1968 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

Let $P$ units be the increase in the circumference of a circle resulting from an increase of $\pi$ units in the diameter. Then $P$ equals:

$\dfrac1\pi$

$\pi$

$\dfrac{\pi^2}{2}$

$\pi^2$

$2\pi$

Answer: D

Concepts:circumference algebraic manipulation

Difficulty rating: 950

Small Hint:

Write circumference as $\pi d$

Big Hint:

Replace $d$ by $d+\pi$ and subtract

Solution:

The original circumference is $\pi d.$ After the diameter increases by $\pi,$ it is $\pi(d+\pi).$ The increase is therefore $\pi(d+\pi)-\pi d=\pi^2.$

Therefore, the correct answer is D.

2.

The real value of $x$ such that $64^{x-1}$ divided by $4^{x-1}$ equals $256^{2x}$ is:

$-\dfrac23$

$-\dfrac13$

$0$

$\dfrac14$

$\dfrac38$

Answer: B

Concepts:exponent linear equation

Difficulty rating: 1320

Small Hint:

Combine the quotient because its powers have the same exponent

Big Hint:

Write both sides as powers of $16$

Solution:

The left side is $(\frac{64}{4})^{x-1}=16^{x-1},$ while $256^{2x}=(16^2)^{2x}=16^{4x}.$ Hence $x-1=4x,$ so $x=-\frac{1}{3}.$

Therefore, the correct answer is B.

3.

A straight line passing through the point $(0,4)$ is perpendicular to the line $x-3y-7=0.$ Its equation is:

$y+3x-4=0$

$y+3x+4=0$

$y-3x-4=0$

$3y+x-12=0$

$3y-x-12=0$

Answer: A

Concepts:coordinate geometry slope linear equation

Difficulty rating: 1290

Small Hint:

The given line has slope $\frac{1}{3}$

Big Hint:

Use the negative reciprocal slope through $(0,4)$

Solution:

The given line has slope $\frac{1}{3},$ so a perpendicular line has slope $-3.$ Through $(0,4)$ its equation is $y=-3x+4,$ or $y+3x-4=0.$

Therefore, the correct answer is A.

4.

Define an operation $*$ for positive real numbers by $a*b=\dfrac{ab}{a+b}.$ Then $4*(4*4)$ equals:

$\dfrac34$

$1$

$\dfrac43$

$2$

$\dfrac{16}{3}$

Answer: C

Concepts:custom operation fraction substitution

Difficulty rating: 1360

Small Hint:

Evaluate the inner operation first

Big Hint:

After finding $4*4,$ substitute it as the second input

Solution:

First, $4*4=\frac{16}{8}=2.$ Therefore $4*(4*4)=4*2,$ which equals $\frac{8}{4+2}=\frac{4}{3}.$

Therefore, the correct answer is C.

5.

If $f(n)=\dfrac13n(n+1)(n+2),$ then $f(r)-f(r-1)$ equals:

$r(r+1)$

$(r+1)(r+2)$

$\dfrac13r(r+1)$

$\dfrac13(r+1)(r+2)$

$\dfrac13r(r+1)(r+2)$

Answer: A

Concepts:function factoring

Difficulty rating: 1260

Small Hint:

Substitute $r$ and $r-1$ separately

Big Hint:

Factor out $\frac13r(r+1)$

Solution:

We have $f(r)=\frac13r(r+1)(r+2)$ and $f(r-1)=\frac13(r-1)r(r+1).$ Factor their difference as $\frac13r(r+1)$ times $(r+2)-(r-1)=3.$ The result is $r(r+1).$

Therefore, the correct answer is A.

6.

Let side $AD$ of convex quadrilateral $ABCD$ be extended through $D,$ and let side $BC$ be extended through $C,$ to meet in point $E.$ Let $S$ represent the degree-sum of angles $CDE$ and $DCE,$ and let $S'$ represent the degree-sum of angles $BAD$ and $ABC.$ If $r=\frac{S}{S'},$ then:

$r=1$ sometimes, $r\gt1$ sometimes

$r=1$ sometimes, $r\lt1$ sometimes

$0\lt r\lt1$

$r\gt1$

$r=1$

Answer: E

Concepts:angle sum angle chasing

Difficulty rating: 1460

Small Hint:

Express both sums using angle $E$

Big Hint:

Apply the triangle angle sum to $ECD$ and $EAB$

Solution:

In triangle $ECD,$ $S=180^\circ-\angle E.$ Since $D$ lies on $AE$ and $C$ lies on $BE,$ triangle $EAB$ gives $S'=180^\circ-\angle E$ as well. Thus $S=S'$ and $r=1.$

Therefore, the correct answer is E.

7.

Let $O$ be the intersection point of medians $AP$ and $CQ$ of triangle $ABC.$ If $OQ$ is $3$ inches, then $OP,$ in inches, is:

$3$

$\dfrac92$

$6$

$9$

undetermined

Answer: E

Concepts:centroid median (geometry)

Difficulty rating: 1360

Small Hint:

A centroid gives a ratio only along each individual median

Big Hint:

Knowing part of $CQ$ does not determine the length of $AP$

Solution:

The centroid divides each median in a $2:1$ ratio, so $OQ=3$ determines $CQ=9.$ It gives no relation between the lengths of the two different medians $CQ$ and $AP.$ Triangles with $CQ=9$ can have different $AP,$ so $OP$ is undetermined.

Therefore, the correct answer is E.

8.

A positive number is mistakenly divided by $6$ instead of being multiplied by $6.$ Based on the correct answer, the error thus committed, to the nearest percent, is:

$100$

$97$

$83$

$17$

$3$

Answer: B

Concepts:percentage fraction

Difficulty rating: 1290

Small Hint:

Let the number be $N$ and compare $\frac{N}{6}$ with $6N$

Big Hint:

Divide the absolute error by the correct result

Solution:

The correct result is $6N,$ while the mistaken result is $\frac{N}{6}.$ Relative to the correct result, the error is $\frac{6N-\frac{N}{6}}{6N}=\frac{35}{36}\approx97.2\%.$

Therefore, the correct answer is B.

9.

The sum of the real values of $x$ satisfying $\lvert x+2\rvert=2\lvert x-2\rvert$ is:

$\dfrac13$

$\dfrac23$

$6$

$6\dfrac13$

$6\dfrac23$

Answer: E

Concepts:absolute value quadratic Vieta’s Formulas

Difficulty rating: 1500

Small Hint:

Square both sides; both sides are nonnegative

Big Hint:

Use the sum of the roots of the resulting quadratic

Solution:

Squaring gives $(x+2)^2=4(x-2)^2,$ or $3x^2-20x+12=0.$ Both roots satisfy the original absolute-value equation, and their sum is $\frac{20}{3}=6\frac23.$

Therefore, the correct answer is E.

10.

Assume that, for a certain school, it is true that

$\mathrm{I}:$ Some students are not honest.

$\mathrm{II}:$ All fraternity members are honest.

A necessary conclusion is:

Some students are fraternity members

Some fraternity members are not students

Some students are not fraternity members

No fraternity member is a student

No student is a fraternity member

Answer: C

Concepts:logical deduction

Difficulty rating: 1220

Small Hint:

Choose a student whose existence is guaranteed by statement $\mathrm{I}$

Big Hint:

Could that dishonest student be a fraternity member under statement $\mathrm{II}?$

Solution:

Statement $\mathrm{I}$ guarantees a dishonest student. Statement $\mathrm{II}$ says every fraternity member is honest, so that dishonest student cannot be a fraternity member. Hence some student is not a fraternity member.

Therefore, the correct answer is C.

11.

If an arc of $60^\circ$ on circle $\mathrm{I}$ has the same length as an arc of $45^\circ$ on circle $\mathrm{II},$ the ratio of the area of circle $\mathrm{I}$ to that of circle $\mathrm{II}$ is:

$16:9$

$9:16$

$4:3$

$3:4$

none of these

Answer: B

Concepts:arc circle area ratio and proportion

Difficulty rating: 1400

Small Hint:

Equate $(\frac{\theta}{360^\circ})2\pi r$ for the two arcs

Big Hint:

Square the resulting radius ratio to get the area ratio

Solution:

Equal arc lengths give $60r_1=45r_2,$ so $\frac{r_1}{r_2}=\frac{3}{4}.$ Circle areas scale as the squares of the radii, giving $A_1:A_2=9:16.$

Therefore, the correct answer is B.

12.

A circle passes through the vertices of a triangle with side-lengths $7\dfrac12,$ $10,$ $12\dfrac12.$ The radius of the circle is:

$\dfrac{15}{4}$

$5$

$\dfrac{25}{4}$

$\dfrac{35}{4}$

$\dfrac{15\sqrt2}{2}$

Answer: C

Concepts:right triangle circumcircle, circumcenter, and circumradius Pythagorean Theorem

Difficulty rating: 1450

Small Hint:

The side lengths are $\frac{5}{2}$ times a familiar Pythagorean triple

Big Hint:

A right triangle’s hypotenuse is the circumcircle’s diameter

Solution:

The sides are $\frac52(3,4,5),$ so the triangle is right with hypotenuse $\frac{25}{2}.$ The hypotenuse is the circumdiameter, making the radius $\frac{25}{4}.$

Therefore, the correct answer is C.

13.

If $m$ and $n$ are the roots of $x^2+mx+n=0,$ $m\ne0,$ $n\ne0,$ then the sum of the roots is:

$-\dfrac12$

$-1$

$\dfrac12$

$1$

undetermined

Answer: B

Concepts:Vieta’s Formulas system of equations

Difficulty rating: 1800

Small Hint:

Use both the sum and product forms of Vieta’s formulas

Big Hint:

The product equation and $n\ne0$ determine $m$

Solution:

Because the roots are $m,$ $n,$ Vieta gives $m+n=-m$ and $mn=n.$ Since $n\ne0,$ the second equation gives $m=1,$ and then $n=-2.$ Their sum is $-1.$

Therefore, the correct answer is B.

14.

If $x$ and $y$ are nonzero numbers such that $x=1+\dfrac1y$ and $y=1+\dfrac1x,$ then $y$ equals:

$x-1$

$1-x$

$1+x$

$-x$

$x$

Answer: E

Concepts:rational equation system of equations algebraic manipulation

Difficulty rating: 1400

Small Hint:

Multiply each equation by its denominator

Big Hint:

Both equations produce an expression for $xy$

Solution:

The equations give $xy=y+1$ and $xy=x+1.$ Therefore $x+1=y+1,$ so $y=x.$

Therefore, the correct answer is E.

15.

Let $P$ be the product of any three consecutive positive odd integers. The largest integer dividing all such $P$ is:

$15$

$6$

$5$

$3$

$1$

Answer: D

Concepts:divisibility greatest common divisor

Difficulty rating: 1630

Small Hint:

Among three consecutive odd integers, inspect residues modulo $3$

Big Hint:

Use two examples to rule out every larger common divisor

Solution:

Three consecutive odd integers occupy three consecutive residues modulo $3,$ so one is divisible by $3.$ Thus every product is divisible by $3.$ The products $1\cdot3\cdot5=15$ and $7\cdot9\cdot11=693$ have greatest common divisor $3,$ so no larger integer always divides the product.

Therefore, the correct answer is D.

16.

If $x$ is such that $\dfrac1x\lt2$ and $\dfrac1x\gt-3,$ then:

$-\dfrac13\lt x\lt\dfrac12$

$-\dfrac12\lt x\lt3$

$x\gt\dfrac12$

$x\gt\dfrac12$ or $-\dfrac13\lt x\lt0$

$x\gt\dfrac12$ or $x\lt-\dfrac13$

Answer: E

Concepts:inequality rational equation casework

Difficulty rating: 1740

Small Hint:

Combine the conditions as $-3\lt\frac{1}{x}\lt2$

Big Hint:

Treat positive and negative $x$ separately when taking reciprocals

Solution:

For $x\gt0,$ $\frac{1}{x}\lt2$ gives $x\gt\frac{1}{2},$ while the other inequality is automatic. For $x\lt0,$ $\frac{1}{x}\gt-3$ gives $x\lt-\frac{1}{3},$ while the first is automatic. Thus $x\gt\frac{1}{2}$ or $x\lt-\frac{1}{3}.$

Therefore, the correct answer is E.

17.

Let $f(n)=\dfrac{x_1+x_2+\cdots+x_n}{n},$ where $n$ is a positive integer. If $x_k=(-1)^k,$ $k=1,$ $2,$ $\ldots,$ $n,$ the set of possible values of $f(n)$ is:

$\{0\}$

$\left\{\dfrac1n\right\}$

$\left\{0,-\dfrac1n\right\}$

$\left\{0,\dfrac1n\right\}$

$\left\{1,\dfrac1n\right\}$

Answer: C

Concepts:parity summation mean

Difficulty rating: 1470

Small Hint:

Pair consecutive terms $-1+1$

Big Hint:

Separate even and odd values of $n$

Solution:

If $n$ is even, the terms pair to give sum $0,$ so $f(n)=0.$ If $n$ is odd, one final $-1$ remains, so $f(n)=-\frac{1}{n}.$ The possible values are therefore $\{0,-\frac{1}{n}\}.$

Therefore, the correct answer is C.

18.

Side $AB$ of triangle $ABC$ has length $8$ inches. Line $DEF$ is drawn parallel to $AB$ so that $D$ is on segment $AC,$ and $E$ is on segment $BC.$ Line $AE$ extended bisects angle $FEC.$ If $DE$ has length $5$ inches, then the length of $CE,$ in inches, is:

$\dfrac{51}{4}$

$13$

$\dfrac{53}{4}$

$\dfrac{40}{3}$

$\dfrac{27}{2}$

Answer: D

Difficulty rating: 2000

Small Hint:

Put $G$ on the extension of $AE$ beyond $E$ , so the bisector condition is $\angle FEG=\angle GEC$

Big Hint:

After showing $BE=AB,$ use similarity of $CDE$ and $CAB$

Solution:

Put $G$ on the extension of $AE$ beyond $E.$ The bisector condition gives $\angle FEG=\angle GEC.$ Since $FE\parallel AB,$ we have $\angle FEG=\angle BAE.$ Also, rays $EG$ and $EA$ are opposite, as are rays $EC$ and $EB,$ so $\angle GEC=\angle AEB.$ Thus triangle $ABE$ is isosceles and $BE=AB=8.$ Similar triangles $CDE$ and $CAB$ give $\frac{CE}{CE+8}=\frac58.$ Hence $8CE=5CE+40,$ so $CE=\frac{40}{3}.$

Therefore, the correct answer is D.

19.

Let $n$ be the number of ways that $10$ dollars can be changed into dimes and quarters, with at least one of each coin being used. Then $n$ equals:

$40$

$38$

$21$

$20$

$19$

Answer: E

Concepts:Diophantine Equation counting integers in a range divisibility

Difficulty rating: 1710

Small Hint:

In cents, solve $10d+25q=1000$

Big Hint:

Use parity to write $q=2k$ , then enforce positive coin counts

Solution:

The equation $10d+25q=1000$ reduces to $2d+5q=200.$ Thus $q$ must be even; write $q=2k,$ giving $d=100-5k.$ Positivity requires $k=1,$ $2,$ $\ldots,$ $19,$ so there are $19$ ways.

Therefore, the correct answer is E.

20.

The measures of the interior angles of a convex polygon of $n$ sides are in arithmetic progression. If the common difference is $5^\circ$ and the largest angle is $160^\circ,$ then $n$ equals:

$9$

$10$

$12$

$16$

$32$

Answer: A

Concepts:angle sum arithmetic sequence quadratic

Difficulty rating: 1800

Small Hint:

Write the smallest angle as $160-5(n-1)$

Big Hint:

Equate the arithmetic-series sum to $180(n-2)$

Solution:

The smallest angle is $160-5(n-1).$ The angle sum is $\frac n2$ times $320-5(n-1),$ and it equals $180(n-2).$ This simplifies to $\begin{aligned} n^2+7n-144&=0,\\ (n-9)(n+16)&=0. \end{aligned}$ Thus $n=9.$

Therefore, the correct answer is A.

21.

If $S=1!+2!+3!+\cdots+99!,$ then the units digit in the value of $S$ is:

$9$

$8$

$5$

$3$

$0$

Answer: D

Concepts:factorial units digit

Difficulty rating: 1130

Small Hint:

Every factorial from $5!$ onward ends in zero

Big Hint:

Only add the units digits of the first four terms

Solution:

For $k\ge5,$ $k!$ is divisible by $10.$ Only the first four terms matter. Their sum is $1+2+6+24=33,$ whose units digit is $3.$

Therefore, the correct answer is D.

22.

A segment of length $1$ is divided into four segments. Then there exists a quadrilateral with the four segments as sides if and only if each segment is:

equal to $\dfrac14$

equal to or greater than $\dfrac18$ and less than $\dfrac12$

greater than $\dfrac18$ and less than $\dfrac12$

greater than $\dfrac18$ and less than $\dfrac14$

less than $\dfrac12$

Answer: E

Concepts:triangle inequality perimeter

Difficulty rating: 1610

Small Hint:

A nondegenerate quadrilateral exists exactly when the longest side is shorter than the other three combined

Big Hint:

The four lengths sum to $1$

Solution:

A simple nondegenerate quadrilateral exists exactly when each side is less than the sum of the other three. Since the total is $1,$ this condition is $s\lt1-s,$ or $s\lt\frac{1}{2},$ for every segment.

Therefore, the correct answer is E.

23.

If all the logarithms are real numbers, the equality

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

is satisfied for:

all real values of $x$

no real values of $x$

all real values of $x$ except $x=0$

no real values of $x$ except $x=0$

all real values of $x$ except $x=1$

Answer: B

Concepts:logarithm rational equation

Difficulty rating: 1800

Small Hint:

Combine the two logarithms on the left

Big Hint:

Solve the resulting algebraic equation, then test its domain

Solution:

Combining logarithms would require $(x+3)(x-1)=x^2-2x-3.$ This gives $x=0.$ But then $x-1=-1,$ so the logarithm on the left is not real. Hence there are no real solutions.

Therefore, the correct answer is B.

24.

A painting $18''\times24''$ is to be placed into a wooden frame with the longer dimension vertical. The wood at the top and bottom is twice as wide as the wood on the sides. If the frame area equals that of the painting itself, the ratio of the smaller to the larger dimension of the framed painting is:

$1:3$

$1:2$

$2:3$

$3:4$

$1:1$

Answer: C

Concepts:rectangle area quadratic ratio and proportion

Difficulty rating: 1690

Small Hint:

Let each side strip have width $x$ , so each top strip has width $2x$

Big Hint:

Set the outside area equal to twice $18\cdot24$

Solution:

The outside dimensions are $18+2x$ and $24+4x.$ Since the frame area equals the painting area, $(18+2x)(24+4x)=2(18)(24).$ This gives $x^2+15x-54=0,$ so $x=3.$ The dimensions are $24$ and $36,$ with ratio $2:3.$

Therefore, the correct answer is C.

25.

Ace runs with constant speed and Flash runs $x$ times as fast, $x\gt1.$ Flash gives Ace a head start of $y$ yards, and, at a given signal, they start off in the same direction. Then the number of yards Flash must run to catch Ace is:

$xy$

$\dfrac{y}{x+y}$

$\dfrac{xy}{x-1}$

$\dfrac{x+y}{x+1}$

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

Answer: C

Concepts:relative speed distance rate and time

Difficulty rating: 1450

Small Hint:

If Ace’s speed is $v,$ the closing speed is $(x-1)v$

Big Hint:

Multiply the catch-up time by Flash’s speed $xv$

Solution:

If Ace runs at speed $v,$ Flash runs at $xv,$ so their closing speed is $(x-1)v.$ The catch-up time is $\frac{y}{(x-1)v},$ during which Flash runs $\frac{xv\cdot y}{(x-1)v}=\frac{xy}{x-1}.$

Therefore, the correct answer is C.

26.

Let $S=2+4+6+\cdots+2N,$ where $N$ is the smallest positive integer such that $S\gt1{,}000{,}000.$ Then the sum of the digits of $N$ is:

$27$

$12$

$6$

$2$

$1$

Answer: E

Concepts:arithmetic sequence inequality digits

Difficulty rating: 1510

Small Hint:

The sum is $N(N+1)$

Big Hint:

Compare $999\cdot1000$ and $1000\cdot1001$ with one million

Solution:

We have $S=N(N+1).$ For $N=999,$ this is $999{,}000\lt1{,}000{,}000,$ while for $N=1000$ it is $1{,}001{,}000.$ Thus $N=1000,$ whose digit sum is $1.$

Therefore, the correct answer is E.

27.

Let

$\begin{aligned} S_n&=1-2+3-4+\cdots\\ &\quad+(-1)^{n-1}n. \end{aligned}$

Here $n=1,$ $2,$ $\ldots.$ Then $S_{17}+S_{33}+S_{50}$ equals:

$0$

$1$

$2$

$-1$

$-2$

Answer: B

Concepts:parity summation pairing and grouping

Difficulty rating: 1600

Small Hint:

Pair terms as $(1-2)+(3-4)+\cdots$

Big Hint:

For even $n,$ $S_n=-\frac{n}{2};$ for odd $n,$ $S_n=\frac{n+1}{2}$

Solution:

Pairing gives $S_n=-\frac{n}{2}$ for even $n$ and $S_n=\frac{n+1}{2}$ for odd $n.$ Hence $S_{17}=9,$ $S_{33}=17,$ $S_{50}=-25,$ and their sum is $1.$

Therefore, the correct answer is B.

28.

If the arithmetic mean of $a$ and $b$ is double their geometric mean, with $a\gt b\gt0,$ then a possible value for the ratio $\frac{a}{b},$ to the nearest integer, is:

$5$

$8$

$11$

$14$

none of these

Answer: D

Concepts:AM-GM Inequality quadratic ratio and proportion

Difficulty rating: 1830

Small Hint:

Set $t=\frac{a}{b}$ and divide the mean equation by $b$

Big Hint:

The equation $t+1=4\sqrt t$ becomes quadratic after squaring

Solution:

Let $t=\frac{a}{b}\gt1.$ The condition gives $\frac{t+1}{2}=2\sqrt t,$ so $t^2-14t+1=0.$ The root greater than $1$ is $t=7+4\sqrt3\approx13.93,$ whose nearest integer is $14.$

Therefore, the correct answer is D.

29.

Given the three numbers $x,$ $y=x^x,$ $z=x^{(x^x)},$ with $0.9\lt x\lt1.0.$ Arranged in order of increasing magnitude, they are:

$x,$ $z,$ $y$

$x,$ $y,$ $z$

$y,$ $x,$ $z$

$y,$ $z,$ $x$

$z,$ $x,$ $y$

Answer: A

Concepts:exponent inequality

Difficulty rating: 2000

Small Hint:

For $0\lt x\lt1,$ compare powers by comparing their positive exponents

Big Hint:

First show $x\lt x^x\lt1$ , then compare $x^{x^x}$

Solution:

Since $0\lt x\lt1$ and $x\lt1,$ raising $x$ to the exponent $x$ gives $x\lt y=x^x\lt1.$ For a base between $0$ and $1,$ a larger exponent gives a smaller value. Because $x\lt y\lt1,$ we have $x=x^1\lt x^y=z\lt x^x=y.$ Thus $x\lt z\lt y.$

Therefore, the correct answer is A.

30.

Convex polygons $P_1$ and $P_2$ are drawn in the same plane with $n_1$ and $n_2$ sides, respectively, $n_1\leq n_2.$ If $P_1$ and $P_2$ do not have any line segment in common, then the maximum number of intersections of $P_1$ and $P_2$ is:

$2n_1$

$2n_2$

$n_1n_2$

$n_1+n_2$

none of these

Answer: A

Concepts:counting intersections extremal argument

Difficulty rating: 2130

Small Hint:

A line segment can enter and leave a convex polygon at most once

Big Hint:

Apply that bound to each side of the polygon with fewer sides

Solution:

Each side of $P_1$ lies on a line, and its intersection with the convex region $P_2$ is a single segment or empty. Thus that side crosses the boundary of $P_2$ at most twice. Across $n_1$ sides there are at most $2n_1,$ and a suitable thin convex $n_1$ -gon crossing a convex $n_2$ -gon attains this bound.

Therefore, the correct answer is A.

31.

In this diagram, not drawn to scale, figures $\mathrm{I}$ and $\mathrm{III}$ are equilateral triangular regions with respective areas of $32\sqrt3$ and $8\sqrt3$ square inches. Figure $\mathrm{II}$ is a square region with area $32$ square inches. Let the length of segment $AD$ be decreased by $12\dfrac12\%$ of itself, while the lengths of $AB$ and $CD$ remain unchanged. The percent decrease in the area of the square is:

$12\dfrac12$

$25$

$50$

$75$

$87\dfrac12$

Answer: D

Concepts:equilateral triangle square (geometry)area percentage

Difficulty rating: 1800

Small Hint:

Convert the three given areas into the side lengths $AB,$ $BC,$ $CD$

Big Hint:

The decrease in $AD$ is absorbed entirely by the square’s side

Solution:

From $(\frac{\sqrt3}{4})s^2=32\sqrt3,$ $AB=8\sqrt2.$ The square has $BC=4\sqrt2,$ and the smaller equilateral triangle also has $CD=4\sqrt2.$ Thus $AD=16\sqrt2.$ Decreasing it by $\frac{1}{8}$ removes $2\sqrt2$ from $BC,$ whose new length is $2\sqrt2.$ The square’s area falls from $32$ to $8,$ a $75\%$ decrease.

Therefore, the correct answer is D.

32.

$A$ and $B$ move uniformly along two straight paths intersecting at right angles in point $O.$ When $A$ is at $O,$ $B$ is $500$ yards short of $O.$ In $2$ minutes they are equidistant from $O,$ and in $8$ minutes more they are again equidistant from $O.$ Then the ratio of $A$ ’s speed to $B$ ’s speed is:

$4:5$

$5:6$

$2:3$

$5:8$

$1:2$

Answer: C

Concepts:distance rate and time system of equations

Difficulty rating: 1900

Small Hint:

Let the speeds be $u,$ $v$ ; at $t=2,$ $2u=500-2v$

Big Hint:

At $t=10,$ $B$ has passed $O,$ so $10u=10v-500$

Solution:

Let the speeds of $A,$ $B$ be $u,$ $v.$ At $t=2,$ $2u=500-2v,$ so $u+v=250.$ At $t=10,$ $B$ has passed $O,$ and $10u=10v-500,$ so $v-u=50.$ Hence $u=100,$ $v=150,$ and $u:v=2:3.$

Therefore, the correct answer is C.

33.

A number $N$ has three digits when expressed in base $7.$ When $N$ is expressed in base $9$ the digits are reversed. Then the middle digit is:

$0$

$1$

$3$

$4$

$5$

Answer: A

Concepts:number base place value Diophantine Equation

Difficulty rating: 2060

Small Hint:

Call the base- $7$ digits $a,$ $b,$ $c$ and equate $49a+7b+c$ to $81c+9b+a$

Big Hint:

Rearrange to $24a-b=40c$ and use the digit bounds

Solution:

Let $N=(abc)_7=(cba)_9.$ Then $49a+7b+c=81c+9b+a,$ so $24a-b=40c.$ With $1\leq a\leq6,$ $0\leq b\leq6,$ and $1\leq c\leq6,$ the only possibility is $a=5,$ $c=3,$ $b=0.$ Indeed, $(503)_7=(305)_9=248.$

Therefore, the correct answer is A.

34.

With $400$ members voting, the House of Representatives defeated a bill. A re-vote, with the same members voting, resulted in passage of the bill by twice the margin by which it was originally defeated. The number voting for the bill on the re-vote was $\dfrac{12}{11}$ of the number voting against it originally. How many more members voted for the bill the second time than voted for it the first time?

$75$

$60$

$50$

$45$

$20$

Answer: B

Concepts:system of equations ratio and proportion

Difficulty rating: 1900

Small Hint:

Let the original and second numbers voting for be $F$ and $G$

Big Hint:

Use $G=\frac{12}{11}(400-F)$ and $2G-400=2(400-2F)$

Solution:

Let $F,$ $G$ be the original and second numbers voting for the bill. The ratio condition gives $G=\frac{12}{11}(400-F).$ The passage margin is twice the original defeat margin, so $2G-400=2(400-2F),$ or $G+2F=600.$ Solving gives $F=180,$ $G=240,$ so $60$ more members voted for it.

Therefore, the correct answer is B.

35.

In this diagram the center of the circle is $O,$ the radius is $a$ inches, chord $EF$ is parallel to chord $CD,$ $O,$ $G,$ $H,$ $J$ are collinear, and $G$ is the midpoint of $CD.$ Let $K$ (square inches) represent the area of trapezoid $CDFE$ and let $R$ (square inches) represent the area of rectangle $ELMF.$ Then, as $CD$ and $EF$ are translated upward so that $OG$ increases toward the value $a,$ while $JH$ always equals $HG,$ the ratio $K:R$ becomes arbitrarily close to:

$0$

$1$

$\sqrt2$

$\dfrac1{\sqrt2}+\dfrac12$

$\dfrac1{\sqrt2}+1$

Answer: D

Concepts:circle chord trapezoid calculus

Difficulty rating: 2650

Small Hint:

Let $JH=HG=x,$ so $OG=a-2x$ and $OH=a-x$

Big Hint:

Express the two half-chords with the Pythagorean theorem, then let $x$ approach $0$

Solution:

Let $JH=HG=x.$ Then $OG=a-2x$ and $OH=a-x.$ The half-chords are $\begin{aligned} GD&=2\sqrt{x(a-x)},\\ HF&=\sqrt{x(2a-x)}. \end{aligned}$ Since both figures have height $x,$ $\begin{aligned} \frac KR&=\frac{x(GD+HF)}{2x(HF)}\\ &=\frac12+\frac{GD}{2HF}. \end{aligned}$ As $x$ approaches $0,$ the latter fraction approaches $\frac{2\sqrt{ax}}{2\sqrt{2ax}}=\frac1{\sqrt2}.$ Thus $K:R$ approaches $\frac{1}{\sqrt2}+\frac{1}{2}.$

Therefore, the correct answer is D.