1962 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

The expression $\dfrac{1^{4y-1}}{5^{-1}+3^{-1}}$ is equal to:

$\dfrac{4y-1}{8}$

$8$

$\dfrac{15}{2}$

$\dfrac{15}{8}$

$\dfrac18$

Answer: D

Concepts:exponent fraction algebraic manipulation

Difficulty rating: 1140

Small Hint:

First simplify the power in the numerator

Big Hint:

Rewrite each negative first power as a reciprocal

Solution:

The numerator is $1,$ while $5^{-1}+3^{-1}=\frac15+\frac13=\frac8{15}.$ Therefore the expression equals $\frac{1}{\frac{8}{15}}=\frac{15}{8}.$

Thus, the correct answer is D.

2.

The expression $\sqrt{\dfrac43}-\sqrt{\dfrac34}$ is equal to:

$\dfrac{\sqrt3}{6}$

$-\dfrac{\sqrt3}{6}$

$\dfrac{\sqrt{-3}}6$

$\dfrac{5\sqrt3}{6}$

$1$

Answer: A

Concepts:radical rationalizing denominator fraction

Difficulty rating: 1320

Small Hint:

Write the two radicals as $\frac{2}{\sqrt3}$ and $\frac{\sqrt3}{2}$

Big Hint:

Use a common denominator before rationalizing

Solution:

We have $\frac2{\sqrt3}-\frac{\sqrt3}{2} =\frac{4-3}{2\sqrt3} =\frac{\sqrt3}{6}.$

Therefore, the correct answer is A.

3.

The first three terms of an arithmetic progression are $x-1,$ $x+1,$ $2x+3,$ in the order shown. The value of $x$ is:

$-2$

$0$

$2$

$4$

undetermined

Answer: B

Concepts:arithmetic sequence linear equation

Difficulty rating: 1030

Small Hint:

Consecutive differences in an arithmetic progression are equal

Big Hint:

Set $(x+1)-(x-1)$ equal to $(2x+3)-(x+1)$

Solution:

Equality of consecutive differences gives $2=(2x+3)-(x+1)=x+2,$ so $x=0.$

Thus, the correct answer is B.

4.

If $8^x=32,$ then $x$ equals:

$4$

$\dfrac53$

$\dfrac32$

$\dfrac35$

$\dfrac14$

Answer: B

Concepts:exponent algebraic manipulation

Difficulty rating: 1400

Small Hint:

Express both sides as powers of $2$

Big Hint:

Equate the exponents in $2^{3x}=2^5$

Solution:

Since $8=2^3$ and $32=2^5,$ the equation becomes $2^{3x}=2^5.$ Hence $3x=5$ and $x=\frac{5}{3}.$

Therefore, the correct answer is B.

5.

If the radius of a circle is increased by $1$ unit, the ratio of the new circumference to the new diameter is:

$\pi+2$

$\dfrac{2\pi+1}{2}$

$\pi$

$\dfrac{2\pi-1}{2}$

$\pi-2$

Answer: C

Concepts:circle circumference ratio and proportion

Difficulty rating: 800

Small Hint:

Call the new radius $R$

Big Hint:

Divide the circumference $2\pi R$ by the diameter $2R$

Solution:

For every circle, regardless of its radius, the circumference divided by the diameter is $\frac{2\pi R}{2R}=\pi.$

Thus, the correct answer is C.

6.

A square and an equilateral triangle have equal perimeters. The area of the triangle is $9\sqrt3$ square inches. Expressed in inches the diagonal of the square is:

$\dfrac92$

$2\sqrt5$

$4\sqrt2$

$\dfrac{9\sqrt2}{2}$

none of these

Answer: D

Concepts:equilateral triangle perimeter diagonal area

Difficulty rating: 1410

Small Hint:

Use $\frac{\sqrt3}{4}s^2$ for the triangle’s area

Big Hint:

Equate the two perimeters, then multiply the square’s side by $\sqrt2$

Solution:

If $s$ is the triangle’s side, then $\frac{\sqrt3}{4}s^2=9\sqrt3,$ so $s=6.$ Its perimeter is $18,$ making the square’s side $\frac{18}{4}=\frac{9}{2}.$ The square’s diagonal is therefore $\frac{9\sqrt2}{2}.$

Thus, the correct answer is D.

7.

Let the bisectors of the exterior angles at $B$ and $C$ of triangle $ABC$ meet at $D.$ Then, if all measurements are in degrees, angle $BDC$ equals:

$\dfrac12(90-A)$

$90-A$

$\dfrac12(180-A)$

$180-A$

$180-2A$

Answer: C

Concepts:angle bisector angle sum angle chasing

Difficulty rating: 1300

Small Hint:

Each exterior-angle bisector makes an angle $90^\circ-\frac{B}{2}$ or $90^\circ-\frac{C}{2}$ with a side

Big Hint:

Apply the angle sum in triangle $BDC$ and use $B+C=180^\circ-A$

Solution:

The angles of triangle $BDC$ at $B$ and $C$ are $90^\circ-\frac{B}{2}$ and $90^\circ-\frac{C}{2}.$ Thus $\begin{aligned} \angle BDC &=180^\circ-\left(90^\circ-\frac B2\right) \\ &\quad-\left(90^\circ-\frac C2\right)\\ &=\frac{B+C}{2} \\ &=\frac{180^\circ-A}{2}. \end{aligned}$

Therefore, the correct answer is C.

8.

Given the set of $n$ numbers, $n\gt1,$ of which one is $1-\dfrac1n$ and all the others are $1.$ The arithmetic mean of the $n$ numbers is:

$1$

$n-\dfrac1n$

$n-\dfrac1{n^2}$

$1-\dfrac1{n^2}$

$1-\dfrac1n-\dfrac1{n^2}$

Answer: D

Concepts:mean algebraic manipulation

Difficulty rating: 1230

Small Hint:

There are $n-1$ copies of $1$

Big Hint:

Add all $n$ numbers, then divide the sum by $n$

Solution:

The sum is $(n-1)+\left(1-\frac1n\right)=n-\frac1n.$ Dividing by $n$ gives $1-\frac{1}{n^2}.$

Thus, the correct answer is D.

9.

When $x^9-x$ is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:

more than $5$

$5$

$4$

$3$

$2$

Answer: B

Concepts:factoring difference of squares polynomial

Difficulty rating: 1280

Small Hint:

Begin with $x(x^8-1)$ and repeatedly use differences of squares

Big Hint:

Over the integers, stop after isolating $x,$ $x-1,$ $x+1,$ $x^2+1,$ and $x^4+1$

Solution:

Factoring over the integers gives $\begin{aligned} x^9-x &=x(x-1)(x+1)\\ &\quad\cdot(x^2+1)(x^4+1). \end{aligned}$ The last two nonconstant factors are irreducible over the integers, so there are $5$ factors.

Thus, the correct answer is B.

10.

A man drives $150$ miles to the seashore in $3$ hours and $20$ minutes. He returns from the shore to the starting point in $4$ hours and $10$ minutes. Let $r$ be the average rate for the entire trip. Then the average rate for the trip going exceeds $r,$ in miles per hour, by:

$5$

$4\dfrac12$

$4$

$2$

$1$

Answer: A

Concepts:distance rate and time unit conversion

Difficulty rating: 1320

Small Hint:

Use total distance divided by total time for $r$

Big Hint:

The outbound time is $\frac{10}{3}$ hours and the round-trip time is $\frac{15}{2}$ hours

Solution:

The outbound rate is $\frac{150}{\frac{10}{3}}=45$ miles per hour. The entire trip covers $300$ miles in $3\frac13+4\frac16=7\frac12$ hours, so $r=\frac{300}{\frac{15}{2}}=40.$ The difference is $45-40=5.$

Thus, the correct answer is A.

11.

The difference between the larger root and the smaller root of $x^2-px+\frac{p^2-1}{4}=0$ is:

$0$

$1$

$2$

$p$

$p+1$

Answer: B

Concepts:quadratic algebraic manipulation

Difficulty rating: 1110

Small Hint:

Compute the discriminant of the quadratic

Big Hint:

The two roots are $\frac{p\pm\sqrt{\Delta}}{2}$

Solution:

The discriminant is $p^2-4\left(\frac{p^2-1}{4}\right)=1.$ The roots are $\frac{p+1}{2}$ and $\frac{p-1}{2},$ whose difference is $1.$

Therefore, the correct answer is B.

12.

When $\left(1-\dfrac1a\right)^6$ is expanded, the sum of the last three coefficients is:

$22$

$11$

$10$

$-10$

$-11$

Answer: C

Concepts:binomial theorem summation

Difficulty rating: 1210

Small Hint:

The last three terms use powers $a^{-4},$ $a^{-5},$ and $a^{-6}$

Big Hint:

Include the alternating signs from $(-\frac{1}{a})^k$

Solution:

The last three terms have coefficients $\binom64,\quad-\binom65,\quad\binom66,$ so their sum is $15-6+1=10.$

Thus, the correct answer is C.

13.

$R$ varies directly as $S$ and inversely as $T.$ When $R=\dfrac43$ and $T=\dfrac9{14},$ $S=\dfrac37.$ Find $S$ when $R=\sqrt{48}$ and $T=\sqrt{75}.$

$28$

$30$

$40$

$42$

$60$

Answer: B

Concepts:ratio and proportion radical algebraic manipulation

Difficulty rating: 1500

Small Hint:

Write the variation as $R=\frac{kS}{T}$

Big Hint:

For fixed $k,$ the quantity $\frac{RT}{S}$ is constant

Solution:

Because $R=\frac{kS}{T},$ we have $S=\frac{RT}{k}.$ Hence $\begin{aligned} \frac{S_2}{S_1} &=\frac{R_2T_2}{R_1T_1}\\ &=\frac{\sqrt{48}\sqrt{75}} {(\frac{4}{3})(\frac{9}{14})}\\ &=\frac{60}{\frac{6}{7}}=70. \end{aligned}$ Therefore $S_2=(\frac{3}{7})\cdot70=30.$

Thus, the correct answer is B.

14.

Let $s$ be the limiting sum of the geometric series $4-\dfrac83+\dfrac{16}{9}-\cdots,$ as the number of terms increases without bound. Then $s$ equals:

a number between $0$ and $1$

$2.4$

$2.5$

$3.6$

$12$

Answer: B

Concepts:geometric sequence summation

Difficulty rating: 1110

Small Hint:

Find the ratio of the second term to the first

Big Hint:

Use $\frac{a}{1-r}$ because the ratio has absolute value less than $1$

Solution:

The first term is $4$ and the common ratio is $-\frac{2}{3}.$ Thus $s=\frac4{1-(-\frac{2}{3})}=\frac{12}{5}=2.4.$

Therefore, the correct answer is B.

15.

Given triangle $ABC$ with base $AB$ fixed in length and position. As the vertex $C$ moves on a straight line, the intersection point of the three medians moves on:

a circle

a parabola

an ellipse

a straight line

a curve here not listed

Answer: D

Concepts:centroid transformation vector

Difficulty rating: 1280

Small Hint:

The centroid lies two-thirds of the way from a vertex to the midpoint of the opposite side

Big Hint:

With $AB$ fixed, express the centroid as a fixed point plus one-third of the position vector of $C$

Solution:

Let $A,$ $B,$ and $C$ also denote their position vectors. The centroid is $G=\frac{A+B+C}{3}.$ Since $A$ and $B$ are fixed, this is a translation and scaling by $\frac{1}{3}$ of the position of $C.$ A straight-line locus therefore maps to a straight-line locus.

Thus, the correct answer is D.

16.

Given rectangle $R_1$ with one side $2$ inches and area $12$ square inches. Rectangle $R_2$ with diagonal $15$ inches is similar to $R_1.$ Expressed in square inches the area of $R_2$ is:

$\dfrac92$

$36$

$\dfrac{135}{2}$

$9\sqrt{10}$

$\dfrac{27\sqrt{10}}4$

Answer: C

Difficulty rating: 1510

Small Hint:

The sides of $R_1$ are $2$ and $6$

Big Hint:

Areas scale as the square of the ratio of corresponding diagonals

Solution:

The diagonal of $R_1$ is $\sqrt{2^2+6^2}=2\sqrt{10}.$ Therefore $\frac{[R_2]}{[R_1]} =\left(\frac{15}{2\sqrt{10}}\right)^2 =\frac{45}{8}.$ Thus $[R_2]=12(\frac{45}{8})=\frac{135}{2}.$

Therefore, the correct answer is C.

17.

If $a=\log_8 225$ and $b=\log_2 15,$ then $a,$ in terms of $b,$ is:

$\dfrac b2$

$\dfrac{2b}{3}$

$b$

$\dfrac{3b}{2}$

$2b$

Answer: B

Concepts:logarithm exponent algebraic manipulation

Difficulty rating: 1180

Small Hint:

Write $225=15^2$ and $8=2^3$

Big Hint:

Use the change-of-base formula with base $2$

Solution:

Changing to base $2$ gives $\begin{aligned} a&=\frac{\log_2(15^2)}{\log_2(2^3)}\\ &=\frac{2\log_2 15}{3} =\frac{2b}{3}. \end{aligned}$

Thus, the correct answer is B.

18.

A regular dodecagon ( $12$ sides) is inscribed in a circle with radius $r$ inches. The area of the dodecagon, in square inches, is:

$3r^2$

$2r^2$

$\dfrac{3r^2\sqrt3}{4}$

$r^2\sqrt3$

$3r^2\sqrt3$

Answer: A

Concepts:regular polygon triangle area trigonometry

Difficulty rating: 1210

Small Hint:

Divide the dodecagon into $12$ triangles with vertex at the center

Big Hint:

Each central angle is $30^\circ$ , so use $\frac12r^2\sin30^\circ$

Solution:

Each of the $12$ central triangles has area $\frac12r^2\sin30^\circ=\frac{r^2}{4}.$ Their total area is $12(\frac{r^2}{4})=3r^2.$

Thus, the correct answer is A.

19.

If the parabola $y=ax^2+bx+c$ passes through the points $(-1,12),$ $(0,5),$ and $(2,-3),$ the value of $a+b+c$ is:

$-4$

$-2$

$0$

$1$

$2$

Answer: C

Concepts:system of equations quadratic substitution

Difficulty rating: 1280

Small Hint:

The requested sum is the value of the parabola at $x=1$

Big Hint:

Use the three given points to solve for $a,$ $b,$ and $c$

Solution:

The point $(0,5)$ gives $c=5.$ The other two points give $a-b=7,\qquad 4a+2b=-8.$ Solving yields $a=1$ and $b=-6.$ Hence $a+b+c=1-6+5=0.$

Thus, the correct answer is C.

20.

The angles of a pentagon are in arithmetic progression. One of the angles, in degrees, must be:

$108$

$90$

$72$

$54$

$36$

Answer: A

Concepts:angle sum arithmetic sequence

Difficulty rating: 1440

Small Hint:

Five terms in arithmetic progression have their middle term equal to their average

Big Hint:

The interior angles of a pentagon sum to $540^\circ$

Solution:

The average of the five angles is $\frac{540^\circ}{5}=108^\circ.$ For five terms in arithmetic progression, the middle term equals the average, so one angle must be $108^\circ.$

Thus, the correct answer is A.

21.

It is given that one root of $2x^2+rx+s=0,$ with $r$ and $s$ real numbers, is $3+2i$ $(i=\sqrt{-1}).$ The value of $s$ is:

undetermined

$5$

$6$

$-13$

$26$

Answer: E

Concepts:complex number Vieta’s Formulas

Difficulty rating: 1210

Small Hint:

A polynomial with real coefficients also has the conjugate root

Big Hint:

Use the product of the roots and Vieta’s formula

Solution:

The other root is $3-2i.$ Their product is $(3+2i)(3-2i)=9+4=13.$ Vieta’s formula gives $\frac{s}{2}=13,$ so $s=26.$

Therefore, the correct answer is E.

22.

The number $121_b,$ written in the integral base $b,$ is the square of an integer, for:

$b=10,$ only

$b=10$ and $b=5,$ only

$2\le b\le10$

$b\gt2$

no value of $b$

Answer: D

Concepts:number base place value perfect square

Difficulty rating: 1440

Small Hint:

Convert $121_b$ to an expression in $b$

Big Hint:

Remember that the digit $2$ requires $b\gt2$

Solution:

In ordinary notation, $121_b=b^2+2b+1=(b+1)^2.$ This is a square for every allowable base. Since digit $2$ occurs, precisely the integral bases $b\gt2$ are allowable.

Thus, the correct answer is D.

23.

In triangle $ABC,$ $CD$ is the altitude to $AB$ and $AE$ is the altitude to $BC.$ If the lengths of $AB,$ $CD,$ and $AE$ are known, the length of $DB$ is:

not determined by the information given

determined only if $A$ is an acute angle

determined only if $B$ is an acute angle

determined only if $ABC$ is an acute triangle

none of these is correct

Answer: E

Concepts:triangle area altitude Pythagorean Theorem

Difficulty rating: 1570

Small Hint:

Compute the triangle’s area in two ways using the two known altitudes

Big Hint:

After finding $BC,$ use right triangle $BCD$

Solution:

Let $AB=c,$ $CD=h,$ and $AE=e.$ Equating two area formulas gives $\frac12ch=\frac12(BC)e,$ so $BC=\frac{ch}{e}.$ Since triangle $BCD$ is right at $D,$ $DB=\sqrt{BC^2-CD^2}.$ This determines $DB$ whether the original triangle is acute, right, or obtuse, so none of the first four choices is correct.

Thus, the correct answer is E.

24.

Three machines $P,$ $Q,$ and $R,$ working together, can do a job in $x$ hours. When working alone, $P$ needs an additional $6$ hours to do the job; $Q,$ one additional hour; and $R,$ $x$ additional hours. The value of $x$ is:

$\dfrac23$

$\dfrac{11}{12}$

$\dfrac32$

$2$

$3$

Answer: A

Concepts:rate rational equation

Difficulty rating: 1710

Small Hint:

The individual completion times are $x+6,$ $x+1,$ and $2x$

Big Hint:

Set the sum of the individual hourly rates equal to $\frac{1}{x}$

Solution:

The rates satisfy $\frac1{x+6}+\frac1{x+1}+\frac1{2x}=\frac1x.$ Thus $\frac1{x+6}+\frac1{x+1}=\frac1{2x},$ which simplifies to $3x^2+7x-6=0,$ or $(3x-2)(x+3)=0.$ A time must be positive, so $x=\frac{2}{3}.$

Therefore, the correct answer is A.

25.

Given square $ABCD$ with side $8$ feet. A circle is drawn through vertices $A$ and $D$ and tangent to side $BC.$ The radius of the circle, in feet, is:

$4$

$4\sqrt2$

$5$

$5\sqrt2$

$6$

Answer: C

Concepts:tangent line perpendicular bisector coordinate geometry circle

Difficulty rating: 1570

Small Hint:

Place $A=(0,0),$ $D=(0,8),$ and $BC$ on the line $x=8$

Big Hint:

The center lies on the perpendicular bisector of $AD$ , and its distance to $BC$ equals the radius

Solution:

Place the center at $(h,4).$ Since the circle passes through $A,$ $r^2=h^2+16.$ Tangency to $x=8$ gives $r=8-h.$ Therefore $h^2+16=(8-h)^2,$ so $h=3$ and $r=5.$

Thus, the correct answer is C.

26.

For any real value of $x$ the maximum value of $8x-3x^2$ is:

$0$

$\dfrac83$

$4$

$5$

$\dfrac{16}{3}$

Answer: E

Concepts:quadratic completing the square optimization

Difficulty rating: 1180

Small Hint:

Complete the square in $8x-3x^2$

Big Hint:

A negative square is largest when it equals $0$

Solution:

Completing the square, $8x-3x^2 =\frac{16}{3}-3\left(x-\frac43\right)^2.$ The square term is nonnegative, so the maximum is $\frac{16}{3}.$

Thus, the correct answer is E.

27.

Let $a\mathbin{@}b$ represent the operation on two numbers, $a$ and $b,$ which selects the larger of the two numbers, with $a\mathbin{@}a=a.$ Let $a\mathbin{!}b$ represent the operation which selects the smaller of the two numbers, with $a\mathbin{!}a=a.$ Which of the following three rules is (are) correct? $\begin{aligned} (1)\quad&a\mathbin{@}b=b\mathbin{@}a,\\ (2)\quad&a\mathbin{@}(b\mathbin{@}c)=(a\mathbin{@}b)\mathbin{@}c,\\ (3)\quad&a\mathbin{!}(b\mathbin{@}c)\\ &\quad=(a\mathbin{!}b)\mathbin{@}(a\mathbin{!}c). \end{aligned}$

$(1)$ only

$(2)$ only

$(1)$ and $(2)$ only

$(1)$ and $(3)$ only

all three

Answer: E

Concepts:custom operation distributive property casework

Difficulty rating: 1500

Small Hint:

Translate $\mathbin{@}$ as maximum and $\mathbin{!}$ as minimum

Big Hint:

For rule $(3),$ compare both sides separately when $a$ is below or above $\max(b,c)$

Solution:

Maximum is commutative and associative, so $(1)$ and $(2)$ hold. Rule $(3)$ is the distributive identity $\begin{aligned} &\min(a,\max(b,c))\\ &\quad=\max(\min(a,b),\min(a,c)). \end{aligned}$ If $a\ge\max(b,c),$ both sides equal $\max(b,c);$ if $a\lt\max(b,c),$ both sides equal $a.$ Thus $(3)$ also holds.

Therefore, the correct answer is E.

28.

The set of $x$ -values satisfying the equation $x^{\log_{10}x}=\frac{x^3}{100}$ consists of:

$\dfrac1{10},$ only

$10,$ only

$100,$ only

$10$ or $100,$ only

more than two real numbers

Answer: D

Concepts:logarithm exponent substitution quadratic

Difficulty rating: 1520

Small Hint:

The logarithm requires $x\gt0$ ; set $y=\log_{10}x$

Big Hint:

Take base- $10$ logarithms to obtain a quadratic in $y$

Solution:

Set $y=\log_{10}x,$ so $x=10^y.$ Taking base- $10$ logarithms gives $y^2=3y-2.$ Hence $(y-1)(y-2)=0,$ so $x=10$ or $x=100.$

Thus, the correct answer is D.

29.

Which of the following sets of $x$ -values satisfy the inequality $2x^2+x\lt6?$

$-2\lt x\lt\dfrac32$

$x\gt\dfrac32$ or $x\lt-2$

$x\lt\dfrac32$

$\dfrac32\lt x\lt2$

$x\lt-2$

Answer: A

Concepts:inequality factoring zero product property

Difficulty rating: 1150

Small Hint:

Move all terms to one side and factor the quadratic

Big Hint:

A product of two linear factors is negative between its roots

Solution:

The inequality is $2x^2+x-6\lt0.$ Factoring gives $(2x-3)(x+2)\lt0.$ The product is negative between its roots, so $-2\lt x\lt\frac{3}{2}.$

Thus, the correct answer is A.

30.

Consider the statements:

$(1)$ $p$ and $q$ are both true
$(2)$ $p$ is true and $q$ is false
$(3)$ $p$ is false and $q$ is true
$(4)$ $p$ is false and $q$ is false.

How many of these imply the negation of the statement “ $p$ and $q$ are both true”?

$0$

$1$

$2$

$3$

$4$

Answer: D

Concepts:logical deduction

Difficulty rating: 1280

Small Hint:

The negation fails only when both statements are true

Big Hint:

Check which of the four listed truth assignments are not case $(1)$

Solution:

The negation of “ $p$ and $q$ are both true” holds whenever at least one of $p$ and $q$ is false. This occurs in cases $(2),$ $(3),$ and $(4),$ for a total of $3.$

Thus, the correct answer is D.

31.

The ratio of the interior angles of two regular polygons with sides of unit length is $3:2.$ How many such pairs are there?

$1$

$2$

$3$

$4$

infinitely many

Answer: C

Concepts:regular polygon angle sum Diophantine Equation

Difficulty rating: 1710

Small Hint:

For an $n$ -gon, an interior angle is $\frac{180^\circ(n-2)}{n}$

Big Hint:

If the smaller polygon has $n$ sides, solve for the larger side count and test the possible integers $3\le n\lt6$

Solution:

Let the smaller and larger polygons have $n$ and $N$ sides. Then $\frac{\frac{N-2}{N}}{\frac{n-2}{n}}=\frac32,$ which gives $N=\frac{4n}{6-n}.$ Positivity and $N\gt n$ require $n=3,$ $n=4,$ or $n=5.$ These give $N=4,$ $N=8,$ and $N=20,$ respectively. Thus there are $3$ pairs.

Therefore, the correct answer is C.

32.

If $x_{k+1}=x_k+\dfrac12$ for $k=1,$ $2,$ $\ldots,$ $n-1$ and $x_1=1,$ find $x_1+x_2+\cdots+x_n.$

$\dfrac{n+1}{2}$

$\dfrac{n+3}{2}$

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

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

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

Answer: E

Concepts:arithmetic sequence summation

Difficulty rating: 1440

Small Hint:

The recurrence defines an arithmetic sequence with common difference $\frac{1}{2}$

Big Hint:

Find $x_n$ , then use $\frac{n(x_1+x_n)}{2}$

Solution:

We have $x_n=1+\frac{n-1}{2}=\frac{n+1}{2}.$ Therefore $\begin{aligned} x_1+\cdots+x_n &=\frac n2\left(1+\frac{n+1}{2}\right)\\ &=\frac{n^2+3n}{4}. \end{aligned}$

Thus, the correct answer is E.

33.

The set of $x$ -values satisfying the inequality $2\le|x-1|\le5$ is:

$-4\le x\le-1$ or $3\le x\le6$

$3\le x\le6$ or $-6\le x\le-3$

$x\le-1$ or $x\ge3$

$-1\le x\le3$

$-4\le x\le6$

Answer: A

Concepts:absolute value inequality

Difficulty rating: 1180

Small Hint:

Interpret the inequality as distances from $1$ between $2$ and $5$

Big Hint:

Solve $2\le x-1\le5$ and $-5\le x-1\le-2$

Solution:

For $x-1\ge0,$ the bounds give $3\le x\le6.$ For $x-1\le0,$ they give $-4\le x\le-1.$ The solution is the union of these two intervals.

Thus, the correct answer is A.

34.

For what real values of $K$ does $x=K^2(x-1)(x-2)$ have real roots?

none

$-2\lt K\lt1$

$-2\sqrt2\lt K\lt2\sqrt2$

$K\gt1$ or $K\lt-2$

all

Answer: E

Concepts:quadratic algebraic manipulation

Difficulty rating: 1550

Small Hint:

Expand and collect terms to obtain a quadratic in $x$

Big Hint:

Its discriminant simplifies to $K^4+6K^2+1$

Solution:

Rearranging gives $K^2x^2-(3K^2+1)x+2K^2=0.$ Its discriminant is $\begin{aligned} &(3K^2+1)^2-8K^4\\ &\qquad=K^4+6K^2+1\gt0 \end{aligned}$ for every real $K.$ This also covers $K=0,$ when the original equation gives $x=0.$

Therefore, the correct answer is E.

35.

A man on his way to dinner shortly after $6{:}00$ p.m. observes that the hands of his watch form an angle of $110^\circ.$ Returning before $7{:}00$ p.m. he notices that again the hands of his watch form an angle of $110^\circ.$ The number of minutes that he has been away is:

$36\dfrac23$

$40$

$42$

$42.4$

$45$

Answer: B

Concepts:clock relative speed

Difficulty rating: 1570

Small Hint:

During the interval, the minute hand gains on the hour hand at $5.5^\circ$ per minute

Big Hint:

Between the two observations, the signed separation changes from $110^\circ$ to $-110^\circ$

Solution:

The two observations lie on opposite sides of the instant when the hands coincide. Their signed angular separation changes by $220^\circ.$ Since the minute hand gains on the hour hand at $6-0.5=5.5^\circ$ per minute, the elapsed time is $\frac{220}{5.5}=40$ minutes.

Thus, the correct answer is B.

36.

If both $x$ and $y$ are integers, how many solutions are there to the equation $(x-8)(x-10)=2^y?$

$0$

$1$

$2$

$3$

more than $3$

Answer: C

Concepts:Diophantine Equation power of 2 difference of squares

Difficulty rating: 1900

Small Hint:

Rewrite the left side as $(x-9)^2-1$

Big Hint:

If $n=x-9,$ then the consecutive even factors $n-1$ and $n+1$ must both be powers of $2$

Solution:

Put $n=x-9.$ Then $2^y=n^2-1=(n-1)(n+1).$ Since the product is a power of $2,$ both factors must have no odd prime divisor. The only consecutive even integers differing by $2$ that are both signed powers of $2$ are $(-4,-2)$ and $(2,4).$ Thus $n=\pm3$ and $y=3,$ giving $x=6$ or $12.$ There are $2$ ordered pairs $(x,y).$

Therefore, the correct answer is C.

37.

$ABCD$ is a square with side of unit length. Points $E$ and $F$ are taken respectively on sides $AB$ and $AD$ so that $AE=AF$ and the quadrilateral $CDFE$ has maximum area. In square units this maximum area is:

$\dfrac12$

$\dfrac9{16}$

$\dfrac{19}{32}$

$\dfrac58$

$\dfrac23$

Answer: D

Concepts:area shoelace formula completing the square optimization

Difficulty rating: 1730

Small Hint:

Let $AE=AF=t$ and use coordinates or the shoelace formula

Big Hint:

The area becomes $\frac12(1+t-t^2)$ ; complete the square

Solution:

Set $A=(0,0),$ $B=(1,0),$ $C=(1,1),$ and $D=(0,1).$ Then $E=(t,0)$ and $F=(0,t).$ The shoelace formula gives $\begin{aligned} [CDFE]&=\frac{1+t-t^2}{2}\\ &=\frac58-\frac12 \left(t-\frac12\right)^2. \end{aligned}$ Its maximum is $\frac{5}{8}.$

Thus, the correct answer is D.

38.

The population of Nosuch Junction at one time was a perfect square. Later, with an increase of $100,$ the population was one more than a perfect square. Now, with an additional increase of $100,$ the population is again a perfect square.

The original population is a multiple of:

$3$

$7$

$9$

$11$

$17$

Answer: B

Concepts:difference of squares perfect square factor systematic listing

Difficulty rating: 1980

Small Hint:

Write the populations as $a^2,$ $b^2+1,$ and $c^2$

Big Hint:

From $b^2-a^2=99,$ test the positive factor pairs of $99$ , then impose $c^2-a^2=200$

Solution:

Let the original population be $a^2.$ Then $\begin{aligned} b^2-a^2&=99,\\ c^2-a^2&=200. \end{aligned}$ From $(b-a)(b+a)=99,$ the positive possibilities for $a$ are $49,$ $15,$ and $1.$ Checking the second condition, only $a=49$ works, since $49^2+200=2601=51^2.$ The population is $49^2=7^4,$ a multiple of $7.$

Thus, the correct answer is B.

39.

The medians $AN$ and $BP$ of a triangle with unequal sides are, respectively, $3$ inches and $6$ inches long. Its area is $3\sqrt{15}$ square inches. The length of the third median, in inches, is:

$4$

$3\sqrt3$

$3\sqrt6$

$6\sqrt3$

$6\sqrt6$

Answer: C

Concepts:median (geometry)law of cosines area

Difficulty rating: 2090

Small Hint:

The three medians form the side lengths of a triangle whose area is three-fourths the original area

Big Hint:

Use the two known median lengths and that area to find the two possible included angles, then reject the case that makes two medians equal

Solution:

The triangle whose sides are the three medians has area $\frac34\left(3\sqrt{15}\right)=\frac{9\sqrt{15}}4.$ If $\theta$ is the included angle between its sides $3$ and $6,$ then $9\sin\theta=\frac{9\sqrt{15}}4,$ so $\cos\theta=\pm\frac{1}{4}.$ By the law of cosines, the third median $m$ satisfies $m^2=3^2+6^2-2(3)(6)\cos\theta,$ giving $m^2=36$ or $54.$ The value $m=6$ would make two medians, and hence two sides, equal. Because the triangle has unequal sides, $m=\sqrt{54}=3\sqrt6.$

Therefore, the correct answer is C.

40.

The limiting sum of the infinite series $\frac1{10}+\frac2{10^2}+\frac3{10^3}+\cdots,$ whose $n$ th term is $\frac{n}{10^n},$ is:

$\dfrac19$

$\dfrac{10}{81}$

$\dfrac18$

$\dfrac{17}{72}$

larger than any finite quantity

Answer: B

Concepts:arithmetico-geometric series calculus summation

Difficulty rating: 1570

Small Hint:

Start from $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$

Big Hint:

Differentiate and then multiply by $x$

Solution:

For $|x|\lt1,$ $\sum_{n=1}^{\infty}nx^n=\frac{x}{(1-x)^2}.$ Taking $x=\frac{1}{10}$ gives $\frac{\frac{1}{10}}{(\frac{9}{10})^2}=\frac{10}{81}.$

Thus, the correct answer is B.