2009 AMC 10A Exam Solutions

1.

One can holds $12$ ounces of soda. What is the minimum number of cans needed to provide a gallon $(128$ ounces $)$ of soda?

$7$

$8$

$9$

$10$

$11$

Concepts:divisibility estimation

Difficulty rating: 560

Solution:

Since $\dfrac{128}{12} = 10\dfrac{2}{3},$ ten cans hold only $120$ ounces, which is not enough.

Therefore $11$ cans are needed.

Thus, the correct answer is E.

2.

Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could not be the total value of the four coins, in cents?

$15$

$25$

$35$

$45$

$55$

Concepts:money divisibility bounding to limit cases

Difficulty rating: 940

Solution:

To get a multiple of $5$ cents, the number of pennies must be a multiple of $5.$ With only four coins, that means using no pennies, but then the four coins are each worth at least $5$ cents, for a total of at least $20$ cents.

So $15$ cents cannot be made. The others can: $25 = 10 + 3(5),$ $35 = 3(10) + 5,$ $45 = 25 + 10 + 2(5),$ and $55 = 25 + 3(10).$

Thus, the correct answer is A.

3.

Which of the following is equal to

$1 + \cfrac{1}{1 + \cfrac{1}{1 + 1}}?$

$\dfrac{5}{4}$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$2$

$3$

Concepts:continued fraction fraction

Difficulty rating: 870

Solution:

Working outward, $1 + \cfrac{1}{1 + \cfrac{1}{1 + 1}} = 1 + \cfrac{1}{1 + \dfrac{1}{2}} = 1 + \dfrac{1}{\frac{3}{2}} = 1 + \dfrac{2}{3} = \dfrac{5}{3}.$

Thus, the correct answer is C.

4.

Eric plans to compete in a triathlon. He can average $2$ miles per hour in the $\tfrac14$ -mile swim and $6$ miles per hour in the $3$ -mile run. His goal is to finish the triathlon in $2$ hours. To accomplish his goal what must his average speed, in miles per hour, be for the $15$ -mile bicycle ride?

$\dfrac{120}{11}$

$11$

$\dfrac{56}{5}$

$\dfrac{45}{4}$

$12$

Concepts:distance rate and time fraction

Difficulty rating: 1030

Solution:

The swim takes $\dfrac{1/4}{2} = \dfrac18$ hour and the run takes $\dfrac{3}{6} = \dfrac12$ hour. This leaves $2 - \dfrac18 - \dfrac12 = \dfrac{11}{8}$ hours for the bicycle ride.

His average speed must be $\dfrac{15}{11/8} = \dfrac{120}{11}$ miles per hour.

Thus, the correct answer is A.

5.

What is the sum of the digits of the square of $111{,}111{,}111?$

$18$

$27$

$45$

$63$

$81$

Concepts:digits palindrome pattern recognition

Difficulty rating: 1070

Solution:

The square of the nine-digit repunit is the palindrome $111{,}111{,}111^2 = 12{,}345{,}678{,}987{,}654{,}321.$

Its digits are $1, 2, \ldots, 9, 8, \ldots, 1,$ so the sum is $2(1 + 2 + \cdots + 8) + 9 = 2 \cdot 36 + 9 = 81.$

Thus, the correct answer is E.

6.

A circle of radius $2$ is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's area is shaded?

$\dfrac{1}{2}$

$\dfrac{\pi}{6}$

$\dfrac{2}{\pi}$

$\dfrac{2}{3}$

$\dfrac{3}{\pi}$

Concepts:circle area tangent circles

Difficulty rating: 1020

Solution:

The inscribed circle rests on the diameter and is tangent to the arc, so the semicircle has radius $4.$ Its area is $\dfrac12 \pi (4)^2 = 8\pi.$

The circle's area is $\pi(2)^2 = 4\pi,$ so the shaded area is $8\pi - 4\pi = 4\pi.$

The shaded fraction is $\dfrac{4\pi}{8\pi} = \dfrac12.$

Thus, the correct answer is A.

7.

A carton contains milk that is $2\%$ fat, an amount that is $40\%$ less fat than the amount contained in a carton of whole milk. What is the percentage of fat in whole milk?

$\dfrac{12}{5}$

$3$

$\dfrac{10}{3}$

$38$

$42$

Concepts:percentage linear equation

Difficulty rating: 960

Solution:

Let whole milk be $x\%$ fat. Since $2$ is $40\%$ less than $x,$ we have $0.6x = 2,$ so $x = \dfrac{2}{0.6} = \dfrac{10}{3}.$

Thus, the correct answer is C.

8.

Three generations of the Wen family are going to the movies, two from each generation. The two members of the youngest generation receive a $50\%$ discount as children. The two members of the oldest generation receive a $25\%$ discount as senior citizens. The two members of the middle generation receive no discount. Grandfather Wen, whose senior ticket costs $\$6.00,$ is paying for everyone. How many dollars must he pay?

$34$

$36$

$42$

$46$

$48$

Concepts:percentage work backwards

Difficulty rating: 1060

Solution:

The senior ticket costs $\$6,$ which is $\dfrac34$ of the full price, so a full ticket costs $\dfrac43 \cdot 6 = \$8,$ and a child ticket costs $\dfrac12 \cdot 8 = \$4.$

The total is $2(6 + 8 + 4) = \$36.$

Thus, the correct answer is B.

9.

Positive integers $a,$ $b,$ and $2009,$ with $a \lt b \lt 2009,$ form a geometric sequence with an integer ratio. What is $a?$

$7$

$41$

$49$

$287$

$2009$

Concepts:geometric sequence prime factorization

Difficulty rating: 1240

Solution:

Let the common ratio be $r.$ Then $a r^2 = 2009 = 7^2 \cdot 41.$

Since $r$ must be an integer greater than $1,$ the only possibility is $r = 7,$ giving $a = 41$ and the sequence $41, 287, 2009.$

Thus, the correct answer is B.

10.

Triangle $ABC$ has a right angle at $B.$ Point $D$ is the foot of the altitude from $B,$ $AD = 3,$ and $DC = 4.$ What is the area of $\triangle ABC?$

$4\sqrt{3}$

$7\sqrt{3}$

$21$

$14\sqrt{3}$

$42$

Concepts:right triangle altitude triangle area

Difficulty rating: 1240

Solution:

For the altitude from the right angle to the hypotenuse, $BD^2 = AD \cdot DC = 3 \cdot 4 = 12,$ so $BD = 2\sqrt{3}.$

The hypotenuse is $AC = 3 + 4 = 7,$ so the area is $\dfrac12 \cdot 7 \cdot 2\sqrt3 = 7\sqrt3.$

Thus, the correct answer is B.

11.

One dimension of a cube is increased by $1,$ another is decreased by $1,$ and the third is left unchanged. The volume of the new rectangular solid is $5$ less than that of the cube. What was the volume of the cube?

$8$

$27$

$64$

$125$

$216$

Concepts:difference of squares volume linear equation

Difficulty rating: 1100

Solution:

Let the cube have side length $x.$ The new solid has volume $x(x+1)(x-1) = x^3 - x.$

Setting this equal to $x^3 - 5$ gives $x^3 - x = x^3 - 5,$ so $x = 5.$

The cube's volume is $5^3 = 125.$

Thus, the correct answer is D.

12.

In quadrilateral $ABCD,$ $AB = 5,$ $BC = 17,$ $CD = 5,$ $DA = 9,$ and $BD$ is an integer. What is $BD?$

$11$

$12$

$13$

$14$

$15$

Concepts:triangle inequality bounding to limit cases

Difficulty rating: 1220

Solution:

In $\triangle BCD,$ the triangle inequality gives $5 + BD \gt 17,$ so $BD \gt 12.$

In $\triangle ABD,$ it gives $5 + 9 \gt BD,$ so $BD \lt 14.$

The only integer with $12 \lt BD \lt 14$ is $BD = 13.$

Thus, the correct answer is C.

13.

Suppose that $P = 2^m$ and $Q = 3^n.$ Which of the following is equal to $12^{mn}$ for every pair of integers $(m, n)?$

$P^2 Q$

$P^n Q^m$

$P^n Q^{2m}$

$P^{2m} Q^n$

$P^{2n} Q^m$

Concepts:exponent prime factorization

Difficulty rating: 1280

Solution:

Since $12 = 2^2 \cdot 3,$ $12^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n} \cdot (3^n)^m = P^{2n} Q^m.$

Thus, the correct answer is E.

14.

Four congruent rectangles are placed as shown. The area of the outer square is $4$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?

$3$

$\sqrt{10}$

$2 + \sqrt{2}$

$2\sqrt{3}$

$4$

Concepts:area ratio rectangle linear equation

Difficulty rating: 1340

Solution:

Let each rectangle have shorter side $x$ and longer side $y.$ The outer square has side length $y + x$ and the inner square has side length $y - x.$

Since the area ratio is $4,$ the side ratio is $2,$ so $y + x = 2(y - x),$ which gives $y = 3x.$

The ratio of longer to shorter side is $\dfrac{y}{x} = 3.$

Thus, the correct answer is A.

15.

The figures $F_1,$ $F_2,$ $F_3,$ and $F_4$ shown are the first in a sequence of figures. For $n \ge 3,$ $F_n$ is constructed from $F_{n-1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n-1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}?$

$401$

$485$

$585$

$626$

$761$

Concepts:recursion arithmetic sequence summation

Difficulty rating: 1400

Solution:

Going from $F_{n-1}$ to $F_n,$ the new outside square carries $4(n-1)$ diamonds. Starting from the single diamond of $F_1,$ $F_n = 1 + 4\big(1 + 2 + \cdots + (n-1)\big) = 1 + 4 \cdot \dfrac{(n-1)n}{2} = 1 + 2n(n-1).$

Therefore $F_{20} = 1 + 2 \cdot 20 \cdot 19 = 761.$

Thus, the correct answer is E.

16.

Let $a,$ $b,$ $c,$ and $d$ be real numbers with $|a - b| = 2,$ $|b - c| = 3,$ and $|c - d| = 4.$ What is the sum of all possible values of $|a - d|?$

$9$

$12$

$15$

$18$

$24$

Concepts:absolute value casework

Difficulty rating: 1400

Solution:

Since $a - d = (a-b) + (b-c) + (c-d) = \pm 2 \pm 3 \pm 4,$ the possible absolute values are $2+3+4 = 9,\quad 2+3-4 = 1,\quad 2-3+4 = 3,\quad -2+3+4 = 5.$

Their sum is $9 + 1 + 3 + 5 = 18.$

Thus, the correct answer is D.

17.

Rectangle $ABCD$ has $AB = 4$ and $BC = 3.$ Segment $EF$ is constructed through $B$ so that $EF \perp DB,$ and $A$ and $C$ lie on $DE$ and $DF,$ respectively. What is $EF?$

$9$

$10$

$\dfrac{125}{12}$

$\dfrac{103}{9}$

$12$

Difficulty rating: 1580

Solution:

The diagonal is $DB = \sqrt{4^2 + 3^2} = 5.$

Right triangles $EBA,$ $DBC,$ and $BFC$ are all similar to $\triangle DBA.$ From $\triangle EBA,$ $\dfrac{EB}{AB} = \dfrac{DB}{BC} \implies \dfrac{EB}{4} = \dfrac{5}{3} \implies EB = \dfrac{20}{3}.$

From $\triangle BFC,$ $\dfrac{BF}{BC} = \dfrac{DB}{AB} \implies \dfrac{BF}{3} = \dfrac{5}{4} \implies BF = \dfrac{15}{4}.$

Therefore $EF = EB + BF = \dfrac{20}{3} + \dfrac{15}{4} = \dfrac{125}{12}.$

Thus, the correct answer is C.

18.

At Jefferson Summer Camp, $60\%$ of the children play soccer, $30\%$ of the children swim, and $40\%$ of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?

$30\%$

$40\%$

$49\%$

$51\%$

$70\%$

Concepts:percentage Venn Diagram conditional probability

Difficulty rating: 1460

Solution:

Take $100$ children: $60$ play soccer, and $40\%$ of them, or $24,$ also swim. So $60 - 24 = 36$ soccer players do not swim.

There are $30$ swimmers and $70$ non-swimmers, so the fraction of non-swimmers who play soccer is $\dfrac{36}{70} \approx 0.514 \approx 51\%.$

Thus, the correct answer is D.

19.

Circle $A$ has radius $100.$ Circle $B$ has an integer radius $r \lt 100$ and remains internally tangent to circle $A$ as it rolls once around the circumference of circle $A.$ The two circles have the same points of tangency at the beginning and end of circle $B$ 's trip. How many possible values can $r$ have?

$4$

$8$

$9$

$50$

$90$

Concepts:circumference divisibility factor counting

Difficulty rating: 1630

Solution:

The circumferences are $200\pi$ and $2\pi r,$ so the initial point of tangency returns after $\dfrac{200\pi}{2\pi r} = \dfrac{100}{r}$ rolls.

For this to be an integer greater than $1,$ $r$ must be a divisor of $100$ less than $100:$ namely $1, 2, 4, 5, 10, 20, 25,$ and $50.$ That is $8$ values.

Thus, the correct answer is B.

20.

Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?

$20$

$30$

$55$

$65$

$80$

Concepts:distance rate and time rate

Difficulty rating: 1510

Solution:

Let Lauren's rate be $r$ km/min. Then $r + 3r = 1,$ so $r = \dfrac14.$

In the first $5$ minutes the gap shrinks by $5$ km, leaving $15$ km. Lauren covers this alone at $\dfrac14$ km/min, taking $\dfrac{15}{1/4} = 60$ minutes.

The total time is $5 + 60 = 65$ minutes.

Thus, the correct answer is D.

21.

Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle. In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?

$3 - 2\sqrt{2}$

$2 - \sqrt{2}$

$4(3 - 2\sqrt{2})$

$\dfrac{1}{2}(3 - \sqrt{2})$

$2\sqrt{2} - 2$

Concepts:tangent circles area ratio rationalizing denominator

Difficulty rating: 1690

Solution:

Let each small circle have radius $1.$ Their centers form a square of side $2,$ whose diagonal is $2\sqrt2.$

The large circle's diameter is $2 + 2\sqrt2,$ so its radius is $1 + \sqrt2.$

The desired ratio is $\dfrac{4 \cdot \pi (1)^2}{\pi (1 + \sqrt2)^2} = \dfrac{4}{3 + 2\sqrt2} = 4(3 - 2\sqrt2).$

Thus, the correct answer is C.

22.

Two cubical dice each have removable numbers $1$ through $6.$ The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is $7?$

$\dfrac{1}{9}$

$\dfrac{1}{8}$

$\dfrac{1}{6}$

$\dfrac{2}{11}$

$\dfrac{1}{5}$

Concepts:dice (probability)basic probability symmetry

Difficulty rating: 1820

Solution:

Randomly attaching the tiles and then rolling is equivalent to choosing two of the twelve numbers at random and adding them.

Suppose the first top face shows $N.$ For a sum of $7,$ the second must be $7 - N,$ and there are exactly $2$ tiles equal to $7 - N$ among the remaining $11.$

So the probability is $\dfrac{2}{11}.$

Thus, the correct answer is D.

23.

Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12.$ Diagonals $AC$ and $BD$ intersect at $E,$ $AC = 14,$ and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE?$

$\dfrac{9}{2}$

$\dfrac{50}{11}$

$\dfrac{21}{4}$

$\dfrac{17}{3}$

$6$

Concepts:triangle area parallel lines similarity

Difficulty rating: 1690

Solution:

Since $[AED] = [BEC],$ adding $[CED]$ to both gives $[ACD] = [BCD].$ These share base $CD,$ so $A$ and $B$ are equidistant from line $CD,$ meaning $AB \parallel CD.$

Then $\triangle ABE \sim \triangle CDE$ with ratio $\dfrac{AB}{CD} = \dfrac{9}{12} = \dfrac34,$ so $\dfrac{AE}{EC} = \dfrac34.$

With $AE + EC = AC = 14,$ we get $AE = \dfrac{3}{7} \cdot 14 = 6.$

Thus, the correct answer is E.

24.

Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{4}{7}$

$\dfrac{5}{7}$

$\dfrac{3}{4}$

Concepts:complementary counting cube geometry combinations

Difficulty rating: 1860

Solution:

Three vertices determine a plane that cuts through the interior unless all three lie on a single face.

Each of the $6$ faces gives $\binom{4}{3} = 4$ triples, so $6 \cdot 4 = 24$ triples lie on a face out of $\binom{8}{3} = 56$ total.

The probability of hitting the interior is $1 - \dfrac{24}{56} = \dfrac{4}{7}.$

Thus, the correct answer is C.

25.

For $k \gt 0,$ let $I_k = 10\ldots064,$ where there are $k$ zeros between the $1$ and the $6.$ Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k.$ What is the maximum value of $N(k)?$

$6$

$7$

$8$

$9$

$10$

Concepts:prime factorization sum and difference of cubes casework

Difficulty rating: 2160

Solution:

Write $I_k = 10^{k+2} + 64 = 2^{k+2} \cdot 5^{k+2} + 2^6.$

If $k \lt 4,$ the first term has fewer than $6$ factors of $2,$ so $N(k) = k + 2 \lt 6.$

If $k \gt 4,$ the first term has at least $7$ factors of $2$ while the second has exactly $6,$ so their sum has exactly $6:$ $N(k) = 6.$

If $k = 4,$ then $I_4 = 2^6(5^6 + 1).$ Since $5^6 + 1 = (5^2 + 1)\big((5^2)^2 - 5^2 + 1\big) = 26 \cdot 601,$ it contributes exactly one more factor of $2.$ Thus $N(4) = 7.$

The maximum value is $N(4) = 7.$

Thus, the correct answer is B.