Aliases

This is really weird, it was from blue book.

Q1. There are 6 red, 6 brown, 6 yellow, and 6 gray scarves packaged in 24 identical, unmarked boxes, 1 scarf per box. What is the least number of boxes that must be selected in order to be sure that among the boxes selected 3 or more contain scarves of the same color?

A. 3
B. 6
C. 7
D. 8
E. 9

Q2. 2x-5y=8, 4x+ky-17
For which of the following values of k will the system of equations above have no solution?

A. -10
B. -5
C. 0
D. 5
E. 10

Q3. RESULTS OF BEANBAG
Number of Throws Number of People
1 7
2 6
3 6
4 4
5 2

In a certain game, each person threw a beanbag at a target until the person missed the target. The table shows the results for the 25 people who played the game. For example, 4 people hit the target on their first 3 throws and missed their 4th throw. Based on the information in the table, which of the following must be true?

I. More than half the people hit the target on their first throw.
II. For all the throws attempted, more hit the target than missed the target.
III. No one hit the target 5 times.

A. I only
B. II only
C. I and III only
D. II and III only
I, II, and III

P.S. I know the answer I want you to tell me what you think the correct answer is and how you got it because I do not get these problems at all.

noocie

Q. 1
The answer is E. 9. Think of the worst case scenario - you could select 8 boxes with only 2 of each color. The next scarf has to give you three of one color.

Q. 2
First, we multiply 2x - 5y = 8 by (-2) and add it on to 4x + ky = 17
4x + ky = 17
-4x+10y = -16
---------------
10y + ky = 1
Then we factor out the y-
(10 + k)y = 1
To get a system with no solution, we need to make y equal 0. To do this, (10 + k) must equal 0. The answer is A. -10.

Q. 3
I. Only 7/25 people missed the target on the first throw, so this statement is true.
II. You can validate this statement by plugging the numbers into an equation to get the number of "true" hits to compare to the 25 misses. The equation I used for each group:
(number of people who missed on that throw)(throw number-1) = number of hits for that throw number. Then you add all the numbers together like so:
(7)(0) + (6)(1) + (6)(2) + (4)(3) + (2)(4)
You get 38 hits, which is greater than the 25 misses.
III. This statement is true because the table only goes up to 5 throws and if you add (7 + 6 + 6 + 4 + 2) you get 25.
The answer is E. I, II, and III.

Please let me know if you need any clarification

fogcity

Q2. 2x-5y=8, 4x+ky=17

Another way of thinking about this question is to observe that two lines have "no" solutions if they are parallel. So write both equations in the form y = mx + b. Recall that m is the slope of the line and b is the y-intercept. Two lines are parallel when their slopes are equal.

So rewrite 2x -5y = 8 as y = (2/5)x - (8/5) ... slope is (2/5)
and rewrite 4x + ky = 17 as y = (-4/k)x + (17/k) ... slope is (-4/k)

The slopes are equal when k = -10

KingAurora

Wow question 1 was exactly like a question on this year's PSAT (may I add that I got it wrong lolol)

DrSteve

Number 2 can be done very quickly by observing that you multiply 2 by 2 to get 4. So (-5)(2) = -10, choice (A).

Remarks: The 2 lines are in the general form ax+by=c. The slope of this line is m =-a/b unless b=0, in which case the line is vertical and has no slope.

Let's take 2 such lines:

ax+by=c
dx+ey=f

If there is a number r such that ra=d, rb=e and rc=f, then the two equations represent the same line, and the systm of equations has infinitely many solutions.

If there is a number r such that ra=d, rb=e but rc is not f, then the two equations represent parallel lines, and the systm of equations has no solution.

Otherwise the system has a unique solution.