2 "Hard" Math Questions -- help please?

Test Preparation SAT Preparation
1

<p>20) If (x+y) X [(x^2)-(y^2)] = 0, which of the following must be true?</p>

<p>A) x = y
B) x = -y
C) x^2 = y^2
D) x^2 = -y^2
E) x^3 = y^3</p>

<p>18) A two digit number, XY, where X and Y are digits, is 3 times the sum of its digits. Which of the following equations could be used to represent the statement above?</p>

<p>A) 3(X)+Y = 10(x)+Y
B) 3(X)+Y = X+Y
C) 3(X+Y) = 10(X+Y)
D) 3(X+Y) = 10(X)+Y
E) 3(X+Y) = X+Y</p>

<p>Thanks in advance, these problems have been raping me in the face...
(note: these are "official" problems from the 2006 PSAT)</p>

2

<p>20: x2-y2=(x+y)(x-y), so it's (x+y)2(x-y)=0, so x=y or x=-y, thus C is correct.</p>

<p>18.well XY=10X+Y...so D is right</p>

3

<p>^ agreed. .</p>

4

<p>thanks DanQ, could someone please elaborate a bit on 18? I don't really get that at all</p>

5

<p>18) A two digit number, XY, where X and Y are digits, is 3 times the sum of its digits. Which of the following equations could be used to represent the statement above?</p>

<p>A) 3(X)+Y = 10(x)+Y
B) 3(X)+Y = X+Y
C) 3(X+Y) = 10(X+Y)
D) 3(X+Y) = 10(X)+Y
E) 3(X+Y) = X+Y</p>

<p>A two digit number has two digits. The unit digit (the digit farthest to the right) is the one's digit, its multiplier is 1. The digit to the left of the ones digit is the tens digit, its multiplier is 10.</p>

<p>so, a number like 23 can be expressed as 2(10) + 3(1).</p>

<p>That goes for any other number in base 10 (what we count in)</p>

<p>2348 = 2(1000)+3(100)+4(10)+8(1)</p>

6

<p>wouldn't number 20 be B?</p>

7

<p>well....actually A and B are both possible. and A+B=C, right? so the correct answer is C.</p>

8

<p>It's not arithmetical addition, if that is what you mean.</p>

<p>(x+y)^2(x-y)=0</p>

<p>The two factors are (x+y) and (x-y). You're basically solving a cubic function for its roots. If either (x+y) or (x-y) equates to zero, then regardless of the other factor, the entire expression equates to zero.</p>

<p>You can try it with numbers:
x= 5 , y = -5</p>

<p>(5+5)^2 * (5-5) = 0</p>

<p>or</p>

<p>x = 5, y = -5
or
x=-5 y = 5</p>

<p>the correct answer, which I believe is C, x^2=y^2 is basically using the exponent as the absolute value function.</p>

<p>|x| = |y|</p>

9

<p>These three are equivalent:
x^2 = y^2
|x| = |y|
x=y or x=-y</p>