2001 AMC 10 Solutions | LIVE by Po-Shen Loh

1.

The median of the list

$\begin{gathered} n,\ n+3,\ n+4,\ n+5,\ n+6, \\ \ n+8,\ n+10,\ n+12,\ n+15 \end{gathered}$

is $10.$ What is the mean?

$4$

$6$

$7$

$10$

$11$

Concepts:median (data)mean

Difficulty rating: 790

Small Hint:

The list has $9$ numbers, so the median is the $5$ th term

Big Hint:

The $5$ th term is $n+6,$ so set $n+6=10$

Solution:

The list has $9$ numbers in increasing order, so the median is the $5$ th term, $n+6.$ Setting $n+6=10$ gives $n=4.$

The sum of the terms is $9n$ $\small {}+(3+4+5+6+8+10+12+15)$ $=9n+63$ $=99,$ so the mean is $\frac{99}{9}=11.$

Thus, the correct answer is E.

2.

A number $x$ is $2$ more than the product of its reciprocal and its additive inverse. In which interval does the number lie?

$-4 \le x \le -2$

$-2 \lt x \le 0$

$0 \lt x \le 2$

$2 \lt x \le 4$

$4 \lt x \le 6$

Concepts:fraction algebraic manipulation

Difficulty rating: 960

Small Hint:

The reciprocal of $x$ is $\dfrac1x$ and the additive inverse is $-x$

Big Hint:

Their product is $\dfrac1x\cdot(-x)=-1$

Solution:

The reciprocal of $x$ is $\dfrac1x$ and its additive inverse is $-x.$ Their product is $\dfrac1x\cdot(-x)=-1.$

So $x=2+(-1)=1,$ which lies in the interval $0\lt x\le2.$

Thus, the correct answer is C.

3.

The sum of two numbers is $S.$ Suppose $3$ is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?

$2S+3$

$3S+2$

$3S+6$

$2S+6$

$2S+12$

Concepts:algebraic manipulation distributive property

Difficulty rating: 870

Small Hint:

Call the numbers $a$ and $b$ with $a+b=S$

Big Hint:

The final sum is $2(a+3)+2(b+3)$

Solution:

Let the numbers be $a$ and $b,$ so $a+b=S.$ After adding $3$ to each and doubling, the sum is $2(a+3)$ $+2(b+3)$ $=2(a+b)+12$ $=2S+12.$

Thus, the correct answer is E.

4.

What is the maximum number of possible points of intersection of a circle and a triangle?

$2$

$3$

$4$

$5$

$6$

Concepts:circle counting intersections

Difficulty rating: 1040

Small Hint:

A line can cross a circle in at most $2$ points

Big Hint:

A triangle has $3$ sides

Solution:

Each side of the triangle is a segment, which can intersect a circle in at most $2$ points. With $3$ sides, the maximum is $3\cdot2=6$ points, and this is achievable.

Thus, the correct answer is E.

5.

How many of the twelve pentominoes pictured below have at least one line of symmetry?

$3$

$4$

$5$

$6$

$7$

Concepts:symmetry systematic listing

Difficulty rating: 1120

Small Hint:

A line of symmetry folds the shape exactly onto itself

Big Hint:

Check each of the twelve shapes for a horizontal, vertical, or diagonal fold line

Solution:

Checking each pentomino for a reflection line, exactly six have at least one: the straight bar, the plus, the $T$ -shape, the $U$ -shape, the $V$ -shape, and the $W$ -shape. The remaining six have no line of symmetry.

Thus, the correct answer is D.

6.

Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits of the integer $n.$ For example, $P(23)=6$ and $S(23)=5.$ Suppose $N$ is a two-digit number such that $N=P(N)+S(N).$ What is the units digit of $N?$

$2$

$3$

$6$

$8$

$9$

Concepts:digits algebraic manipulation

Difficulty rating: 1100

Small Hint:

Write $N=10a+b,$ where $a$ and $b$ are its digits

Big Hint:

Then $10a+b=ab+(a+b),$ which simplifies to $9a=ab$

Solution:

Write $N=10a+b.$ Then $P(N)=ab$ and $S(N)=a+b,$ so $10a+b=ab+a+b.$ This reduces to $9a=ab,$ and since $a\ne0,$ we get $b=9.$

The units digit is $9.$ Thus, the correct answer is E.

7.

When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?

$0.0002$

$0.002$

$0.02$

$0.2$

$2$

Concepts:place value decimal algebraic manipulation

Difficulty rating: 1170

Small Hint:

Moving the decimal point four places right multiplies by $10000$

Big Hint:

So $10000x=\dfrac{4}{x},$ giving $x^2=\dfrac{4}{10000}$

Solution:

Moving the decimal four places right multiplies $x$ by $10000.$ So $10000x=4\cdot\dfrac1x,$ giving $x^2=\dfrac{4}{10000}.$

Since $x\gt0,$ $x=\dfrac{2}{100}=0.02.$

Thus, the correct answer is C.

8.

Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab?

$42$

$84$

$126$

$178$

$252$

Concepts:least common multiple prime factorization

Difficulty rating: 960

Small Hint:

Find the least common multiple of $3,$ $4,$ $6,$ and $7$

Big Hint:

The least common multiple needs the factors $2^2,$ $3,$ and $7$

Solution:

They meet again after $\text{lcm}(3,4,6,7)$ days. Since $4=2^2$ and $6=2\cdot3,$ the least common multiple is $2^2\cdot3\cdot7=84.$

Thus, the correct answer is B.

9.

The state income tax where Kristin lives is levied at the rate of $p\%$ of the first $\$28000$ of annual income plus $(p+2)\%$ of any amount above $\$28000.$ Kristin noticed that the state income tax she paid amounted to $(p+0.25)\%$ of her annual income. What was her annual income?

$\$28000$

$\$32000$

$\$35000$

$\$42000$

$\$56000$

Concepts:percentage linear equation

Difficulty rating: 1370

Small Hint:

Let her income be $x\gt28000$ and write the tax two different ways

Big Hint:

$\dfrac{p}{100}(28000)$ $+\dfrac{p+2}{100}(x-28000)$ $=\dfrac{p+0.25}{100}x$

Solution:

Let her income be $x\ge28000.$ Then

$\begin{aligned} &\tfrac{p}{100}(28000) \\ &\quad {}+\tfrac{p+2}{100}(x-28000) \\ &\quad =\tfrac{p+0.25}{100}x. \end{aligned}$

Multiplying by $100$ and expanding, all the $p$ terms cancel, leaving $2x-56000=0.25x.$ So $1.75x=56000$ and $x=32000.$

Thus, the correct answer is B.

10.

If $x,$ $y,$ and $z$ are positive with $xy=24,$ $xz=48,$ and $yz=72,$ then $x+y+z$ is

$18$

$19$

$20$

$22$

$24$

Concepts:system of equations algebraic manipulation

Difficulty rating: 1240

Small Hint:

Divide $xz=48$ by $xy=24$ to relate $y$ and $z$

Big Hint:

From $z=2y$ and $yz=72,$ solve for $y$

Solution:

Dividing $xz=48$ by $xy=24$ gives $z=2y.$ Then $yz=2y^2=72,$ so $y=6,$ $z=12,$ and $x=\frac{24}{y}=4.$

Hence $x+y+z=22.$ Thus, the correct answer is D.

11.

Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains $8$ unit squares. The second ring contains $16$ unit squares. If we continue this process, the number of unit squares in the $100$ th ring is

$396$

$404$

$800$

$10{,}000$

$10{,}404$

Concepts:difference of squares pattern recognition

Difficulty rating: 1070

Small Hint:

The $n$ th ring is a $(2n+1)\times(2n+1)$ square with the $(2n-1)\times(2n-1)$ square removed

Big Hint:

$(2n+1)^2-(2n-1)^2=8n$

Solution:

The $n$ th ring is the border of a $(2n+1)\times(2n+1)$ square surrounding a $(2n-1)\times(2n-1)$ square, so it contains $(2n+1)^2-(2n-1)^2=8n$ unit squares.

For $n=100,$ that is $800.$ Thus, the correct answer is C.

12.

Suppose that $n$ is the product of three consecutive integers and that $n$ is divisible by $7.$ Which of the following is not necessarily a divisor of $n?$

$6$

$14$

$21$

$28$

$42$

Concepts:divisibility prime factorization counterexample

Difficulty rating: 1370

Small Hint:

Among three consecutive integers, one is even and one is a multiple of $3$

Big Hint:

Look for a choice requiring two factors of $2;$ three consecutive integers need not provide both

Solution:

Among three consecutive integers, at least one is even and one is a multiple of $3,$ so $n$ is divisible by $6.$ With the given factor of $7,$ it is divisible by $6,$ $14,$ $21,$ and $42.$

But $28=2^2\cdot7$ requires two factors of $2,$ which is not guaranteed: $5\cdot6\cdot7=210$ is divisible by $7$ but not by $28.$

Thus, the correct answer is D.

13.

A telephone number has the form $ABC-DEF-GHIJ,$ where each letter represents a different digit. The digits in each part of the number are in decreasing order; that is, $A\gt B\gt C,$ $D\gt E\gt F,$ and $G\gt H\gt I\gt J.$ Furthermore, $D,$ $E,$ and $F$ are consecutive even digits; $G,$ $H,$ $I,$ and $J$ are consecutive odd digits; and $A+B+C=9.$ Find $A.$

$4$

$5$

$6$

$7$

$8$

Concepts:digits logical deduction casework

Difficulty rating: 1550

Small Hint:

The four consecutive odd digits $GHIJ$ must be $9753$ or $7531$

Big Hint:

Since $A+B+C=9$ uses the leftover odd digit, that odd digit must be $1$

Solution:

The consecutive odd digits $GHIJ$ are $9753$ or $7531,$ leaving one odd digit ( $1$ or $9$ ) for $A,$ $B,$ $C.$ Since $A+B+C=9,$ the odd digit there must be $1,$ so the two even digits in $ABC$ sum to $8.$

The consecutive even digits $DEF$ are $864,$ $642,$ or $420,$ leaving even-digit pairs $\{2,0\},$ $\{8,0\},$ or $\{8,6\}$ for $ABC.$ Only $\{8,0\}$ sums to $8,$ so $ABC=810$ and $A=8.$

Thus, the correct answer is E.

14.

A charity sells $140$ benefit tickets for a total of $\$2001.$ Some tickets sell for full price (a whole dollar amount), and the rest sell for half price. How much money is raised by the full-price tickets?

$\$782$

$\$986$

$\$1158$

$\$1219$

$\$1449$

Concepts:Diophantine Equation prime factorization bounding to limit cases

Difficulty rating: 1490

Small Hint:

Let $n$ tickets sell at full price $p;$ then $np+(140-n)\dfrac p2=2001$

Big Hint:

This gives $p(n+140)$ $=4002$ $=2\cdot3\cdot23\cdot29,$ with $140\le n+140\le280$

Solution:

Let $n$ full-price tickets sell at $p$ dollars each. Then $np+(140-n)\dfrac p2=2001,$ so $p(n+140)$ $=4002$ $=2\cdot3\cdot23\cdot29.$

Since $140\le n+140\le280,$ the only factor of $4002$ in range is $174=2\cdot3\cdot29.$ So $n+140=174,$ giving $n=34$ and $p=23.$ The full-price tickets raise $34\cdot23=782$ dollars.

Thus, the correct answer is A.

15.

A street has parallel curbs $40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $15$ feet and each stripe is $50$ feet long. Find the distance, in feet, between the stripes.

$9$

$10$

$12$

$15$

$25$

Concepts:parallelogram area

Difficulty rating: 1410

Small Hint:

The crosswalk is a parallelogram; compute its area using the curb as base

Big Hint:

The same area equals stripe length times the distance between the stripes

Solution:

The crosswalk is a parallelogram. Using the curb ( $15$ ft) as base and the street width ( $40$ ft) as height, its area is $15\cdot40=600$ square feet.

Using a stripe ( $50$ ft) as base, the area equals $50$ times the distance $d$ between the stripes, so $d=\frac{600}{50}=12.$

Thus, the correct answer is C.

16.

The mean of three numbers is $10$ more than the least of the numbers and $15$ less than the greatest. The median of the three numbers is $5.$ What is their sum?

$5$

$20$

$25$

$30$

$36$

Concepts:mean median (data)linear equation

Difficulty rating: 1280

Small Hint:

Let $m$ be the mean; then the least is $m-10$ and the greatest is $m+15$

Big Hint:

The median is $5,$ so $\dfrac13\big((m-10)+5+(m+15)\big)$ $=m$

Solution:

Let $m$ be the mean. The least number is $m-10$ and the greatest is $m+15,$ with median $5.$

Since the mean of the three is $m,$ $\begin{aligned} &\tfrac13\big((m-10)+5+(m+15)\big) \\ &\quad =m, \end{aligned}$ which gives $m=10.$

The sum is $3m=30.$ Thus, the correct answer is D.

17.

Which of the cones below can be formed from a $252^\circ$ sector of a circle of radius $10$ by aligning the two straight sides?

Concepts:cone sector circumference

Difficulty rating: 1490

Small Hint:

The sector’s radius becomes the slant height of the cone

Big Hint:

The arc length $\dfrac{252}{360}\cdot2\pi\cdot10$ becomes the base circumference $2\pi r$

Solution:

When rolled into a cone, the sector’s radius $10$ becomes the slant height, and the arc length becomes the base circumference.

The arc length is $\dfrac{252}{360}\cdot2\pi\cdot10=14\pi,$ so $2\pi r=14\pi$ gives base radius $r=7.$ The cone has slant height $10$ and base radius $7.$

Thus, the correct answer is C.

18.

The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to

$50$

$52$

$54$

$56$

$58$

Concepts:area ratio tiling percentage

Difficulty rating: 1530

Small Hint:

Work within a repeating $3\times3$ block of small squares

Big Hint:

Of the $9$ small squares, $4$ lie outside the pentagons

Solution:

Consider a repeating $3\times3$ block of nine small squares. Four of these nine squares are not part of the pentagons, so the pentagons cover $1-\dfrac49=\dfrac59\approx55.6\%$ of the area.

This is closest to $56.$ Thus, the correct answer is D.

19.

Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?

$6$

$9$

$12$

$15$

$18$

Concepts:stars and bars combinations

Difficulty rating: 1340

Small Hint:

Count nonnegative integer solutions of $g+c+p=4$

Big Hint:

Use stars and bars: $\dbinom{4+2}{2}$

Solution:

The number of selections is the number of nonnegative integer solutions of $g+c+p=4.$ By stars and bars, this is $\dbinom{4+2}{2}=\dbinom62=15.$

Thus, the correct answer is D.

20.

A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length $2000.$ What is the length of each side of the octagon?

$\dfrac13(2000)$

$2000(\sqrt2-1)$

$2000(2-\sqrt2)$

$1000$

$1000\sqrt2$

Concepts:regular polygon special right triangle rationalizing denominator

Difficulty rating: 1580

Small Hint:

Let each octagon side be $x;$ each cut triangle has legs $\dfrac{x}{\sqrt2}$

Big Hint:

Two legs plus a side span one edge of the square: $2\cdot\dfrac{x}{\sqrt2}+x=2000$

Solution:

Let each octagon side be $x.$ It is the hypotenuse of each cut isosceles right triangle, whose legs are $\dfrac{x}{\sqrt2}.$

Along one side of the square, two legs and one octagon side give $2\cdot\dfrac{x}{\sqrt2}+x=2000,$ so $x(\sqrt2+1)=2000$ and $x=\dfrac{2000}{\sqrt2+1}=2000(\sqrt2-1).$

Thus, the correct answer is B.

21.

A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12,$ and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

$\dfrac83$

$\dfrac{30}{11}$

$3$

$\dfrac{25}{8}$

$\dfrac72$

Concepts:similarity cone cylinder

Difficulty rating: 1680

Small Hint:

In an axial cross-section, the cylinder has radius $r$ and height $2r$

Big Hint:

Similar triangles give $\dfrac{12-2r}{r}=\dfrac{12}{5}$

Solution:

Take an axial cross-section. The cone has base radius $5$ and height $12;$ the cylinder appears as a rectangle of width $2r$ and height $2r.$

By similar triangles, $\dfrac{12-2r}{r}=\dfrac{12}{5},$ so $5(12-2r)=12r,$ giving $60=22r$ and $r=\dfrac{30}{11}.$

Thus, the correct answer is B.

22.

In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v,$ $w,$ $x,$ $y,$ and $z.$ Find $y+z.$

$43$

$44$

$45$

$46$

$47$

Concepts:magic square linear equation

Difficulty rating: 1530

Small Hint:

The first row, first column, and main diagonal all contain $v$

Big Hint:

So the other two entries of each match: $25+18=24+w=21+x$

Solution:

Since $v$ sits in the first row, first column, and main diagonal, the remaining two entries of each of those lines have equal sums: $25+18=24+w=21+x.$ So $w=19$ and $x=22.$

The anti-diagonal $25,$ $22,$ $19$ sums to $66,$ so the magic sum is $66.$ Then $v=66-24-19=23,$ $y=66-18-22=26,$ and $z=66-25-21=20.$

Hence $y+z=46.$ Thus, the correct answer is D.

23.

A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?

$\dfrac{3}{10}$

$\dfrac25$

$\dfrac12$

$\dfrac35$

$\dfrac{7}{10}$

Concepts:sampling without replacement basic probability symmetry

Difficulty rating: 1690

Small Hint:

Imagine drawing all five chips; the game stops when one color is exhausted

Big Hint:

The last drawn chip is white exactly when the very last of all five chips is red

Solution:

Imagine drawing all five chips in a random order. The drawing stops on a white chip exactly when both white chips come out before all three reds, which happens precisely when the very last chip in the full ordering is red.

That probability is $\dfrac{3}{5}.$ Thus, the correct answer is D.

24.

In trapezoid $ABCD,$ $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD},$ with $AB+CD=BC,$ $AB\lt CD,$ and $AD=7.$ What is $AB\cdot CD?$

$12$

$12.25$

$12.5$

$12.75$

$13$

Concepts:trapezoid Pythagorean Theorem difference of squares

Difficulty rating: 1810

Small Hint:

Drop a perpendicular from $B$ to $CD$ at $E;$ then $BE=AD=7$ and $CE=CD-AB$

Big Hint:

$BC^2=BE^2+CE^2$ with $BC=CD+AB$

Solution:

Drop a perpendicular from $B$ to $CD,$ meeting it at $E.$ Then $BE=AD=7$ and $CE=CD-AB.$ By the Pythagorean theorem, $BC^2=BE^2+CE^2.$

Since $BC=CD+AB,$ $\begin{aligned} &(CD+AB)^2 \\ &\quad {}-(CD-AB)^2 \\ &\quad =BE^2=49. \end{aligned}$

The left side equals $4\cdot AB\cdot CD,$ so $AB\cdot CD=\dfrac{49}{4}=12.25.$

Thus, the correct answer is B.

25.

How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5?$

$768$

$801$

$934$

$1067$

$1167$

Concepts:inclusion-exclusion counting integers in a range multiple

Difficulty rating: 1530

Small Hint:

Count the multiples of $3$ or $4,$ then remove those that are also multiples of $5$

Big Hint:

Use inclusion-exclusion with multiples of $3,$ $4,$ $12$ and then $15,$ $20,$ $60$

Solution:

Multiples of $3$ or $4$ up to $2001:$ $\left\lfloor\tfrac{2001}{3}\right\rfloor$ $+\left\lfloor\tfrac{2001}{4}\right\rfloor$ $-\left\lfloor\tfrac{2001}{12}\right\rfloor$ $=667+500-166$ $=1001.$

Among these, remove the multiples of $5:$ there are $133$ multiples of $15$ and $100$ multiples of $20,$ re-adding the $33$ multiples of $60:$ $133+100-33=200.$

So the count is $1001-200=801.$ Thus, the correct answer is B.