1971 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

The number of digits in the number $N=2^{12}\times5^8$ is:

$9$

$10$

$11$

$12$

$20$

Answer: B

Concepts:exponent place value pairing and grouping

Difficulty rating: 1200

Small Hint:

Pair as many factors of $2$ and $5$ as possible

Big Hint:

Rewrite the number as $2^4\cdot10^8$

Solution:

We have $2^{12}5^8=2^4(2^85^8)$ and $16\cdot10^8=1{,}600{,}000{,}000,$ which has $10$ digits.

Therefore, the correct answer is B.

2.

If $b$ men take $c$ days to lay $f$ bricks, then the number of days it will take $c$ men working at the same rate to lay $b$ bricks is:

$fb^2$

$\frac{b}{f^2}$

$\frac{f^2}{b}$

$\frac{b^2}{f}$

$\frac{f}{b^2}$

Answer: D

Concepts:rate ratio and proportion algebraic manipulation

Difficulty rating: 1430

Small Hint:

Find the number of bricks laid by one man in one day

Big Hint:

The rate is $\frac{f}{bc}$ bricks per man-day

Solution:

The rate is $\frac{f}{bc}$ bricks per man-day. Thus $c$ men lay $\frac{f}{b}$ bricks per day, so laying $b$ bricks takes $\frac{b}{\frac{f}{b}}=\frac{b^2}{f}$ days.

Therefore, the correct answer is D.

3.

If the point $(x,-4)$ lies on the straight line joining the points $(0,8)$ and $(-4,0)$ in the $xy$ -plane, then $x$ is equal to:

$-2$

$2$

$-8$

$6$

$-6$

Answer: E

Concepts:coordinate geometry slope linear equation

Difficulty rating: 1310

Small Hint:

Find the slope through the two given fixed points

Big Hint:

The line has equation $y=2x+8$

Solution:

The slope through $(0,8)$ and $(-4,0)$ is $2,$ so the line is $y=2x+8.$ Substituting $y=-4$ gives $-4=2x+8,$ hence $x=-6.$

Therefore, the correct answer is E.

4.

After simple interest for two months at $5\%$ per annum was credited, a Boy Scout Troop had a total of $\$255.31$ in the Council Treasury. The interest credited was a number of dollars plus the following number of cents:

$11$

$12$

$13$

$21$

$31$

Answer: A

Concepts:percentage money rate

Difficulty rating: 1310

Small Hint:

Two months is one-sixth of a year

Big Hint:

If $P$ is the principal, solve $255.31=P(1+\frac{0.05}{6})$

Solution:

If $P$ is the principal, then $\begin{aligned} 255.31&=P\left(1+\frac{0.05}{6}\right)\\ &=P\frac{121}{120}. \end{aligned}$ Hence $P=253.20,$ so the interest is $\$2.11$ and its cents part is $11.$

Therefore, the correct answer is A.

5.

Points $A,$ $B,$ $Q,$ $D,$ and $C$ lie on the circle shown, and the measures of arcs $BQ$ and $QD$ are $42^\circ$ and $38^\circ,$ respectively. The sum of the measures of angles $P$ and $Q$ is:

$80^\circ$

$62^\circ$

$40^\circ$

$46^\circ$

None of these

Answer: C

Concepts:circle inscribed angle arc angle chasing

Difficulty rating: 1780

Small Hint:

Use the external-secant formula for angle $P$

Big Hint:

Adding angle $Q$ cancels the unknown intercepted arc

Solution:

Let $m\overset{\frown}{AC}=u.$ The external-secant formula gives $m\angle P=\frac12\left(m\overset{\frown}{BD}-u\right),$ while the inscribed-angle theorem gives $m\angle Q=\frac{u}{2}.$ Therefore $\begin{aligned} m\angle P+m\angle Q &=\frac12m\overset{\frown}{BD}\\ &=\frac12(42^\circ+38^\circ)\\ &=40^\circ. \end{aligned}$

Therefore, the correct answer is C.

6.

Let $*$ be a symbol denoting the binary operation on the set $S$ of all nonzero real numbers as follows: for any $a,$ $b\in S,$ $a*b=2ab.$ Which statement is not true?

$*$ is commutative over $S$

$*$ is associative over $S$

$\frac{1}{2}$ is an identity element for $*$ in $S$

Every element of $S$ has an inverse for $*$

$\frac{1}{2a}$ is an inverse for $*$ of the element $a$ of $S$

Answer: E

Concepts:custom operation algebraic manipulation

Difficulty rating: 1740

Small Hint:

First determine the identity $e$ from $a*e=a$

Big Hint:

An inverse $x$ of $a$ must satisfy $a*x=\frac{1}{2}$

Solution:

The operation is commutative, and $(a*b)*c=4abc=a*(b*c),$ so it is associative. Its identity is $\frac{1}{2}.$ The inverse of $a$ must satisfy $2ax=\frac{1}{2},$ so it is $x=\frac{1}{4a},$ not $\frac{1}{2a}.$

Therefore, the correct answer is E.

7.

$2^{-(2k+1)}-2^{-(2k-1)}+2^{-2k}$ is equal to:

$2^{-2k}$

$2^{-(2k-1)}$

$-2^{-(2k+1)}$

$0$

$2$

Answer: C

Concepts:exponent factoring algebraic manipulation

Difficulty rating: 1590

Small Hint:

Factor out the term with the smallest power of $2$

Big Hint:

Use $2^{-(2k-1)}=4\cdot2^{-(2k+1)}$

Solution:

Factoring gives $2^{-(2k+1)}(1-4+2)=-2^{-(2k+1)}.$

Therefore, the correct answer is C.

8.

The solution set of $6x^2+5x\lt4$ is the set of all values of $x$ such that:

$-2\lt x\lt1$

$-\dfrac43\lt x\lt\dfrac12$

$-\dfrac12\lt x\lt\dfrac43$

$x\lt\dfrac12$ or $x\gt-\dfrac43$

$x\lt-\dfrac43$ or $x\gt\dfrac12$

Answer: B

Concepts:inequality factoring quadratic

Difficulty rating: 1640

Small Hint:

Move $4$ to the left and factor

Big Hint:

The product $(3x+4)(2x-1)$ is negative between its roots

Solution:

The inequality is $\begin{aligned} 6x^2+5x-4&\lt0,\\ (3x+4)(2x-1)&\lt0. \end{aligned}$ The roots are $-\frac{4}{3}$ and $\frac{1}{2},$ and the upward-opening quadratic is negative between them.

Therefore, the correct answer is B.

9.

An uncrossed belt is fitted without slack around two circular pulleys with radii of $14$ inches and $4$ inches. If the distance between the points of contact of the belt with the pulleys is $24$ inches, then the distance between the centers of the pulleys in inches is:

$24$

$2\sqrt{119}$

$25$

$26$

$4\sqrt{35}$

Answer: D

Concepts:circle tangent line Pythagorean Theorem

Difficulty rating: 1760

Small Hint:

Join the centers and use the difference of the radii

Big Hint:

The center distance is the hypotenuse of a right triangle with legs $24$ and $10$

Solution:

The common external tangent and the radii to its contact points form a right triangle whose legs are $24$ and $14-4=10.$ Thus the center distance is $\sqrt{24^2+10^2}=\sqrt{676}=26.$

Therefore, the correct answer is D.

10.

Each of a group of $50$ girls is blonde or brunette and is blue-eyed or brown-eyed. If $14$ are blue-eyed blondes, $31$ are brunettes, and $18$ are brown-eyed, then the number of brown-eyed brunettes is:

$5$

$7$

$9$

$11$

$13$

Answer: E

Concepts:Venn Diagram basic counting

Difficulty rating: 1360

Small Hint:

First find the total number of blondes

Big Hint:

Subtract the brown-eyed blondes from all brown-eyed girls

Solution:

There are $50-31=19$ blondes, of whom $19-14=5$ are brown-eyed. Hence the number of brown-eyed brunettes is $18-5=13.$

Therefore, the correct answer is E.

11.

The numeral $47$ in base $a$ represents the same number as $74$ in base $b.$ Assuming both bases are positive integers, the least possible value of $a+b,$ written as a Roman numeral, is:

$\mathrm{XIII}$

$\mathrm{XV}$

$\mathrm{XXI}$

$\mathrm{XXIV}$

$\mathrm{XVI}$

Answer: D

Concepts:number base Diophantine Equation modular arithmetic

Difficulty rating: 1850

Small Hint:

Translate the numerals into $4a+7=7b+4$

Big Hint:

Both bases exceed $7$ ; solve $4a-7b=-3$ for the least valid pair

Solution:

The equality is $4a+7=7b+4,$ or $4a-7b=-3.$ Since both digits must be valid, $a,b\gt7.$ Reducing modulo $7$ gives $a\equiv1\pmod7.$ The first valid value is $a=15,$ which gives $b=9.$ Thus $a+b=24,$ written $\mathrm{XXIV}.$

Therefore, the correct answer is D.

12.

For each integer $N\gt1,$ define positive integers to be congruent if they leave the same nonnegative remainder when divided by $N.$ If $69,$ $90,$ and $125$ are congruent in one such system, then in that same system, $81$ is congruent to:

$3$

$4$

$5$

$7$

$8$

Answer: B

Concepts:modular arithmetic greatest common divisor divisibility

Difficulty rating: 1520

Small Hint:

The modulus divides the difference of any two congruent integers

Big Hint:

Use the differences $21$ and $35$

Solution:

The modulus $N$ divides both $90-69=21$ and $125-90=35.$ Since $N\gt1,$ this forces $N=7.$ Then $81\equiv4\pmod7.$

Therefore, the correct answer is B.

13.

If $(1.0025)^{10}$ is evaluated correct to $5$ decimal places, then the digit in the fifth decimal place is:

$0$

$1$

$2$

$5$

$8$

Answer: E

Concepts:binomial theorem decimal estimation

Difficulty rating: 1900

Small Hint:

Write $1.0025=1+0.0025$

Big Hint:

In the binomial expansion, retain terms large enough to affect the fifth decimal place

Solution:

The binomial expansion gives $\begin{gathered} (1+0.0025)^{10}\\ =1+10(0.0025)\\ \quad+45(0.0025)^2\\ \quad+120(0.0025)^3+\cdots\\ =1.025283125\ldots \end{gathered}$ which rounds to $1.02528.$ The fifth decimal digit is $8.$

Therefore, the correct answer is E.

14.

The number $2^{48}-1$ is exactly divisible by two numbers between $60$ and $70.$ These numbers are:

$61,$ $63$

$61,$ $65$

$63,$ $65$

$63,$ $67$

$67,$ $69$

Answer: C

Concepts:factoring exponent divisibility

Difficulty rating: 1990

Small Hint:

Use the fact that $2^m-1$ divides $2^n-1$ whenever $n$ is divisible by $m$

Big Hint:

Recognize $63=2^6-1$ and that $2^{12}-1$ is divisible by $65$

Solution:

Because $48$ is divisible by $6,$ the number $2^{48}-1$ is divisible by $63=2^6-1.$ Also $2^{12}-1=4095=63\cdot65,$ and $48$ is divisible by $12,$ so $2^{48}-1$ is divisible by $65.$ Thus the two numbers are $63$ and $65.$

Therefore, the correct answer is C.

15.

An aquarium on a level table has rectangular faces and is $10$ inches wide and $8$ inches high. When it was tilted, the water in it just covered an $8$ -inch by $10$ -inch end but only three-fourths of the rectangular bottom. The depth of the water when the bottom was again made level was:

$2\tfrac12$ inches

$3$ inches

$3\tfrac14$ inches

$3\tfrac12$ inches

$4$ inches

Answer: B

Concepts:volume triangle area ratio and proportion

Difficulty rating: 1760

Small Hint:

Compare the water volume in the tilted and level positions

Big Hint:

The tilted side view is a triangle with height $8$ and base three-fourths of the aquarium length

Solution:

Let the aquarium length be $L$ and the level-water depth be $h.$ In the tilted position the longitudinal cross-section of the water is a triangle with base $\frac{3L}{4}$ and height $8.$ Therefore $10\left(\frac12\cdot\frac{3L}{4}\cdot8\right)=10Lh,$ so $h=3$ inches.

Therefore, the correct answer is B.

16.

After finding the average of $35$ scores, a student carelessly included the average with the $35$ scores and found the average of these $36$ numbers. The ratio of the second average to the true average was:

$1:1$

$35:36$

$36:35$

$2:1$

None of these

Answer: A

Concepts:mean algebraic manipulation

Difficulty rating: 1200

Small Hint:

Call the original average $x$

Big Hint:

The original sum is $35x$ , and the extra number is also $x$

Solution:

If the true average is $x,$ then the original sum is $35x.$ Including $x$ itself gives a sum of $36x$ over $36$ numbers, so the new average is still $x.$ The ratio is $1:1.$

Therefore, the correct answer is A.

17.

A circular disk is divided by $2n$ equally spaced radii $(n\gt0)$ and one secant line. The maximum number of nonoverlapping areas into which the disk can be divided is:

$2n+1$

$2n+2$

$3n-1$

$3n$

$3n+1$

Answer: E

Concepts:counting regions counting intersections circle

Difficulty rating: 2060

Small Hint:

Begin with the $2n$ sectors made by the radii

Big Hint:

A secant can cross at most $n$ of the radii, so count the pieces of the secant chord

Solution:

The radii first make $2n$ sectors. A secant not through the center can meet at most one radius in each of $n$ opposite pairs, hence at most $n$ radii. Those intersections divide the secant chord into $n+1$ pieces, each of which adds one region. The maximum is therefore $2n+(n+1)=3n+1.$

Therefore, the correct answer is E.

18.

The current in a river flows steadily at $3$ miles per hour. A motorboat traveling at a constant rate in still water goes downstream $4$ miles and then returns to its starting point. The trip takes one hour, excluding turning time. The ratio of the downstream rate to the upstream rate is:

$4:3$

$3:2$

$5:3$

$2:1$

$5:2$

Answer: D

Concepts:relative speed distance rate and time quadratic

Difficulty rating: 1760

Small Hint:

Let $v$ be the boat’s still-water speed

Big Hint:

Solve $\frac{4}{v+3}+\frac{4}{v-3}=1$

Solution:

If $v$ is the still-water speed, then $\frac4{v+3}+\frac4{v-3}=1.$ This simplifies to $v^2-8v-9=0,$ so the positive admissible solution is $v=9.$ The downstream and upstream rates are $12$ and $6,$ whose ratio is $2:1.$

Therefore, the correct answer is D.

19.

If the line $y=mx+1$ intersects the ellipse $x^2+4y^2=1$ exactly once, then the value of $m^2$ is:

$\frac{1}{2}$

$\frac{2}{3}$

$\frac{3}{4}$

$\frac{4}{5}$

$\frac{5}{6}$

Answer: C

Concepts:ellipse tangent line quadratic

Difficulty rating: 2020

Small Hint:

Substitute the line equation into the ellipse

Big Hint:

Exactly one intersection means the resulting quadratic has discriminant zero

Solution:

Substitution gives $(1+4m^2)x^2+8mx+3=0.$ Tangency requires its discriminant to vanish: $64m^2-12(1+4m^2)=0.$ Thus $16m^2=12,$ so $m^2=\frac{3}{4}.$

Therefore, the correct answer is C.

20.

The sum of the squares of the roots of the equation $x^2+2hx=3$ is $10.$ The absolute value of $h$ is equal to:

$-1$

$\frac{1}{2}$

$\frac{2}{3}$

$2$

None of these

Answer: E

Concepts:Vieta’s Formulas quadratic algebraic manipulation

Difficulty rating: 1640

Small Hint:

Use the sum and product of the two roots

Big Hint:

If the roots are $r,s,$ then $r+s=-2h$ and $rs=-3$

Solution:

For roots $r,s,$ Vieta’s formulas give $r+s=-2h$ and $rs=-3.$ Hence $\begin{aligned} 4h^2=(r+s)^2 &=r^2+s^2+2rs\\ &=10-6=4. \end{aligned}$ Therefore $|h|=1,$ which is not among choices A-D because choice A is $-1.$

Therefore, the correct answer is E.

21.

If $\begin{gathered} \log_2(\log_3(\log_4x))=0,\\ \log_3(\log_4(\log_2y))=0,\\ \log_4(\log_2(\log_3z))=0, \end{gathered}$ then $x+y+z$ is equal to:

$50$

$58$

$89$

$111$

$1296$

Answer: C

Concepts:logarithm exponent function

Difficulty rating: 1850

Small Hint:

Undo each logarithm from the outside inward

Big Hint:

For the first chain, the successive inner values are $1,3,$ and then $x=4^3$

Solution:

Undoing the logarithms from the outside inward gives $\begin{gathered} x=4^3=64,\\ y=2^4=16,\\ z=3^2=9. \end{gathered}$ Therefore $x+y+z=64+16+9=89.$

Therefore, the correct answer is C.

22.

If $w$ is one of the imaginary roots of $x^3=1,$ then $(1-w+w^2)(1+w-w^2)$ is equal to:

$4$

$w$

$2$

$w^2$

$1$

Answer: A

Concepts:complex number roots of unity factoring

Difficulty rating: 1700

Small Hint:

Use $w^2+w+1=0$

Big Hint:

Replace $1+w^2$ by $-w$ and $1+w$ by $-w^2$

Solution:

Because $w\ne1$ and $w^3=1,$ we have $1+w+w^2=0.$ Thus $\begin{gathered} 1-w+w^2=-2w,\\ 1+w-w^2=-2w^2. \end{gathered}$ Their product is $4w^3=4.$

Therefore, the correct answer is A.

23.

Teams $A$ and $B$ are playing a series of games. If either team has an equal chance to win any game, and Team $A$ must win two games while Team $B$ must win three games to win the series, then the odds favoring Team $A$ to win the series are:

$11$ to $5$

$5$ to $2$

$8$ to $3$

$3$ to $2$

$13$ to $6$

Answer: A

Concepts:basic probability complementary probability casework

Difficulty rating: 1990

Small Hint:

It is shorter to enumerate the ways Team $A$ can lose

Big Hint:

Before $A$ ’s second win, $B$ can finish as BBB, ABBB, BABB, or BBAB

Solution:

Team $A$ loses in the sequences BBB, ABBB, BABB, and BBAB. Their total probability is $\frac18+3\left(\frac1{16}\right)=\frac5{16}.$ Thus Team $A$ wins with probability $\frac{11}{16},$ so the odds in its favor are $11:5.$

Therefore, the correct answer is A.

24.

Pascal’s triangle is an array of positive integers, shown below, in which the first row is $1,$ the second row is two $1$ ’s, each row begins and ends with $1,$ and each other entry is the sum of the two entries above it.

The quotient of the number of entries in the first $n$ rows which are not $1$ ’s and the number of $1$ ’s is:

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

$\dfrac{n^2-n}{4n-2}$

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

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

None of these

Answer: D

Concepts:Pascal’s Triangle arithmetic sequence basic counting

Difficulty rating: 1760

Small Hint:

Count all entries and then subtract the boundary $1$ ’s

Big Hint:

The first $n$ rows contain $\frac{n(n+1)}{2}$ entries and $2n-1$ boundary $1$ ’s

Solution:

The first $n$ rows contain $1+2+\cdots+n=\frac{n(n+1)}{2}$ entries. There are $2n-1$ boundary $1$ ’s, so the number of other entries is $\begin{gathered} \frac{n(n+1)}2-(2n-1)\\ =\frac{n^2+n-4n+2}{2}\\ =\frac{n^2-3n+2}{2}. \end{gathered}$ Dividing by $2n-1$ gives $\frac{n^2-3n+2}{4n-2}.$

Therefore, the correct answer is D.

25.

A teenage boy wrote his own age after his father’s. From this new four-place number, he subtracted the absolute value of the difference of their ages to get $4{,}289.$ The sum of their ages was:

$48$

$52$

$56$

$59$

$64$

Answer: D

Concepts:ages digits Diophantine Equation modular arithmetic

Difficulty rating: 1880

Small Hint:

Let the father’s age be $f$ and the boy’s age be $b$

Big Hint:

The concatenated number is $100f+b$ , so use $99f+2b=4289$

Solution:

Let the father and boy be $f$ and $b$ years old. Since $f\gt b,$ $\begin{aligned} 100f+b-(f-b)&=4289,\\ 99f+2b&=4289. \end{aligned}$ The boy is a teenager, so $13\le b\le19.$ Reducing modulo $9$ gives $2b\equiv5\pmod9,$ hence $b=16.$ Then $f=\frac{4289-32}{99}=43,$ and $f+b=59.$

Therefore, the correct answer is D.

26.

In triangle $ABC,$ point $F$ divides side $AC$ in the ratio $1:2.$ Let $E$ be the point where side $BC$ meets $AG,$ where $G$ is the midpoint of $BF.$ Then $E$ divides $BC$ in the ratio:

$1:4$

$1:3$

$2:5$

$4:11$

$3:8$

Answer: B

Concepts:mass points midpoint ratio and proportion

Difficulty rating: 2060

Small Hint:

Assign endpoint masses so that $AF:FC=1:2$

Big Hint:

The midpoint condition makes the masses at $B$ and $F$ equal

Solution:

Use mass points. Since $AF:FC=1:2,$ assign masses $2$ and $1$ to $A$ and $C,$ so the mass at $F$ is $3.$ Because $G$ is the midpoint of $BF,$ the mass at $B$ is also $3.$ Therefore $BE:EC=m_C:m_B=1:3.$

Therefore, the correct answer is B.

27.

A box contains chips, each of which is red, white, or blue. The number of blue chips is at least half the number of white chips and at most one-third the number of red chips. The number which are white or blue is at least $55.$ The minimum number of red chips is:

$24$

$33$

$45$

$54$

$57$

Answer: E

Concepts:inequality optimization bounding to limit cases

Difficulty rating: 1880

Small Hint:

Let the counts be $r,w,b$ and translate every condition into an inequality

Big Hint:

From $w\le2b$ and $w+b\ge55$ , find the least possible integer $b$

Solution:

Let the counts be $r,w,b.$ The conditions give $\begin{gathered} w\le2b,\\ r\ge3b,\\ w+b\ge55. \end{gathered}$ Hence $3b\ge w+b\ge55,$ so $b\ge19$ and $r\ge57.$ Equality is possible with $(w,b,r)=(36,19,57),$ so the minimum is $57.$

Therefore, the correct answer is E.

28.

Nine lines parallel to the base of a triangle divide the other sides each into $10$ equal segments and the area into $10$ distinct parts. If the area of the largest of these parts is $38,$ then the area of the original triangle is:

$180$

$190$

$200$

$210$

$240$

Answer: C

Difficulty rating: 1780

Small Hint:

The largest part is the bottom strip

Big Hint:

The smaller triangle above that strip has linear scale $\frac{9}{10}$

Solution:

If the whole area is $K,$ the triangle above the bottom strip is similar to the original with scale $\frac{9}{10},$ so its area is $\frac{81K}{100}.$ Thus the largest strip has area $K-\frac{81K}{100}=\frac{19K}{100}=38,$ giving $K=200.$

Therefore, the correct answer is C.

29.

Given the progression $\begin{gathered} 10^{\frac{1}{11}},10^{\frac{2}{11}},10^{\frac{3}{11}},\\ 10^{\frac{4}{11}},\ldots,10^{\frac{n}{11}}, \end{gathered}$ the least positive integer $n$ such that the product of the first $n$ terms exceeds $100{,}000$ is:

$7$

$8$

$9$

$10$

$11$

Answer: E

Concepts:geometric sequence exponent triangular number inequality

Difficulty rating: 1760

Small Hint:

Add the exponents when multiplying the terms

Big Hint:

Require $\frac{n(n+1)}{22}\gt5$ , noting that equality is not enough

Solution:

The product is $10^{\frac{1+2+\cdots+n}{11}}=10^{\frac{n(n+1)}{22}}.$ It exceeds $100{,}000=10^5$ exactly when $n(n+1)\gt110.$ For $n=10$ there is equality, while $n=11$ works.

Therefore, the correct answer is E.

30.

Given the linear fractional transformation $f_1(x)=\frac{2x-1}{x+1},$ define $f_{n+1}(x)=f_1(f_n(x))$ for $n=1,$ $2,$ $3,$ $\ldots.$ Assuming $f_{35}(x)=f_5(x),$ it follows that $f_{28}(x)$ is equal to:

$x$

$\frac{1}{x}$

$\frac{x-1}{x}$

$\frac{1}{1-x}$

None of these

Answer: D

Concepts:function recursion transformation

Difficulty rating: 2340

Small Hint:

Cancel five iterates by composing with the inverse transformation

Big Hint:

If $g=f_1^{-1}$ , then $f_{28}=g^2$ once $f_{30}$ is the identity

Solution:

The transformation is invertible. From $f_{35}=f_5,$ composing with $f_5^{-1}$ shows that $f_{30}$ is the identity. Solving $y=\frac{2x-1}{x+1}$ for $x$ gives $g(y)=f_1^{-1}(y)=\frac{y+1}{2-y}.$ Since iterates have period $30,$ $f_{28}=f_{-2}=g^2.$ Direct composition gives $g(g(x))=\frac{1}{1-x}.$

Therefore, the correct answer is D.

31.

Quadrilateral $ABCD$ is inscribed in a circle with side $AD,$ a diameter of length $4.$ If sides $AB$ and $BC$ each have length $1,$ then side $CD$ has length:

$\frac{7}{2}$

$\frac{5\sqrt2}{2}$

$\sqrt{11}$

$\sqrt{13}$

$2\sqrt3$

Answer: A

Concepts:circle chord trigonometric identity

Difficulty rating: 2190

Small Hint:

Equal chords $AB$ and $BC$ subtend equal central angles

Big Hint:

If half of either central angle is $t$ , then $4\sin t=1$ and $CD=4\cos(2t)$

Solution:

The circle has radius $2.$ Let the central angles subtending the equal chords $AB$ and $BC$ each be $2t.$ Then $1=4\sin t,$ so $\sin t=\frac{1}{4}.$ The remaining central angle from $C$ to $D$ along the semicircle is $\pi-4t,$ hence $\begin{aligned} CD&=4\sin\left(\frac{\pi-4t}{2}\right)\\ &=4\cos(2t)\\ &=4\left(1-2\cdot\frac1{16}\right)\\ &=\frac72. \end{aligned}$

Therefore, the correct answer is A.

32.

If $\begin{aligned} s={}&(1+2^{-\frac{1}{32}})(1+2^{-\frac{1}{16}})\\ &\cdot(1+2^{-\frac{1}{8}})(1+2^{-\frac{1}{4}})\\ &\cdot(1+2^{-\frac{1}{2}}), \end{aligned}$ then $s$ is equal to:

$\dfrac12(1-2^{-\frac{1}{32}})^{-1}$

$(1-2^{-\frac{1}{32}})^{-1}$

$1-2^{-\frac{1}{32}}$

$\dfrac12(1-2^{-\frac{1}{32}})$

$\frac{1}{2}$

Answer: A

Concepts:difference of squares telescoping exponent

Difficulty rating: 2210

Small Hint:

Set $x=2^{-\frac{1}{32}}$

Big Hint:

Apply the difference-of-squares identity repeatedly through the factor $1+x^{16}$

Solution:

Let $x=2^{-\frac{1}{32}}.$ Then $\begin{aligned} (1-x)s&=1-x^{32}\\ &=1-\frac12=\frac12. \end{aligned}$ Therefore $s=\frac12(1-2^{-\frac{1}{32}})^{-1}.$

Therefore, the correct answer is A.

33.

If $P$ is the product of $n$ quantities in geometric progression, $S$ their sum, and $S'$ the sum of their reciprocals, then $P$ in terms of $S,$ $S',$ and $n$ is:

$(SS')^{\frac{n}{2}}$

$(\frac{S}{S'})^{\frac{n}{2}}$

$(SS')^{n-2}$

$(\frac{S}{S'})^n$

$(\frac{S}{S'})^{\frac{n-1}{2}}$

Answer: B

Concepts:geometric sequence algebraic manipulation

Difficulty rating: 2300

Small Hint:

Write the progression as $a,ar,\ldots,ar^{n-1}$

Big Hint:

Show that $\frac{S}{S'}=a^2r^{n-1}$ , whose $\frac{n}{2}$ power is the product

Solution:

Write the terms as $a,ar,\ldots,ar^{n-1}.$ Reversing the reciprocal sum gives $S'=\frac{S}{a^2r^{n-1}},$ so $\frac{S}{S'}=a^2r^{n-1}.$ Meanwhile $\begin{aligned} P&=a^nr^{\frac{n(n-1)}{2}}\\ &=\left(a^2r^{n-1}\right)^{\frac{n}{2}}\\ &=\left(\frac{S}{S'}\right)^{\frac{n}{2}}. \end{aligned}$

Therefore, the correct answer is B.

34.

An ordinary clock in a factory is running slow so that the minute hand passes the hour hand at the usual dial positions ( $12$ o’clock, etc.) but only every $69$ minutes. At time and one-half for overtime, the extra pay to which a $\$4.00$ -per-hour worker should be entitled after working a normal $8$ -hour day by that slow-running clock is:

$\$2.30$

$\$2.60$

$\$2.80$

$\$3.00$

$\$3.30$

Answer: B

Concepts:clock rate money ratio and proportion

Difficulty rating: 1850

Small Hint:

A normal clock’s hands pass every $\frac{720}{11}$ minutes

Big Hint:

Eleven slow-clock intervals total $12$ displayed hours but $12$ hours $39$ minutes of real time

Solution:

Successive hand-overlaps are $\frac{720}{11}$ real minutes apart, while this slow clock displays $69$ minutes between them. Thus $12$ displayed hours correspond to $759$ real minutes, since $11\cdot69=759=12\cdot60+39.$ Eight displayed hours therefore take $8$ hours $26$ minutes, so the overtime is $26$ minutes. At $\$6$ per hour, that pays $\$2.60.$

Therefore, the correct answer is B.

35.

Each circle in an infinite sequence with decreasing radii is tangent externally to the one following it and to both sides of a given right angle. The ratio of the area of the first circle to the sum of the areas of all the other circles in the sequence is:

$(4+3\sqrt2):4$

$9\sqrt2:2$

$(16+12\sqrt2):1$

$(2+2\sqrt2):1$

$(3+2\sqrt2):1$

Answer: C

Concepts:tangent circles geometric sequence area ratio radical

Difficulty rating: 2380

Small Hint:

The centers lie on the angle bisector; find the ratio of consecutive radii

Big Hint:

The radius ratio is $3-2\sqrt2$ , so the area ratio is its square

Solution:

If consecutive radii are $r\gt r',$ their centers lie on the angle bisector at distances $r\sqrt2$ and $r'\sqrt2$ from the vertex. External tangency gives $\sqrt2(r-r')=r+r',$ so $\frac{r'}{r}=3-2\sqrt2.$ The ratio of successive areas is $q=(3-2\sqrt2)^2.$ Therefore the first area divided by the sum of all later areas is $\begin{aligned} \frac{1-q}{q} &=\frac1q-1\\ &=(3+2\sqrt2)^2-1\\ &=16+12\sqrt2. \end{aligned}$

Therefore, the correct answer is C.