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1993 AMC 8 真题及答案详细解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月4日下午3:09

1993 AMC 8 真题

答案详细解析请参考文末

Problem 1

Which pair of numbers does NOT have a product equal to $36$ ?

$text{(A)} { -4,-9} qquad text{(B)} { -3,-12} qquad text{(C)} left{ dfrac{1}{2},-72right} qquad text{(D)} { 1,36} qquad text{(E)} left{dfrac{3}{2},24right}$

Problem 2

When the fraction $dfrac{49}{84}$ is expressed in simplest form, then the sum of the numerator and the denominator will be

$text{(A)} 11 qquad text{(B)} 17 qquad text{(C)} 19 qquad text{(D)} 33 qquad text{(E)} 133$

Problem 3

Which of the following numbers has the largest prime factor?

$text{(A)} 39 qquad text{(B)} 51 qquad text{(C)} 77 qquad text{(D)} 91 qquad text{(E)} 121$

Problem 4

$1000times 1993 times 0.1993 times 10 =$

$text{(A)} 1.993times 10^3 qquad text{(B)} 1993.1993 qquad text{(C)} (199.3)^2 qquad text{(D)} 1,993,001.993 qquad text{(E)} (1993)^2$

Problem 5

Which one of the following bar graphs could represent the data from the circle graph?

$[asy] unitsize(36); draw(circle((0,0),1),gray); fill((0,0)--arc((0,0),(0,-1),(1,0))--cycle,gray); fill((0,0)--arc((0,0),(1,0),(0,1))--cycle,black); [/asy]$

$[asy] unitsize(4); fill((1,0)--(1,15)--(5,15)--(5,0)--cycle,gray); fill((6,0)--(6,15)--(10,15)--(10,0)--cycle,black); draw((11,0)--(11,20)--(15,20)--(15,0)); fill((26,0)--(26,15)--(30,15)--(30,0)--cycle,gray); fill((31,0)--(31,15)--(35,15)--(35,0)--cycle,black); draw((36,0)--(36,15)--(40,15)--(40,0)); fill((51,0)--(51,10)--(55,10)--(55,0)--cycle,gray); fill((56,0)--(56,10)--(60,10)--(60,0)--cycle,black); draw((61,0)--(61,20)--(65,20)--(65,0)); fill((76,0)--(76,10)--(80,10)--(80,0)--cycle,gray); fill((81,0)--(81,15)--(85,15)--(85,0)--cycle,black); draw((86,0)--(86,20)--(90,20)--(90,0)); fill((101,0)--(101,15)--(105,15)--(105,0)--cycle,gray); fill((106,0)--(106,10)--(110,10)--(110,0)--cycle,black); draw((111,0)--(111,20)--(115,20)--(115,0)); for(int a = 0; a < 5; ++a) { draw((25*a,21)--(25*a,0)--(25*a+16,0)); } label("(A)",(8,21),N); label("(B)",(33,21),N); label("(C)",(58,21),N); label("(D)",(83,21),N); label("(E)",(108,21),N); [/asy]$

Problem 6

A can of soup can feed $3$ adults or $5$ children. If there are $5$ cans of soup and $15$ children are fed, then how many adults would the remaining soup feed?

$text{(A)} 5 qquad text{(B)} 6 qquad text{(C)} 7 qquad text{(D)} 8 qquad text{(E)} 25$

Problem 7

$3^3+3^3+3^3 =$

$text{(A)} 3^4 qquad text{(B)} 9^3 qquad text{(C)} 3^9 qquad text{(D)} 27^3 qquad text{(E)} 3^{27}$

Problem 8

To control her blood pressure, Jill's grandmother takes one half of a pill every other day. If one supply of medicine contains $60$ pills, then the supply of medicine would last approximately

$text{(A)} 1text{ month} qquad text{(B)} 4text{ months} qquad text{(C)} 6text{ months} qquad text{(D)} 8text{ months} qquad text{(E)} 1text{ year}$

Problem 9

Consider the operation $*$ defined by the following table:

$[begin{tabular}{c|cccc} * & 1 & 2 & 3 & 4 hline 1 & 1 & 2 & 3 & 4 2 & 2 & 4 & 1 & 3 3 & 3 & 1 & 4 & 2 4 & 4 & 3 & 2 & 1 end{tabular}]$

For example, $3*2=1$ . Then $(2*4)*(1*3)=$

$text{(A)} 1 qquad text{(B)} 2 qquad text{(C)} 3 qquad text{(D)} 4 qquad text{(E)} 5$

Problem 10

This line graph represents the price of a trading card during the first $6$ months of $1993$ .

$[asy] unitsize(18); for (int a = 0; a <= 6; ++a) { draw((4*a,0)--(4*a,10)); } for (int a = 0; a <= 5; ++a) { draw((0,2*a)--(24,2*a)); } draw((0,5)--(4,4)--(8,8)--(12,3)--(16,9)--(20,6)--(24,2),linewidth(1.5)); label("$Jan$",(2,0),S); label("$Feb$",(6,0),S); label("$Mar$",(10,0),S); label("$Apr$",(14,0),S); label("$May$",(18,0),S); label("$Jun$",(22,0),S); label("$textbf{1993 PRICES FOR A TRADING CARD}$",(12,10),N); label("$begin{tabular}{c}textbf{P} textbf{R} textbf{I} textbf{C} textbf{E} end{tabular}$",(-2,5),W); label("$1$",(0,2),W); label("$2$",(0,4),W); label("$3$",(0,6),W); label("$4$",(0,8),W); label("$5$",(0,10),W); [/asy]$

The greatest monthly drop in price occurred during

$text{(A)} text{January} qquad text{(B)} text{March} qquad text{(C)} text{April} qquad text{(D)} text{May} qquad text{(E)} text{June}$

Problem 11

Consider this histogram of the scores for $81$ students taking a test:

$[asy] unitsize(12); draw((0,0)--(26,0)); draw((1,1)--(25,1)); draw((3,2)--(25,2)); draw((5,3)--(23,3)); draw((5,4)--(21,4)); draw((7,5)--(21,5)); draw((9,6)--(21,6)); draw((11,7)--(19,7)); draw((11,8)--(19,8)); draw((11,9)--(19,9)); draw((11,10)--(19,10)); draw((13,11)--(19,11)); draw((13,12)--(19,12)); draw((13,13)--(17,13)); draw((13,14)--(17,14)); draw((15,15)--(17,15)); draw((15,16)--(17,16)); draw((1,0)--(1,1)); draw((3,0)--(3,2)); draw((5,0)--(5,4)); draw((7,0)--(7,5)); draw((9,0)--(9,6)); draw((11,0)--(11,10)); draw((13,0)--(13,14)); draw((15,0)--(15,16)); draw((17,0)--(17,16)); draw((19,0)--(19,12)); draw((21,0)--(21,6)); draw((23,0)--(23,3)); draw((25,0)--(25,2)); for (int a = 1; a < 13; ++a) { draw((2*a,-.25)--(2*a,.25)); } label("$40$",(2,-.25),S); label("$45$",(4,-.25),S); label("$50$",(6,-.25),S); label("$55$",(8,-.25),S); label("$60$",(10,-.25),S); label("$65$",(12,-.25),S); label("$70$",(14,-.25),S); label("$75$",(16,-.25),S); label("$80$",(18,-.25),S); label("$85$",(20,-.25),S); label("$90$",(22,-.25),S); label("$95$",(24,-.25),S); label("$1$",(2,1),N); label("$2$",(4,2),N); label("$4$",(6,4),N); label("$5$",(8,5),N); label("$6$",(10,6),N); label("$10$",(12,10),N); label("$14$",(14,14),N); label("$16$",(16,16),N); label("$12$",(18,12),N); label("$6$",(20,6),N); label("$3$",(22,3),N); label("$2$",(24,2),N); label("Number",(4,8),N); label("of Students",(4,7),N); label("$textbf{STUDENT TEST SCORES}$",(14,18),N); [/asy]$

The median is in the interval labeled

$text{(A)} 60 qquad text{(B)} 65 qquad text{(C)} 70 qquad text{(D)} 75 qquad text{(E)} 80$

Problem 12

If each of the three operation signs, $+$ , $-$ , $times$ , is used exactly ONCE in one of the blanks in the expression

$[5hspace{1 mm}underline{hspace{4 mm}}hspace{1 mm}4hspace{1 mm}underline{hspace{4 mm}}hspace{1 mm}6hspace{1 mm}underline{hspace{4 mm}}hspace{1 mm}3]$

then the value of the result could equal

$text{(A)} 9 qquad text{(B)} 10 qquad text{(C)} 15 qquad text{(D)} 16 qquad text{(E)} 19$

Problem 13

The word "HELP" in block letters is painted in black with strokes $1$ unit wide on a $5$ by $15$ rectangular white sign with dimensions as shown. The area of the white portion of the sign, in square units, is

$[asy] unitsize(12); fill((0,0)--(0,5)--(1,5)--(1,3)--(2,3)--(2,5)--(3,5)--(3,0)--(2,0)--(2,2)--(1,2)--(1,0)--cycle,black); fill((4,0)--(4,5)--(7,5)--(7,4)--(5,4)--(5,3)--(7,3)--(7,2)--(5,2)--(5,1)--(7,1)--(7,0)--cycle,black); fill((8,0)--(8,5)--(9,5)--(9,1)--(11,1)--(11,0)--cycle,black); fill((12,0)--(12,5)--(15,5)--(15,2)--(13,2)--(13,0)--cycle,black); fill((13,3)--(14,3)--(14,4)--(13,4)--cycle,white); draw((0,0)--(15,0)--(15,5)--(0,5)--cycle); label("$5left{ begin{tabular}{c} end{tabular}right.$",(1,2.5),W); label(rotate(90)*"${$",(0.5,0.1),S); label("$1$",(0.5,-0.6),S); label(rotate(90)*"${$",(3.5,0.1),S); label("$1$",(3.5,-0.6),S); label(rotate(90)*"${$",(7.5,0.1),S); label("$1$",(7.5,-0.6),S); label(rotate(90)*"${$",(11.5,0.1),S); label("$1$",(11.5,-0.6),S); label(rotate(270)*"$left{ begin{tabular}{c} end{tabular}right.$",(1.5,4),N); label("$3$",(1.5,5.8),N); label(rotate(270)*"$left{ begin{tabular}{c} end{tabular}right.$",(5.5,4),N); label("$3$",(5.5,5.8),N); label(rotate(270)*"$left{ begin{tabular}{c} end{tabular}right.$",(9.5,4),N); label("$3$",(9.5,5.8),N); label(rotate(270)*"$left{ begin{tabular}{c} end{tabular}right.$",(13.5,4),N); label("$3$",(13.5,5.8),N); label("$left. begin{tabular}{c} end{tabular}right} 2$",(14,1),E); [/asy]$

$text{(A)} 30 qquad text{(B)} 32 qquad text{(C)} 34 qquad text{(D)} 36 qquad text{(E)} 38$

Problem 14

The nine squares in the table shown are to be filled so that every row and every column contains each of the numbers $1,2,3$ . Then $A+B=$

$[begin{tabular}{|c|c|c|} hline 1 & & hline & 2 & A hline & & B hline end{tabular}]$

$text{(A)} 2 qquad text{(B)} 3 qquad text{(C)} 4 qquad text{(D)} 5 qquad text{(E)} 6$

Problem 15

The arithmetic mean (average) of four numbers is $85$ . If the largest of these numbers is $97$ , then the mean of the remaining three numbers is

$text{(A)} 81.0 qquad text{(B)} 82.7 qquad text{(C)} 83.0 qquad text{(D)} 84.0 qquad text{(E)} 84.3$

Problem 16

$dfrac{1}{1+dfrac{1}{2+dfrac{1}{3}}} =$

$text{(A)} dfrac{1}{6} qquad text{(B)} dfrac{3}{10} qquad text{(C)} dfrac{7}{10} qquad text{(D)} dfrac{5}{6} qquad text{(E)} dfrac{10}{3}$

Problem 17

Square corners, $5$ units on a side, are removed from a $20$ unit by $30$ 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

$[asy] fill((0,0)--(20,0)--(20,5)--(0,5)--cycle,lightgray); fill((20,0)--(20+5*sqrt(2),5*sqrt(2))--(20+5*sqrt(2),5+5*sqrt(2))--(20,5)--cycle,lightgray); draw((0,0)--(20,0)--(20,5)--(0,5)--cycle); draw((0,5)--(5*sqrt(2),5+5*sqrt(2))--(20+5*sqrt(2),5+5*sqrt(2))--(20,5)); draw((20+5*sqrt(2),5+5*sqrt(2))--(20+5*sqrt(2),5*sqrt(2))--(20,0)); draw((5*sqrt(2),5+5*sqrt(2))--(5*sqrt(2),5*sqrt(2))--(5,5),dashed); draw((5*sqrt(2),5*sqrt(2))--(15+5*sqrt(2),5*sqrt(2)),dashed); [/asy]$

$text{(A)} 300 qquad text{(B)} 500 qquad text{(C)} 550 qquad text{(D)} 600 qquad text{(E)} 1000$

Problem 18

The rectangle shown has length $AC=32$ , width $AE=20$ , and $B$ and $F$ are midpoints of $overline{AC}$ and $overline{AE}$ , respectively. The area of quadrilateral $ABDF$ is

$[asy] pair A,B,C,D,EE,F; A = (0,20); B = (16,20); C = (32,20); D = (32,0); EE = (0,0); F = (0,10); draw(A--C--D--EE--cycle); draw(B--D--F); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); label("$A$",A,NW); label("$B$",B,N); label("$C$",C,NE); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,W); [/asy]$

$text{(A)} 320 qquad text{(B)} 325 qquad text{(C)} 330 qquad text{(D)} 335 qquad text{(E)} 340$

Problem 19

$(1901+1902+1903+cdots + 1993) - (101+102+103+cdots + 193) =$

$text{(A)} 167,400 qquad text{(B)} 172,050 qquad text{(C)} 181,071 qquad text{(D)} 199,300 qquad text{(E)} 362,142$

Problem 20

When $10^{93}-93$ is expressed as a single whole number, the sum of the digits is

$text{(A)} 10 qquad text{(B)} 93 qquad text{(C)} 819 qquad text{(D)} 826 qquad text{(E)} 833$

Problem 21

If the length of a rectangle is increased by $20%$ and its width is increased by $50%$ , then the area is increased by

$text{(A)} 10% qquad text{(B)} 30% qquad text{(C)} 70% qquad text{(D)} 80% qquad text{(E)} 100%$

Problem 22

Pat Peano has plenty of 0's, 1's, 3's, 4's, 5's, 6's, 7's, 8's and 9's, but he has only twenty-two 2's. How far can he number the pages of his scrapbook with these digits?

$text{(A)} 22 qquad text{(B)} 99 qquad text{(C)} 112 qquad text{(D)} 119 qquad text{(E)} 199$

Problem 23

Five runners, $P$ , $Q$ , $R$ , $S$ , $T$ , have a race, and $P$ beats $Q$ , $P$ beats $R$ , $Q$ beats $S$ , and $T$ finishes after $P$ and before $Q$ . Who could NOT have finished third in the race?

$text{(A)} Ptext{ and }Q qquad text{(B)} Ptext{ and }R qquad text{(C)} Ptext{ and }S qquad text{(D)} Ptext{ and }T qquad text{(E)} P,Stext{ and }T$

Problem 24

What number is directly above $142$ in this array of numbers?

$[begin{array}{cccccc} & & & 1 & & & & 2 & 3 & 4 & & 5 & 6 & 7 & 8 & 9 10 & 11 & 12 & cdots & & end{array}]$

$text{(A)} 99 qquad text{(B)} 119 qquad text{(C)} 120 qquad text{(D)} 121 qquad text{(E)} 122$

Problem 25

A checkerboard consists of one-inch squares. A square card, $1.5$ inches on a side, is placed on the board so that it covers part or all of the area of each of $n$ squares. The maximum possible value of $n$ is

$text{(A)} 4text{ or }5 qquad text{(B)} 6text{ or }7qquad text{(C)} 8text{ or }9 qquad text{(D)} 10text{ or }11 qquad text{(E)} 12text{ or more}$

1993 AMC8真答案详细解析

A. The ordered pair ${-4,-9}$ has a product of $-4*-9=36$

B. The ordered pair ${-3,-12}$ has a product of $-3*-12=36$

C. The ordered pair ${1/12, -72}$ has a product of $-36$

D. The ordered pair ${1, 36}$ has a product of $1*36=36$

E. The ordered pair ${3/2, 24}$ has a product of $3/2*24=36$

Since C is the only ordered pair which doesn't equal 36, $boxed{text{(C)}}$ is the answer.

$begin{align*} dfrac{49}{84} &= dfrac{7^2}{2^2cdot 3cdot 7} &= dfrac{7}{2^2cdot 3} &= dfrac{7}{12}. end{align*}$

The sum of the numerator and denominator is $7+12=19rightarrow boxed{text{C}}$ .

A. The prime factors of $39$ are $3$ and $13$ . Therefore, the largest prime factor is $13$ .

B. The prime factors of $51$ are $3$ and $17$ . Therefore, the largest prime factor is $17$ .

C. The prime factors of $77$ are $7$ and $11$ . Therefore, the largest prime factor is $11$ .

D. The prime factors of $91$ are $7$ and $13$ . Therefore, the largest prime factor is $13$ .

E. The only prime factor of $121$ is $11$ . This makes $11$ the largest prime factor of this number.

Since we are looking for the number with the largest prime factor, the correct answer would be $boxed{text{(B)}}$ .

4. $1000times10=10^4 0.1993=1993times10^{-4} 1993times1993times10^{-4}times10^4= boxed{textbf{(E)} (1993)^2}$

5.Since the white portion of the graph is twice as much as the black and gray, its bar in the bar graph will need to be twice as much as the black and gray. Note that the black and gray portions are equal. This cancels choices $A, B, D, E$ . The final answer is now $boxed{C}$ .

6.A can of soup will feed $5$ children so $15$ children are feed by $3$ cans of soup. Therefore, there are $5-3=2$ cans for adults, so $3 times 2 =boxed{textbf{(B)} 6}$ adults are fed.

7. $3^3+3^3+3^3 = 3 times 3^3 = boxed{text{(A)} 3^4}$

8.If Jill's grandmother takes one half of a pill every other day, she takes a pill every $4$ days. Since she has $60$ pills, the supply will last $60 times 4=240$ days which is about $boxed{text{(D)} 8text{ months}}$ .

9.Using the chart, $(2*4)=3$ and $(1*3)=3$ . Therefore, $(2*4)*(1*3)=3*3=boxed{text{(D)} 4}$ .

10.

The graph shows the following price changes:

Jan : Change:-0.50

Feb : Change:+2.00

Mar : Change:-2.50

Apr : Change:+3.00

May : Change:-0.50

Jun : Change:-2.00

Therefore, $boxed{text{(B)} text{March}}$ has the largest price drop.

11.Since $81$ students took the test, the median is the score of the $41^{st}$ student. The five rightmost intervals include $2+3+6+12+16=39$ students, so the $41^{st}$ one must lie in the next interval, which is $boxed{text{(C)} 70}$ .

12.There are a reasonable number of ways to place the operation signs, so guess and check to find that $5-4+6 times 3 = boxed{text{(E)} 19}$ .

13.Count the number of black squares in each letter. H has 11, E has 11, L has 7, and P has 10, giving the number of black squares to be $11+11+7+10=39$ . The total number of squares is $(15)(5)=75$ and the number of white squares is $75-39=boxed{text{(D)} 36}$ .

14.

The square connected both to 1 and 2 cannot be the same as either of them, so must be 3.

$[begin{tabular}{|c|c|c|}hline 1 & 3 & hline & 2 & A hline & & B hlineend{tabular}]$

The last square in the top row cannot be either 1 or 3, so it must be 2.

$[begin{tabular}{|c|c|c|}hline 1 & 3 & 2 hline & 2 & hline & & B hlineend{tabular}]$

The other two squares in the rightmost column with A and B cannot be two, so they must be 1 and 3 and therefore have a sum of $1+3=boxed{text{(C)} 4}$ .

15.Say that the four numbers are $a, b, c,$ & $97$ . Then $frac{a+b+c+97}{4} = 85$ . What we are trying to find is $frac{a+b+c}{3}$ . Solving, $[frac{a+b+c+97}{4} = 85]$ $[a+b+c+97 = 340]$ $[a+b+c = 243]$ $[frac{a+b+c}{3} = boxed{mathrm{(A)} 81}]$

16. $[frac{1}{1+frac{1}{2+frac13}} = frac{1}{1+frac{1}{frac73}} = frac{1}{1+frac37} = frac{1}{frac{10}{7}} = boxed{text{(C)} frac{7}{10}}]$

17.If the sides of the open box are folded down so that a flat sheet with four corners cur out remains, then the revealed surface would have the same area as the interior of the box. This is equal to the area of the four corners subtracted from the area of the original sheet, which is $(20)(30)-4(5)(5) = 600-100 = boxed{text{(B)} 500}$ .

18.

The area of the quadrilateral $ABDF$ is equal to the areas of the two right triangles $triangle BCD$ and $triangle EFD$ subtracted from the area of the rectangle $ABCD$ . Because $B$ and $F$ are midpoints, we know the dimensions of the two right triangles.

$[(20)(32)-frac{(16)(20)}{2}-frac{(10)(32)}{2} = 640-160-160 = boxed{text{(A)} 320}]$

19.

We see that $1901=1800+101$ , $1902=1800+102$ , etc. Each term in the first set of numbers is $1800$ more than the corresponding term in the second set; Because there are $93$ terms in the first set, the expression can be paired up as follows and simplified:

$[(1901-101) + (1902-102) + (1903-103) + cdots + (1993-193) =1800 + 1800 + 1800 + cdots + 1800 =(1800)(93) =boxed{text{(A)} 167,400}]$

20.

$begin{align*} 10^2-93&=7 10^3-93&=907 10^4-93&=9907 end{align*}$

This can be generalized into $10^n-93$ is equal is $n-2$ nines followed by the digits $07$ . Then $10^{93}-93$ is equal to $91$ nines followed by $07$ . The sum of the digits is equal to $9(91)+7=819+7=boxed{text{(D)} 826}$ .

21.

If a rectangle had dimensions $10 times 10$ and area $100$ , then its new dimensions would be $12 times 15$ and area $180$ . The area is increased by $180-100=80$ or $80/100 = boxed{text{(D)} 80%}$ .

Alternate Solution

Let the dimensions of the rectangle be $x times y$ . This rectangle has area $xy$ . The new dimensions would be $1.2x times 1.5y$ , so the area is $=1.8xy$ , which is $boxed{80%}$ more than the original area.

22.

There is $1$ two in the one-digit numbers.

The number of two-digit numbers with a two in the tens place is $10$ and the number with a two in the ones place is $9$ . Thus the digit two is used $10+9=19$ times for the two digit numbers.

Now, Pat Peano only has $22-1-19=2$ remaining twos. The last numbers with a two that he can write are $102$ and $112$ . He can continue numbering the last couple pages without a two until $120$ , with the last number he writes being $boxed{text{(D)} 119}$ .

23.First, note that $P$ must beat $Q$ , $R$ , $T$ , and by transitivity, $S$ . Thus $P$ is in first, and not in 3rd. Similarly, $S$ is beaten by $P$ , $Q$ , and by transitivity, $T$ , so $S$ is in fourth or fifth, and not in third. All of the others can be in third, as all of the following sequences show. Each follows all of the assumptions of the problem, and they are in order from first to last: PTQRS, PTRQS, PRTQS. Thus the answer is $boxed{text{(C)} P text{and} S}$ .

24.

Solution 1

Notice that a number in row $k$ is $2k$ less than the number directly below it. For example, $5$ , which is in row $3$ , is $(2)(3)=6$ less than the number below it, $11$ .

From row 1 to row $k$ , there are $k left(frac{1+(-1+2k)}{2} right) = k^2$ numbers in those $k$ rows. Because there are $12^2=144$ numbers up to the 12th row, $142$ is in the $k^{th}$ row. The number directly above is in the 11th row, and is $22$ less than $142$ . Thus the number directly above $142$ is $142-22=boxed{text{(C)} 120}$ .

Solution 2

Writing a couple more rows, the last number in each row ends in a perfect square. Thus $142$ is two left from the last number in its row, $144$ . One left and one up from $144$ is the last number of its row, also a perfect square, and is $121$ . This is one right and one up from $142$ , so the number directly above $142$ is one less than $121$ , or $boxed{text{(C)} 120}$ .

25.

Using Pythagorean Theorem, the diagonal of the square $sqrt{(1.5)^2+(1.5)^2}=sqrt{4.5}>2$ . Because this is longer than $2$ , the length of the sides of two adjacent squares, the card can be placed like so, covering $12$ squares. $rightarrow boxed{text{(E)} 12 text{or more}}$ .

$[asy] for (int a = -2; a <= 2; ++a) { draw((-2,a)--(2,a)); draw((a,-2)--(a,2)); } pair A,B,C,D; A=(0,sqrt(2.25)); B=(sqrt(2.25),0); C=(0,-sqrt(2.25)); D=(-sqrt(2.25),0); draw(A--B--C--D--cycle); fill(A--B--C--D--cycle,lightgray); [/asy]$