Consolidated List of Solutions to Blue Book Math Problems

<p>598, #13</p>

<p>Primary equation: x [bulbs]<em>[$1/bulb] + y [bulbs]</em>[$2/bulb] = $600 or
x + 2y = 600</p>

<p>Secondary equation: "If twice as many $1 bulbs as $2 bulbs were ordered..." means 2y = x.</p>

<p>Substitute...</p>

<p>2y + 2y = 600
4y = 600
y = 150
so, x = 2(150) = 300</p>

<p>Total # of bulbs = 150 + 300 = 450</p>

<p>Can someone help me understand these questions?? They are...</p>

<p>If K(2X+3)(X-1) = 0 and x > 1, what is the value of K?
(A) -3/2
(B) 0
(C) 2/3
(D) 1
(E) 2
The answer is B.</p>

<p>If n and p are integers greater than 1 and if p is a factor of both n + 3 and n + 10, what is the value of p?
(A) 3
(B) 7
(C) 10
(D) 13
(E) 30
The answer is D.</p>

<p>Why are these answers correct????</p>

<p>At the risk of sounding completely clueless, I must ask-- what is "the blue book?"</p>

<p>Caesar:</p>

<p>For the first one :</p>

<p>we have three different numbers:k,(2x+3) and (x-1) . The given equation gives their product to be 0 . </p>

<p>So , one or more of these three numbers has to be zero in order to satisfy the equation.</p>

<p>But,it is also given that x>1.By this condition,(2x+3) becomes strictly positive in all cases and the number (x-1) is ALSO strictly positive because since x>1; x-1>0 .</p>

<p>So,the only possible number that can be zero is k .Hence,k=0.</p>

<p>For the second one :</p>

<p>we have n,p>1 . </p>

<p>I'll give you an alternative method for this.Suppose you have 5 which is a factor of both 20 and 25 . Now,since 5 is a factor of both 25 and 20 , 5 is also a factor of (25-20) . </p>

<p>Applying this, since p is a factor of both n+3 and n+10 ,it is also a factor of (n+10) - (n+3) .Which comes out to 7 . Now,since 7 is prime , you cannot go further than this and hence p=7. </p>

<p>Ohh,the answer is B ,not D .Check your book ;) </p>

<p>Sparkshooter : The infamous blue book is The Official SAT Study Guide.</p>

<p>Caesar (or anyone else), can you post page and problem numbers from the blue book for the problems you posted?</p>

<p>thanks iceman!!!!!!!</p>

<p>tanman,
The first one is on pg. 793 (question #9).
The second one is on pg. 807 (question #14).</p>

<p>this one wasnt listed yet...pg 491 # 15:</p>

<p>the lengths of the sides of a right triangle are consecutive even integers, and the length of the shortest side is x. which of the following equations could be used to find x?</p>

<p>b. x^2+(x+1)^2=(x+2)^2
c. x^2+(x+1)^2=(x+4)^2</p>

<p>i got b, but they're saying the answer is c and i dont see how...</p>

<p>thanks</p>

<p>p. 491, #15</p>

<p>Consecutive even integers where x = smallest one...
x, x + 2, x + 4</p>

<p>In a right triangle, the hypotenuse is always the longest side, and the sum of the squares of the two legs = the square of the hypotenuse (the Pythagorean Theorem)...</p>

<p>So, we have
x^2 + (x + 2)^2 = (x + 4)^2, choice (C). [You accidentally typed part of choice (C) incorrectly.]</p>

<p>thanks. i realized what i did wrong - i forgot that the numbesr had to be even integers. its so ironic how most of my mistakes on math come from my inability to read the question correctly (ie my critical reading skill) altho i do well on critreading...</p>

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<p>Please use the new</a> solutions thread</p>