SAT Math Bible Help

Test Preparation SAT Preparation
1

<ol>
<li>
If 1 < p < 3, then which of the following could be true?
(I) p2 < 2p
(II) p2 = 2p
(III) p2 > 2p
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III</li>
</ol>

<p>In their answer explanation, they used three different values for P to show that all of the roman numerals are true. My question is can you do that? Because I had just used p=1.5, and I found that only roman numeral number 1 is true.</p>

<p>The actual answer explanation is: </p>

<p>If p = 3/2, then p2 = (3/2)2 = 9/4 = 2.25 and 2p = 2 ⋅ 3/2 = 3. Hence, p2 < 2p, I is true, and clearly II
(p2 = 2p) and III (p2 > 2p) are both false. This is true for all 1 < p < 2.
If p = 2, then p2 = 22 = 4 and 2p = 2 ⋅ 2 = 4. Hence, p2 = 2p, II is true, and clearly I (p2 < 2p) and III
(p2 > 2p) are both false.
18 SAT Math Bible
If p = 5/2, then p2 = (5/2)2 = 25/4 = 6.25 and 2p = 2 ⋅ 5/2 = 5. Hence, p2 > 2p, III is true, and clearly I
(p2 < 2p) and II(p2 = 2p) are both false. This is true for any 2 < p < 3.
Hence, exactly one of the three choices I, II, and III is true simultaneously (for a given value of p). The answer is (E).</p>

<ol>
<li>
If 42.42 = k(14 + m/50), where k and m are positive integers and m < 50, then what is the value of
k + m ?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10</li>
</ol>

<p>My question is that they gues three random numbers, and then proved that E is the correct answer. However, is there any other way I can do this problem without guessing? Maybe an algebraic equation or a way to know WHAT numbers to guess?</p>

<p>Answer explanation from the book is:</p>

<p>We are given that k is a positive integer and m is a positive integer less than 50. We are also given that
42.42 = k(14 + m/50).
Suppose k = 1. Then k(14 + m/50) = 14 + m/50 = 42.42. Solving for m yields m = 50(42.42 – 14) =
50 </p>

2

<p>Hm, well, if I were you, I’d go with what the book says, unless you can really prove that there are alternate ways of doing things. If it says to try different numbers, then do so. You are free to try weird numbers, just in case, as long as they fit the criteria of the problem.</p>

3

<ol>
<li><p>Yes, plugging in numbers is your best option. It’s not particularly slow, but you have to make sure you satisfy all cases. For the domain -2<x<2 for="" example,="" you="" need="" to="" test="" in="" between="" -1<x<1="" and="" 1<x<2="" -1="">x>-2</x<2></p></li>
<li><p>While solutions are usually the best answer, this solution is WAY TOO LONG.
If 42.42 = k(14 + m/50), where k and m are positive integers and m < 50, then what is the value of
k + m ?
if m<50, then (14+m/50) must be greater than 14 (assuming m does not equal 0) and less than 15. Since 14x3=42 and 15x3=45, you can immediately deduce that k=3, since 42.42 is in between 42 and 45. Knowing that k must be 3, solve for m and simply add k and m.
P.S. Please use the carot sign to indicate powers. p2=2p is really annoying because that’s not what you really mean. it should be p^2=2p</p></li>
</ol>

4

<p>For#1, the trouble you have with their explanation is that you are misunderstanding what it means when they say “could be”. That doesn’t mean that the options could all be true using the same p value but rather that there are p values which fall in the permitted range which make each option possible. </p>

<p>Let me give you an easier example to illustrate what I mean: </p>

<p>I’m thinking of an integer between one and ten. Which of the following could be true?
i. The number is even
ii. The number is odd</p>

<p>The correct answer is that both i and ii COULD be true depending on what # you picked, even though there is no SINGLE number you could have picked to make them both simultaneously true.</p>

<p>(I hope this is what you were asking!)</p>