1951 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

The percent that $M$ is greater than $N$ is:

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

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

$\dfrac{M-N}{N}$

$\dfrac{M-N}{M}$

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

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:

$\dfrac{x^2}{2}$

$2x^2$

$\dfrac{2x^2}{9}$

$\dfrac{x^2}{18}$

$\dfrac{x^2}{72}$

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

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:

$360$

$460$

$490$

$590$

$720$

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

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

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 $(D)$

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

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

$\left(5\dfrac14\right)a$

$2a$

$6a$

$\left(4\dfrac12\right)a$

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.

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

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:

$30^\circ$

$5^\circ$

$22\dfrac12^\circ$

$7\dfrac12^\circ$

$28^\circ$

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

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.

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:

$2$

$3$

$4$

$5$

$6$

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

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 $x$ :

$x+y=0$

$3xy=10$

$x=5y$

$3x+y=10$

$\dfrac{x}{y}=\sqrt3$

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

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

$101$

$1001$

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:

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

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

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

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

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

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:

$x+z\gt y+z$

$x-z\gt y-z$

$xz\gt yz$

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

$xz^2\gt yz^2$

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

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

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:

$2^{n+1}-\dfrac18$

$-2^{n+1}$

$1-2^n$

$\dfrac78$

$\dfrac74$

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

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:

$m=1$

$m=\dfrac12$

$m=0$

$m=-1$

$m=-\dfrac12$

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

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

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

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

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:

$14$

$28$

$56$

$8$

$7$

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 $\overline{AB}\sqrt2$

$\ge \overline{AB}\sqrt2$

$\le \overline{AB}\sqrt2$

$\overline{AB}^{\,2}$

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:

$y=x^2,$ $y=2x$

$y=x^2-2x,$ $y=0$

$y=x,$ $y=x-2$

$y=x^2-2x+1,$ $y=1$

$y=x^2-1,$ $y=2x-1$

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:

$7$

$1$

$10$

$\log_{10}7$

$\log_7 10$

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:

$xy=qz$

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

$x+y=q+z$

$x-y=q-z$

$x^y=q^z$

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.

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

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

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

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:

$(x+1)^4$

$(x^3+1)^4$

$1$

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

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

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:

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

$y=2x-4$

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

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

$y=x^2-4x$

$y=x^2-4$

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

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.

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:

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

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

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

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

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

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:

$\log 17$

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

$\log 15$

$\log 600$

$\log 0.4$

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 $C$

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

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:

$1:2$

$2:3$

$2:5$

$3:4$

$3:5$

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

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

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.