AMC 8真题

1993年AJHSME真题

这是一个包含25个问题的单项选择题测试。每个问题后面都有标记为A、B、C、D和E的答案。其中只有一个是正确的。每个正确答案得1分。错误的答案不会扣分。
除普通纸、书写工具、尺子和橡皮外，不允许使用任何辅助工具。特别是，不允许使用方格纸、指南针、量角器、计算器、电脑、智能手表和智能手机。
您将有40分钟的工作时间来完成测试。

1993年AJHSME真题+答案在线试做！

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1993年AJHSME 真题：

Problem 1

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

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

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

$1000\times 1993 \times 0.1993 \times 10 =$

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

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?