第一次做错的题目

终于考好了，CD给了我蛮多帮助。贴上我第一次做错的题和讲解。解释是从网上找的。希望对大家有一点帮助。－Math

1. Before being simplified, the instructions for computing income tax in Country R were to add 2 percent of one's annual income to the average(arithmetic mean)of 100units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax, in Country R's currency, for a person in that country whose annual income is I?
   
   A 50+I/200 B 50+3I/100 C 50+I/40 D 100+I/50 E100+3I/100
   
   answer is C

Total income is I.

take average of:
100 and 1% of income I/100

average is

(100 + I/100) / 2

50 + I/200

result is

I/50 + 50 + I/200
=50 + 5I/200
=50 + I/40

2. For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n. If p is the smallest prime factor of h(100)+1,then p is ?
   
   
   A between 2 and 10 B between 10 and 20 C between 20 and 30 D between 30 and 40 E greater than 40
   
   Answer is E

h(100) will be 2 * 4 * 6 * 8 .... 100

= 2^50 (1 * 2 * 3 ... 50)

= 2^50 * 50!

This h(100) will be divisible by all numbers from 1 to 50

adding 1
h(100) + 1 will not be divisible by any number from 1 to 50

hence smallest prime factor will be greater than 50
so answer is E

3. In the XY plane, does the line with equation y = 3x + 2 contain the point (r,s)?
   
   1) (3r + 2 - s)(4r + 9 - s) = 0
   2) (4r - 6 - s)(3r + 2 - s) = 0
   
   In case 1, one of 2 brackets can be 0. If 3r+2-s = 0 then yes, line contains the point. Is 2nd bracket value is 0 , there is one valuse of r and s on which it does, but not for each value of r and s.
   
   Same for case 2.
   
   Coombined together, if both are true, 4r+9-s and 4r-6-S, can not both be 0 at the same time. So 3r+2-S = 0. This is same as y=3x+2. So if 3r+2-s = 0, this line will contain every (R,S) satisfying this equation.
   
   So C is the answer.

4. If x is positive which of the following could be correct ordering of 1/x, 2x, x^2.
   1) x^2 < 2x < 1/x
   
   2) x^2 < 1/x < 2x
   
   3) 2x < x^2 < 1/x
   
   Choices are:
   
   1) None
   2) 1 Only
   3) 3 Only
   4) I and 2 only
   5) 1, 2, 3
   
   
   

My answer would be 4) 1 & 2 only.

If x = 0.1, then x^2 < 2x < 1/x (so 1 is possible)

If x = 0.9, then x^2 < 1/x < 2x (so 2 is possible)

I don't think there is a positive value for x to make 3 possible.

Are x and y both positive?

1) 2x - 2y = 1

2) x/y > 1

I agree (1) or (2) are INSUFFICIENT. but consider both of them together

(1) 2x-2y = 1
(2) x/y >1

Rephrasing (1)

x-y = 1/2
x = y + 1/2

Now substituting value of x in (2)

(y+1/2)/y >1

Solving:

1+ 1/2y > 1

so, 1/2y > 0 it means y >0 since x = y+1/2 s0 X >0 as well.

For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative. What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6?

a. one
b. two
c. three
d. four
e. five

The number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative.

Our sequence is 1, -3, 2, 5, -4, -6?

We are looking for the number of pairs of consecutive terms for which the product of the two terms is negative.

The first consecutive terms of our sequence is (1,-3).

1*-3 = -3. Therefore, the product of these two terms is negative. This equals the firstvariation in sign.

The second consecutive terms of our sequence is (-3,2)

-3*2 = -6. Once again, the product of these two terms is negative. This equals the second variation in sign.

The third consecutive terms of our sequence is (2,5)
2*5 = 10. Since the product is positive it is NOT a variation in sign.

The fourth consecutive term in out sequence is (5-4)
5*-4 = -20. The product of these two terms is negative. This equals the third variation in sign.


The final term in our sequence is (-4,-6). -4 *-6 = 24. Since the product is positive it is NOT a variation in sign.

When you count them up, we have 3 variations in sign.

This question is misleading because it makes you assume that the test makers want to know how many negative numbers you can get in a row. If that was the case then the answer would be (B).

What they actually want to know is how many negative numbers you can get total. Therefore, the answer is C.

Of the 75 houses in a certain community, 48 have a patio. How many of the houses in the community have a swimming pool?

1) 38 of the houses in the community have a patio but do not have a swimming pool.

2) The number of houses in the community that have a patio and a swimming pool is equal to the number of houses in the community that have neither a swimming pool nor a patio.

from stmt 2

48 + x(swimming pool) -(y(common house)) + y(neither ) = 75
48 + x = 75
x = 27

If Bob produces 36 or fewer in a week, he is paid X dollars per item. If Bob produces more than 36 items, he is paid X dollars per item for the first 36 items, and 3/2 times that amount for each additional item. How many items did Bob produce last week?

1). Last week Bob was paid total of $480 for the items that he produced that week.

2). This week produced 2 items more than last week and was paid a total of $510 for the item that he produced this week.

When reproducing questions, please make sure that you do so exactly. I think I got everything here, but there are some words missing, which makes me wonder if something important was left out...

36 or fewer: X per item
> 36: X per item for first 36 + 1.5X for each item over 36.

How many items last week?

Some upfront thought: For exactly 36 items, Bob would be paid $36*X. For, say, 40 items, Bob would be paid 36*X + 4*1.5*X. I don't know what X is.

(1) Last week, Bob was paid $480. Maybe he made one item and is paid 480 per item. Maybe he made two items and is paid 240 per item. ??? Not sufficient. Elim A and D.

(2) Bob made Y items last week and Y+2 items this week. Bob make $510 this week. Maybe Bob made 1 item last week and 3 this week, and was paid 170 per item. Maybe Bob made 3 items last week and 5 this week and was paid 102 per item. ??? Not suff. Elim B.

(1) + (2) Bob was paid 480 for Y items last week. Bob was paid 510 for Y+2 items this week. Y = 480. Y + 2 = 510. So two items added an additional $30, of $15 per item. But wait - is this the regular X rate, or the "overtime" 1.5X rate?

If it's the regular rate, then he produced 480/15 items last week, or 32 items. That's one possibility. If it's the 1.5X rate, then this week he produced some number of items at $15 per and $10 per. He had to have made at least 36 items at the $10 per rate (in order to get to the point of making $15 per item). So that's 36*10 = 360. He made 510 this week, so he still has another 150 to make, at 15 per. That's 10 items. So Bob made 36 + 10 = 46 this week. Then according to statement 2, he made 44 last week (46-2).

Two answers. Not sufficient. Elim C. E is the answer.

Notice that I checked the work for (1) + (2) even once it became apparent that there are two possibilities. You can go with it or keep checking - I checked b/c I figured there was a possibility that one of the options wouldn't be acceptable given the parameters of the question, so I wanted to make sure. But if I was behind on time on the test, I might not have checked.

Six countries in a certain region sent a total of 75 representatives to an international congress, and no two countries sent the same number of representatives. Of the six countries, if Country A sent the second greatest number of representatives, did Country A send at least 10 representatives?

(1) One of the six countries sent 41 representatives to the congress
(2) Country A sent fewer than 12 representatives to the congress

I selected B on this one when the answer is E. I can kind of see why I am not correct, but am not completely sure. Can someone prove this to me mathematically? Thanks!

so this problem is basically concerned with EXTREMES: you're trying to figure the least, or greatest, number of representatives (or both) that could be sent in certain situations, in order to determine the range of possibilities. remember, then, if you want to figure extreme values, you have to consider extreme situations. here's one way you can progress through the problem:

-- (1) alone --

clearly 41 representatives = greatest number sent by any one country.
therefore, the countries from second place on down sent a total of 75 - 41 = 34 representatives.

EXTREME CASE 1: smallest possible # for the second country
in this case, you want to spread the remaining 34 representatives out as evenly as possible, so that the 2nd, 3rd, 4th, 5th, and 6th place countries are as near each other as possible.
34/5 = 6.8, so try to cluster the numbers around this average: the distribution with the least possible amount of variation is 9, 8, 7, 6, 4 (you can't get consecutive integers - try it for yourself)
therefore, the second greatest number of representatives must be at least 9
   

EXTREME CASE 2: largest possible # for the second country
in this case, you want to make the 3rd, 4th, 5th, and 6th values as small as possible. this is straightforward: make them 4, 3, 2, and 1 respectively.
this means that the 2nd place country sent 34 - 4 - 3 - 2 - 1 = 24 representatives
therefore, the second greatest number of representatives must be 24 or less
   

9 < second highest number < 24

insufficient


-- (2) alone --

in this case, there are no further restrictions on the numbers of representatives.

the highest number of representatives that country a could send is clearly 11.
therefore, the second greatest number of representatives must be 11 or less
   

to make the number as small as possible, just let the 2nd, 3rd, 4th, 5th, 6th place numbers be 5, 4, 3, 2, 1 respectively, and give all the rest of the representatives to the first place country.
therefore, the second greatest number of representatives must be at least 5
   

5 < second highest number < 11

insufficient


-- together --

we have
9 < second highest number < 24
AND
5 < second highest number < 11

therefore
9 < second highest number < 11

still insufficient

answer = e

The positive integers x, y and z are such that x is a factor of y and y is a factor of z. Is z even.

1) xz is even
2) y is even.

Answer is D.

I think the best way to solve this problem is to plug in different numbers.

1) xz = even.

Take x=3 y=6 and z=12. xz=even.. z is even
Take x=2 y=4 and z=8. xz=even.. z is even

Say xz is not even. Take x=1, y=3 and z=9.. z is not even.

So, sufficient.

2) y is even

An even integer will have all its multiples as even. Since z is a multiple of y, z has to be even.

So, sufficient.

D is the right answer

if the terms of a sequence are t1, t2, t3... tn, then what is n? Given information...?

is:

1) The sum of the terms is 3,124
2) Average of the n terms is 4

Given the first piece of information, we know that
t1 + t2 + t3 + .... + tn = 3124
The average of the n terms would be:
(t1 + t2 + t3 + .... + tn)/n
It is equal to 4
So,
(t1 + t2 + t3 + .... + tn)/n = 4
(t1 + t2 + t3 + .... + tn) = 4n
But, t1 + t2 + t3 .... + tn = 3124
So,
4n = 3124
n = 781
We need both pieces of information to find n. Each individual piece of information by itself has too many variables and we cannot isolate n.

On the number line, the distance between x and y is greater than the distance between x and z. Does z lie between x and y on the number line?

1) xyz <0

2) xy<0

The answer is E I belive.

Scenario A ) where XYZ < 0.

2 positives 1 negative
3 negatives.

For three negatives on the number the order can be z, x, y, 0...or x, z, y, 0.

So A is insufficient.

Scenario B) where XY < 0.

One negative one positive. So the order can go. x 0 z y...or z x 0 y. So B is insufficient.

Scenario A+B.

If X and Y are both of oppositie signs (XY<0)..then Z has to be postive in order for XYZ < 0.

So One Negative and two positives.

Order can be.....x 0 z y...or y 0 x z.

我是9号考

以下是引用must700在2009-7-13 13:29:00的发言：
这些都是哪里的题目啊？prep or jj?

基本是Prep

我的理解就象100RMB