2023 AMC 8 — Problems & Online Timed Exam

Timed

40:00

1.

What is the value of the following expression?

$(8 \times 4 + 2) - (8 + 4 \times 2)$

$0$

$6$

$10$

$18$

$24$

Answer: D

Concepts:order of operations

Difficulty rating: 370

Small Hint:

Follow order of operations before subtracting

Big Hint:

Compute the two parenthesized expressions separately

Video solution:

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Written solution:

We can simplify this as follows.

$\begin{align*} (32 + 2) - (8 + 8) &= 34 - 16 \\ &= 18 \end{align*}$

Thus, D is the correct answer.

2.

A square piece of paper is folded twice into four equal quarters, as shown below, then cut along the dashed line. When unfolded, the paper will match which of the following figures?

Answer: E

Concepts:paper folding symmetry

Difficulty rating: 660

Small Hint:

Unfold the cut by reflecting it across each fold line

Big Hint:

The single cut appears four times after both folds are undone

Video solution:

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Written solution:

We can unfold the cut up paper to achieve the following figure.

Thus, E is the correct answer.

3.

Wind chill is a measure of how cold people feel when exposed to wind outside. A good estimate for wind chill can be found using this calculation:

$W=T-0.7\cdot S$

Here $W$ represents the wind chill, $T$ represents air temperature measured in degrees Fahrenheit $(^{\circ}F),$ and $S$ represents wind speed measured in miles per hour (mph).

Suppose the air temperature is $36^{\circ}F$ and the wind speed is $18$ mph. Which of the following is closest to the approximate wind chill?

$18$

$23$

$28$

$32$

$35$

Answer: B

Concepts:substitution decimal estimation

Difficulty rating: 450

Small Hint:

Substitute $T=36$ and $S=18$

Big Hint:

$0.7\cdot18$ is a little more than $12$

Video solution:

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Written solution:

Using the formula, the wind chill is

$\begin{align*} 36 - 0.7 \cdot 18 &= 36 - 12.6 \\ &=23.4. \end{align*}$

The closest choice is $23.$

Thus, B is the correct answer.

4.

The numbers from $1$ to $49$ are arranged in a spiral pattern on a square grid, beginning at the center. The first few numbers have been entered into the grid below. Consider the four numbers that will appear in the shaded squares, on the same diagonal as the number $7.$ How many of these four numbers are prime?

$0$

$1$

$2$

$3$

$4$

Answer: D

Concepts:prime pattern recognition

Difficulty rating: 960

Small Hint:

Continue the spiral until the shaded diagonal is filled

Big Hint:

Check each shaded number for divisibility by small primes

Video solution:

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Written solution:

We can fill in the other numbers to get the complete grid.

From this, we can see that the only prime numbers in the shaded boxes are $19,$ $23,$ and $47.$

Thus, D is the correct answer.

5.

A lake contains $250$ trout, along with a variety of other fish. When a marine biologist catches and releases a sample of $180$ fish from the lake, $30$ are identified as trout. Assume that the ratio of trout to the total number of fish is the same in both the sample and the lake. How many fish are there in the lake?

$1250$

$1500$

$1750$

$1800$

$2000$

Answer: B

Concepts:ratio and proportion

Difficulty rating: 720

Small Hint:

The sample has $30$ trout out of $180$ fish

Big Hint:

Trout are $\frac16$ of all fish in the lake

Video solution:

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Written solution:

Note that $\dfrac{30}{180} = \dfrac{1}{6}.$ This means that a sixth of the fish in the lake are trout, or in other words, the total number of fish is $6$ times the number of trout.

Therefore, there are $6 \cdot 250 = 1500$ fish in the lake.

Thus, B is the correct answer.

6.

The digits $2,$ $0,$ $2,$ and $3$ are placed in the expression below, one digit per box. What is the maximum possible value of the expression?

$0$

$8$

$9$

$16$

$18$

Answer: C

Concepts:exponent optimization

Difficulty rating: 900

Small Hint:

Avoid putting $0$ in a base

Big Hint:

Use $0$ as an exponent so one factor becomes $1$

Video solution:

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Written solution:

Note that we do not want $0$ as a base, since that would make the expression equal to $0.$

This means that $0$ must be an exponent. The number whose exponent is $0$ will automatically evaluate to $1.$

Using one of the two copies of $2$ in the factor $2^0$ leaves $2$ and $3$ for the other factor.

The other factor can then be $2^3$ or $3^2.$ Since $3^2$ is larger, we use $3^2.$

This gives us a final value of $2^0 \times 3^2 = 1 \times 9 = 9.$

Thus, C is the correct answer.

7.

A rectangle, with sides parallel to the $x$ -axis and $y$ -axis, has opposite vertices located at $(15, 3)$ and $(16, 5).$ A line is drawn through points $A(0, 0)$ and $B(3, 1).$ Another line is drawn through points $C(0, 10)$ and $D(2, 9).$ How many points on the rectangle lie on at least one of the two lines?

$0$

$1$

$2$

$3$

$4$

Answer: B

Concepts:coordinate geometry linear equation

Difficulty rating: 1070

Small Hint:

Find the equations of the two lines

Big Hint:

The rectangle has $15\le x\le16$ and $3\le y\le5$

Video solution:

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Written solution:

We can graph the two lines.

From this, we see that only the top left corner of the rectangle intersects either line.

Thus, B is the correct answer.

8.

Lola, Lolo, Tiya, and Tiyo participated in a ping pong tournament. Each player competed against each of the other three players exactly twice. Shown below are the win-loss records for the players. The numbers $1$ and $0$ represent a win or loss, respectively. For example, Lola won five matches and lost the fourth match. What was Tiyo’s win-loss record?

$000101$

$001001$

$010000$

$010101$

$011000$

Answer: A

Concepts:logical deduction parity

Difficulty rating: 1020

Small Hint:

Each match contributes one win and one loss

Big Hint:

In each column of the records, two entries must equal $1$ and two must equal $0$

Video solution:

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Written solution:

Each column represents one round of two matches, so every column must contain exactly two $1$ ’s and two $0$ ’s.

Completing each column to contain two $1$ entries and two $0$ entries gives Tiyo’s record $000101.$

Thus, A is the correct answer.

9.

Malaika is skiing on a mountain. The graph below shows her elevation, in meters, above the base of the mountain as she skis along a trail. In total, how many seconds does she spend at an elevation between $4$ and $7$ meters?

$6$

$8$

$10$

$12$

$14$

Answer: B

Concepts:data and graph interpretation

Difficulty rating: 1020

Small Hint:

Count the time intervals where the graph is between the two heights

Big Hint:

Split the count into the three separate intervals visible on the graph

Video solution:

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Written solution:

The first time that she hits an elevation of $7$ meters is at $2$ seconds.

She then dips below $4$ meters after $4$ seconds. This adds $4 - 2 = 2$ seconds to the total answer.

Malaika then goes above $4$ meters at $6$ seconds. She hits $7$ meters again at $10$ seconds.

This adds $10 - 6 = 4$ more seconds to the total. She finally dips below $7$ meters for the last time at $12$ seconds.

She then falls below $4$ meters at $14$ seconds, finally adding $14 - 12 = 2$ seconds to the total time.

The desired total is $2 + 4 + 2 = 8$ Thus, B is the correct answer.

10.

Harold made a plum pie to take on a picnic. He was able to eat only $\frac{1}{4}$ of the pie, and he left the rest for his friends. A moose came by and ate $\frac{1}{3}$ of what Harold left behind. After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind. How much of the original pie still remained after the porcupine left?

$\dfrac{1}{12}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{5}{12}$

Answer: D

Concepts:fraction

Difficulty rating: 900

Small Hint:

Work with the fraction of the original pie remaining

Big Hint:

After Harold, $\frac34$ remains; after the moose, multiply by $\frac23$

Video solution:

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Written solution:

Harold left $1 - \frac{1}{4} = \frac{3}{4}$ of the pie for his friends.

The moose ate $\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}$ of the pie, leaving $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the pie.

Finally, the porcupine ate $\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$ of the pie. This leaves $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the pie.

Thus, D is the correct answer.

11.

NASA’s Perseverance Rover was launched on July $30,$ $2020.$ After traveling $292{,}526{,}838$ miles, it landed on Mars in Jezero Crater about $6.5$ months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?

$6{,}000$

$12{,}000$

$60{,}000$

$120{,}000$

$600{,}000$

Answer: C

Concepts:distance rate and time estimation unit conversion

Difficulty rating: 1100

Small Hint:

Approximate $6.5$ months as about $200$ days

Big Hint:

Convert miles per day to miles per hour by dividing by $24$

Video solution:

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Written solution:

We can round the distance to $300{,}000{,}000$ miles and approximate $6.5$ months as $30 \times 6.5 = 195 \approx 200$ days.

This gives $300{,}000{,}000 \div 200 = 1{,}500{,}000$ miles per day.

Dividing by $24$ gives about $62{,}500$ miles per hour, which is closest to $60{,}000.$

Thus, C is the correct answer.

12.

The figure below shows a large unshaded circle with a number of smaller unshaded and shaded circles in its interior. What fraction of the interior of the large unshaded circle is shaded?

$\dfrac{1}{4}$

$\dfrac{11}{36}$

$\dfrac{1}{3}$

$\dfrac{19}{36}$

$\dfrac{5}{9}$

Answer: B

Concepts:circle area area ratio

Difficulty rating: 1370

Small Hint:

Circle areas are proportional to the square of the radius

Big Hint:

Count shaded area in units of the smallest circle’s area

Video solution:

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Written solution:

Let each smallest circle have radius $1.$

This means that there are $3$ shaded unit circles, which total to $3 \cdot 1^2 \pi = 3 \pi$ area.

There is also a shaded circle with radius $4$ with two unshaded circles of radius $2$ inside.

This gives us an extra shaded area of $\begin{align*} 4^2 \pi - 2 \cdot 2^2 \pi &= 16 \pi - 8 \pi \\ &= 8 \pi. \end{align*}$

The total shaded area is $8\pi + 3\pi = 11\pi.$ The area of the large unshaded circle is $6^2\pi = 36\pi.$ Therefore, the desired fraction is $\dfrac{11\pi}{36\pi} = \dfrac{11}{36}.$

Thus, B is the correct answer.

13.

Along the route of a bicycle race, $7$ water stations are evenly spaced between the start and finish lines, as shown in the figure below. There are also $2$ repair stations evenly spaced between the start and finish lines. The $3$ rd water station is located $2$ miles after the $1$ st repair station. How long is the race in miles?

$8$

$16$

$24$

$48$

$96$

Answer: D

Concepts:fraction ratio and proportion

Difficulty rating: 1240

Small Hint:

Seven water stations divide the race into $8$ equal gaps

Big Hint:

Two repair stations divide the race into $3$ equal gaps

Video solution:

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Written solution:

The $3$ rd water station is located $\frac{3}{8}$ of the way along the race (the water stations split the race up into $8$ equal spaces).

The first repair station is located $\frac{1}{3}$ of the way along the race. The distance between the stations is $\dfrac{3}{8} - \dfrac{1}{3} = \dfrac{1}{24}$ of the race length. This distance is $2$ miles, so the race is $24 \cdot 2 = 48$ miles long.

Thus, D is the correct answer.

14.

Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $5$ -cent, $10$ -cent, and $25$ -cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $\$7.10$ in postage?

(Note: The amount $\$7.10$ corresponds to $7$ dollars and $10$ cents. One dollar is worth $100$ cents.)

$45$

$46$

$51$

$54$

$55$

Answer: E

Concepts:optimization modular arithmetic

Difficulty rating: 1480

Small Hint:

Use as many low-value stamps as possible

Big Hint:

All $5$ - and $10$ -cent stamps total $300$ cents, leaving a nonmultiple of $25$

Video solution:

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Written solution:

Let $a,$ $b,$ $c$ be the numbers of $5$ -, $10$ -, and $25$ -cent stamps, and let $n=a+b+c.$ Dividing the value equation by $5$ gives $a+2b+5c=142,$ so $n+b+4c=142.$

Suppose $n\ge56.$ Then $b+4c\le86.$ Also $a=142-2b-5c\le20,$ so $2b+5c\ge122.$ Combining these inequalities gives $c\le16,$ but then $2b+5c\le2(20)+5(16)=120,$ a contradiction. Thus at most $55$ stamps can be used.

The bound is attainable with $a=19,$ $b=19,$ $c=17:$ $19(5)+19(10)+17(25)=710.$ Hence the greatest possible number of stamps is $19+19+17=55.$

Thus, E is the correct answer.

15.

Viswam walks half a mile to get to school each day. His route consists of $10$ city blocks of equal length and he takes one minute to walk each block. Today, after walking $5$ blocks, Viswam discovers that he has to make a detour, walking $3$ blocks of equal length instead of $1$ block to reach the next corner. From the time he starts his detour, at what speed, in miles per hour, must Viswam walk in order to arrive at school at his usual time?

$4$

$4.2$

$4.5$

$4.8$

$5$

Answer: B

Concepts:distance rate and time unit conversion

Difficulty rating: 1410

Small Hint:

One block is $\frac1{20}$ mile

Big Hint:

From the detour point, he must walk $7$ blocks in the usual $5$ minutes

Video solution:

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Written solution:

If half a mile is the same as $10$ blocks, then one block is $\dfrac{1}{2} \div 10 = \dfrac{1}{20}$ miles.

Starting from the detour, Viswam has to walk $7 \cdot \dfrac{1}{20} = \dfrac{7}{20}$ miles.

Normally, from this spot Viswam would take $5$ minutes to walk to school. Now he has to travel $\dfrac{7}{20}$ miles in $5$ minutes.

Note that $5$ minutes is $\dfrac{5}{60} = \dfrac{1}{12}$ hours. This means his speed must be $\begin{align*} \dfrac{\frac{7}{20}}{\frac{1}{12}} &= 12 \cdot \dfrac{7}{20} \\&= \dfrac{21}{5} \\ &= 4.2 \end{align*}$ miles per hour.

Thus, B is the correct answer.

16.

The letters $P,$ $Q,$ and $R$ are entered into a $20 \times 20$ table according to the pattern shown below. How many $P$ s, $Q$ s, and $R$ s will appear in the completed table?

$132$ $P$ s, $134$ $Q$ s, $134$ $R$ s

$133$ $P$ s, $133$ $Q$ s, $134$ $R$ s

$133$ $P$ s, $134$ $Q$ s, $133$ $R$ s

$134$ $P$ s, $132$ $Q$ s, $134$ $R$ s

$134$ $P$ s, $133$ $Q$ s, $133$ $R$ s

Answer: C

Concepts:pattern recognition modular arithmetic

Difficulty rating: 1540

Small Hint:

The pattern repeats every $3$ entries along each row and column

Big Hint:

Since $20=3\cdot6+2$ , two letters get one extra appearance in each line

Video solution:

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Written solution:

Since $20 = 3 \cdot 6 + 2,$ the bottom $2$ letters in each column will occur one more time than the third letter.

This means that the third letter in each column will occur $6$ times, whereas the other $2$ will appear $7$ times.

Across the $20$ columns, $P$ is the letter that appears $6$ times in $7$ columns and appears $7$ times in the other $13$ columns. The same is true of $R.$

Thus $P$ and $R$ each appear $7\cdot6+13\cdot7=42+91=133$ times.

$Q$ will therefore appear $\begin{align*} 20 \cdot 20 - 2 \cdot 133 &=400 - 266 \\ &= 134 \end{align*}$ times.

Thus, C is the correct answer.

17.

A regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown in the figure by folding the piece of paper below. Which numbered face will end up to the right of the shaded region $Q?$

$1$

$2$

$3$

$4$

$5$

Answer: A

Concepts:net (3D geometry)polyhedron

Difficulty rating: 1580

Small Hint:

Track which faces fold around the shaded face $Q$

Big Hint:

Faces $2,$ $3,$ $4,$ $5$ form the opposite half from $Q$

Video solution:

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Written solution:

Begin by observing that when folded, the faces labelled $2,$ $3,$ $4,$ and $5$ form the bottom half of the octahedron. As such, the remaining four faces must make up the top half of the octahedron.

From here, we have narrowed down our possibilities to $1,$ $6,$ and $7.$ We can see that $6$ will be the face to the left of the shaded region $Q.$ This also gives us that $7$ is to the left of $6.$

Therefore, we know that the only remaining face, $1,$ must be to the right of the shaded region $Q.$

Thus, A is the correct answer.

18.

Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting position?

$405$

$407$

$409$

$411$

$413$

Answer: D

Concepts:Diophantine Equation modular arithmetic optimization

Difficulty rating: 1740

Small Hint:

Start with $404$ jumps to the right, which reaches $2020$

Big Hint:

The extra jumps must have net displacement $3$

Video solution:

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Written solution:

Let $r$ be the number of right jumps and $l$ be the number of left jumps. We need $5r-3l=2023.$ Reducing modulo $5$ gives $-3l\equiv3\pmod5$ , so $l\equiv4\pmod5$ .

To minimize the total number of jumps, use the smallest possible $l$ , namely $l=4$ . Then $5r-12=2023$ , so $r=407$ .

The fewest number of jumps is $407+4=411$ .

Thus, D is the correct answer.

19.

An equilateral triangle is placed inside a larger equilateral triangle so that the region between them can be divided into three congruent trapezoids, as shown below. The side length of the inner triangle is $\frac{2}{3}$ the side length of the larger triangle. What is the ratio of the area of one trapezoid to the area of the inner triangle?

$1 : 3$

$3 : 8$

$5 : 12$

$7 : 16$

$4 : 9$

Answer: C

Concepts:area ratio similarity

Difficulty rating: 1410

Small Hint:

Areas of similar triangles scale as the square of side lengths

Big Hint:

The inner triangle has $\left(\frac23\right)^2$ of the large triangle’s area

Video solution:

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Written solution:

Since the inner triangle’s side length is $\frac{2}{3}$ the side length of the outer triangle, its area is $\left(\frac{2}{3}\right)^2 = \frac{4}{9}$ the area of the outer triangle.

This means that the three trapezoids are $1 - \frac{4}{9} = \frac{5}{9}$ the area of the outer triangle.

Therefore, one trapezoid is $\frac{5}{9} \div 3 = \frac{5}{27}$ the area of the outer triangle.

This makes the ratio of the areas of one trapezoid and the inner triangle $\dfrac{\frac{5}{27}}{\frac{4}{9}} = \dfrac{5}{27} \cdot \dfrac{9}{4} = \dfrac{5}{12}.$

Thus, C is the correct answer.

20.

Two integers are inserted into the list $3,$ $3,$ $8,$ $11,$ $28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?

$56$

$57$

$58$

$60$

$61$

Answer: D

Concepts:range median (data)mode optimization

Difficulty rating: 1600

Small Hint:

The original range is $25$ , so the new range must be $50$

Big Hint:

To keep the median $8$ , add one number below $8$ and one above $8$

Video solution:

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Written solution:

The original range is $28-3=25,$ so the new range must be $50.$ To keep the median $8,$ one added number $x$ must be less than $8$ and the other, $y,$ must be greater than $8.$

If $x\ge3,$ the minimum remains $3,$ so the new maximum must be $53.$ The value $x=3$ would change the mode, so $x\le7.$ Hence $x+y\le7+53=60.$

If $x<3$ and $y>28,$ then $y-x=50,$ so $x+y=2x+50\le54.$ If $y\le28,$ the maximum remains $28,$ forcing $x=-22$ and giving an even smaller sum. Therefore no case exceeds $60.$

The values $x=7$ and $y=53$ preserve the mode and median and give range $50,$ so the maximum sum is $60.$

Thus, D is the correct answer.

21.

Alina writes the numbers $1,$ $2,$ $\cdots,$ $9$ on separate cards, one number per card. She wishes to divide the cards into $3$ groups of $3$ cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?

$0$

$1$

$2$

$3$

$4$

Answer: C

Concepts:partitions and compositions casework

Difficulty rating: 1690

Small Hint:

Each group must have sum $15$

Big Hint:

Look at the group containing $9$

Video solution:

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Written solution:

The sum of all the numbers is $\dfrac{9 \cdot 10}{2} = 45.$ This means that the sum of each group is $45 \div 3 = 15.$

Consider the group with $9$ in it. The other two numbers must add to $6.$ Therefore, the other cards in this group are $1$ and $5$ or $2$ and $4.$

Case $1:$ One group is $1,$ $5,$ and $9.$

Consider the group with $8$ in it. The other numbers must add to $7.$ The only option is $3$ and $4$ with the remaining cards.

The other group is then $2,$ $6,$ and $7.$ This adds to $15,$ so this case contributes one possibility.

Case $2:$ One group is $2,$ $4,$ and $9.$

Consider the group with $8$ in it. As above, the other numbers have to add to $7.$ The only option is $1$ and $6.$

The final group is $3,$ $5,$ and $7,$ which adds to $15.$ This is another configuration.

We have gone through all the cases, which revealed that there are only $2$ possible groupings.

Thus, C is the correct answer.

22.

In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term in the sequence is $4000.$ What is the first term?

$1$

$2$

$4$

$5$

$10$

Answer: D

Concepts:recursion prime factorization

Difficulty rating: 1790

Small Hint:

Write the first several terms using first term $x$ and second term $y$

Big Hint:

The sixth term is $x^3y^5$

Video solution:

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Written solution:

Let $x$ be the first term and $y$ be the second.

Then we get the following sequence: $x, y, xy, xy^2, x^2y^3, x^3y^5, \cdots$ Thus $x^3y^5 = 4000.$

Factoring $4000,$ gives $4000 = 2^5 \cdot 5^3.$ Because $x$ and $y$ are positive integers, the only fifth powers that divide $4000$ are $1$ and $32.$

If $y = 1,$ then $x^3 = 4000,$ which is impossible because $4000$ is not a perfect cube. Therefore, $y^5 = 32$ and $y = 2.$

This forces $x$ to equal $5.$

Thus, D is the correct answer.

23.

Each square in a $3 \times 3$ grid is randomly filled with one of the $4$ shaded-and-unshaded tiles shown below on the right.

What is the probability that the tiling will contain a large shaded diamond in one of the smaller $2 \times 2$ grids? Below is an example of such a tiling.

$\dfrac{1}{1024}$

$\dfrac{1}{256}$

$\dfrac{1}{64}$

$\dfrac{1}{16}$

$\dfrac{1}{4}$

Answer: C

Concepts:basic probability multiplication principle

Difficulty rating: 1840

Small Hint:

Choose which $2\times2$ grid contains the large diamond

Big Hint:

Once a $2\times2$ grid is chosen, its four tile orientations are forced

Video solution:

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Written solution:

There are $4^9$ possible tilings. There are $4$ possible $2\times2$ grids where a large shaded diamond could appear.

After one of these $2\times2$ grids is chosen, the four tile orientations inside it are forced, and the other $5$ squares can be filled in any way. This gives $4^5$ tilings for each chosen $2\times2$ grid.

Two different $2\times2$ grids cannot both contain a large shaded diamond, because their overlapping squares would require incompatible tile orientations. Therefore the number of favorable tilings is $4\cdot4^5=4^6.$

The desired probability is then $\dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \dfrac{1}{64}.$

Thus, C is the correct answer.

24.

Isosceles triangle $ABC$ has equal side lengths $AB$ and $BC.$ In the figures below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are $11$ and $5$ units, respectively. What is the height $h$ of $\triangle ABC?$

$14.6$

$14.8$

$15$

$15.2$

$15.4$

Answer: A

Concepts:similarity area ratio

Difficulty rating: 1930

Small Hint:

Equal shaded areas give an equation between the two unshaded triangle areas

Big Hint:

Similar triangle areas scale as the square of their heights

Video solution:

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Written solution:

Let $a$ be the area of $\triangle ABC.$ The unshaded triangle in the left figure has height $11$ and is similar to the full triangle, so its area is $a\left(\frac{11}{h}\right)^2.$ Therefore, the shaded area there is $a\left(1-\left(\frac{11}{h}\right)^2\right).$

In the right figure, the shaded triangle has height $h-5$ and is similar to the full triangle, so its area is $a\left(\frac{h-5}{h}\right)^2.$ Equating the shaded areas and canceling $a$ gives $1-\left(\dfrac{11}{h}\right)^2 =\left(\dfrac{h-5}{h}\right)^2.$

Simplifying yields $h^2 - 121 = h^2 - 10h + 25.$ This simplifies to $10h = 146,$

so $h=14.6.$

Thus, A is the correct answer.

25.

Fifteen integers $a_1,$ $a_2,$ $a_3,$ $\cdots,$ $a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that

$1 \leq a_1 \leq 10,$

$13 \leq a_2 \leq 20,$

and

$241 \leq a_{15} \leq 250.$

What is the sum of the digits of $a_{14}?$

$8$

$9$

$10$

$11$

$12$

Answer: A

Concepts:arithmetic sequence bounding to limit cases digits

Difficulty rating: 1950

Small Hint:

Let $d$ be the common difference

Big Hint:

Use the widest possible bounds for $a_1$ and $a_{15}$ to force $d$

Video solution:

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Written solution:

Let $d$ be the common difference. Using the largest possible value $10$ for $a_1$ and the smallest possible value $241$ for $a_{15},$ we have $d \geq \dfrac{241 - 10}{14} = 16.5.$ Since all the numbers are integers, $d$ must be at least $17.$

Using the smallest possible value $1$ for $a_1$ and the largest possible value $250$ for $a_{15},$ we have $d \leq \dfrac{250 - 1}{14} \approx 17.8.$ Since all the numbers are integers, $d$ is at most $17.$ Therefore, $d=17.$

Note that $17 \cdot 14 = 238.$ Since $a_{15}$ is at least $241,$ $a_1$ must be at least $3.$

On the other hand, if $a_1$ were greater than $3,$ then $a_2=a_1+17$ would be greater than $20,$ which is not allowed.

Now we know that $a_1 = 3$ and $d = 17.$ This tells us that $a_{14} = 3 + 13 \cdot 17 = 224.$

Therefore, sum of the digits is $2 + 2 + 4 = 8.$

Thus, A is the correct answer.