2015 AMC 10A Exam Solutions

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1.

What is the value of (201+520)1×5?(2^0-1+5^2-0)^{-1} \times 5?

125-125

120-120

15\dfrac{1}{5}

524\dfrac{5}{24}

2525

Solution(s):

We can evaluate it as follows. (201+520)1×5=111+25×5=525=15\begin{align*} &(2^0-1+5^2-0)^{-1} \times 5 \\&=\dfrac{1}{1 - 1 + 25} \times 5 \\ &= \dfrac{5}{25}\\ &= \dfrac{1}{5} \end{align*}

Thus, C is the correct answer.

2.

A box contains a collection of triangular and square tiles. There are 2525 tiles in the box, containing 8484 edges total. How many square tiles are there in the box?

33

55

77

99

1111

Solution(s):

Let xx be the number of triangular tiles and yy be the number of square tiles. We then have that x+y=25 x + y = 25 and 3x+4y=84. 3x + 4y = 84. Multiplying the first equation by 33 gives us 3x+3y=75 3x + 3y = 75 and subtracting this from the other equation gives us y=9.y = 9.

Thus, D is the correct answer.

3.

Ann made a 33-step staircase using 1818 toothpicks as shown in the figure. How many toothpicks does she need to add to complete a 55-step staircase?

99

1818

2020

2222

2424

Solution(s):

Let us try to find a pattern between the number of toothpicks needed for the staircases.

For a 11-step staircase, we would only need 44 toothpicks (just a square).

For a 22-step staircase, we would need 1010 toothpicks according to the diagram.

Similarly, we would need 1818 toothpicks for a 33-step staircase.

A 22-step staircase needs 104=610 - 4 = 6 more toothpicks than a 11-step staircase. A 33-step staircase needs 1810=818 - 10 = 8 more toothpicks than a 22-step staircase.

Following this pattern, we can see that a 44-step staircase will need 18+10=2818 + 10 = 28 toothpicks, and a 55-step staircase will need 28+12=4028 + 12 = 40 toothpicks.

This means that Ann would need to add 4018=2240 - 18 = 22 more toothpicks.

Thus, D is the correct answer.

4.

Pablo, Sofia, and Mia got some candy eggs at a party. Pablo had three times as many eggs as Sofia, and Sofia had twice as many eggs as Mia. Pablo decides to give some of his eggs to Sofia and Mia so that all three will have the same number of eggs. What fraction of his eggs should Pablo give to Sofia?

112\dfrac{1}{12}

16\dfrac{1}{6}

14\dfrac{1}{4}

13\dfrac{1}{3}

12\dfrac{1}{2}

Solution(s):

Let mm be the number of candy eggs that Mia had. Then SofiaSofia had 2m2m eggs and PabloPablo had 6m6m eggs.

The total number of eggs is then m+2m+6m=9m. m + 2m + 6m = 9m. For all of them to have the same number of eggs, they each must have 9m÷3=3m9m \div 3 = 3m eggs.

Sofia needs 3m2m=m3m - 2m = m more eggs. This means Pablo must give m6m=16\dfrac{m}{6m} = \dfrac{1}{6} of his eggs to Sofia.

Thus, B is the correct answer.

5.

Mr. Patrick teaches math to 1515 students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was 80.80. After he graded Payton's test, the test average became 81.81. What was Payton's score on the test?

8181

8585

9191

9494

9595

Solution(s):

The class's total score with Payton's test is 1480=1120. 14 \cdot 80 = 1120. Including Payton's test, the total score goes up to 1581=1215. 15 \cdot 81 = 1215. This means that Payton got a 12151120=95 1215 - 1120 = 95 on his test.

Thus, E is the correct answer.

6.

The sum of two positive numbers is 55 times their difference. What is the ratio of the larger number to the smaller number?

54\dfrac{5}{4}

32\dfrac{3}{2}

95\dfrac{9}{5}

22

52\dfrac{5}{2}

Solution(s):

Let xx and yy be the two numbers. Then we have that x+y=5(xy). x + y = 5(x - y). Note that we are assuming x>y.x \gt y. This gives us x+y=5x5y x + y = 5x - 5y 6y=4x. 6y = 4x.

Dividing through yields xy=64=32.\dfrac{x}{y} = \dfrac{6}{4} = \dfrac{3}{2}.

Thus, B is the correct answer.

7.

How many terms are in the arithmetic sequence 13,13, 16,16, 19,19, ,\dotsc, 70,70, 73?73?

2020

2121

2424

6060

6161

Solution(s):

Recall that the nn th term of an arithmetic sequence is a+d(n1),a + d(n - 1), where aa is the first term and dd is the common difference.

For us, a=13a = 13 and d=3.d = 3. Plugging these in, we get that 73=13+3(n1)20=n1n=21.\begin{align*} 73 &= 13 + 3(n - 1) \\ 20&= n - 1 \\ n &= 21. \end{align*} Thus, B is the correct answer.

8.

Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be 22 : 11 ?

22

44

55

66

88

Solution(s):

Let pp and cc be Pete's and Claire's current ages respectively.

Then we have that p2=3(c2) p - 2 = 3(c - 2) and p4=4(c4). p - 4 = 4(c - 4). Simplifying both equations gives us p=3c4 p = 3c - 4 and p=4c12. p = 4c - 12. Setting them equal, we have 3c4=4c12 3c - 4 = 4c - 12 c=8. c = 8.

This means that p=384=20. p = 3 \cdot 8 - 4 = 20.

Now, we need to find the number of years (yy) until 20+y=2(8+y), 20 + y = 2(8 + y), which gives us 20+y=16+2y 20 + y = 16 + 2y y=4. y = 4. Thus, B is the correct answer.

9.

Two right circular cylinders have the same volume. The radius of the second cylinder is 10%10\% more than the radius of the first. What is the relationship between the heights of the two cylinders?

The second height is 10%10\% less than the first.

The first height is 10%10\% more than the second.

The second height is 21%21\% less than the first.

The first height is 21%21\% more than the second.

The second height is 80%80\% of the first.

Solution(s):

Let r1r_1 and h1h_1 be the radius and height of the first cylinder and similarly define r2r_2 and h2h_2 for the second cylinder.

We know that r2=1110r1 r_2 = \dfrac{11}{10}r_1 and πr12h1=πr22h2. \pi r_1^2h_1 = \pi r_2^2h_2.

Substituting and simplifying gives us r12h1=121100r12h2, r_1^2h_1 = \dfrac{121}{100}r_1^2h_2, which tells us that h1=121100h2. h_1 = \dfrac{121}{100}h_2.

Thus, D is the correct answer.

10.

How many rearrangements of abcdabcd are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either abab or ba.ba.

00

11

22

33

44

Solution(s):

The initial step would be to choose a letter that could work and build from there.

For instance, if we begin with a,a, the only letters that can be placed next to it are cc or d.d.

From here, we cannot proceed further since the combinations acbdacbd and acdbacdb are not allowed due to the presence of cbcb and cd,cd, respectively.

The same issue arises if we start with d,d, as a bb would have to be placed in the middle and end up being adjacent to either an aa or a c.c.

On the other hand, if we start with a b,b, the next letter would have to be d,d, and after that, an aa and a cc can be placed, making this arrangement possible.

The same methodology holds true for starting with a c.c. This gives us a total of 22 rearrangements.

Thus, C is the correct answer.

11.

The ratio of the length to the width of a rectangle is 44 : 3.3. If the rectangle has diagonal of length d,d, then the area may be expressed as kd2kd^2 for some constant k.k. What is k?k?

27\dfrac{2}{7}

37\dfrac{3}{7}

1225\dfrac{12}{25}

1625\dfrac{16}{25}

34\dfrac{3}{4}

Solution(s):

Let the side lengths be 4x4x and 3x.3x. Then the diagonal has length (4x)2+(3x)2=25x2=5x. \sqrt{(4x)^2 + (3x)^2} = \sqrt{25x^2} = 5x.

The area of the rectangle is 4x3x=12x2. 4x \cdot 3x = 12x^2. Then we get that kd2=12x2 kd^2 = 12x^2 k=12x225x2=1225. k = \dfrac{12x^2}{25x^2} = \dfrac{12}{25}.

Thus, C is the correct answer.

12.

Points (π,a)(\sqrt{\pi}, a) and (π,b)(\sqrt{\pi}, b) are distinct points on the graph of y2+x4=2x2y+1.y^2 + x^4 = 2x^2 y + 1. What is ab?|a-b|?

11

π2\dfrac{\pi}{2}

22

1+π\sqrt{1+\pi}

1+π1 + \sqrt{\pi}

Solution(s):

Plugging in π\sqrt{\pi} into the equation, we get that y2+π2=2πy+1, y^2 + \pi^2 = 2\pi y + 1, which can be rearranged to get y22πy+π2=1(yπ)2=1yπ=±1y=π±1. \begin{gather*} y^2 - 2\pi y + \pi^2 = 1 \\ (y - \pi)^2 = 1 \\ y - \pi = \pm 1 \\ y = \pi \pm 1. \end{gather*} This gives us the values for aa and b.b. We then get that ab=π+1π+1=2. |a - b| = |\pi + 1 - \pi + 1| = 2.

Thus, C is the correct answer.

13.

Claudia has 1212 coins, each of which is a 55-cent coin or a 1010-cent coin. There are exactly 1717 different values that can be obtained as combinations of one or more of his coins. How many 1010-cent coins does Claudia have?

33

44

55

66

77

Solution(s):

Let the number of 55-cent coins be xx and the number of 1010-cent coins be 12x.12 - x.

Then we have that any multiple of 55 between 55 and 5x+10(12x)=1205x 5x + 10(12 - x) = 120 - 5x can be achieved by a combination of coins.

There are 24x24 - x such multiples of 5,5, which means that x=7x = 7 to get 1717 possible different values.

The number of 1010-cent coins is therefore 127=5.12 - 7 = 5.

Thus, C is the correct answer.

14.

The diagram below shows the circular face of a clock with radius 2020 cm and a circular disk with radius 1010 cm externally tangent to the clock face at 1212 o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?

22 o' clock

33 o' clock

44 o' clock

66 o' clock

88 o' clock

Solution(s):

Note that the circumference of the clock is twice the circumference of the disk.

This means that at 3,3, the disk would have traveled half way around, which means that the arrow would be pointing left.

This is because the tip of the arrow is half way around, and that is the point that is tangent to the clock.

Now, we see that in 33 hours the disk rotated 34\dfrac{3}{4} of the way.

To get the final quarter rotation, the disk needs to travel 33=1\dfrac{3}{3} = 1 more hour.

Thus, C is the correct answer.

15.

Consider the set of all fractions xy,\dfrac{x}{y}, where xx and yy are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by 1,1, the value of the fraction is increased by 10%?10\%?

00

11

22

33

infinitely many\text{infinitely many}

Solution(s):

We get that x+1y+1=11x10y. \dfrac{x + 1}{y + 1} = \dfrac{11x}{10y}. Cross-multiplying and simplifying yields xy+11x10y=0. xy + 11x - 10y = 0. We can add 110-110 to add side and then factor to get (x1)(y+11)=110. (x - 1)(y + 11) = -110. We are restricted by x>0x \gt 0 and y>0y \gt 0 to get x10>10x - 10 \gt 10 and y+11>11.y + 11 \gt 11.

The only factor pairs of 110-110 that satisfy the above constraints are (1,110),(2,55), (-1, 110), (-2, 55), and (5,22). (-5, 22).

Solving these get the following (x,y)(x, y) pairs: (9,99),(8,44),(5,11).(9, 99), (8, 44), (5, 11).

Only one of these pairs gives values for xx and yy that are relatively prime.

Thus, B is the correct answer.

16.

If we have that y+4=(x2)2,y+4 = (x-2)^2,x+4=(y2)2, x+4 = (y-2)^2, and xy,x \neq y, what is the value of x2+y2?x^2+y^2?

1010

1515

2020

2525

30\text{30}

Solution(s):

Adding the two equations gives us x2+y24x4y+8 x^2 + y^2 - 4x - 4y + 8 =x+y+8. = x + y + 8. We can rearrange this equation to get x2+y2=5(x+y). x^2 + y^2 = 5(x + y). We can then subtract them to get x2y24x+4y=yx. x^2 - y^2 - 4x + 4y = y - x. Once again rearranging, we can find x2y2=3(xy). x^2 - y^2 = 3(x - y). We have that xy,x \neq y, which means that we can divide both sides by xy.x - y. This gives us x+y=3 x + y = 3 x2+y2=15. x^2 + y^2 = 15.

Thus, B is the correct answer.

17.

A line that passes through the origin intersects both the line x=1x = 1 and the line y=1+33x.y=1+ \dfrac{\sqrt{3}}{3} x. The three lines create an equilateral triangle. What is the perimeter of the triangle?

262\sqrt{6}

2+232 + 2\sqrt{3}

66

3+233 + 2\sqrt{3}

6+336 + \dfrac{\sqrt{3}}{3}

Solution(s):

Since one of the sides of the equilateral triangle is a vertical line, the line of symmetry perpendicular to this side must be horizontal.

This means that the slope of the third side must be opposite the slope of the second side, which would be 33.-\dfrac{\sqrt{3}}{3}.

To find the perimeter, we only need to find the length of one of the sides of the triangle.

We can plug in x=1x = 1 into the two other equations to get the two vertices on the vertical line.

The two yy-values are then 1+33 and 33. 1 + \dfrac{\sqrt{3}}{3} \text{ and } -\dfrac{\sqrt{3}}{3}. The distance between the two is 1+233,1 + \dfrac{2\sqrt{3}}{3}, which makes the perimeter 3(1+233)=3+23. 3 \cdot \left(1 + \dfrac{2\sqrt{3}}{3}\right) = 3 + 2\sqrt{3}. Thus, D is the correct answer.

18.

Hexadecimal (base-16) numbers are written using numeric digits 00 through 99 as well as the letters AA through FF to represent 1010 through 15.15. Among the first 10001000 positive integers, there are nn whose hexadecimal representation contains only numeric digits. What is the sum of the digits of n?n?

1717

1818

1919

2020

2121

Solution(s):

Note that 10001000 converted to hexadecimal is 3E8.3E8. Now we need to count the number of numbers that have only numerical digits in their hexadecimal.

The first digit can be 0,1,20, 1, 2 or 3.3. The second and third digits can be any number from 09.0 - 9. This gives us 41010=400 4 \cdot 10 \cdot 10 = 400 numbers. This, however, includes 0,0, which is not a positive integer so we have to subtract one.

The sum of the digits in 399399 is 21.21.

Thus, E is the correct answer.

19.

The isosceles right triangle ABCABC has right angle at CC and area 12.5.12.5. The rays trisecting ACB\angle ACB intersect ABAB at DD and E.E. What is the area of CDE?\triangle CDE?

523\dfrac{5\sqrt{2}}{3}

503754\dfrac{50\sqrt{3}-75}{4}

1538\dfrac{15\sqrt{3}}{8}

502532\dfrac{50-25\sqrt{3}}{2}

256\dfrac{25}{6}

Solution(s):

Note that ACD=30\angle ACD = 30^{\circ} since it trisects a right angle. This means that we can split ACD\triangle ACD into a 45459045-45-90 right triangle and a 3060=9030-60=90 right triangle.

We have that AF=DFAF = DF since AFD\triangle AFD is isosceles. Using this, we have that CFDF=5hh=31. \dfrac{CF}{DF} = \dfrac{5 - h}{h} = \dfrac{\sqrt{3}}{1}. Cross-multiplying gives us 5h=h3 5 - h = h\sqrt{3} h=51+3. h = \dfrac{5}{1 + \sqrt{3}}.

Rationalizing the denominator gives us h=51+33131 h = \dfrac{5}{1 + \sqrt{3}} \cdot \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}=5352. = \dfrac{5\sqrt{3} - 5}{2}.

The area of ACD\triangle ACD is then 1255352=253254. \dfrac{1}{2} \cdot 5 \cdot \dfrac{5\sqrt{3} - 5}{2} = \dfrac{25\sqrt{3} - 25}{4}.

Note that the area of ABC\triangle ABC is 252.\dfrac{25}{2}. We also have that [ACD]=[BCE].[ACD] = [BCE].

Finally, we get that [CDE]=[ABC]2[ACD] [CDE] = [ABC] - 2 \cdot [ACD] =252253252 = \dfrac{25}{2} - \dfrac{25\sqrt{3} - 25}{2}=502532. = \dfrac{50 - 25\sqrt{3}}{2}.

Thus, D is the correct answer.

20.

A rectangle with positive integer side lengths in cm\text{cm} has area AA cm2\text{cm}^2 and perimeter PP cm.\text{cm}. Which of the following numbers cannot equal A+P?A+P?

100100

102102

104104

106106

108108

Solution(s):

Let xx and yy be the side lengths of the rectangle. Then we have that A=xy and P=2(x+y). A = xy \text{ and } P = 2(x + y).

Adding, we get A+P=xy+2(x+y) A + P = xy + 2(x + y)=(x+2)(y+2)4. = (x + 2)(y + 2) - 4.

This means that A+P4A + P - 4 must be the product of two numbers which are both greater than 2.2.

The only answer choice that cannot be expressed as such is 102.102.

Thus, B is the correct answer.

21.

Tetrahedron ABCDABCD has AB=5,AB=5, AC=3,AC=3, BC=4,BC=4, BD=4,BD=4, AD=3,AD=3, and CD=1252.CD=\tfrac{12}5\sqrt2. What is the volume of the tetrahedron?

323\sqrt2

252\sqrt5

245\dfrac{24}5

333\sqrt3

2452\dfrac{24}5\sqrt2

Solution(s):

We claim that triangles ABCABC and ABDABD are perpendicular to each other.

We can show this be dropping the altitudes from CC to ABAB and from DD to ABAB in each triangle.

Since AC=ADAC = AD and BC=BD,BC = BD, we have that the feet of these altitudes will coincide at point P.P.

Then we have that CP=DP=345=125. CP = DP = \dfrac{3 \cdot 4}{5} = \dfrac{12}{5}. We then have that CD=CP2,CD = CP\sqrt{2}, which shows that CPDCPD is an isosceles right triangle.

This proves the above claim. Finally, the volume of the tetrahedron is 13[ABD]CP=63125=245. \dfrac{1}{3}[ABD] \cdot CP = \dfrac{6}{3} \cdot \dfrac{12}{5} = \dfrac{24}{5}.

Thus, C is the correct answer.

22.

Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?

47256\dfrac{47}{256}

316\dfrac{3}{16}

49256\dfrac{49}{256}

25128\dfrac{25}{128}

51256\dfrac{51}{256}

Solution(s):

We can case on the number of people standing.

Case 1:01: 0 people standing

There is only 11 way for this to happen: everyone sits.

Case 2:12: 1 person stands

There are 88 choices for which person stands.

Case 3:23: 2 people stand

There are (82)=28\binom{8}{2} = 28 ways to choose the pair of people that stand. There are, however, 88 pairs of adjacent people, so we have to subtract those out for a total of 288=2028 - 8 = 20 configurations.

Case 4:34: 3 people stand

There are 88 choices for the first person. This rules out the person standing and their two neighbors, leaving 55 people to choose from.

There are (52)=10\binom{5}{2} = 10 ways to choose the other 22 people, but 44 of these pairs involve adjacent people standing.

Therefore, there are a total of 86=488 \cdot 6 = 48 arrangements for this case. Note, however, that we have divide by 33 since we overcounted which person we choose first.

This case only has 48÷3=1648 \div 3 = 16 configurations then.

Case 5:45: 4 people stand

There are only 22 choices depending on which half of the people stand up.

These cases together contribute 1+8+20+16+2=47 1 + 8 + 20 + 16 + 2 = 47 configurations. There are 28=2562^8 = 256 possibilities for the coin flips, which makes the desired probability 47256.\dfrac{47}{256}.

Thus, A is the correct answer.

23.

The zeroes of the function f(x)=x2ax+2af(x)=x^2-ax+2a are integers. What is the sum of the possible values of a?a?

77

88

1616

1717

1818

Solution(s):

Let the zeroes be rr and s.s. Using Vieta's formulas, we have that a=r+sa = r + s and 2a=rs.2a = rs.

Then we get that rs=2(r+s), rs = 2(r + s), which rearranges to rs2r2s=0 rs - 2r - 2s = 0 rs2r2s+4=4 rs - 2r - 2s + 4 = 4 (r2)(s2)=4. (r - 2)(s - 2) = 4.

The only possible pairs (r2,s2)(r - 2, s - 2) that work are (1,4),(1,4),(4,1),(4,1), (1, 4), (-1, -4), (4, 1), (-4, -1), (2,2),(2,2). (2, 2), (-2, -2). For any of these pairs, we have that a=r2+s2+4. a = r - 2 + s - 2 + 4. We want all the such unique values of a.a. We get that they are 1,0,8,9. -1, 0, 8, 9. The sum of these values is 1+8+9=16.-1 + 8 + 9 = 16.

Thus, C is the correct answer.

24.

For some positive integers p,p, there is a quadrilateral ABCDABCD with positive integer side lengths, perimeter p,p, right angles at BB and C,C, AB=2,AB=2, and CD=AD.CD=AD. How many different values of p<2015p < 2015 are possible?

3030

3131

6161

6262

6363

Solution(s):

Drop the altitude from AA down to CD.\overline{CD}. Then we have that (y2)2+x2=y2, (y - 2)^2 + x^2 = y^2, which simplifies to x2=4(y1). x^2 = 4(y - 1). Since xx and yy are both integers, we have that yy must be one more than a perfect square.

We know that p=2+2y+2y1. p = 2 + 2y + 2\sqrt{y - 1}.

We have that p<2015.p \lt 2015. Guessing and checking tells us that y=312+!y = 31^2 + ! is the max value, whereas y=11+1y = 1^1 + 1 is the minimum value.

This means that there are 3131 values of yy for which all the problem constraints are satisfied.

Thus, B is the correct answer.

25.

Let SS be a square of side length 1.1. Two points are chosen independently at random on the sides of S.S. The probability that the straight-line distance between the points is at least 12\dfrac{1}{2} is abπc,\dfrac{a-b\pi}{c}, where a,a, b,b, and cc are positive integers with gcd(a,b,c)=1.\gcd(a,b,c)=1. What is a+b+c?a+b+c?

5959

6060

6161

6262

6363

Solution(s):

Fix one of the points. Then the probability the other point is on the same side is 14.\dfrac{1}{4}.

The probability that the other point is on an adjacent side is 12\dfrac{1}{2} and 14\dfrac{1}{4} for the opposite side.

Case 1:1: the other point is one the same side

Let the two points be aa and b.b. If we view them as points in the unit square, then we see that the area such that ab>12|a - b| \gt \dfrac{1}{2} forms a triangle with area (12)2=14.\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}.

Case 2:2: the points are on adjacent sides

WLOG, let the two points be on the bottom and left sides. Then the two points are (0,a)(0, a) and (b,0).(b, 0).

The distance between the two points is a2+b2.\sqrt{a^2 + b^2}. We want this to be greater than 12.\dfrac{1}{2}.

When graphed as above, we get that the points fill out the unit square except for a quarter circle of radius 12.\dfrac{1}{2}.

This means that the probability that the distance is greater than 12\dfrac{1}{2} is 114π122=1π16. 1 - \dfrac{1}{4} \cdot \pi \cdot \dfrac{1}{2}^2 = 1 - \dfrac{\pi}{16}.

Case 3:3: the points are on opposite sides

The length will always be greater than 12,\dfrac{1}{2}, which means that the probability is 1.1.

The total probability is therefore 1414+12(1π16)+141 \dfrac{1}{4} \cdot \dfrac{1}{4} + \dfrac{1}{2} \cdot \left(1 - \dfrac{\pi}{16}\right) + \dfrac{1}{4} \cdot 1 =26π32. = \dfrac{26 - \pi}{32}. The desired sum is 26+1+32=59. 26 + 1 + 32 = 59. Thus, A is the correct answer.