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标题: 曼哈顿模考数学两道错题求助 [打印本页]

作者: nengneng1120 时间: 2014-9-30 16:14
标题: 曼哈顿模考数学两道错题求助
撸了一发曼哈顿模考，免费的，数学部分果如前辈所说比PREP难。而且字小到完全不能忍。VERBAL部分还是挺有参考价值的。
发个模考链接：
http://www.manhattangmat.com/free-gmat.cfm

有两道数学题看解释死活没懂....求助..

作者: nengneng1120 时间: 2014-9-30 16:17
A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

10(根号3 – 1)

5

10( 根号2– 1)

5(根号3 – 1)

5( 根号2– 1)

作者: nengneng1120 时间: 2014-9-30 16:18
x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT

x = w
x > w
x/y is an integer
w/z is an integer
x/z is an integer

作者: nengneng1120 时间: 2014-9-30 16:19
答案是 D 和E

作者: 苏鹿 时间: 2014-9-30 16:52
第一个  正方体对角线=10根号3 减去圆球的直径10  除以2 =答案

第二题我举得反例选的C 错了我也不晓得：当y=2时，x=1+2=3 时候  不就不是整数了吗

还有若是x/z都可以不是这个数那么它的2倍--x/y咋回事整数  真是要流眼泪了

作者: pippy1108 时间: 2014-9-30 16:55

nengneng1120 发表于 2014-9-30 16:17
A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one ...

第一题蛮简单的，就是说边长为10的立方体里面内切了一个球，问立方体顶点到球面的最短距离是多少。
球面的这个点在立方体的体对角线上，所以最短距离等于体对角线一半减去球半径，就是5（根号3-1）

作者: pippy1108 时间: 2014-9-30 17:05

苏鹿发表于 2014-9-30 16:52
第一个正方体对角线=10根号3 减去圆球的直径10 除以2 =答案

第二题我举得反例选的C 错了我也不晓 ...

第二题我也觉得答案有问题

设第一个序列为a，a+1，a+2,...a+2y-1,那么x=（a+a+2y-1)*y/2=(2a+4z-1)*z,即x一定被Z整除的

作者: nengneng1120 时间: 2014-9-30 17:28

我想我是把正方体当作正方形了，第一题可以无视掉惹..

第二题，问 then each of the following could be true EXCEPT，也就是问哪个选项肯定是错的。

对于E. 假设Y=4, Z=2, X=1+2+3+4=10, W=1+2=3, 酱紫X/Z=5
不懂.... 咋肥事儿

作者: nengneng1120 时间: 2014-9-30 17:31
抱歉....我发现答案是C ... 撸主今天脑抽了

For any set of consecutive integers with an odd number of terms, the sum of the integers is always a multiple of the number of terms. For example, the sum of 1, 2, and 3 (three consecutives -- an odd number) is 6, which is a multiple of 3. For any set of consecutive integers with an even number of terms, the sum of the integers is never a multiple of the number of terms. For example, the sum of 1, 2, 3, and 4 (four consecutives -- an even number) is 10, which is not a multiple of 4.

The question tells us that y = 2z, which allows us to deduce that y is even. Since y is even, then the sum of y integers, x, cannot be a multiple of y. Therefore, x/y cannot be an integer; choice C is the correct answer. We can verify this by showing that the other choices could indeed be true:

(A) The sum x can equal the sum w: 4 + 5 + 6 + 7 + 8 + 9 = 12 + 13 + 14 = 39, for example.

(B) The sum x can be greater than the sum w: 1 + 2 + 3 + 4 > 1 + 2, for example.

(D) z could be odd (the question does not restrict this), making the sum w a multiple of z. Thus, w/z could be an integer. For example, if z = 3, then we are dealing with three consecutive integers. We can choose any three: 2, 3, and 4, for example. 2 + 3 + 4 = 9 and 9/3 = 3, which is an integer.

(E) x/z could be an integer. If z = 2 and if x is an even sum, then x/z would be an integer. For example, if z = 2, then y = 4. We can choose any four consecutive integers: 1 + 2 + 3 + 4, for example. So the sum x of these four integers is 10. 10/2 = 5, which is an integer.

The correct answer is C.

作者: 苏鹿 时间: 2014-9-30 17:50
差一点斯巴达儿了

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