AMC 10 Daily Practice Round 4

Complete problem set with solutions and individual problem pages

Problem 1 Easy

Define a※b=\frac{a-b}{a+b} for all real numbers a and b. Find the value of 5※(4※3).

  • A.

    \frac{16}{17}

  • B.

    \frac{17}{18}

  • C.

    \frac{18}{19}

  • D.

    \frac{19}{20}

  • E.

    \frac{21}{22}

Answer:B

4※3=\frac{4-3}{4+3}=\frac{1}{7},

5※\frac{1}{7}=\frac{5-\dfrac{1}{7}}{5+\dfrac{1}{7}}=\frac{17}{18}.

∴ 5※（4※3）=\frac{17}{18}.

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Problem 2 Easy

Real numbers x,y, and z satisfy the inequalities 0<x<1,-1<y<0, and 1<z<2. Which of the following numbers is necessarily positive?

  • A.

    y+x^2

  • B.

    y+xz

  • C.

    y+y^2

  • D.

    y+2y^2

  • E.

    xy+z^2

Answer:E

Notice that xy+z^2 must be positive because \left| z^2\right|\gt 1\gt \left| xy\right|. Therefore the answer is \rm E.

The other choices:

\rm (A) As x grows closer to 0, x^2 decreases and thus becomes less than y.

\rm (B) x can be as small as possible (x>0), so xz grows close to 0 as x approaches 0.

\rm (C)  For all -1<y<0, y>y^2, and thus it is always negative.

\rm (D) The same logic as above, but when -\frac 12<y<0 this time.

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Problem 3 Easy

Jason walks from his home to school. If he walks 75 meters per minute, he will be 8 minutes late; if he walks 80 meters per minute, he will be 6 minutes late. In order to arrive on time, how fast should Jason walk in meters per minute?

  • A.

    90

  • B.

    100

  • C.

    110

  • D.

    120

  • E.

    130

Answer:B

The required time is \left( 75\times 8-80\times 6 \right)\div \left( 80-75 \right)=24 minutes. 75\times \left( 24+8 \right)\div 24=100 meters per minute. Alternatively, we can also set up a rational equation to solve the problem.

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Problem 4 Easy

A rectangle A B C D has A B=8 and B C=4. Points P and Q lie on sides \overline{A B} and \overline{B C}, respectively, such that A P=C Q and the area of \triangle B P Q is 6. What is P Q^2?

  • A.

    32

  • B.

    34

  • C.

    36

  • D.

    38

  • E.

    40

Answer:E

Since the area of \triangle B P Q is 6, we get that \frac{B P \cdot B Q}{2}=6. Thus, B P \cdot B Q=12. Let A P=Q C=x. Then B P=4-x and B Q=8-x, so (4-x)(8-x)=12. Expanding and factoring gives (x-2)(x-10)=0, so either x=2 or x=10.

If x=10, then B P=-6 and B Q=-2, which is impossible, so thus x=2. This gives B P=2 and B Q=6. Since A B C D is a rectangle, \angle B=90^{\circ}, so applying the Pythagorean Theorem on \triangle B P Q gives 2^2+6^2=P Q^2. Thus, P Q^2=4+36=40.

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Problem 5 Medium

In the coordinate system, the x and y coordinates of the intersection point between lines y=x-k and y=kx+2 are both integers. How many possible values of k are there?

  • A.

    4

  • B.

    5

  • C.

    6

  • D.

    7

  • E.

    8

Answer:A

Construct the system \begin{cases}y=x-k ①\\ y=kx+2② \\ \end{cases}.

Plug the first equation into the second equation, we have x\left( 1-k \right)=k+2.

It is clear that 1-k\ne 0, so we have x=\frac{k+2}{1-k}=-1+\frac{3}{1-k}.

When 1-k=\pm 3 or \pm 1, x is an integer, and now y is also an integer.

Therefore, k=\{4,2,0,-2\}, four of them.

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Problem 6 Easy

A bicycle has two wheels, while a tricycle has three wheels. There are a 17 such vehicles with 41 wheels. Find the number of bicycles.

  • A.

    10

  • B.

    11

  • C.

    12

  • D.

    13

  • E.

    14

Answer:A

Suppose that all vehicles are tricycles, then there are (3\times17-41)\div(3-2)=10 bicycles. Alternatively, we can also set up system of linear equations to solve the problem.

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Problem 7 Easy

Two squares of side lengths 2 and 3 lie within a third square of side length 5, as shown below. What is the area of the shaded region?

  • A.

    11

  • B.

    11.5

  • C.

    12

  • D.

    12.5

  • E.

    13

Answer:D

Each of the 4 shaded triangles can be paired with a congruent but unshaded triangle, and vice versa. Hence, the shaded region and the combined unshaded regions have the same area, so the shaded region has half the area of the entire square. The requested answer is therefore \frac{5^2}{2}=12.5.

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Problem 8 Easy

Which of the following numbers is a perfect square?

  • A.

    \frac{23!24!}{3}

  • B.

    \frac{24!25!}{3}

  • C.

    \frac{25!26!}{3}

  • D.

    \frac{26!27!}{3}

  • E.

    \frac{27!28!}{3}

Answer:D

Note that for all positive n, we have \dfrac{n!\left( n+1\right)!}{3}=\dfrac{\left( n!\right)^{2}\cdot\left( n+1\right)}{3}=\left( n!\right)^{2}\cdot\dfrac{n+1}{3}.

We must find a  value of n such that \left( n!\right)^{2}\cdot\dfrac{n+1}{3} is a perfect  square. Since(n!)^2 is a perfect square, we must also have \dfrac{n+1}{3} be a perfect square. In order for \dfrac{n+1}{3} to be a perfect square, n+1 must be twice a perfect square. From the answer choices, n+1=27 works, thus, n=26 and our desired answer is (\rm D)\dfrac{26!27!}{3}.

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Problem 9 Easy

Cards labeled from 1 to 9 are distributed to students A, B, and C, three cards each person without repetition. The following are the conversations between the students.

A: The three numbers on my cards form an arithmetic sequence.

B: Me too.

C: Only my cards do not form an arithmetic sequence.

Suppose that everyone is telling the truth, find the minimum of the sum of the numbers in three cards of C.

  • A.

    6

  • B.

    7

  • C.

    8

  • D.

    9

  • E.

    10

Answer:D

As the numbers of A and B form arithmetic sequences, their sums must be multiples of 3. The sum of 9 cards is 45, which is also a multiple of 3. Therefore, the sum of C is also a multiple of 3. The sum of C can not be 6=(1+2+3), so the sum is at least 9. Check and find 9 is the correct answer.

A: (9,8,7)

B: (5,4,3)

C: (1,2,6)

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Problem 10 Medium

Let ABCD be a convex quadrilateral such that \angle DAB=\angle DBC=90{}^\circandDA=AB=6. Let E be the intersection of the diagonals AC and BD. If BE=2\sqrt{2}, find the area of ABCD.

  • A.

    30

  • B.

    36

  • C.

    48

  • D.

    54

  • E.

    60

Answer:D

Construct AH\bot BD.

∵ \angle DAB=90{}^\circ, DA=AB=6,

∴ BD=\sqrt{D{{A}^{2}}+A{{B}^{2}}}=6\sqrt{2}.

∵ AH\bot BD,

∴ AH=BH=DH=\frac{1}{2}BD=3\sqrt{2}, \angle AHE=90{}^\circ,

∵ BE=2\sqrt{2},

∴ EH=BH-BE=3\sqrt{2}-2\sqrt{2}=\sqrt{2},

∵ \angle AHE=\angle CBE=\angle DBC=90{}^\circ, \angle AEH=\angle CEB,

∴ \triangle AHE \sim \triangle CEB,

∴ \frac{AH}{BC}=\frac{EH}{BE}, or \frac{3\sqrt{2}}{BC}=\frac{\sqrt{2}}{2\sqrt{2}}, BC=6\sqrt{2},

∴ {{A}_{\triangle BCD}}=\frac{1}{2}BD\times BC=\frac{1}{2}\times 6\sqrt{2}\times 6\sqrt{2}=36,

{{A}_{\triangle ABD}}=\frac{1}{2}AD\times AB=\frac{1}{2}\times 6\times 6=18,

∴ {{A}_{ABCD}}={{A}_{\triangle BCD}}+{{A}_{\triangle ABD}}=36+18=54.

Therefore, A_{ABCD}=54.

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Problem 11 Easy

A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Determine the number of palindromes between 1000 and 2023.

  • A.

    10

  • B.

    11

  • C.

    12

  • D.

    13

  • E.

    14

Answer:B

Enumerate based on patterns: 1001, 1111, 1221, 1331, 1441, 1551, 1661, 1771, 1881, 1991, 2002.

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Problem 12 Easy

Consider the Fibonacci Sequence 1, 1, 2, 3, 5, 8, \cdots. Starting the third number, each succeeding number is the sum of the previous two numbers. Among the first 2023 numbers in the sequence, how many of them are multiples of 5?

  • A.

    403

  • B.

    404

  • C.

    405

  • D.

    505

  • E.

    506

Answer:B

We can determine the pattern of remainders when each term in the sequence is divided by 5: 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0, 1, 1, 2, 3, \cdots. We find that a remainder of 0 appears once for every 5 numbers if the divisor is 5. As 2023\div 5=404 \ R \ 3, there are 404 such numbers.

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Problem 13 Medium

Consider positive integers a,b,c,d such that ab=cd, which of the following is a possible value of a+b+c+d?

  • A.

    97

  • B.

    101

  • C.

    301

  • D.

    401

  • E.

    None of the above

Answer:C

Since 97, 101, and 401 are both prime and 301=7\times 43.

Let a=mn, b=pq, c=mp, d=nq, m, n, p, q\in {{\mathbb{N}}^{*}},

Then a+b+c+d=mn+pq+mp+nq=(m+q)(n+p)，

So we are good as long as a+b+c+d is not prime.

For example, 301=7\times 43=(1+6)\cdot (1+42).

This is true when a=1, b=252, c=42, d=6.

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Problem 14 Medium

Circles A, B, and C each have radius 1. Circles A and B share one point of tangency. Circle C has a point of tangency with the midpoint of \overline{AB}. One side of rectangle D is tangent to both circles A and B, and its opposite side is tangent to circle C. The other two sides are tangent to circles A and B, respectively. What is the area of the part that is inside the rectangle but not inside the circle?

  • A.

    12-3\pi

  • B.

    16-3\pi

  • C.

    10-2\pi

  • D.

    14-2\pi

  • E.

    10-3\pi

Answer:C

By the principle of inclusion-exclusion, it can be deduced that the desired area is equal to the area of the rectangle D minus the sum of the areas of the three circles A,B, and C, plus the overlapping area between circle C and circles A and B.

Then, we can compute the shaded area as the area of half of C plus the area of the rectangle minus the area of the two sectors created by A and B. This is \dfrac{\pi(1)^{2}}{2}+(2)(1)-2 \cdot \dfrac{\pi(1)^{2}}{4}=2.

The area of the rectangle is 4\times 3=12, so the area is 12-2\pi-2=10-2\pi.

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Problem 15 Easy

In the figure, a semicircle with diameter EF intersects each side of rectangle ABCD at exactly one point. The lengths of segments EF and FC are 20 centimeters and 2 centimeters, respectively. The area of the shaded region is a- b \cdot \pi square centimeters. Find the value of a+b. (Assuming the value of \pi is 3.14.)

  • A.

    326

  • B.

    338

  • C.

    348

  • D.

    444

  • E.

    500

Answer:B

Let EF be a line segment with midpoint O. Draw a perpendicular line OP from O to CD at point P. Connect OM and ON.

Then, OM is perpendicular to AB, and ON is perpendicular to BC. Given that EF has a length of 20\text{ cm}, we can deduce that OE = OF = OM = ON = 10\text{ cm}. Therefore, PF can be calculated as 10 - 2 = 8\text{ cm}.

The length of OP is 6\text{ cm}, and since OP bisects CD, we have DP = PF = 8\text{ cm}. Consequently, AD = MP = 16\text{ cm}.

The length of BC is given as 8 + 8 + 2 = 18\text{ cm}.

Now, we want to find the area of the shaded region, denoted as S_{\text{shaded}}. This can be obtained by subtracting the area of the semicircle from the area of quadrilateral ABCD. The area of quadrilateral ABCD can be calculated as AD \times BC = 16 \times 18. The area of the semicircle can be calculated as \frac{1}{2}\pi \times r^2, where the radius r is \frac{1}{2} \times EF = 10\text{ cm}. Thus, the area of the shaded region is:

S_{\text{shaded}} = 16 \times 18 - \frac{1}{2} \times \pi \times 10^2 = 288 - 50\pi.

Therefore, the answer is 288 + 50 =\boxed{\textbf{(B)} 338}.

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Problem 16 Easy

Let a be a real number. If \left| a \right|=-a, what is the simplified value of \left| a-1 \right|-\left| a-2 \right|?

  • A.

    -3

  • B.

    -1

  • C.

    1

  • D.

    2a-3

  • E.

    3-2a

Answer:B

Given \left| a \right| = -a, we can deduce that a \leqslant 0 (since the absolute value of a number is non-negative, and here it equals the negative value of a).

Now, let's simplify the expression \left| a-1 \right|-\left| a-2 \right|:

\left| a-1 \right| - \left| a-2 \right| = - (a-1) + (a-2) = -1.

Hence, the simplified value of the expression is -1.

The correct answer is \boxed{\text{B}}.

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Problem 17 Easy

Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no children? (2004 AMC 10A Problems, Problem #6)

  • A.

    22

  • B.

    23

  • C.

    24

  • D.

    25

  • E.

    26

Answer:E

Solution 1: Since Bertha has 6 daughters, she has 30-6=24 granddaughters, of which none have daughters. Of Bertha's daughters, \dfrac{24}{6}=4 have daughters, so 6-4=2 do not have daughters. Therefore, of Bertha's daughters and granddaughters, 24+2=26 do not have daughters. \boxed{ (\text{E}) 26}.

Solution 2: Bertha has 30-6=24 granddaughters, none of whom have any daughters. The granddaughters are the children of \dfrac{24}{6}=4 of Bertha's daughters, so the number of women having no daughters is 30-4=\boxed {26}.

Draw a tree diagram and see that the answer can be found in the sum of 6+6 granddaughters, 5+5 daughters, and 4 more daughters. Adding them together gives the answer of \boxed{ (\text{E}) 26}.

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Problem 18 Easy

In the figure, the area of triangle ABC is 1. Connecting the points that divide segments AB and AC into five equal parts, what is the area of the shaded region in the figure?

  • A.

    \dfrac{13}{25}

  • B.

    \dfrac{3}{5}

  • C.

    \dfrac{16}{25}

  • D.

    \dfrac{18}{25}

  • E.

    \dfrac{4}{5}

Answer:B

The areas of the shaded regions in the figure have a ratio of 1:3:5:7:9 from top to bottom. To find the area of the shaded region, we sum up the areas of the three shaded regions from top to bottom, which correspond to the areas with ratios 1:5:9. Then, we divide this sum by the total sum of all the areas (which corresponds to the total sum of the ratios 1:3:5:7:9 in the figure).

The shaded region's area is \frac{1+5+9}{1+3+5+7+9} = \frac{15}{25} = \frac{3}{5}.

Hence, the area of the shaded region is \boxed{\frac{3}{5}}.

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Problem 19 Easy

In rectangle ABCD, AD =1. Point P lies on side \overline{AB}, and segments \overline{DB} and \overline{DP} trisect \angle ADC. What is the perimeter of \triangle BDP?

  • A.

    3+ \frac{ \sqrt{3}}{3}

  • B.

    2+ \frac{ 4\sqrt{3}}{3}

  • C.

    2+ 2\sqrt{2}

  • D.

    \frac{ 3+3\sqrt{5}}{2}

  • E.

    2+ \frac{ 5\sqrt{3}}{3}

Answer:B

Since \angle A D C is trisected, \angle A D P=\angle P D B=\angle B D C=30^{\circ}. Thus, P D=\frac{2 \sqrt{3}}{3}, D B=2, and B P=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2 \sqrt{3}}{3}. Adding them, we get \boxed{\text{(B) }2+\frac{4 \sqrt{3}}{3}}.

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Problem 20 Medium

In a candy distribution scenario, three individuals, A, B, and C, each have a positive integer number of candies. If A gives B 20 candies, B's candy count will be twice the sum of A and C's candy counts. If A gives C 30 candies, C's candy count will be three times the sum of A and B's candy counts. How many candies do A, B, and C have in total?

  • A.

    48

  • B.

    60

  • C.

    80

  • D.

    90

  • E.

    100

Answer:A

Let x, y, and z represent the numbers of candies for A, B, and C, respectively. From the given conditions, we have the following system of equations:

\begin{cases} y+20 = 2(x-20+z) \quad (1) \\ z+30 = 3(x-30+y) \quad (2) \end{cases}

Solving and simplifying, we get:

\begin{cases} 2x+2z-y = 60 \quad (3) \\ 3x+3y-z = 120 \quad (4) \end{cases}

From equation (3), we have y=2x+2z-60. Substituting this into equation (4), we get:

3x+3(2x+2z-60)-z = 120

Hence, 9x+5z=300. Since x and z are positive integers, we have:

9 \leq 9x \leq 300

Therefore, 1 \leq x \leq 33\frac{1}{3}. After checking, we find that x=30 satisfies the equation, and this implies that 5z=30. Thus, z=6 and y=2 \times 30 + 2 \times 6 - 60 = 12.

So, the number of candies for A, B, and C respectively are:

A: 30 candies B: 12 candies C: 6 candies

The total number of candies for A, B, and C is 30+12+6 = 48.

Therefore, the answer is \boxed{\textbf{(A) } 48}.

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Related Paper
AMC 10 Daily Practice - Sequences AMC 10 Daily Practice - Exponent AMC 10 Daily Practice Round 3 AMC 10 Daily Practice Round 2
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