1997 AMC 8 Exam Problems

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Used with permission of the Mathematical Association of America.

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1.

110+9100+91000+710000= \dfrac{1}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + \dfrac{7}{10000} =

0.00260.0026

0.01970.0197

0.19970.1997

0.260.26

1.9971.997

Answer: C
Solution(s):

Converting all the fractions to decimals, we get .1+.09+.009+.0007=.1997. .1 + .09 + .009 + .0007 = .1997.

Thus, C is the correct answer.

2.

Ahn chooses a two-digit integer, subtracts it from 200,200, and doubles the result. What is the largest number Ahn can get?

200200

202202

220220

380380

398398

Answer: D
Solution(s):

To get the largest number, we would want to subtract the smallest number possible from 200.200.

The smallest two-digit number is 10.10. Having Ahn choose this number will give us 2(20010)=2190=380 2 (200 - 10) = 2 \cdot 190 = 380 as the final result.

Thus, D is the correct answer.

3.

Which of the following numbers is the largest?

0.970.97

0.9790.979

0.97090.9709

0.9070.907

0.90890.9089

Answer: B
Solution(s):

We have that all the tenths place digits are the same, so we then look at the hundredths digit.

The largest hundredths digit is 7,7, so we can limit the answer choices to the ones with this value.

The largest thousandths digit is 9,9, which is achieved by .979..979.

Thus, B is the correct answer.

4.

Julie is preparing a speech for her class. Her speech must last between one-half hour and three-quarters of an hour. The ideal rate of speech is 150150 words per minute. If Julie speaks at the ideal rate, which of the following number of words would be an appropriate length for her speech?

22502250

30003000

42004200

43504350

56505650

Answer: E
Solution(s):

One-half hour is 3030 minutes, in which Julie can speak 15030=4500 150 \cdot 30 = 4500 words. Three-quarters of an hour is 4545 minutes, in which Julie can speak 15045=6750 150 \cdot 45 = 6750 words. The only answer choice in between these two values is 5650.5650.

Thus, E is the correct answer.

5.

There are many two-digit multiples of 7,7, but only two of the multiples have a digit sum of 10.10. The sum of these two multiples of 77 is

119119

126126

140140

175175

189189

Answer: A
Solution(s):

Listing out all the two-digit multiples of 7,7, we get 14,21,28,35,42,49,56,63,70, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77,84,91, and 98. 77, 84, 91, \text{ and } 98. We have that 2828 and 9191 are the two multiples whose digits add to 10.10.

The sum of these numbers is 28+91=119.28 + 91 = 119.

Thus, A is the correct answer.

6.

In the number 74982.103574982.1035 the value of the place occupied by the digit 99 is how many times as great as the value of the place occupied by the digit 3?3?

1,0001,000

10,00010,000

100,000100,000

1,000,0001,000,000

10,000,00010,000,000

Answer: C
Solution(s):

Note that each digit has ten times the value of the digit to its right.

The place occupied by 99 is 55 spaces to the right of the place occupied by 3.3.

This means that it is 105=100,000 10^5 = 100,000 times as a great.

Thus, C is the correct answer.

7.

The area of the smallest square that will contain a circle of radius 44 is

88

1616

3232

6464

128128

Answer: D
Solution(s):

We can inscribe the circle inside the square so that it is tangent to the midpoints of each side of the square.

This means that the side length of the square is two times the radius of the circle, making it 24=8.2 \cdot 4 = 8.

Then the area of the square is 82=64. 8^2 = 64.

Thus, D is the correct answer.

8.

Walter gets up at 6:306:30 a.m., catches the school bus at 7:307:30 a.m., has 66 classes that last 5050 minutes each, has 3030 minutes for lunch, and has 22 hours additional time at school. He takes the bus home and arrives at 4:004:00 p.m. How many minutes has he spent on the bus?

3030

6060

7575

9090

120120

Answer: B
Solution(s):

There are 8.58.5 hours between the time Walter catches the school bus and arrives at home.

This is a total of 608.5=510 60 \cdot 8.5 = 510 minutes.

The total time Walter spends at school is 650+30+260= 6 \cdot 50 + 30 + 2 \cdot 60 =300+30+120=450 300 + 30 + 120 = 450 minutes.

This means that Walter spends 510450=60 510 - 450 = 60 minutes on the bus.

Thus, B is the correct answer.

9.

Three students, with different names, line up single file. What is the probability that they are in alphabetical order from front-to-back?

112\dfrac{1}{12}

19\dfrac{1}{9}

16\dfrac{1}{6}

13\dfrac{1}{3}

23\dfrac{2}{3}

Answer: C
Solution(s):

There are 33 options for the person in front. Then, there are 22 options for the person in the middle.

This leaves 11 choice for the person at the end. There are 321=63 \cdot 2 \cdot 1 = 6 ways for the people to line up.

Only one of these lines is correct, and the probability it occurs is 16.\dfrac{1}{6}.

Thus, C is the correct answer.

10.

What fraction of this square region is shaded? Stripes are equal in width, and the figure is drawn to scale.

512\dfrac{5}{12}

12\dfrac{1}{2}

712\dfrac{7}{12}

23\dfrac{2}{3}

56\dfrac{5}{6}

Answer: C
Solution(s):

We can split the square into 62=366^2 = 36 unit squares. The number of black squares is 3+7+11=21. 3 + 7 + 11 = 21.

The fraction of the square that is shaded is 2136=712. \dfrac{21}{36} = \dfrac{7}{12}.

Thus, C is the correct answer.

11.

Let N\boxed{N} mean the number of whole number divisors of N.N. For example, 3=2\boxed{3}=2 because 33 has two divisors, 11 and 3.3. Find the value of 11×20.\boxed{\boxed{11}\times\boxed{20}}.

66

88

1212

1616

2424

Answer: A
Solution(s):

We know that 1111 is prime, which means that it only has 22 divisors.

The prime factorization of 2020 is 225. 2^2 \cdot 5. Recall that the number of divisors a number has is the product of all the exponents plus one in the prime factorization.

Here, that product would be (2+1)(1+1)=32=6. (2 + 1) (1 + 1) = 3 \cdot 2 = 6.

Then 26=12.2 \cdot 6 = 12. We have the prime factorization of 1212 is 223. 2^2 \cdot 3. This also has (2+1)(1+1)=32=6 (2 + 1) (1 + 1) = 3 \cdot 2 = 6 divisors.

Thus, A is the correct answer.

12.

1+2=180\angle 1 + \angle 2 = 180^\circ 3=4\angle 3 = \angle 4 Find 4\angle 4

2020^\circ

2525^\circ

3030^\circ

3535^\circ

4040^\circ

Answer: D
Solution(s):

We have that 1=1807040=70 \angle 1 = 180^{\circ} - 70^{\circ} - 40^{\circ} = 70^{\circ} using the fact that the interior angles of a triangle add to 180.180^{\circ}.

This tells us that 2=1801=110 \angle 2 = 180^{\circ} - \angle 1 = 110^{\circ} since 1\angle 1 and 2\angle 2 are supplementary.

Finally, 3+4+110=180 \angle 3 + \angle 4 + 110^{\circ} = 180^{\circ} \Rightarrow24=704=35. 2 \angle 4 = 70^{\circ} \Rightarrow \angle 4 = 35^{\circ}.

Thus, D is the correct answer.

13.

Three bags of jelly beans contain 26,28,26, 28, and 3030 beans. The ratios of yellow beans to all beans in each of these bags are 50%,50\%, 25%,25\%, and 20%,20\%, respectively. All three bags of candy are dumped into one bowl. Which of the following is closest to the ratio of yellow jelly beans to all beans in the bowl?

31%31\%

32%32\%

33%33\%

35%35\%

95%95\%

Answer: A
Solution(s):

There are 26.5=26÷2=13 26 \cdot .5 = 26 \div 2 = 13 yellow jelly beans in the first bag, 28.25=28÷4=7 28 \cdot .25 = 28 \div 4 = 7 yellow jelly beans in the second bag, and 30.2=30÷5=6 30 \cdot .2 = 30 \div 5 = 6 yellow jelly beans in the third bag.

The total number of yellow jelly beans is 13+7+6=26 13 + 7 + 6 = 26 and the total number of jelly beans is 26+28+30=84. 26 + 28 + 30 = 84. The ratio of yellow jelly beans to all the beans is 2684100%=1342100% \dfrac{26}{84} \cdot 100 \% = \dfrac{13}{42} \cdot 100 \%30.9%. \approx 30.9 \%.

Thus, A is the correct answer.

14.

There is a set of five positive integers whose average (mean) is 5,5, whose median is 5,5, and whose only mode is 8.8. What is the difference between the largest and smallest integers in the set?

33

55

66

77

88

Answer: D
Solution(s):

The sum of all the numbers in the list is 55=25.5 \cdot 5 = 25.

The only mode is 8,8, which means that there are guaranteed two 88 s.

The sum of the numbers is up to 28+5=16+5=21. 2 \cdot 8 + 5 = 16 + 5 = 21.

The other two numbers must then add to 4.4. They are both positive and not equal, leaving the only possible two numbers as 1 and 3. 1 \text{ and } 3.

The desired difference is then 81=7. 8 - 1 = 7.

Thus, D is the correct answer.

15.

Each side of the large square in the figure is trisected (divided into three equal parts). The corners of an inscribed square are at these trisection points, as shown. The ratio of the area of the inscribed square to the area of the large square is

33\dfrac{\sqrt{3}}{3}

59\dfrac{5}{9}

23\dfrac{2}{3}

53\dfrac{\sqrt{5}}{3}

79\dfrac{7}{9}

Answer: B
Solution(s):

Let 3x3x be the side length of the large square. Then we can find the side length of the inner square via (2x)2+x2=5x2=x5 \sqrt{(2x)^2 + x^2} = \sqrt{5x^2} = x\sqrt{5} from the Pythagorean Theorem.

The area of the larger square is (3x)2=9x2 (3x)^2 = 9x^2 and that of the inner square is (x5)2=5x2. (x\sqrt{5})^2 = 5x^2.

The ratio of the areas is then 59.\dfrac{5}{9}.

Thus, B is the correct answer.

16.

Penni Precisely buys $ 100 worth of stock in each of three companies: Alabama Almonds, Boston Beans, and California Cauliflower. After one year, AA was up 20%,20 \%, BB was down 25%,25 \%, and CC was unchanged. For the second year, AA was down 20%20 \% from the previous year, BB was up 25%25 \% from the previous year, and CC was unchanged. If A,B,A, B, and CC are the final values of the stock, then

A=B=CA = B = C

A=B<CA = B \lt C

C<B=AC \lt B = A

A<B<CA \lt B \lt C

B<A<CB \lt A \lt C

Answer: E
Solution(s):

After the first year, AA's stock's worth goes up to $ 100 \cdot 1.2 = $ 120 and BB's goes down to $ 100 \cdot .75 = $ 75.

After the second year, AA's stock is worth $ 120 \cdot .8 = $ 96 and BB's is worth $ 75 \cdot 1.25 = $ 93.75.

CC's stock worth remains the same, so the ordering of the stock worths is now B<A<C. B \lt A \lt C.

Thus, E is the correct answer.

17.

A cube has eight vertices (corners) and twelve edges. A segment, such as x,x, which joins two vertices not joined by an edge is called a diagonal. Segment yy is also a diagonal. How many diagonals does a cube have?

66

88

1212

1414

1616

Answer: E
Solution(s):

Each face has two diagonals connecting each of the two pairs of opposite vertices.

Also, for each vertex, there is one corresponding vertex that lies opposite it on the cube.

There are then 8÷2=48 \div 2 = 4 interior space diagonals in the cube.

The total number of diagonals is then 62+4=12+4=16. 6 \cdot 2 + 4 = 12 + 4 = 16.

Thus, E is the correct answer.

18.

At the grocery store last week, small boxes of facial tissue were priced at 44 boxes for $5. This week they are on sale at 55 boxes for $4. The percent decrease in the price per box during the sale was closest to

30%30\%

35%35\%

40%40\%

45%45\%

65%65\%

Answer: B
Solution(s):

Originally, each box is worth $ 5 \div 4 = $ 1.25.

Now, each box is worth $ 4 \div 5 = $ .8.

The percent decrease is then 1.25.81.25100%=45125100% \dfrac{1.25 - .8}{1.25} \cdot 100 \% = \dfrac{45}{125} \cdot 100 \% =925100%=36%.= \dfrac{9}{25} \cdot 100 \% = 36 \%.

Thus, B is the correct answer.

19.

If the product 32435465ab=9,\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \ldots\cdot \dfrac{a}{b} = 9, what is the sum of aa and b?b?

1111

1313

1717

3535

3737

Answer: D
Solution(s):

Note that the numerator of each fraction cancels with the denominator of the fraction to its right.

We can then cancel out all these terms to get a final equation of a2=9a=18b=17. \dfrac{a}{2} = 9 \Rightarrow a = 18 \Rightarrow b = 17.

The desired sum is then 18+17=35. 18 + 17 = 35.

Thus, D is the correct answer.

20.

A pair of 88-sided dice have sides numbered 11 through 8.8. Each side has the same probability (chance) of landing face up. The probability that the product of the two numbers that land face-up exceeds 3636 is

532\dfrac{5}{32}

1164\dfrac{11}{64}

316\dfrac{3}{16}

14\dfrac{1}{4}

12\dfrac{1}{2}

Answer: A
Solution(s):

We can case on the value of the first die. If its value is 14,1 - 4, then it is impossible for the product be greater than 36.36.

If it is a 5,5, then the other dice has to roll an 8,8, otherwise the product is less than 36.36.

If the first die is a 66 or 7,7, then other die has to be at least a 6,6, giving 23=62 \cdot 3 = 6 possibilities.

Finally, if the first roll is an 8,8, the other die must roll at least a 5,5, giving us 33 more possibilities.

The total number of working pairs is 1+6+3=10, 1 + 6 + 3 = 10, and the total number of pairs is 82=64.8^2 = 64. The desired probability is then 1064=532. \dfrac{10}{64} = \dfrac{5}{32}.

Thus, A is the correct answer.

21.

Each corner cube is removed from this 3 cm×3 cm×3 cm3\text{ cm}\times 3\text{ cm}\times 3\text{ cm} cube. The surface area of the remaining figure is

19 sq.cm19\text{ sq.cm}

24 sq.cm24\text{ sq.cm}

30 sq.cm30\text{ sq.cm}

54 sq.cm54\text{ sq.cm}

72 sq.cm72\text{ sq.cm}

Answer: D
Solution(s):

Note that there is one unit cube between any pair of corner cubes, so the removal of each does not affect the others.

When we remove a corner, we are losing three unit squares. We, however, gain these back from the three faces that get uncovered.

This means that removing a corner cube does not change the surface area. The surface area of the original cube is 632=69=54 6 \cdot 3^2 = 6 \cdot 9 = 54 square centimeters.

Thus, D is the correct answer.

22.

A two-inch cube (2×2×2)(2\times 2\times 2) of silver weighs 33 pounds and is worth $ 200. How much is a three-inch cube of silver worth?

$300

$375

$450

$560

$675

Answer: E
Solution(s):

The two-inch cube consists of 23=82^3 = 8 unit cubes. Each of these unit cubes is worth 200÷8=25 200 \div 8 = 25 dollars. To form a three-inch cube, you need 33=273^3 = 27 unit cubes. This means that it is worth 2527=675 25 \cdot 27 = 675 dollars.

Thus, E is the correct answer.

23.

There are positive integers that have these properties:

I. the sum of the squares of their digits is 50,50, and

II. each digit is larger than the one to its left.

The product of the digits of the largest integer with both properties is

77

2525

3636

4848

6060

Answer: C
Solution(s):

Note that if the number has 55 digits, then the sum of the squares of the digits is at least 12+22+32+42+52=55, 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55, which violates the first condition.

This means that we should aim to find the largest four digit number that satisfies the properties.

Let the four digit number be abcdabcd where 0<a<b<c<d. 0 \lt a \lt b \lt c \lt d. We must have that d<8d \lt 8 since otherwise d8d264>50. d \geq 8 \Rightarrow d^2 \geq 64 \gt 50.

If d=7,d = 7, then a2+b2+c212+22+32=14, a^2 + b^2 + c^2 \geq 1^2 + 2^2 + 3^2 = 14, which means that the sum of the digits is greater than 50.50.

Then if d=6,d = 6, we have that a=1,a = 1, b=2,b = 2, and c=3c = 3 work. If d=5,d = 5, then a2+b2+c2=25. a^2 + b^2 + c^2 = 25. The only option is (0,4,5)(0, 4, 5) which causes us to end up with a 33-digit number.

If d=4,d = 4, then a2+b2+c2+d2=14, a^2 + b^2 + c^2 + d^2 = 14, which does not satisfy the first condition. The only viable option is then 1236.1236. The product of the digits is then 1236=36. 1 \cdot 2 \cdot 3 \cdot 6 = 36.

Thus, C is the correct answer.

24.

Diameter ACEACE is divided at CC in the ratio 2:3.2:3. The two semicircles, ABCABC and CDE,CDE, divide the circular region into an upper (shaded) region and a lower region. The ratio of the area of the upper region to that of the lower region is

2:32:3

1:11:1

3:23:2

9:49:4

5:25:2

Answer: C
Solution(s):

WLOG, let AE=10.AE = 10. Then AC=2510=4 AC = \dfrac{2}{5} \cdot 10 = 4 and CE=3510=6. CE = \dfrac{3}{5} \cdot 10 = 6. Then the area of the semicircle ABCABC is 12π(4÷2)2=2π. \dfrac{1}{2} \cdot \pi (4 \div 2)^2 = 2 \pi.

We also have that the area of semicircle CDECDE is 12π(6÷2)2=92π. \dfrac{1}{2} \cdot \pi (6 \div 2)^2 = \dfrac{9}{2}\pi.

Then the area of the upper shaded region is 12(10÷2)22π+92π=15π. \dfrac{1}{2} \cdot (10 \div 2)^2 - 2 \pi + \dfrac{9}{2}\pi = 15 \pi.

Subtracting this from the area of the total circle gives us that the area of the lower region is π5215π=10π. \pi 5^2 - 15 \pi = 10\pi.

The desired ratio is then 15π10π=32=3:2. \dfrac{15\pi}{10\pi} = \dfrac{3}{2} = 3:2.

Thus, C is the correct answer.

25.

All of the even numbers from 22 to 9898 inclusive, excluding those ending in 0,0, are multiplied together. What is the rightmost digit (the units digit) of the product?

00

22

44

66

88

Answer: D
Solution(s):

We only care about the units digit, which means that the tens digits don't matter.

Then we have 1010 groups of 2468=384. 2 \cdot 4 \cdot 6 \cdot 8 = 384.

The units digit of each group is 4.4. We now need to find the units digit of 410.4^{10}.

The units digit of 42=164^2 = 16 is 6.6. This means we only need to find the units digit of 65.6^5.

Note that every power of 66 always ends in a 66 (e.g. 6,36,216,6, 36, 216, \cdots).

Thus, D is the correct answer.