2011 AMC 10B Exam Problems

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1.

What is $\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} ?$

$-1$

$\dfrac{5}{36}$

$\dfrac{7}{12}$

$\dfrac{147}{60}$

$\dfrac{43}{3}$

Answer: C

Solution:

Simply solving directly: $\begin{align*} &\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} \\ &= \dfrac{12}9 - \dfrac 9{12} \\ &= \dfrac {16}{12} - \dfrac 9{12} \\ &=\dfrac 7{12}.\end{align*}$

Thus, the correct answer is C .

2.

Josanna's test scores to date are $90, 80, 70, 60,$ and $85.$ Her goal is to raise her test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?

$80$

$82$

$85$

$90$

$95$

Answer: E

Solution:

Her current average is $\dfrac{90+80+70+60+85}5 = 77,$ and the sum of her scores is $385.$ The desired average is then $80,$ so the sum of scores required is $80\cdot 6.$ Therefore, the answer is $80\cdot 6-385 = 95.$

Thus, the correct answer is E .

3.

At a store, when a length is reported as $x$ inches that means it is at least $x - 0.5$ inches and at most $x + 0.5$ inches. Suppose the dimensions of a rectangular tile are reported as $2$ inches by $3$ inches. In square inches, what is the minimum area for the rectangle?

$3.75$

$4.5$

$5$

$6$

$8.75$

Answer: A

Solution:

The smallest possible dimensions are $1.5\times 2.5,$ so the area is $1.5\cdot 2.5=3.75.$

Thus, the correct answer is A .

4.

LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid $A$ dollars and Bernardo had paid $B$ dollars, where $A < B.$ How many dollars must LeRoy give to Bernardo so that they share the costs equally?

$\dfrac{A + B}{2}$

$\dfrac{A - B}{2}$

$\dfrac{B - A}{2}$

$B - A$

$A + B$

Answer: C

Solution:

The amount they each would have to pay is $\dfrac{A+B}2,$ and LeRoy paid $A.$ Thus, he has to pay $\dfrac{A+B}2-A = \dfrac{B-A}2$ more.

Thus, the correct answer is C .

5.

In multiplying two positive integers $a$ and $b,$ Ron reversed the digits of the two-digit number $a.$ His erroneous product was $161.$ What is the correct value of the product of $a$ and $b?$

$116$

$161$

$204$

$214$

$224$

Answer: E

Solution:

The number $161$ is equal to $7\cdot 23.$ There are no other pairs of numbers that multiply to $161$ besides $1\cdot 161,$ so $23$ is the only two digit factor. Thus, $23$ is the number reversed, so he meant to get $32\cdot 7$ which is $224.$

Thus, the correct answer is E .

6.

On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?

$30$

$39$

$48$

$57$

$66$

Answer: A

Solution:

Work backward. Before giving $4$ candies to his sister, Casper had $12$ candies, because he ended the second day with $8$ .

Those $12$ candies were $\dfrac23$ of what he had after the first day, so after the first day he had $18$ candies.

Before giving $2$ candies to his brother, he had $20$ candies. This was $\dfrac23$ of his original amount, so he began with $30$ candies.

Thus, A is the correct answer.

7.

The sum of two angles of a triangle is $\frac{6}{5}$ of a right angle, and one of these two angles is $30^{\circ}$ larger than the other. What is the degree measure of the largest angle in the triangle?

$69$

$72$

$90$

$102$

$108$

Answer: B

Solution:

The two angles add to $\frac 65 \cdot 90 = 108.$ This makes the other angle $180-108=72.$ Then, if the larger of the two angles is $x$ then the smaller of them is $x-30$ so their sum is $2x-30=108,$ making $x=69.$

This means no angle is larger than $72,$ making the largest equal to $72.$

Thus, the correct answer is B .

8.

At a certain beach if it is at least $80^{\circ} F$ and sunny, then the beach will be crowded. On June 10 the beach was not crowded. What can be concluded about the weather conditions on June 10?

The temperature was cooler than $80^{\circ}$ F and it was not sunny.

The temperature was cooler than $80^{\circ}$ F or it was not sunny.

If the temperature was at least $80^{\circ}$ F, then it was sunny.

If the temperature was cooler than $80^{\circ}$ F, then it was sunny.

If the temperature was cooler than $80^{\circ}$ F, then it was not sunny.

Answer: B

Solution:

The statement says that if the weather was at least $80^\circ F$ and sunny, then the beach was crowded.

Because the beach was not crowded, the two conditions could not both have been true. Therefore the temperature was cooler than $80^\circ F$ , or it was not sunny, or both.

Thus, B is the correct answer.

9.

The area of $\triangle EBD$ is one third of the area of the $3$ - $4$ - $5$ triangle $ABC.$ Segment $DE$ is perpendicular to segment $AB.$ What is $BD?$

$\dfrac{4}{3}$

$\sqrt{5}$

$\dfrac{9}{4}$

$\dfrac{4\sqrt{3}}{3}$

$\dfrac{5}{2}$

Answer: D

Solution:

By angle angle similarity, we have $BDE \sim BCA .$

Then, since the ratio of the areas is $\frac 13,$ the ratio of the sidelengths is $\frac{1}{\sqrt 3}.$

As such, $\dfrac{BD}{BC} = \dfrac{BD}4 = \dfrac{1}{\sqrt 3},$ making $BD = \dfrac 4{ \sqrt 3} = \dfrac{4\sqrt{3}}{3} .$

Thus, the correct answer is D .

10.

Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}.$ The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?

$1$

$9$

$10$

$11$

$101$

Answer: B

Solution:

The largest number is $10^{10}.$ The rest of the number have a sum of $S=\sum_{i=0}^9 10^i.$ Then, $10S = \sum_{i=0}^9 10^{i+1},$ making $9S = 10^{10}-1.$ This means that $\dfrac{10^{10}}{9S}$ is close to one, so the ratio between $10^{10}$ and the sum is close to $9.$

Thus, the correct answer is B .

11.

There are $52$ people in a room. What is the largest value of $n$ such that the statement "At least $n$ people in this room have birthdays falling in the same month" is always true?

$2$

$3$

$4$

$5$

$12$

Answer: D

Solution:

It isn't necessarily true for $n\geq 6$ as we could have $5$ people born in the first $4$ months and $4$ people born in the subsequent months.

However, one month must be greater than or equal to $5$ as the average of the number of people born in each month is $\frac{52}{12}$ which is greater than $4,$ and some month must be above average.

Thus, the correct answer is D .

12.

Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?

$\dfrac{\pi}{3}$

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

$\pi$

$\dfrac{4\pi}{3}$

$\dfrac{5\pi}{3}$

Answer: A

Solution:

Let the inner semicircle radius be $r$ . The straight parts of the inside and outside paths have the same total length, so only the semicircular ends change the distance.

The two inner semicircles have total length $2\pi r$ , while the two outer semicircles have total length $2\pi(r+6)$ . The outside path is therefore $12\pi$ meters longer.

Keiko takes $36$ more seconds to walk $12\pi$ more meters, so her speed is $12\pi/36=\dfrac\pi3$ meters per second.

Thus, A is the correct answer.

13.

Two real numbers are selected independently at random from the interval $[-20, 10].$ What is the probability that the product of those numbers is greater than zero?

$\dfrac{1}{9}$

$\dfrac{1}{3}$

$\dfrac{4}{9}$

$\dfrac{5}{9}$

$\dfrac{2}{3}$

Answer: D

Solution:

There is $0$ probability that our number is $0,$ so we need to just find the probability that the product isn't less than $0.$ The number product is less than zero if one of the numbers is less than $0$ and one of them is greater than $0.$

First, there are $2$ ways to choose the designated lower number. Then, the probability that the designated lower number is less than $0$ is $\frac 23$ and the probability that the designated higher number is greater than $0$ is $\frac 13.$

This makes the probability that the product is less than $0$ equal to $2 \cdot \dfrac 23 \cdot \dfrac 13 = \dfrac 49.$ As such, the probability that the product is greater than $0$ equal to $\dfrac 59.$

Thus, the correct answer is D .

14.

A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?

$52$

$58$

$62$

$68$

$70$

Answer: C

Solution:

Let the side lengths be $l,w.$ We wish to find $2(l+w).$ From the Pythagorean Theorem, we get $25 = \sqrt{l^2+w^2}$ $l^2+w^2 = 625$ We also know $lw = 168.$

As such $\begin{align*}l^2+2lw+ w^2 &= (l+w)^2 \\&= 961 \\&= 31^2.\end{align*}$ This makes $l+w = 31,$ and as such, our answer is $31\cdot 2=62.$

Thus, the correct answer is C .

15.

Let $@$ denote the "averaged with" operation: $a @ b = \frac{a+b}{2}.$ Which of the following distributive laws hold for all numbers $x, y,$ and $z?$

I. $x @ (y + z) = (x @ y) + (x @ z)$

II. $x + (y @ z) = (x + y) @ (x + z)$

III. $x @ (y @ z) = (x @ y) @ (x @ z)$

I only

II only

III only

I and III only

II and III only

Answer: E

Solution:

In text 1, the left hand side equals $\dfrac{x+y+z}2$ and the right hand side equals $x+ \dfrac{y+z}2,$ so they aren't equal.

In text 2, the left hand side equals $x+ \dfrac{y+z}2$ and the right hand side equals $x+ \dfrac{y+z}2,$ so they are equal.

In text 3, the left hand side equals $\dfrac{2x+y+z}4$ and the right hand side equals $\dfrac{2x+y+z}4,$ so they are equal.

Thus, the correct answer is E .

16.

A dart board is a regular octagon divided into regions as shown. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?

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

$\dfrac{1}{4}$

$\dfrac{2 - \sqrt{2}}{2}$

$\dfrac{\sqrt{2}}{4}$

$2 - \sqrt{2}$

Answer: A

Solution:

Let the side length be $1.$ Then, the area of the center is $1.$

Then, we must find the area of the octagon. It can be found as a square with $4$ isosceles right triangles taken out. The side length of this square is $1 + 2\cdot \dfrac 1{\sqrt 2} = 1 + \sqrt 2 .$ It has an area of $(1+\sqrt 2)^2 = 3 + 2 \sqrt 2.$

Then, the side length of the right triangles is $\dfrac {1}{\sqrt 2} ,$ making the area of one equal to $\dfrac {\frac {1}{\sqrt 2}^2}{2} = \dfrac 14 .$ This makes them have a total combined area of $1,$ so the area of the octagon is $2+ 2 \sqrt 2.$

Thus, the ratio is $\begin{align*}&\dfrac 1{2+ 2 \sqrt 2} \\&= \dfrac {-2+ 2 \sqrt 2}{(2+ 2 \sqrt 2)(-2+ 2 \sqrt 2)}\\ &=\dfrac {2(\sqrt 2-1)}{4} \\&=\dfrac {\sqrt 2-1}{2}. \end{align*}$

Thus, the correct answer is A .

17.

In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC},$ and $\overline{AB}$ is parallel to $\overline{ED}.$ The angles $AEB$ and $ABE$ are in the ratio $4 : 5.$ What is the degree measure of angle $BCD?$

$120$

$125$

$130$

$135$

$140$

Answer: C

Solution:

Since $EB$ is a diameter, $\angle EAB=90^\circ$ . The ratio $\angle AEB:\angle ABE=4:5$ then gives $\angle AEB=40^\circ$ and $\angle ABE=50^\circ$ .

Because $AB\parallel ED$ , $\angle DEB=50^\circ$ . Since $EB\parallel DC$ , quadrilateral $EBCD$ is an isosceles trapezoid, so $\angle BCD=\angle CDE$ .

Angles $DEB$ and $CDE$ are supplementary, so $\angle CDE=180^\circ-50^\circ=130^\circ$ . Hence $\angle BCD=130^\circ$ .

Thus, C is the correct answer.

18.

Rectangle $ABCD$ has $AB = 6$ and $BC = 3.$ Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD.$ What is the degree measure of $\angle AMD?$

$15$

$30$

$45$

$60$

$75$

Answer: E

Solution:

The angles $\angle AMD$ and $\angle MDC$ are equal since $AB \mid \mid DC.$

As such, $\angle MDC = \angle DMC ,$ making $MDC$ isosceles and $MC = DC = 6.$

As we can see, $\sin (CMB) = \frac 12,$ making $\angle CMB = 30^\circ .$

Therefore, $\angle AMC = 150^\circ .$ Since $\angle AMD$ is half of that, $\angle AMD = 75^\circ .$

Thus, the correct answer is E .

19.

What is the product of all the roots of the equation $\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}?$

$-64$

$-24$

$-9$

$24$

$576$

Answer: A

Solution:

The equation is equal to $\sqrt{5 | x | + 8} = \sqrt{|x|^2 - 16}.$ Solving, we get that: $\begin{align*} 5|x|+8 &= |x|^2-16 \\ |x|^2-5|x|-24&=0 \\ (|x|-8)(|x|+3)&=0 \end{align*}$ This makes $|x|=-3,8,$ making $|x|=8$ the only possible value. Thus, $x =8,-8$ with a product of $-64.$

Thus, the correct answer is A .

20.

Rhombus $ABCD$ has side length $2$ and $\angle B = 120^\circ$ . Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R?$

$\dfrac{\sqrt{3}}{3}$

$\dfrac{\sqrt{3}}{2}$

$\dfrac{2\sqrt{3}}{3}$

$1 + \dfrac{\sqrt{3}}{3}$

$2$

Answer: C

Solution:

The points closer to $B$ than to another vertex are bounded by the perpendicular bisectors of $BA$ , $BC$ , and $BD$ . The bisector of $BD$ is diagonal $AC$ , so the desired region lies in $\triangle ABC$ .

Triangle $ABC$ has area $\dfrac12\cdot2\cdot2\sin120^\circ=\sqrt3$ . The perpendicular bisectors of $BA$ and $BC$ cut off two congruent $30$ - $60$ - $90$ triangles, each with area $\dfrac{\sqrt3}{6}$ .

Therefore the desired area is $\sqrt3-2\cdot\dfrac{\sqrt3}{6}=\dfrac{2\sqrt3}{3}$ .

Thus, C is the correct answer.

21.

Brian writes down four integers $w > x > y > z$ whose sum is $44.$ The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9.$ What is the sum of the possible values for $w?$

$16$

$31$

$48$

$62$

$93$

Answer: B

Solution:

The largest difference is $9$ , so $w-z=9$ . For either middle number $n$ , the two differences $w-n$ and $n-z$ must add to $9$ .

The available pairs of differences that add to $9$ are $3+6$ and $4+5$ , and the remaining difference between the two middle numbers is $1$ .

One possible set is ${w,w-5,w-6,w-9}$ , giving $4w-20=44$ and $w=16$ . The other is ${w,w-3,w-4,w-9}$ , giving $4w-16=44$ and $w=15$ .

The sum of the possible values of $w$ is $16+15=31$ .

Thus, B is the correct answer.

22.

A pyramid has a square base with sides of length $1$ and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?

$5\sqrt{2} - 7$

$7 - 4\sqrt{3}$

$\dfrac{2\sqrt{2}}{27}$

$\dfrac{\sqrt{2}}{9}$

$\dfrac{\sqrt{3}}{9}$

Answer: A

Solution:

Let the cube have side length $x$ . Take a vertical diagonal cross-section of the pyramid through opposite vertices of the square base.

This cross-section is an isosceles right triangle with hypotenuse $\sqrt2$ . The cube appears as a rectangle of height $x$ and width $\sqrt2x$ , leaving two congruent right isosceles triangles of leg $x$ .

Thus $\sqrt2=\sqrt2x+2x$ , so $x=\sqrt2-1$ . The cube volume is $x^3=(\sqrt2-1)^3=5\sqrt2-7$ .

Thus, A is the correct answer.

23.

What is the hundreds digit of $2011^{2011}?$

$1$

$4$

$5$

$6$

$9$

Answer: D

Solution:

Because $2011\equiv11\pmod{1000}$ , it is enough to find $11^{2011}=(10+1)^{2011}$ modulo $1000$ .

All terms with $10^3$ or higher are divisible by $1000$ , so only the first three terms matter: $1+2011\cdot10+\binom{2011}{2}10^2.$

Modulo $1000$ , this is $1+11\cdot10+\dfrac{2011\cdot2010}{2}\cdot100\equiv1+110+500=611$ .

The hundreds digit is therefore $6$ .

Thus, D is the correct answer.

24.

A lattice point in an $xy$ -coordinate system is any point $(x, y)$ where both $x$ and $y$ are integers. The graph of $y = mx +2$ passes through no lattice point with $0 < x \le 100$ for all $m$ such that $\frac{1}{2} < m < a.$ What is the maximum possible value of $a?$

$\dfrac{51}{101}$

$\dfrac{50}{99}$

$\dfrac{51}{100}$

$\dfrac{52}{101}$

$\dfrac{13}{25}$

Answer: B

Solution:

Shift the graph down by $2$ . The problem is equivalent to finding the smallest slope $m\gt\dfrac12$ for which $y=mx$ passes through a lattice point with $0\lt x\le100$ .

For a fixed integer $x$ , the smallest integer $y$ with $y/x\gt1/2$ is $x/2+1$ when $x$ is even, and $(x+1)/2$ when $x$ is odd.

Thus the candidate slopes are $\dfrac12+\dfrac1x$ for even $x$ , minimized at $x=100$ as $\dfrac{51}{100}$ , and $\dfrac12+\dfrac1{2x}$ for odd $x$ , minimized at $x=99$ as $\dfrac{50}{99}$ .

The smaller of these is $\dfrac{50}{99}$ , so every $m$ with $\dfrac12\lt m\lt\dfrac{50}{99}$ avoids such lattice points, and this upper endpoint is best possible.

Thus, B is the correct answer.

25.

Let $T_1$ be a triangle with side lengths $2011, 2012,$ and $2013.$ For $n \ge 1,$ if $T_n = \triangle ABC$ and $D, E,$ and $F$ are the points of tangency of the incircle of $\triangle ABC$ to the sides $AB, BC,$ and $AC,$ respectively, then $T_{n+1}$ is a triangle with side lengths $AD, BE,$ and $CF,$ if it exists. What is the perimeter of the last triangle in the sequence $( T_n )?$

$\dfrac{1509}{8}$

$\dfrac{1509}{32}$

$\dfrac{1509}{64}$

$\dfrac{1509}{128}$

$\dfrac{1509}{256}$

Answer: D

Solution:

For a triangle with side lengths $a=BC$ , $b=CA$ , and $c=AB$ , equal tangents from the same vertex give the next side lengths $\dfrac{b+c-a}{2},\quad \dfrac{a+c-b}{2},\quad \dfrac{a+b-c}{2}.$

If the current side lengths are $s-1,s,s+1$ , then the next side lengths are $\dfrac{s}{2}-1,\dfrac{s}{2},\dfrac{s}{2}+1$ . Thus the same form persists while the middle side halves each time.

For $T_n$ , the middle side is $2012/2^{n-1}$ . A triangle of the form $s-1,s,s+1$ exists exactly when $s>2$ .

The last valid triangle has $2012/2^{n-1}>2$ , but the next one does not. This gives $n=10$ , with middle side $2012/2^9=503/128$ .

The perimeter is $3\cdot\dfrac{503}{128}=\dfrac{1509}{128}$ .

Thus, D is the correct answer.