2003年AMC8中⽂版答案
1.(E)A cube has12edges,8corners and6faces.The sum is26.
2.(C)The smallest prime is2,which is a factor of every even number.Because
58is the only even number,it has the smallest prime factor.
3.(D)Since30of the120grams are?ller,30
120=25%of the burger is?ller.So
100%?25%=75%of the burger is not?ller.
OR
There are120?30=90grams that are not?ller.So90
120
=75%is not?ller.
4.(C)The following chart shows that the answer must be5tricycles.
Bicycles Tricycles Wheels
0721
1620
2519
3418
OR
Let b equal the number of bicycles and t equal the number of tricycles.Then the number of vehicles is b+t=7,and the number of wheels is2b+3t=19.
Because b=7?t,it follows that
2(7?t)+3t=19
14?2t+3t=19
14+t=19
t=5.
OR
If each child had a bicycle,there would be14wheels.Since there are19 wheels,5of the vehicles must be tricycles.
5.(B)If20%of the number is12,the number must be60.Then30%of60is
0.30×60=18.
OR
Since20%of the number is12,it follows that10%of the number is6.So 30%of the number is18.
5
6.(B)
A =
12(√
144)(√25)A =1
2
·12·5
A =30square units 7.(A)Blake scored a total of 4×78=312points on the four tests.Jenny scored 10?10+20+20=40more points than Blake,so her average was 352
4=88,or 10points higher than Blake’s.
OR
The total point di ?erence between Jenny’s and Blake’s tests was 10?10+20+20=40points.The avera
ge di ?erence is
404=10points.8.(A)Because all of the cookies have the same thickness,only the surface area of their shapes needs to be considered.The surface area of each of Art’s
trapezoid cookies is 1
2·3·8=12in 2.Since he makes 12cookies,the surface area of the dough is 12×12=144in 2.Roger’s rectangle cookies each have surface area 2·4=8in 2;therefore,he makes 144÷8=18cookies.
Paul’s parallelogram cookies each have surface area 2·3=6in 2.He makes 144÷6=24cookies.
Trisha’s triangle cookies each have surface area 1
2
·4·3=6in 2.She makes
144÷6=24cookies.
So Art makes the fewest cookies.
9.(C)Art’s 12cookies sell for 12×$0.60=$7.20.Roger’s 18cookies should cost $7.20÷18=$.40each.
OR
The trapezoid’s area is 12in 2and the rectangle’s area is 8in 2.So the cost of a rectangle cookie should be 812 60c |=40c |.
10.(E)The triangle’s area is 6in 2,or half that of the trapezoid.So Trisha will
make twice as many cookies as Art,or 24.
11.(B)Thursday’s price of$40is increased10%or$4,so on Friday the shoes are
marked$44.Then10%of$44or$4.40is taken o?,so the price on Monday is$44?$4.40=$39.60.
OR
Marking a price10%higher multiplies the original price by1.1,and reducing
a price by10%multiplies the price by0.9.So the price of a pair of shoes that
was originally$40will be$40×1.1×0.9=$39.60.
12.(E)If6is one of the visible faces,the product will be divisible by6.If6is
not visible,the product of the visible faces will be1×2×3×4×5=120, which is also divisible by6.Because the product is always divisible by6,the probability is1.
13.(B)A cube has four red faces if it is attached to exactly two other cubes.
The four top cubes are each attached to only one other cube,so they have?ve red faces.The four bottom corner cubes are each attached to three others, so they have three red faces.The remaining six each have four red faces. 14.(D)As
given,T=7.This implies that F=1and that O equals either4or
5.Since O is even,O=4.Therefore,R=8.Replacing letters with numerals
gives
7W4
+7W4
14U8
W+W must be less than10;otherwise,a1would be carried to the next column,and O would be5.So W<5.W=0because
W=U,W=1 because F=1,W=2because if W=2then U=4=O,and W=4 because O=4.So W=3.
The addition problem is
734
+734
1468
15.(B)There are only two ways to construct a solid from three cubes so that
each cube shares a face with at least one
other:
and
Neither of these con?gurations has both the front and side views shown.The four-cube con?guration has the required front and side views.Thus at least four cubes are necessary.
FRONT SIDE
1
32
1
Question:Is it possible to construct a ?ve-cube con?guration with these front and side views?
16.(D)There are 2choices for the driver.The other three can seat themselves
in 3×2×1=6di ?erent ways.So the number of seating arrangements is 2×6=12.17.(E)Because Jim has brown eyes and blond hair,none of his siblings can
have both blue eyes and black hair.Therefore,neither Benjamin nor Tevyn can be Jim’s sibling.Consequently,there are only three possible pairs for Jim’s siblings –Nadeen and Austin,Nadeen and Sue,or Austin and Sue.Since Nadeen has di ?erent hair color and eye color from both Austin and Sue,neither can be Nadeen’s sibling.So Austin and Sue are Jim’s siblings.Benjamin,Nadeen and Tevyn are siblings in the other family.
18.(D)In the graph below,the six classmates who are not friends with Sarah or
with one of Sarah’s friends are circled.Consequently,six classmates will not be invited to the party.
19.(C)A number with 15,20and 25as factors must be divisible by their least
common multiple (LCM).Because 15=3×5,20=22×5,and 25=52,the LCM of 15,20and 25is 22×3×52=300.There are three multiples of 300between 1000and 2000:1200,1500and 1800.
20.(D)Note that in the same period of time,the hour hand moves 1
12as far as
the minute hand.At ,the minute hand is at 12and the hour hand is
at 4.By ,the minute hand has moved 1
3of way around the clock to
4,and the hour hand has moved 112×13=1
36of the way around the clock from
4.Therefore,the angle formed by the hands at is 1
36·360?=10?.
OR
As the minute hand moves 13of the way around the clock face from 12to 4,the hour hand will move 1
3of the way from 4to 5.So the hour hand will
move 13of 112of 360?,or 10?
.
21.(B)Label the feet of the altitudes from B and C as E and F respectively.
Considering right triangles AEB and DF C ,AE =√102?82=√36=6cm,and F D =√172?82=√225=15cm.So the area of △AEB is 12(6)(8)=24cm 2
,and the area of △DF C is 12 (15)(8)=60cm 2
.Rectangle BCF E has area 164?(24+60)=80cm 2.Because BE =CF =8cm,it follows that BC =10cm.
A
blondB C D
10
8
17
8
E F
OR
Let BC =EF =x .From the ?rst solution we know that AE =6and F D =15.Therefore,AD =x +21,and the area of the trapezoid ABCD is (8) 1
2[x +(x +21)] =164.So
4(2x +21)=164,2x +21=41,
2x =20,
and x =10.
22.(C)For Figure A,the area of the square is 22=4cm 2.The diameter of the
circle is 2cm,so the radius is 1cm and the area of the circle is πcm 2.So the area of the shaded region is 4?πcm 2.
For Figure B,the area of the square is also 4cm 2.The radius of each of
the four circles is 12cm,and the area of each circle is 12 2
π=14πcm 2
.The combined area of all four circles is πcm 2.So the shaded regions in A and B have the same area.
For Figure C,the radius of the circle is 1cm,so the area of the circle is πcm 2.Because the diagonal of the inscribed square is the hypotenuse of a right triangle with legs of equal lengths,use the Pythagorean Theorem to determine the length s of one side of the inscribed square.That is,s 2+s 2=22=4.So s 2=2cm 2,the area of the square.Therefore,the area of the shaded region is π?2cm 2.Because π?2>1and 4?π<1,the shaded region in Figure C has the largest area.
Note that the second ?gure consists of four small copies of the ?rst ?gure.Because each of the four small squares has sides half the length of the sides
of the big square,the area of each of the four small ?gures is 1
4
the area of Figure A.Because there are four such small ?gures in Figure B,the shaded regions in A and B have the same area.
23.(A)There are four di ?erent positions for the cat in the 2×2array,so after
every fourth move,the cat will be in the same location.Because 247=4×61+3,the cat will be in the 3rd position clockwise from the ?rst,or the lower right quadrant.There are eight possible positions for the mouse.Because 247=8×30+7,the mouse will be in the 7th position counterclockwise from the ?rst,or the left-hand side of the lower left
quadrant.
B
24.(B)All points along the semicircular part of the
course are the same distance from X ,so the ?rst part of the graph is a horizontal line.As the ship mo
ves from B to D ,its distance from X decreases,then it increases as the ship moves from D to C .Only graph B has these features.

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