2002 AMC 10A Solutions | LIVE by Po-Shen Loh

1.

The ratio $\dfrac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ is closest to which of the following numbers?

$0.1$

$0.2$

$1$

$5$

$10$

Concepts:exponent estimation

Difficulty rating: 960

Small Hint:

Factor $10^{2000}$ out of the numerator

Big Hint:

The denominator equals $2\cdot 10^{2001}$

Solution:

Factoring gives $\dfrac{10^{2000}(1+100)}{2\cdot 10^{2001}}=\dfrac{101}{20}=5.05,$ which is closest to $5.$

Thus, the correct answer is D.

2.

For the nonzero numbers $a,$ $b,$ and $c,$ define $(a,b,c)=\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}.$ Find $(2,12,9).$

$4$

$5$

$6$

$7$

$8$

Concepts:custom operation fraction

Difficulty rating: 960

Small Hint:

Substitute $a=2,$ $b=12,$ $c=9$

Big Hint:

Add $\dfrac{2}{12}+\dfrac{12}{9}+\dfrac{9}{2}$ over a common denominator

Solution:

$\begin{aligned} (2,12,9) &= \dfrac{2}{12}+\dfrac{12}{9}+\dfrac{9}{2} \\ &= \dfrac{1}{6}+\dfrac{4}{3}+\dfrac{9}{2}. \end{aligned}$ Over a denominator of $6,$ this is $\dfrac{1+8+27}{6}=\dfrac{36}{6}=6.$

Thus, the correct answer is C.

3.

According to the standard convention for exponentiation, $2^{2^{2^{2}}}=2^{\left(2^{\left(2^{2}\right)}\right)}=2^{16}=65{,}536.$ If the order in which the exponentiations are performed is changed, how many other values are possible?

$0$

$1$

$2$

$3$

$4$

Concepts:exponent order of operations systematic listing

Difficulty rating: 1190

Small Hint:

List the five ways to parenthesize a tower of four $2$ ’s

Big Hint:

Every grouping evaluates to either $2^{16}$ or $2^{8}$

Solution:

There are five ways to parenthesize the tower. Three of them, $(2^2)^{\left(2^2\right)},$ $\left(2^{\left(2^2\right)}\right)^2,$ and $\left(\left(2^2\right)^2\right)^2,$ all equal $2^{8}=256.$ The other two both give the standard value $2^{16}=65{,}536.$

So exactly one other value, $256,$ is possible.

Thus, the correct answer is B.

4.

For how many positive integers $m$ does there exist at least one positive integer $n$ such that $m\cdot n\le m+n?$

$4$

$6$

$9$

$12$

infinitely many

Concepts:inequality small cases

Difficulty rating: 980

Small Hint:

Try a very small positive value of $n$

Big Hint:

See whether the resulting inequality places any upper bound on $m$

Solution:

Take $n=1.$ Then $m\cdot 1\le m+1$ becomes $m\le m+1,$ which holds for every positive integer $m.$

So every positive integer $m$ works, giving infinitely many.

Thus, the correct answer is E.

5.

Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.

$\pi$

$1.5\pi$

$2\pi$

$3\pi$

$3.5\pi$

Concepts:tangent circles circle area

Difficulty rating: 1060

Small Hint:

The large radius spans the center circle plus two more radii

Big Hint:

Subtract the seven unit circles from the large circle

Solution:

The center of a surrounding circle is $2$ from the center (two radii), and adding its own radius $1$ gives a large radius of $3.$

The large circle has area $9\pi,$ and the seven unit circles have total area $7\pi,$ so the shaded region is $9\pi-7\pi=2\pi.$

Thus, the correct answer is C.

6.

Cindy was asked by her teacher to subtract $3$ from a certain number and then divide the result by $9.$ Instead, she subtracted $9$ and then divided the result by $3,$ giving an answer of $43.$ What would her answer have been had she worked the problem correctly?

$15$

$34$

$43$

$51$

$138$

Concepts:linear equation work backwards

Difficulty rating: 1120

Small Hint:

Recover the original number from Cindy’s incorrect steps

Big Hint:

Solve $\dfrac{x-9}{3}=43$

Solution:

Let $x$ be the number. Cindy computed $\dfrac{x-9}{3}=43,$ so $x-9=129$ and $x=138.$

The correct computation is $\dfrac{138-3}{9}=\dfrac{135}{9}=15.$

Thus, the correct answer is A.

7.

If an arc of $45^\circ$ on circle $A$ has the same length as an arc of $30^\circ$ on circle $B,$ then the ratio of the area of circle $A$ to the area of circle $B$ is

$\dfrac{4}{9}$

$\dfrac{2}{3}$

$\dfrac{5}{6}$

$\dfrac{3}{2}$

$\dfrac{9}{4}$

Concepts:arc area ratio ratio and proportion

Difficulty rating: 1190

Small Hint:

Arc length is $\dfrac{\theta}{360}\cdot 2\pi r$

Big Hint:

Equal arc lengths give the ratio $\dfrac{r_A}{r_B}$

Solution:

Equal arc lengths give $\dfrac{45}{360}\cdot 2\pi r_A=\dfrac{30}{360}\cdot 2\pi r_B,$ so $45 r_A=30 r_B$ and $\dfrac{r_A}{r_B}=\dfrac{2}{3}.$

The ratio of areas is $\left(\dfrac{r_A}{r_B}\right)^2=\dfrac{4}{9}.$

Thus, the correct answer is A.

8.

Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let $B$ be the total area of the blue triangles, $W$ the total area of the white squares, and $R$ the area of the red square. Which of the following is correct?

$B=W$

$W=R$

$B=R$

$3B=2R$

$2R=W$

Concepts:area decomposition congruence (geometry)

Difficulty rating: 1270

Small Hint:

Cut the whole flag into congruent right triangles

Big Hint:

Count how many of those congruent triangles fill each color

Solution:

Divide the flag into congruent right triangles by drawing the grid lines and diagonals. Counting gives $24$ triangles in the blue region, $24$ in the white region, and $16$ in the red region.

Hence $B=W.$

Thus, the correct answer is A.

9.

Suppose $A,$ $B,$ and $C$ are three numbers for which $1001C-2002A=4004$ and $1001B+3003A=5005.$ The average of the three numbers $A,$ $B,$ and $C$ is

$1$

$3$

$6$

$9$

not uniquely determined

Concepts:system of equations mean

Difficulty rating: 1170

Small Hint:

Add the two equations together

Big Hint:

The sum collapses to a multiple of $A+B+C$

Solution:

Adding the equations, $\begin{gathered} 1001C-2002A+1001B \\ {}+3003A \\ = 1001A+1001B+1001C \\ = 9009. \end{gathered}$

So $A+B+C=9$ and the average is $\dfrac{9}{3}=3.$

Thus, the correct answer is B.

10.

Compute the sum of all the roots of $(2x+3)(x-4)$ $+(2x+3)(x-6)=0.$

$\dfrac{7}{2}$

$4$

$5$

$7$

$13$

Concepts:quadratic factoring

Difficulty rating: 1120

Small Hint:

Factor out the common $(2x+3)$

Big Hint:

Combine $(x-4)+(x-6)$ inside the bracket

Solution:

Factoring, $\begin{gathered} (2x+3)\left[(x-4)+(x-6)\right] \\ = (2x+3)(2x-10) \\ = 0. \end{gathered}$

The roots are $-\dfrac{3}{2}$ and $5,$ which sum to $\dfrac{7}{2}.$

Thus, the correct answer is A.

11.

Jamal wants to store $30$ computer files on floppy disks, each of which has a capacity of $1.44$ megabytes (mb). Three of his files require $0.8$ mb of memory each, $12$ more require $0.7$ mb each, and the remaining $15$ require $0.4$ mb each. No file can be split between floppy disks. What is the minimal number of floppy disks that will hold all the files?

$12$

$13$

$14$

$15$

$16$

Concepts:optimization bounding to limit cases

Difficulty rating: 1420

Small Hint:

A $0.8$ mb file leaves room for only one $0.4$ mb file on its disk

Big Hint:

Account for the wasted space on the three $0.8$ mb disks, then build a packing that reaches the bound

Solution:

The files need $3(0.8)+12(0.7)+15(0.4)$ $=16.8$ mb. On any disk holding a $0.8$ mb file, only one $0.4$ mb file fits alongside it (since $0.8+0.7\gt 1.44$ ), leaving at least $0.24$ mb wasted. Across the three such disks that is at least $0.72$ mb, so the effective demand is at least $16.8+0.72=17.52$ mb, requiring at least $\left\lceil\dfrac{17.52}{1.44}\right\rceil=13$ disks.

This is achievable: $3$ disks each hold one $0.8$ file and one $0.4$ file, $6$ disks each hold two $0.7$ files, and $4$ disks each hold three $0.4$ files.

Thus, the correct answer is B.

12.

Mr. Earl E. Bird leaves his house for work at exactly $8{:}00$ A.M. every morning. When he averages $40$ miles per hour, he arrives at his workplace three minutes late. When he averages $60$ miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?

$45$

$48$

$50$

$55$

$58$

Concepts:distance rate and time linear equation

Difficulty rating: 1410

Small Hint:

Let $t$ be the on-time travel time in hours; three minutes is $0.05$ hours

Big Hint:

The distance is the same both ways: $40(t+0.05)=60(t-0.05)$

Solution:

Let $t$ hours be the on-time travel time. Since $3$ minutes is $0.05$ hours, $40(t+0.05)=60(t-0.05).$ Then $40t+2=60t-3,$ so $t=0.25.$

The distance is $40(0.30)=12$ miles, so the required speed is $\dfrac{12}{0.25}=48$ mph.

Thus, the correct answer is B.

13.

The sides of a triangle have lengths of $15,$ $20,$ and $25.$ Find the length of the shortest altitude.

$6$

$12$

$12.5$

$13$

$15$

Concepts:right triangle altitude triangle area

Difficulty rating: 1280

Small Hint:

Check whether $15,$ $20,$ $25$ is a right triangle

Big Hint:

The shortest altitude is drawn to the longest side

Solution:

Since $15^2+20^2=225+400$ $=625=25^2,$ the triangle is right with legs $15$ and $20,$ and area $\dfrac{1}{2}(15)(20)=150.$

The shortest altitude falls to the longest side $25,$ and equals $\dfrac{2\cdot 150}{25}=12.$

Thus, the correct answer is B.

14.

Both roots of the quadratic equation $x^2-63x+k=0$ are prime numbers. The number of possible values of $k$ is

$0$

$1$

$2$

$4$

more than four

Concepts:Vieta’s Formulas prime parity

Difficulty rating: 1310

Small Hint:

The roots sum to $63$ and multiply to $k$

Big Hint:

An odd sum of two primes forces one of them to be $2$

Solution:

If the roots are primes $p$ and $q,$ then $p+q=63$ and $pq=k.$ Because $63$ is odd, one prime must be $2,$ making the other $61,$ which is prime.

So $k=2\cdot 61=122$ is the only possible value.

Thus, the correct answer is B.

15.

The digits $1,$ $2,$ $3,$ $4,$ $5,$ $6,$ $7,$ and $9$ are used to form four two-digit prime numbers, with each digit used exactly once. What is the sum of these four primes?

$150$

$160$

$170$

$180$

$190$

Concepts:prime digits place value

Difficulty rating: 1390

Small Hint:

A two-digit prime cannot end in $2,$ $4,$ $5,$ or $6$

Big Hint:

Those four digits must therefore be the tens digits

Solution:

A two-digit prime cannot end in $2,$ $4,$ $5,$ or $6,$ so these four are the tens digits and $1,$ $3,$ $7,$ and $9$ are the units digits.

The sum is $10(2+4+5+6)$ $+(1+3+7+9)$ $=170+20=190.$ One valid set is $\{23,47,59,61\}.$

Thus, the correct answer is E.

16.

If $a+1$ $=b+2$ $=c+3$ $=d+4$ $=a+b+c+d+5,$ then $a+b+c+d$ is

$-5$

$-\dfrac{10}{3}$

$-\dfrac{7}{3}$

$\dfrac{5}{3}$

$5$

Concepts:system of equations substitution

Difficulty rating: 1330

Small Hint:

Set every expression equal to a single value $k$

Big Hint:

Write $a,$ $b,$ $c,$ $d$ in terms of $k$ and add them

Solution:

Let the common value be $k.$ Then $a=k-1,$ $b=k-2,$ $c=k-3,$ $d=k-4,$ so $a+b+c+d=4k-10.$

Since $a+b+c+d+5=k,$ we get $4k-10+5=k,$ so $3k=5$ and $k=\dfrac{5}{3}.$ Then $a+b+c+d=k-5$ $=\dfrac{5}{3}-5=-\dfrac{10}{3}.$

Thus, the correct answer is B.

17.

Sarah pours four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then transfers half the coffee from the first cup to the second and, after stirring thoroughly, transfers half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

$\dfrac{2}{5}$

$\dfrac{1}{2}$

Concepts:mixture process simulation

Difficulty rating: 1450

Small Hint:

Track ounces of coffee and cream after each transfer

Big Hint:

The second cup is uniformly mixed before the second transfer

Solution:

After transferring $2$ oz of coffee, cup $1$ has $2$ oz coffee and cup $2$ has $2$ oz coffee plus $4$ oz cream, a total of $6$ oz.

Transferring back half of cup $2$ (that is $3$ oz, consisting of $1$ oz coffee and $2$ oz cream) leaves cup $1$ with $3$ oz coffee and $2$ oz cream. The fraction that is cream is $\dfrac{2}{2+3}=\dfrac{2}{5}.$

Thus, the correct answer is D.

18.

A $3\times 3\times 3$ cube is formed by gluing together $27$ standard cubical dice. (On a standard die, the sum of the numbers on any pair of opposite faces is $7.$ ) The smallest possible sum of all the numbers showing on the surface of the $3\times 3\times 3$ cube is

$60$

$72$

$84$

$90$

$96$

Concepts:cube geometry optimization casework

Difficulty rating: 1540

Small Hint:

Classify the $27$ dice as corner, edge, face-center, and interior dice

Big Hint:

Minimize each die’s showing faces: corners show $1+2+3,$ edges show $1+2,$ face-centers show $1$

Solution:

The $8$ corner dice show $3$ faces each, minimized at $1+2+3=6,$ contributing $8\cdot 6=48.$ The $12$ edge dice show $2$ faces, minimized at $1+2=3,$ contributing $12\cdot 3=36.$

The $6$ face-center dice show $1$ face, minimized at $1,$ contributing $6,$ and the hidden interior die contributes $0.$ The total is $48+36+6=90.$

Thus, the correct answer is D.

19.

Spot’s doghouse has a regular hexagonal base that measures one yard on each side. He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside the doghouse that Spot can reach?

$\dfrac{2}{3}\pi$

$2\pi$

$\dfrac{5}{2}\pi$

$\dfrac{8}{3}\pi$

$3\pi$

Concepts:sector regular polygon circle area

Difficulty rating: 1600

Small Hint:

The interior angle of a regular hexagon is $120^\circ,$ so Spot sweeps $240^\circ$ at radius $2$

Big Hint:

Around each adjacent vertex the leftover $1$ yard of rope sweeps a $60^\circ$ sector

Solution:

At the tether vertex the hexagon blocks its $120^\circ$ interior angle, leaving a $240^\circ$ sector of radius $2:$ area $\dfrac{240}{360}\pi(2)^2=\dfrac{8\pi}{3}.$

Wrapping around each of the two adjacent vertices, $1$ yard of rope remains and sweeps a $60^\circ$ sector: $2\cdot\dfrac{60}{360}\pi(1)^2=\dfrac{\pi}{3}.$ The total is $\dfrac{8\pi}{3}+\dfrac{\pi}{3}=3\pi.$

Thus, the correct answer is E.

20.

Points $A,$ $B,$ $C,$ $D,$ $E,$ and $F$ lie, in that order, on $\overline{AF},$ dividing it into five segments, each of length $1.$ Point $G$ is not on line $AF.$ Point $H$ lies on $\overline{GD},$ and point $J$ lies on $\overline{GF}.$ The line segments $\overline{HC},$ $\overline{JE},$ and $\overline{AG}$ are parallel. Find $\frac{HC}{JE}.$

$\dfrac{5}{4}$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$2$

Concepts:similarity parallel lines

Difficulty rating: 1460

Small Hint:

$\triangle DHC\sim\triangle DGA$ gives $HC$ in terms of $AG$

Big Hint:

$\triangle FJE\sim\triangle FGA$ gives $JE$ in terms of $AG$

Solution:

Since $HC\parallel AG,$ $\triangle DHC\sim\triangle DGA,$ so $\dfrac{HC}{AG}=\dfrac{DC}{DA}=\dfrac{1}{3},$ giving $HC=\dfrac{AG}{3}.$

Since $JE\parallel AG,$ $\triangle FJE\sim\triangle FGA,$ so $\dfrac{JE}{AG}=\dfrac{FE}{FA}=\dfrac{1}{5},$ giving $JE=\dfrac{AG}{5}.$

Therefore $\dfrac{HC}{JE}=\dfrac{\frac{AG}{3}}{\frac{AG}{5}}=\dfrac{5}{3}.$

Thus, the correct answer is D.

21.

The mean, median, unique mode, and range of a collection of eight integers are all equal to $8.$ The largest integer that can be an element of this collection is

$11$

$12$

$13$

$14$

$15$

Concepts:mean median (data)mode extremal argument

Difficulty rating: 1660

Small Hint:

The eight integers sum to $64,$ and a range of $8$ ties the largest and smallest together

Big Hint:

Test the largest candidate values while keeping both the median and unique mode equal to $8$

Solution:

The sum is $8\cdot 8=64.$ The collection $6,$ $6,$ $6,$ $8,$ $8,$ $8,$ $8,$ $14$ has mean, median, unique mode, and range all equal to $8,$ so $14$ is attainable.

If the largest were at least $16,$ the range condition would make the smallest at least $8.$ A mean of $8$ would then force all eight integers to equal $8,$ contradicting the range.

If the largest were $15,$ the range $8$ forces the smallest to be $7,$ so all eight integers are at least $7.$ The other seven then sum to $64-15=49=7\cdot 7,$ forcing every one of them to equal $7.$ But then the median and mode would be $7,$ not $8,$ a contradiction.

Thus, the correct answer is D.

22.

A set of tiles numbered $1$ through $100$ is modified repeatedly by the following operation: remove all tiles numbered with a perfect square, and renumber the remaining tiles consecutively starting with $1.$ How many times must the operation be performed to reduce the number of tiles in the set to one?

$10$

$11$

$18$

$19$

$20$

Concepts:perfect square recursion pattern recognition

Difficulty rating: 1790

Small Hint:

Removing the perfect squares from $n^2$ tiles leaves $n^2-n$ tiles

Big Hint:

Two operations take $n^2$ tiles down to $(n-1)^2$

Solution:

Starting from $n^2$ tiles, one operation removes the $n$ perfect squares, leaving $n^2-n.$ The next operation removes $n-1$ perfect squares, leaving $n^2-n-(n-1)=(n-1)^2.$

So every two operations reduce $n^2$ to $(n-1)^2.$ Going from $10^2=100$ down to $1^2=1$ takes $2(10-1)=18$ operations.

Thus, the correct answer is C.

23.

Points $A,$ $B,$ $C,$ and $D$ lie on a line, in that order, with $AB=CD$ and $BC=12.$ Point $E$ is not on the line, and $BE=CE=10.$ The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC.$ Find $AB.$

$\dfrac{15}{2}$

$8$

$\dfrac{17}{2}$

$9$

$\dfrac{19}{2}$

Concepts:isosceles triangle Pythagorean Theorem perimeter

Difficulty rating: 1660

Small Hint:

Let $M$ be the midpoint of $BC;$ then $EM\perp AD$ and $EM=8$

Big Hint:

With $AB=CD=x$ and $AE=ED=y,$ use the perimeter condition and the Pythagorean theorem

Solution:

Let $M$ be the midpoint of $BC.$ Since $BE=CE,$ $EM\perp BC$ and $EM=\sqrt{10^2-6^2}=8.$ By symmetry $AE=ED;$ write $AB=CD=x$ and $AE=ED=y.$

The perimeter condition gives $2y+(2x+12)$ $=2(10+10+12)=64,$ so $x+y=26.$ Also $y^2=EM^2+(x+6)^2$ $=64+(x+6)^2.$

Substituting $y=26-x,$ $(26-x)^2=64+(x+6)^2,$ which simplifies to $676-52x=100+12x,$ so $64x=576$ and $x=9.$

Thus, the correct answer is D.

24.

Tina randomly selects two distinct numbers from the set $\{1,2,3,4,5\},$ and Sergio randomly selects a number from the set $\{1,2,\ldots,10\}.$ The probability that Sergio’s number is larger than the sum of the two numbers chosen by Tina is

$\dfrac{2}{5}$

$\dfrac{9}{20}$

$\dfrac{1}{2}$

$\dfrac{11}{20}$

$\dfrac{24}{25}$

Concepts:basic probability casework

Difficulty rating: 1900

Small Hint:

List the ten possible sums Tina can make and how often each occurs

Big Hint:

For a sum $s,$ Sergio’s number exceeds it with probability $\dfrac{10-s}{10}$

Solution:

Tina’s $10$ equally likely pairs give sums $3,$ $4,$ $5,$ $5,$ $6,$ $6,$ $7,$ $7,$ $8,$ and $9.$ For a sum $s,$ Sergio’s number exceeds it with probability $\dfrac{10-s}{10}.$

Averaging the winning probability over the ten pairs, the total is $\small\dfrac{7+6+5+5+4+4+3+3+2+1}{100}$ $=\dfrac{40}{100}=\dfrac{2}{5}.$

Thus, the correct answer is A.

25.

In trapezoid $ABCD$ with bases $\overline{AB}$ and $\overline{CD},$ we have $AB=52,$ $BC=12,$ $CD=39,$ and $DA=5.$ The area of $ABCD$ is

$182$

$195$

$210$

$234$

$260$

Concepts:trapezoid similarity Pythagorean Triple

Difficulty rating: 1790

Small Hint:

Extend the legs $DA$ and $CB$ until they meet at a point $P$

Big Hint:

$\triangle PDC\sim\triangle PAB$ with ratio $\dfrac{39}{52}=\dfrac{3}{4}$

Solution:

Extend $DA$ and $CB$ to meet at $P.$ Since $DC\parallel AB,$ $\triangle PDC\sim\triangle PAB$ with ratio $\dfrac{39}{52}=\dfrac{3}{4}.$ From $\dfrac{PD}{PD+5}=\dfrac{3}{4}$ we get $PD=15,$ and similarly $PC=36.$

Then $PD:PC:DC=15:36:39$ $=3\cdot(5:12:13),$ so $\angle P$ is a right angle. The area of $ABCD$ is $\begin{gathered} \dfrac{1}{2}(PA)(PB) \\ {}-\dfrac{1}{2}(PD)(PC) \\ = \dfrac{1}{2}(20)(48) \\ {}-\dfrac{1}{2}(15)(36) \\ = 480-270=210. \end{gathered}$

Thus, the correct answer is C.