Real Math IIC questions - Come test your knowledge!

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<p>Answer and explain your methods for the following old Math IIC questions. You’ll be much loved by the CC community.</p>

<li>If the magnitudes of vectors a and b are 5 and 12 respectively, then the magnitude of vector (b-a) could NOT be</li>
</ol>

<p>a. 5
b. 7
c. 10
d. 12
e. 17</p>

<li><p>if the 20th term of an arithmetic sequence is 100 and the 40th term of the sequence is 250, what is the first term of the sequence?
a. -50
b. -42.5
c. 5
d. 42.5
e. 50</p></li>
<li><p>an indirect proof of the statement “If x = 2, then radical x is not a rational number” could begin with the assumption that
a. x = radical 2
b. x^2 = 2
c. radical x is rational
d. radical x is not rational
e. x is nonnegative</p></li>
<li><p>In how many ways can 10 people be divided into two groups, one with 7 people and the other with 3 people?
a. 120
b. 210
c. 240
d. 5,040
e. 14,400</p></li>
<li><p>Which of the following has an element that is less than any other element in that set?
I. The set of positive rational numbers
II. The set of positive rational numbers r such that r^2 => (equal to or greater than) 2
III. The set of positive rational numbers r such that r^2 > 4</p></li>
</ol>

<p>A. None
B. I only
C. II only
D. III only
E. I and III</p>

<p>Major round of applause to whoever can explain the last 2 questions - those were the hardest on the test.</p>

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<ol>
<li>In how many ways can 10 people be divided into two groups, one with 7 people and the other with 3 people?
a. 120
b. 210
c. 240
d. 5,040
e. 14,400</li>
</ol>

<p>This question is basically asking hm ways can make a 7 out of 10 group (b/c the left over people are obviously all going to be in the other group)</p>

<p>So 10!/7!3!= 120</p>

<p>There is a thread for #48 somewhere.</p>

3

<p><em>Applauds Skeet Daddy</em>! I get it now. One more question:</p>

<p>What if it they asked: how many ways can 10 people be divided into two groups, one with 6 people and the other with 5 people?</p>

<p>would that be </p>

<p>10 choose 6 * 10 choose 5</p>

<p>or 10 choose 6 + 10 choose 5?</p>

4

<p>Eh...6+5 = 11</p>

<p>Multinomial theorem!!!</p>

5

<ol>
<li> When subtracting vectors b-a, you draw vector a so that its tail touches the head of vector b. Vector b-a is the vector drawn between the tail of b and the head of a. So this is essentially a triangle! Yay! Since the magnitude, or length, of b is 12 and the magnitude of a is 5, we know that the length of the remaining side of the triangle must be less than a + b (17). Answer E, 17, is the answer because it cannot be the magnitude.</li>
</ol>

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<ol>
<li>if the 20th term of an arithmetic sequence is 100 and the 40th term of the sequence is 250, what is the first term of the sequence?
a. -50
b. -42.5
c. 5
d. 42.5
e. 50</li>
</ol>

<p>We can find d, or the amount by which this arithmetic series increments, by plugging stuff into the formula:</p>

<p>an = a1 + (n-1)d</p>

<p>But we don't have a1!, you say. What if we simply shift our frame of reference over a bit? Set a20 to be a1 by subtracting 19 from n and do the same to a40 to get a21.</p>

<p>Now we can find d:</p>

<p>250 = 100 - 20d
d = 7.5</p>

<p>Now use the formula but with the original references to get a1.</p>

<p>100 = a1 + 19(7.5)
-42.5 = a1</p>

<p>Thus B is the answer.</p>

7

<p>I get confused on the types of probs presented in #47. In the same vein of obsessedAndre's question, what if they had said:</p>

<p>In how many ways can 10 people be divided into two groups, one with 5 people and the other with 4 people? (Yah know, with a random person left over...) </p>

<p>Is it (10 choose 5) * (5 choose 4)???</p>

8

<p>Aniviel, you're right on for #40. However, the answer to number 37 is actually A. Only 24% of people got it right on the test.</p>

<p>And your answer makes sense to me..</p>

9

<p>Agh! God! Okay, I see now. I was basing this on triangle properties, but we are working with vectors, so....AGHH! Okay. The reason that 17 works is because vectors a and b could be pointing in exactly opposite directions. Reverse the direction of a because you're subtracting. When you draw a line from the tail of b to the tip of a, you get a line that is the length of the sum of the two vectors. 5 could not be the answer because, first of all, you couldn't form a triangle with sides 5, 5, 12. The sum of the shorter sides must be equal to or greater than the longer side to form at least a line of the same length, or a triangle if they are greater than. </p>

<p>Yurg.</p>

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<p>Here's a practice question:</p>

<p>Factor (x^7 - y^7)</p>

11

<p>A simpler way for doing # 47
10 nCr 3 = 120
or
10 nCr 7 = 120</p>

<p>the nCr function is on any TI calculator</p>

12

<p>can anyone do 48 and 45? I have guesses but not sure</p>

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<p>Whats The Answer To Number 37? I Know Its Not (e)!!!</p>

14

<p>45: Proof by contradiction, assume what is to be proven false. Thus assume rad2 is rational.
48: No rational number is smaller than any other rational number. You can keep making the denominator a larger integer. Thus you can keep adding modicums of fractions to 0 and same for 2. For radical 2, since it never terminates, you can always add fraction that is smaller as the digits get further to the right.</p>

<p>37: Triangle inequality. </p>

<p>The vectors form a triangle or line (degenerate triangle) with sides 5, 12, and x. 12-5 <= x, thus the third side (|a-b|) is equal to any number >= 7. 5 isn't.</p>

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<p>wait, so what's the answer to 48? C, right?</p>

16

<p>48's answer is A none.</p>

<p>also factoring xˆ7 - yˆ7 you get:
(x-y)*
(x^6+x^5y+x^4y^2+x^3y^3+x^2y^4+xy^5+y^6)</p>