1993 AMC 8 Solutions

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All problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

Which pair of numbers does not have a product equal to 36?36?
{−4,−9}\{-4, -9\}
{−3,−12}\{-3, -12\}
{12,−72}\left\{\dfrac12, -72\right\}
{1,36}\{1, 36\}
{32,24}\left\{\dfrac32, 24\right\}
Concepts:fraction
Small hint:
Multiply the two numbers in each pair and check which product is not 3636
Big hint:
A negative times a negative is positive, so watch the signs of 12×(−72)\dfrac12 \times (-72)
Solution:
Checking each pair: (−4)(−9)=36,(-4)(-9) = 36, (−3)(−12)=36,(-3)(-12) = 36, 12×(−72)=−36,\dfrac12 \times (-72) = -36, (1)(36)=36,(1)(36) = 36, and 32×24=36.\dfrac32 \times 24 = 36.
Only 12×(−72)=−36\dfrac12 \times (-72) = -36 fails to equal 36.36.
Thus, the correct answer is C .

2.

When the fraction 4984\dfrac{49}{84} is expressed in simplest form, then the sum of the numerator and the denominator will be
1111
1717
1919
3333
133133
Difficulty rating: 450
Small hint:
Find the greatest common factor of 4949 and 8484
Big hint:
49=7249 = 7^2 and 84=7×12,84 = 7 \times 12, so divide both by 77
Solution:
Since 49=7×749 = 7 \times 7 and 84=7×12,84 = 7 \times 12, the fraction reduces to 712.\dfrac{7}{12}.
The sum of numerator and denominator is 7+12=19.7 + 12 = 19.
Thus, the correct answer is C .

3.

Which of the following numbers has the largest prime factor?
3939
5151
7777
9191
121121
Difficulty rating: 660
Small hint:
Factor each number completely into primes
Big hint:
Compare the largest prime in each factorization; note 51=3×1751 = 3 \times 17
Solution:
Factoring: 39=3×13,39 = 3 \times 13, 51=3×17,51 = 3 \times 17, 77=7×11,77 = 7 \times 11, 91=7×13,91 = 7 \times 13, and 121=11×11.121 = 11 \times 11.
The largest prime factor among these is 17,17, which is a factor of 51.51.
Thus, the correct answer is B .

4.

1000×1993×0.1993×10=1000 \times 1993 \times 0.1993 \times 10 =
1.993×1031.993 \times 10^3
1993.19931993.1993
(199.3)2(199.3)^2
1,993,001.9931{,}993{,}001.993
(1993)2(1993)^2
Difficulty rating: 730
Small hint:
Group the powers of ten: 1000×10=100001000 \times 10 = 10000
Big hint:
10000×0.1993=1993,10000 \times 0.1993 = 1993, so the product is 1993×19931993 \times 1993
Solution:
Regroup as (1000×10)(1000 \times 10) ×0.1993\times 0.1993 ×1993\times 1993 =10000= 10000 ×0.1993\times 0.1993 ×1993.\times 1993.
Since 10000×0.1993=1993,10000 \times 0.1993 = 1993, the product is 1993×1993=(1993)2.1993 \times 1993 = (1993)^2.
Thus, the correct answer is E .

5.

Which one of the following bar graphs could represent the data from the circle graph shown?
Difficulty rating: 660
Small hint:
Decide what fraction of the whole circle each of the three regions takes up
Big hint:
Each shaded region is a quarter and the unshaded region is a half, so the three bars should have heights in the ratio 1:1:21 : 1 : 2
Solution:
The two shaded regions are each one quarter of the circle, and the unshaded region is one half. So the three quantities are in the ratio 14:14:12,\tfrac14 : \tfrac14 : \tfrac12, or 1:1:2.1 : 1 : 2.
A matching bar graph must have the two shaded bars equal in height and the unshaded bar exactly twice as tall. Only one bar graph has two equal shaded bars with the white bar double their height.
Thus, the correct answer is C .

6.

A can of soup can feed 33 adults or 55 children. If there are 55 cans of soup and 1515 children are fed, then how many adults would the remaining soup feed?
55
66
77
88
1010
Concepts:rate
Difficulty rating: 730
Small hint:
1515 children need 15÷5=315 \div 5 = 3 cans, so find how many cans remain
Big hint:
Each remaining can feeds 33 adults
Solution:
Feeding 1515 children uses 15÷5=315 \div 5 = 3 cans, leaving 5−3=25 - 3 = 2 cans.
Those 22 cans feed 2×3=62 \times 3 = 6 adults.
Thus, the correct answer is B .

7.

33+33+33=3^3 + 3^3 + 3^3 =
343^4
939^3
393^9
27327^3
3273^{27}
Concepts:exponent
Difficulty rating: 660
Small hint:
Adding three copies of 333^3 is the same as 3×333 \times 3^3
Big hint:
3×33=31+33 \times 3^3 = 3^{1+3}
Solution:
Adding three equal terms, 33+33+33=3×33=34=81.3^3 + 3^3 + 3^3 = 3 \times 3^3 = 3^4 = 81.
Thus, the correct answer is A .

8.

To control her blood pressure, Jill’s grandmother takes one half of a pill every other day. If one supply of medicine contains 6060 pills, then the supply of medicine will last approximately
11 month
44 months
66 months
88 months
11 year
Difficulty rating: 860
Small hint:
Half a pill every other day means one whole pill covers 44 days
Big hint:
Find the total number of days, then convert to months using about 3030 days per month
Solution:
She takes half a pill every two days, so one pill lasts 44 days. Then 6060 pills last 60×4=24060 \times 4 = 240 days.
At about 3030 days per month, that is roughly 240÷30=8240 \div 30 = 8 months.
Thus, the correct answer is D .

9.

Consider the operation ∗* defined by the following table:
∗123411234224133314244321\begin{array}{c|cccc} * & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{array}
For example, 3∗2=1.3 * 2 = 1. Then (2∗4)∗(1∗3)=(2 * 4) * (1 * 3) =
11
22
33
44
55
Difficulty rating: 730
Small hint:
Read 2∗42 * 4 from row 2,2, column 4,4, and 1∗31 * 3 from row 1,1, column 33
Big hint:
Both inner values equal 3,3, so compute 3∗33 * 3 from the table
Solution:
From the table, 2∗4=32 * 4 = 3 and 1∗3=3.1 * 3 = 3.
Then (2∗4)∗(1∗3)=3∗3=4.(2 * 4) * (1 * 3) = 3 * 3 = 4.
Thus, the correct answer is D .

10.

This line graph represents the price of a trading card during the first 66 months of 1993.1993. The greatest monthly drop in price occurred during which month?
January
March
April
May
June
Difficulty rating: 660
Small hint:
A drop happens where the line goes down from one month to the next; measure how far it falls each time
Big hint:
Compare only the downward segments and find the steepest one
Solution:
The price changes month to month are: January $2.50→$2.00\$2.50 \to \$2.00 (drop $0.50\$0.50), February $2.00→$4.00\$2.00 \to \$4.00 (rise), March $4.00→$1.50\$4.00 \to \$1.50 (drop $2.50\$2.50), April $1.50→$4.50\$1.50 \to \$4.50 (rise), May $4.50→$3.00\$4.50 \to \$3.00 (drop $1.50\$1.50), and June $3.00→$1.00\$3.00 \to \$1.00 (drop $2.00\$2.00).
The largest drop is $2.50,\$2.50, which occurred during March.
Thus, the correct answer is B .

11.

Consider this histogram of the scores for 8181 students taking a test. The median is in the interval labeled which value?
6060
6565
7070
7575
8080
Difficulty rating: 800
Small hint:
With 8181 scores in order, the median is the middle one: the 4141st score counting from the lowest
Big hint:
Add the bar heights from left to right until the running total first reaches 4141
Solution:
Since 8181 students took the test, the median is the 4141st score counting up from the lowest.
Adding the bar heights from the left gives running totals 1,1, 3,3, 7,7, 12,12, 18,18, 28,28, 42,42, …\ldots The total first passes 4141 at the interval labeled 70,70, which contains the 2929th through 4242nd scores. So the 4141st score lies in the interval labeled 70.70.
Thus, the correct answer is C .

12.

If each of the three operation signs, +,+, −,-, ×,\times, is used exactly once in one of the blanks in the expression
5 x‾ 4 x‾ 6 x‾ 35 \, \underline{\phantom{x}} \, 4 \, \underline{\phantom{x}} \, 6 \, \underline{\phantom{x}} \, 3
then the value of the result could equal
99
1010
1515
1616
1919
Difficulty rating: 890
Small hint:
Multiplication happens before addition and subtraction, so where you place ×\times matters most
Big hint:
Work through each way of assigning +,+, −,-, ×\times to the three blanks and compute the result
Solution:
The six arrangements give 5×4+6−3=23,5 \times 4 + 6 - 3 = 23, 5×4−6+3=17,5 \times 4 - 6 + 3 = 17, 5+4×6−3=26,5 + 4 \times 6 - 3 = 26, 5−4×6+3=−16,5 - 4 \times 6 + 3 = -16, 5+4−6×3=−9,5 + 4 - 6 \times 3 = -9, and 5−4+6×3=19.5 - 4 + 6 \times 3 = 19.
The only value among the choices is 19.19.
Thus, the correct answer is E .

13.

The word “HELP” in block letters is painted as a shaded region with strokes 11 unit wide on a 55 by 1515 rectangular sign. Each letter is 33 units wide with a 11-unit gap between letters, as shown. The area of the unshaded portion of the sign, in square units, is
3030
3232
3434
3636
3838
Difficulty rating: 960
Small hint:
The whole sign is 5×15=755 \times 15 = 75 square units
Big hint:
Count the shaded unit squares making up each letter, then subtract the total from 7575
Solution:
The full sign has area 5×15=755 \times 15 = 75 square units. Counting the shaded unit squares in each letter gives H=11,H = 11, E=11,E = 11, L=7,L = 7, and P=10,P = 10, for a shaded total of 11+11+7+10=39.11 + 11 + 7 + 10 = 39.
The unshaded area is 75−39=36.75 - 39 = 36.
Thus, the correct answer is D .

14.

The nine squares in the table shown are to be filled so that every row and every column contains each of the numbers 1,1, 2,2, 3.3. Then A+B=A + B =
1XXX2AXXB\begin{array}{|c|c|c|} \hline 1 & \phantom{X} & \phantom{X} \\ \hline \phantom{X} & 2 & A \\ \hline \phantom{X} & \phantom{X} & B \\ \hline \end{array}
22
33
44
55
66
Difficulty rating: 930
Small hint:
Each row and each column must contain 1,1, 2,2, 33 exactly once
Big hint:
Once two entries of a row or column are known, the third is forced; fill in step by step
Solution:
Filling the grid so each row and column has 1,1, 2,2, 3,3, the top row becomes 1,1, 3,3, 2,2, the middle row 3,3, 2,2, A,A, and the bottom row 2,2, 1,1, B.B. The middle row forces A=1,A = 1, and the last column 2,2, 1,1, BB forces B=3.B = 3.
So A+B=1+3=4.A + B = 1 + 3 = 4.
Thus, the correct answer is C .

15.

The arithmetic mean (average) of four numbers is 85.85. If the largest of these numbers is 97,97, then the mean of the remaining three numbers is
81.081.0
82.782.7
83.083.0
84.084.0
84.384.3
Concepts:mean
Difficulty rating: 660
Small hint:
The four numbers sum to 4×854 \times 85
Big hint:
Subtract the largest number, then divide the remaining sum by 33
Solution:
The four numbers sum to 4×85=340,4 \times 85 = 340, so the remaining three sum to 340−97=243.340 - 97 = 243.
Their mean is 243÷3=81.243 \div 3 = 81.
Thus, the correct answer is A .

16.

What is the value of the following expression?
11+12+13\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{3}}}
16\dfrac16
310\dfrac{3}{10}
710\dfrac{7}{10}
56\dfrac56
103\dfrac{10}{3}
Difficulty rating: 860
Small hint:
Work from the bottom up: first simplify 2+132 + \dfrac13
Big hint:
Then 1+173=1+37,1 + \dfrac{1}{\frac{7}{3}} = 1 + \dfrac37, and take the reciprocal
Solution:
Starting inside, 2+13=73,2 + \dfrac13 = \dfrac73, so 173=37.\dfrac{1}{\frac{7}{3}} = \dfrac37.
Then 1+37=107,1 + \dfrac37 = \dfrac{10}{7}, and the whole expression is 1107=710.\dfrac{1}{\frac{10}{7}} = \dfrac{7}{10}.
Thus, the correct answer is C .

17.

Square corners, 55 units on a side, are removed from a 2020 unit by 3030 unit rectangular sheet of cardboard. The sides are then folded to form an open box. The surface area, in square units, of the interior of the box is
300300
500500
550550
600600
10001000
Difficulty rating: 980
Small hint:
Folding does not change area; the interior surface equals the sheet minus the four corner squares
Big hint:
The sheet is 20×30;20 \times 30; subtract four 5×55 \times 5 corner squares
Solution:
The interior surface is exactly one face of the cardboard after the corners are removed. The sheet has area 20×30=600,20 \times 30 = 600, and each removed corner has area 5×5=25.5 \times 5 = 25.
So the interior surface area is 600−4×25=500.600 - 4 \times 25 = 500.
Thus, the correct answer is B .

18.

The rectangle shown has length AC=32,AC = 32, width AE=20,AE = 20, and BB and FF are midpoints of AC‾\overline{AC} and AE‾,\overline{AE}, respectively. The area of the quadrilateral ABDFABDF is
320320
325325
330330
335335
340340
Difficulty rating: 1090
Small hint:
The rectangle ACDEACDE has area 32×20;32 \times 20; remove the parts outside ABDFABDF
Big hint:
Subtract triangle BCDBCD and triangle DEFDEF from the full rectangle
Solution:
Rectangle ACDEACDE has area 32×20=640.32 \times 20 = 640. Triangle BCDBCD has area 16×202=160,\dfrac{16 \times 20}{2} = 160, and triangle DEFDEF has area 10×322=160.\dfrac{10 \times 32}{2} = 160.
The remaining region ABDFABDF has area 640−(160+160)=320.640 - (160 + 160) = 320.
Thus, the correct answer is A .

19.

What is the value of the following expression?
(1901+1902+1903+⋯+1993)−(101+102+103+⋯+193) \begin{gathered} \small (1901 + 1902 + 1903 + \cdots + 1993) \\ \small {}- (101 + 102 + 103 + \cdots + 193) \end{gathered}
167,400167{,}400
172,050172{,}050
181,071181{,}071
199,300199{,}300
362,142362{,}142
Difficulty rating: 960
Small hint:
Pair each term with the one below it: 1901−101,1901 - 101, 1902−102,1902 - 102, …\ldots
Big hint:
Each of the 9393 pairs has difference 18001800
Solution:
Each number in the first sum is exactly 18001800 more than the matching number in the second sum, and there are 9393 such pairs.
So the difference is 93×1800=167,400.93 \times 1800 = 167{,}400.
Thus, the correct answer is A .

20.

When 1093−9310^{93} - 93 is expressed as a single whole number, the sum of the digits is
1010
9393
819819
826826
833833
Difficulty rating: 1140
Small hint:
109310^{93} is a 11 followed by 9393 zeros; subtracting 9393 changes only the right end
Big hint:
The result is a block of nines ending in 07;07; determine how many nines there are
Solution:
Subtracting 9393 from 109310^{93} (a 11 followed by 9393 zeros) gives a number that is 9191 nines followed by 07.07.
The digit sum is 91×9+0+7=819+7=826.91 \times 9 + 0 + 7 = 819 + 7 = 826.
Thus, the correct answer is D .

21.

If the length of a rectangle is increased by 20%20\% and its width is increased by 50%,50\%, then the area is increased by
10%10\%
30%30\%
70%70\%
80%80\%
100%100\%
Concepts:percentage
Difficulty rating: 820
Small hint:
New area == (new length) ×\times (new width) =1.2×1.5= 1.2 \times 1.5 times the old area
Big hint:
Multiply the two growth factors 1.21.2 and 1.51.5 to get the factor by which the area grows
Solution:
The new length is 1.21.2 times the old and the new width is 1.51.5 times the old, so the new area is 1.2×1.5=1.81.2 \times 1.5 = 1.8 times the old area.
That is an increase of 80%.80\%.
Thus, the correct answer is D .

22.

Pat Peano has plenty of 00’s, 11’s, 33’s, 44’s, 55’s, 66’s, 77’s, 88’s and 99’s, but he has only twenty-two 22’s. How far can he number the pages of his scrapbook with these digits?
2222
9999
112112
119119
199199
Difficulty rating: 1200
Small hint:
Count how many 22’s appear when numbering pages 11 through 9999
Big hint:
Ten 22’s appear in the units place and ten in the tens place through 99;99; then track the remaining twos past 100100
Solution:
Numbering 11 through 9999 uses ten 22’s in the units place and ten in the tens place, a total of twenty 22’s. Pages 100100 and 101101 use none.
The remaining two 22’s are used on pages 102102 and 112.112. After that, pages 113113 through 119119 need no 2,2, but 120120 would require another 2,2, so he can number up to 119.119.
Thus, the correct answer is D .

23.

Five runners, P,P, Q,Q, R,R, S,S, T,T, have a race, and PP beats Q,Q, PP beats R,R, QQ beats S,S, and TT finishes after PP and before Q.Q. Who could not have finished third in the race?
PP and QQ
PP and RR
PP and SS
PP and TT
P,P, SS and TT
Difficulty rating: 1070
Small hint:
PP finishes ahead of Q,Q, R,R, and T,T, while QQ finishes ahead of S;S; determine PP’s place
Big hint:
If several runners must all finish ahead of a given runner, that runner cannot be as high as third
Solution:
Since PP beats QQ and R,R, finishes ahead of T,T, and QQ beats S,S, runner PP finishes first and so cannot be third.
The clues give the chain PP before TT before QQ before S.S. So P,P, T,T, and QQ all finish ahead of S,S, meaning SS is no better than fourth and cannot be third either.
Each of Q,Q, R,R, TT can finish third: for example P,P, T,T, Q,Q, R,R, SS puts QQ third; P,P, R,R, T,T, Q,Q, SS puts TT third; and P,P, T,T, R,R, Q,Q, SS puts RR third. So only PP and SS cannot be third.
Thus, the correct answer is C .

24.

What number is directly above 142142 in this array of numbers?
123456789101112⋯\begin{array}{ccccccccc} & & & & 1 & & & & \\ & & & 2 & 3 & 4 & & & \\ & & 5 & 6 & 7 & 8 & 9 & & \\ & 10 & 11 & 12 & \cdots & & & & \end{array}
9999
119119
120120
121121
122122
Difficulty rating: 1140
Small hint:
Each row ends at a perfect square: row nn ends at n2n^2
Big hint:
142142 is in the row ending at 144=122;144 = 12^2; the row above ends at 121=112,121 = 11^2, and the rows line up at their right edges
Solution:
Each row ends at a perfect square, so the row containing 142142 ends at 144=122,144 = 12^2, and the row above it ends at 121=112.121 = 11^2.
Since the rows are aligned at their right edges, 121121 sits directly above 143,143, and therefore 120120 sits directly above 142.142.
Thus, the correct answer is C .

25.

A checkerboard consists of one-inch squares. A square card, 1.51.5 inches on a side, is placed on the board so that it covers part or all of the area of each of nn squares. The maximum possible value of nn is
44 or 55
66 or 77
88 or 99
1010 or 1111
1212 or more
Difficulty rating: 1270
Small hint:
A tilted card can cross more grid lines than a card lined up with the squares; the card’s diagonal has length 1.52+1.52=4.5≈2.1\sqrt{1.5^2 + 1.5^2} = \sqrt{4.5} \approx 2.1
Big hint:
Try tilting the card 45∘45^\circ with its center on a grid corner, and count how many squares each corner of the card pokes into
Solution:
Tilt the card 45∘45^\circ and center it on a corner where four grid squares meet, as shown. Because the card’s diagonal, 1.52+1.52=4.5≈2.1,\sqrt{1.5^2 + 1.5^2} = \sqrt{4.5} \approx 2.1, is longer than 2,2, each of the four corners of the card reaches past a grid line into the next square.
The card covers the central 2×22 \times 2 block of 44 squares and pokes into 22 more squares on each of its four sides, giving 4+4×2=124 + 4 \times 2 = 12 squares.
This is also the most possible. The card is only 1.51.5 inches wide, so its overall width and height are each at most 2.12.1 inches; it therefore lies within a 4×44 \times 4 block of 1616 squares. Its four pointed corners are the only parts that reach the edge of that block, so it can never reach the four corner squares of the block, leaving at most 12.12. Since 1212 is achievable, the maximum is 12,12, which falls in the range “1212 or more.”
Thus, the correct answer is E .
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