)(a+5)\\ &\quad=0. \end{aligned}

Both

a=20

and

a=-5

make the logarithm’s argument

100,

so both are valid.

Thus, the correct answer is B.

23.

The radius of a cylindrical box is

8

inches and the height is

3

inches. The number of inches that may be added to either the radius or the height to give the same nonzero increase in volume is:

1

5\dfrac13

Any number

Non-existent

None of these

Answer: B

Concepts:cylinder volume quadratic

Difficulty rating: 1530

Small hint:

Compare the volume after adding

x

to the radius with the volume after adding

x

to the height

Big hint:

Set

\pi(8+x)^2(3)=\pi(8)^2(3+x)

and discard

x=0

Solution:

For the two new volumes to be equal,

3\pi(8+x)^2=64\pi(3+x).

Cancelling

\pi

and expanding gives

3x^2-16x=0,

so

x=0

or

x=\frac{16}{3}=5\dfrac13.

The problem requires a nonzero increase.

Thus, the correct answer is B.

24.

When simplified,

\dfrac{2^{n+4}-2(2^n)}{2(2^{n+3})}

is:

2^{n+1}-\dfrac18

-2^{n+1}

1-2^n

\dfrac78

\dfrac74

Answer: D

Concepts:exponent algebraic manipulation

Difficulty rating: 1320

Small hint:

Factor

2^n

from the numerator

Big hint:

Rewrite the denominator as

2^{n+4}

Solution:

Factoring and combining powers gives

\frac{2^n(2^4-2)}{2^{n+4}} =\frac{14}{16} =\frac78.

Thus, the correct answer is D.

25.

The apothem of a square having its area numerically equal to its perimeter is compared with the apothem of an equilateral triangle having its area numerically equal to its perimeter. The first apothem will be:

Equal to the second

\dfrac43

times the second

\dfrac{2}{\sqrt3}

times the second

\dfrac{\sqrt2}{\sqrt3}

times the second

Indeterminately related to the second

Answer: A

Concepts:regular polygon area perimeter

Difficulty rating: 1610

Small hint:

For a regular polygon, area equals one half the apothem times the perimeter

Big hint:

If a nonzero perimeter equals the area, cancel the perimeter from

A=\frac12rP

Solution:

Every regular polygon with apothem

r

and perimeter

P

has area

A=\frac12rP.

If its numerical area equals its nonzero perimeter, then

\frac12rP=P,

so

r=2.

This applies to both the square and the equilateral triangle, so their apothems are equal.

Thus, the correct answer is A.

26.

In the equation

\frac{x(x-1)-(m+1)}{(x-1)(m-1)}=\frac{x}{m},

the roots are equal when:

m=1

m=\dfrac12

m=0

m=-1

m=-\dfrac12

Answer: E

Concepts:quadratic rational equation

Difficulty rating: 1830

Small hint:

Clear denominators and collect the resulting quadratic in

x

Big hint:

The equation reduces to

x^2-x-m(m+1)=0

; equal roots require zero discriminant

Solution:

Clearing denominators and simplifying yields

x^2-x-m(m+1)=0.

Equal roots require

\begin{aligned} 1+4m(m+1) &=(2m+1)^2\\ &=0, \end{aligned}

hence

m=-\frac12.

This value does not violate the original denominators.

Thus, the correct answer is E.

27.

Through a point inside a triangle, three lines are drawn from the vertices to the opposite sides, forming six triangular sections. Then:

The triangles are similar in opposite pairs

The triangles are congruent in opposite pairs

The triangles are equal in area in opposite pairs

Three similar quadrilaterals are formed

None of the above relations is true

Answer: E

Concepts:triangle area counterexample

Difficulty rating: 1470

Small hint:

The interior point and the three cevians are arbitrary

Big hint:

Move the point very close to one side to test the claimed opposite-area relation

Solution:

No angle or length condition forces opposite sections to be similar or congruent. Their areas also need not match: placing the interior point very close to one side makes the sections adjoining that side arbitrarily small without forcing their opposite sections to be small. No quadrilaterals are among the six sections.

Thus, the correct answer is E.

28.

The pressure

P

of wind on a sail varies jointly as the area

A

of the sail and the square of the velocity

V

of the wind. The pressure on a square foot is

1

pound when the velocity is

16

miles per hour. The velocity of the wind when the pressure on a square yard is

36

pounds is:

10\dfrac23

mph

96

mph

32

mph

1\dfrac13

mph

16

mph

Answer: C

Concepts:ratio and proportion unit conversion

Difficulty rating: 1520

Small hint:

Use

P=kAV^2

and remember that one square yard is nine square feet

Big hint:

The first condition gives

1=k(1)(16^2)

Solution:

Write

P=kAV^2.

From

1=k(1)(16^2),

we get

k=\frac{1}{256}.

A square yard has area

9

square feet, so

36=\frac1{256}\cdot9V^2.

Thus

V^2=1024

and the positive speed is

V=32

mph.

Thus, the correct answer is C.

29.

Of the following sets of data, the only one that does not determine the shape of a triangle is:

The ratio of two sides and the included angle

The ratios of the three altitudes

The ratios of the three medians

The ratio of the altitude to the corresponding base

Two angles

Answer: D

Concepts:similarity triangle area

Difficulty rating: 1530

Small hint:

Determining shape means determining the triangle up to similarity

Big hint:

A single altitude-to-base ratio fixes an area ratio but can occur in differently shaped triangles

Solution:

Choices A and E determine the angles up to similarity. Ratios of all three altitudes determine reciprocal side ratios, and ratios of all three medians determine side ratios. But a single ratio

\frac{h}{b}

does not determine the remaining side lengths or angles; many non-similar triangles can share it.

Thus, the correct answer is D.

30.

If two poles

20''

and

80''

high are

100''

apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:

50''

40''

16''

60''

None of these

Answer: C

Concepts:coordinate geometry linear equation

Difficulty rating: 1580

Small hint:

Place the pole bases at

x=0

and

x=100

Big hint:

The cross-lines can be written

y=\frac{4x}{5}

and

y=20-\frac{x}{5}

Solution:

Put the bases at

(0,0)

and

(100,0),

with tops

(0,80)

and

(100,20).

The cross-lines are

y=\frac45x

and

y=20-\frac15x.

Equating them gives

x=20,

and hence

y=16.

Thus, the correct answer is C.

31.

A total of

28

handshakes was exchanged at the conclusion of a party. Assuming that each participant was equally polite toward all the others, the number of people present was:

14

28

56

8

7

Answer: D

Concepts:counting pairs quadratic

Difficulty rating: 1290

Small hint:

With

n

people, each unordered pair shakes hands once

Big hint:

Solve

\binom n2=28

Solution:

With

n

people the number of handshakes is

\binom n2=\frac{n(n-1)}2.

Thus

n(n-1)=56,

and the positive solution is

n=8.

Thus, the correct answer is D.

32.

If

\triangle ABC

is inscribed in a semicircle whose diameter is

\overline{AB},

then

\overline{AC}+\overline{BC}

must be:

Equal to

\overline{AB}

Equal to

\overline{AB}\sqrt2

\ge \overline{AB}\sqrt2

\le \overline{AB}\sqrt2

\overline{AB}^{\,2}

Answer: D

Concepts:right triangle Cauchy-Schwarz Inequality

Difficulty rating: 1610

Small hint:

The angle subtending the diameter is a right angle

Big hint:

For legs

a,b

, compare

(a+b)^2

with

2(a^2+b^2)

Solution:

By Thales’ theorem,

AC

and

BC

are the legs of a right triangle with hypotenuse

AB.

Therefore

\begin{aligned} (AC+BC)^2 &\le2(AC^2+BC^2)\\ &=2AB^2. \end{aligned}

Taking positive square roots gives

AC+BC\le AB\sqrt2,

with equality for an isosceles right triangle.

Thus, the correct answer is D.

33.

The roots of the equation

x^2-2x=0

can be obtained graphically by finding the abscissas of the points of intersection of each of the following pairs of equations except the pair:

y=x^2,

y=2x

y=x^2-2x,

y=0

y=x,

y=x-2

y=x^2-2x+1,

y=1

y=x^2-1,

y=2x-1

Answer: C

Concepts:function quadratic algebraic manipulation

Difficulty rating: 1320

Small hint:

Set the two right-hand sides in each pair equal

Big hint:

Four pairs reduce to

x^2-2x=0

; one pair reduces to an impossible constant equation

Solution:

Choices A, B, D, and E all reduce their intersection condition to

x^2-2x=0.

Choice C instead requires

x=x-2,

which has no solution and therefore cannot produce the desired roots.

Thus, the correct answer is C.

34.

The value of

10^{\log_{10}7}

is:

7

1

10

\log_{10}7

\log_7 10

Answer: A

Concepts:logarithm exponent

Difficulty rating: 1260

Small hint:

Exponentiation by

10

and the base-

10

logarithm are inverse operations

Big hint:

Use the identity

b^{\log_b x}=x

Solution:

Since the base-

10

exponential and logarithm are inverse functions,

10^{\log_{10}7}=7.

Thus, the correct answer is A.

35.

If

a^x=c^q=b

and

c^y=a^z=d,

then:

xy=qz

\dfrac{x}{y}=\dfrac{q}{z}

x+y=q+z

x-y=q-z

x^y=q^z

Answer: A

Concepts:logarithm system of equations

Difficulty rating: 1580

Small hint:

Take logarithms of both chains of equal powers

Big hint:

From

x\log a=q\log c

and

y\log c=z\log a

, eliminate the logarithms

Solution:

Taking logarithms gives

\begin{gathered} x\log a=q\log c,\\ y\log c=z\log a. \end{gathered}

Multiplying these equations and cancelling the nonzero logarithmic factors under the historical nondegenerate-base convention yields

xy=qz.

Thus, the intended correct answer is A.

36.

Which of the following methods of proving a geometric figure a locus is not correct?

Every point on the locus satisfies the conditions and every point not on the locus does not satisfy the conditions.

Every point not satisfying the conditions is not on the locus and every point on the locus does satisfy the conditions.

Every point satisfying the conditions is on the locus and every point on the locus satisfies the conditions.

Every point not on the locus does not satisfy the conditions and every point not satisfying the conditions is not on the locus.

Every point satisfying the conditions is on the locus and every point not satisfying the conditions is not on the locus.

Answer: B

Concepts:logical deduction

Difficulty rating: 1610

Small hint:

A locus proof needs both implications between “on the locus” and “satisfies the conditions”

Big hint:

Use contrapositives to see whether each choice establishes both directions

Solution:

Let

L

mean “on the locus” and

C

mean “satisfies the conditions.” A complete proof needs both

L\Rightarrow C

and

C\Rightarrow L.

Choice B states

\neg C\Rightarrow\neg L,

which is merely the contrapositive of

L\Rightarrow C,

and then repeats

L\Rightarrow C.

It never proves

C\Rightarrow L.

Thus, the correct answer is B.

37.

A number which when divided by

10

leaves a remainder of

9,

when divided by

9

leaves a remainder of

8,

by

8

leaves a remainder of

7,

etc., down to where, when divided by

2,

it leaves a remainder of

1,

is:

59

419

1259

2519

None of these answers

Answer: D

Concepts:least common multiple modular arithmetic

Difficulty rating: 1580

Small hint:

Adding

1

to the desired number removes every listed remainder

Big hint:

Find

\operatorname{lcm}(2,3,\ldots,10)

Solution:

If the number is

N,

then

N+1

is divisible by every integer from

2

through

10.

Their least common multiple is

2^3\cdot3^2\cdot5\cdot7=2520.

Thus the least positive number fitting all the conditions is

N=2520-1=2519.

Thus, the correct answer is D.

38.

A rise of

600

feet is required to get a railroad line over a mountain. The grade can be kept down by lengthening the track and curving it around the mountain peak. The additional length of track required to reduce the grade from

3\%

to

2\%

is approximately:

10{,}000

ft.

20{,}000

ft.

30{,}000

ft.

12{,}000

ft.

None of these

Answer: A

Concepts:percentage ratio and proportion

Difficulty rating: 1410

Small hint:

Grade is rise divided by track length

Big hint:

Compare

\frac{600}{0.03}

with

\frac{600}{0.02}

Solution:

A

3\%

grade requires about

\frac{600}{0.03}=20{,}000

feet of track, while a

2\%

grade requires

\frac{600}{0.02}=30{,}000

feet. The additional length is

10{,}000

feet.

Thus, the correct answer is A.

39.

A stone is dropped into a well and the report of the stone striking the bottom is heard

7.7

seconds after it is dropped. Assume that the stone falls

16t^2

feet in

t

seconds and that the velocity of sound is

1120

feet per second. The depth of the well is:

784

ft.

342

ft.

1568

ft.

156.8

ft.

None of these

Answer: A

Concepts:distance rate and time quadratic

Difficulty rating: 1670

Small hint:

The total time is the falling time plus the sound’s return time

Big hint:

If the falling time is

t

, solve

t+\dfrac{16t^2}{1120}=7.7

Solution:

If the stone falls for

t

seconds, the depth is

16t^2,

and the sound takes

\frac{16t^2}{1120}=\frac{t^2}{70}

seconds to return. Hence

t+\frac{t^2}{70}=7.7,

or

t^2+70t-539=0.

The positive root is

t=7,

giving depth

16(7^2)=784

feet.

Thus, the correct answer is A.

40.

The expression

\begin{aligned} &\left(\frac{(x+1)^2(x^2-x+1)^2}{(x^3+1)^2}\right)^2\\ &\quad{}\cdot \left(\frac{(x-1)^2(x^2+x+1)^2}{(x^3-1)^2}\right)^2 \end{aligned}

equals:

(x+1)^4

(x^3+1)^4

1

\big[(x^3+1)(x^3-1)\big]^2

\big[(x^3-1)^2\big]^2

Answer: C

Concepts:sum and difference of cubes algebraic manipulation

Difficulty rating: 1400

Small hint:

Factor

x^3+1

and

x^3-1

Big hint:

Use

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

and the analogous difference formula

Solution:

Because

\begin{aligned} x^3+1&=(x+1)(x^2-x+1),\\ x^3-1&=(x-1)(x^2+x+1), \end{aligned}

each fraction inside parentheses equals

1

wherever the original expression is defined. Their squared product is therefore

1.

Thus, the correct answer is C.

41.

The formula expressing the relationship between

x

and

y

in the table is:

$x$	$2$	$3$	$4$	$5$	$6$
$y$	$0$	$2$	$6$	$12$	$20$

y=2x-4

y=x^2-3x+2

y=x^3-3x^2+2x

y=x^2-4x

y=x^2-4

Answer: B

Concepts:data and graph interpretation quadratic pattern recognition

Difficulty rating: 1390

Small hint:

The first differences

2,4,6,8

suggest a quadratic rule

Big hint:

The values are

(x-1)(x-2)

Solution:

The

y

-values factor naturally as

\begin{gathered} 0=1\cdot0,\quad 2=2\cdot1,\quad 6=3\cdot2,\\ 12=4\cdot3,\quad 20=5\cdot4. \end{gathered}

Thus

y=(x-1)(x-2)

and hence

y=x^2-3x+2.

Thus, the correct answer is B.

42.

If

x

equals

\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}},

then:

x=1

0\lt x\lt1

1\lt x\lt2

x

is infinite

x\gt2

but finite

Answer: C

Concepts:radical recursion quadratic

Difficulty rating: 1530

Small hint:

The expression under the first radical contains another copy of

x

Big hint:

Solve

x=\sqrt{1+x}

and keep the nonnegative root

Solution:

The repeating tail gives

x=\sqrt{1+x},

so

x^2-x-1=0.

Since

x\ge0,

x=\frac{1+\sqrt5}{2},

which lies strictly between

1

and

2.

Thus, the correct answer is C.

43.

Of the following statements, the only one that is incorrect is:

An inequality will remain true after each side is increased, decreased, multiplied, or divided (zero excluded) by the same positive quantity.

The arithmetic mean of two unequal positive quantities is greater than their geometric mean.

If the sum of two positive quantities is given, their product is largest when they are equal.

If

a

and

b

are positive and unequal,

\dfrac12(a^2+b^2)

is greater than

\left[\dfrac12(a+b)\right]^2.

If the product of two positive quantities is given, their sum is greatest when they are equal.

Answer: E

Concepts:AM-GM Inequality optimization

Difficulty rating: 1450

Small hint:

For a fixed positive product, compare the sum at equality with sums from increasingly unequal factors

Big hint:

AM-GM gives a minimum, not a maximum, for the sum when the product is fixed

Solution:

If

uv=P\gt0,

then

u+v\ge2\sqrt P,

with equality at

u=v.

Thus equality gives the least possible sum. The sum can grow without bound by taking

u

large and

v=\frac{P}{u}

small, so it is not greatest at equality.

Thus, the correct answer is E.

44.

If

\dfrac{xy}{x+y}=a,

\dfrac{xz}{x+z}=b,

and

\dfrac{yz}{y+z}=c,

where

a,

b,

c

are other than zero, then

x

equals:

\dfrac{abc}{ab+ac+bc}

\dfrac{2abc}{ab+bc+ac}

\dfrac{2abc}{ab+ac-bc}

\dfrac{2abc}{ab+bc-ac}

\dfrac{2abc}{ac+bc-ab}

Answer: E

Concepts:system of equations algebraic manipulation

Difficulty rating: 1830

Small hint:

Invert all three given equations to make them linear in

\frac{1}{x},\frac{1}{y},\frac{1}{z}

Big hint:

Compute

\dfrac1b+\dfrac1a-\dfrac1c=\dfrac2x

Solution:

Inverting gives

\begin{gathered} \frac1x+\frac1y=\frac1a,\\ \frac1x+\frac1z=\frac1b,\\ \frac1y+\frac1z=\frac1c. \end{gathered}

Adding the first two and subtracting the third yields

\begin{aligned} \frac2x &=\frac1a+\frac1b-\frac1c\\ &=\frac{ac+bc-ab}{abc}. \end{aligned}

Hence

x=\dfrac{2abc}{ac+bc-ab}.

Thus, the correct answer is E.

45.

If you are given

\log 8=0.9031

and

\log 9=0.9542,

then the only logarithm that cannot be found without the use of tables is:

\log 17

\log(\frac{5}{4})

\log 15

\log 600

\log 0.4

Answer: A

Concepts:logarithm prime factorization

Difficulty rating: 1470

Small hint:

The two given values determine

\log2

and

\log3

Big hint:

Use

\log5=1-\log2

, then see which choice has a prime factor other than

2,3,5

Solution:

From the data,

\begin{gathered} \log2=\frac13\log8,\\ \log3=\frac12\log9, \end{gathered}

and

\log5=1-\log2.

Therefore logarithms of products and quotients of powers of

2,3,5

can be computed. The number

17

has none of those prime factors, so its logarithm is not determined by the data.

Thus, the correct answer is A.

46.

\overline{AB}

is a fixed diameter of a circle whose center is

O.

From

C,

any point on the circle, a chord

\overline{CD}

is drawn perpendicular to

\overline{AB}.

Then, as

C

moves over a semicircle, the bisector of angle

OCD

cuts the circle in a point that always:

Bisects the arc

AB

Trisects the arc

AB

Varies

Is as far from

\overline{AB}

as from

D

Is equidistant from

B

and

C

Answer: A

Concepts:circle angle bisector arc

Difficulty rating: 1880

Small hint:

Extend

\overline{CO}

through the center to the opposite point

E

of the circle

Big hint:

Because

CE

is a diameter and

CD\perp AB,

the angle bisector at

C

bisects arc

DE

Solution:

Extend

CO

to meet the circle again at

E.

Since

CE

is a diameter,

\angle CDE=90^\circ.

Also

CD\perp AB,

so

DE\parallel AB.

If the bisector of

\angle OCD=\angle ECD

meets the circle at

P,

equal inscribed angles give equal arcs

EP=PD.

Because the parallel chord

DE

has endpoints symmetrically placed relative to the fixed diameter

AB,

their arc midpoint

P

is the midpoint of arc

AB.

Thus, the correct answer is A.

47.

If

r

and

s

are the roots of the equation

ax^2+bx+c=0,

the value of

\dfrac1{r^2}+\dfrac1{s^2}

is:

b^2-4ac

\dfrac{b^2-4ac}{2a}

\dfrac{b^2-4ac}{c^2}

\dfrac{b^2-2ac}{c^2}

None of these

Answer: D

Concepts:Vieta’s Formulas algebraic manipulation

Difficulty rating: 1530

Small hint:

Write the expression over the common denominator

r^2s^2

Big hint:

Use

r+s=-\frac{b}{a}

and

rs=\frac{c}{a}

Solution:

By Vieta’s formulas,

r+s=-\frac ba,\qquad rs=\frac ca.

Therefore

\begin{aligned} \frac1{r^2}+\frac1{s^2} &=\frac{(r+s)^2-2rs}{(rs)^2}\\ &=\frac{b^2-2ac}{c^2}. \end{aligned}

Thus, the correct answer is D.

48.

The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:

1:2

2:3

2:5

3:4

3:5

Answer: C

Concepts:circle square (geometry)area ratio

Difficulty rating: 1470

Small hint:

Let the circle have radius

R

and the semicircle-square have side

s

Big hint:

For the semicircle-square, a top vertex gives

(\frac{s}{2})^2+s^2=R^2

Solution:

For the square in the semicircle, put its base on the diameter. A top vertex has horizontal distance

\frac{s}{2}

from the center and vertical distance

s,

so

\left(\frac s2\right)^2+s^2=R^2,

giving

s^2=\frac{4R^2}{5}.

A square inscribed in the full circle has diagonal

2R,

hence area

2R^2.

The ratio is

\frac{\frac{4R^2}{5}}{2R^2}=\frac25.

Thus, the correct answer is C.

49.

The medians of a right triangle which are drawn from the vertices of the acute angles are

5

and

\sqrt{40}.

The value of the hypotenuse is:

10

2\sqrt{40}

\sqrt{13}

2\sqrt{13}

None of these

Answer: D

Concepts:median (geometry)right triangle system of equations

Difficulty rating: 1930

Small hint:

Let the legs be

a,b

and place the right angle at their common endpoint

Big hint:

The two median lengths give

a^2+\frac{b^2}{4}=25

and

\frac{a^2}{4}+b^2=40

Solution:

Let the legs be

a,b.

The medians from their opposite acute vertices have squared lengths

a^2+\frac{b^2}{4}=25,\qquad \frac{a^2}{4}+b^2=40.

Solving gives

a^2=16

and

b^2=36.

Thus the hypotenuse

c

satisfies

c^2=a^2+b^2=52,

so

c=2\sqrt{13}.

Thus, the correct answer is D.

50.

Tom, Dick, and Harry started out on a

100

-mile journey. Tom and Harry went by automobile at the rate of

25

mph, while Dick walked at the rate of

5

mph. After a certain distance, Harry got off and walked on at

5

mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived. The number of hours required for the trip was:

5

6

7

8

None of these answers

Answer: D

Concepts:distance rate and time system of equations

Difficulty rating: 2100

Small hint:

Separate the car’s motion into the first forward trip, the return for Dick, and the final forward trip

Big hint:

If those times are

t_1,t_2,t_3

, write one

100

-mile equation for the car, Dick, and Harry

Solution:

Let

t_1,t_2,t_3

be the car’s times before Harry leaves it, while it returns for Dick, and while it carries Dick forward. The car, Dick, and Harry each cover

100

miles, giving

\begin{aligned} 25t_1-25t_2+25t_3&=100,\\ 5t_1+5t_2+25t_3&=100,\\ 25t_1+5t_2+5t_3&=100. \end{aligned}

Dividing by

5

and solving yields

t_1=3,\ t_2=2,\ t_3=3.

Therefore the common travel time is

t_1+t_2+t_3=8

hours.

Thus, the correct answer is D.

1951 AMC 12 Problems

Answer: B

Difficulty rating: 1020

Small hint:

Big hint:

Solution:

Answer: D

Difficulty rating: 1180

Small hint:

Big hint:

Solution:

Answer: B

Difficulty rating: 1270

Small hint:

Big hint:

Solution:

Answer: D

Difficulty rating: 1220

Small hint:

Big hint:

Solution:

Answer: B

Difficulty rating: 1180

Small hint:

Big hint:

Solution:

Answer: D

Difficulty rating: 1270

Small hint:

Big hint:

Solution:

Answer: B

Difficulty rating: 1240

Small hint:

Big hint:

Solution:

Answer: C

Difficulty rating: 1320

Small hint:

Big hint:

Solution:

Answer: D

Difficulty rating: 1400

Small hint:

Big hint:

Solution:

Answer: C

Difficulty rating: 1180

Small hint:

Big hint:

Solution:

Answer: C

Difficulty rating: 1470

Small hint:

Big hint:

Solution:

Answer: C

Difficulty rating: 1200

Small hint:

Big hint:

Solution:

Answer: C

Difficulty rating: 1150

Small hint:

Big hint:

Solution:

Answer: C

Difficulty rating: 1260

Small hint:

Big hint:

Solution:

Answer: E

Difficulty rating: 1390

Small hint:

Big hint:

Solution:

Answer: C

Difficulty rating: 1400

Small hint:

Big hint: