<p>1)The graph above is a parabole whose equation is y = ax^2 + 2,where a is a constant If y = a/3x^2 + 2 is graphed on the same axes,which of the following best describes the resulting graph as compared with the graph above?
1)it will be narrower
2)it will be wider
3)it will be moved to the left
4)it will be moved to the right
5)it will be moved 3 units downard.</p>
<p>According to information given only 1 and 2 are possible,but how to solve this equation fast I had to solve it by plotting some points,which is not efficient way of doing it.</p>
<p>2)A box contains wood beads,red glass beads,and blue glass beads. The number of glass beads is 4 times the number of wood beads. If one bead is to be chosen at random from the box, the probability that a red glass bead will be chosen is 3 times the probability that a blue glas bead will be chosen. If there are 12 red glass beads in the box, what is the total number of beads in the box?</p>
<p>1)20
2)45
3)48
4)60
5)90</p>
<p>On the square gameboard that is divided into n rows of n squares each, k of these squares lie along the boundary of the gameboard. Which of the following is a possible value for k?
1)10
2)25
3)34
4)42
5)52</p>
<p>How to solve this problem and your not even given how much rows and how much column are there??!</p>
<p>1) You just need one piece of information to solve this problem quickly:</p>
<p>A controls stretch, and the smaller a is, the wider the graph. The bigger a is, the narrower the graph.</p>
<p>Counter-intuitive? Yes. </p>
<p>2) There are 20 beads total. First, the number of wood beads can be represented as x. The number of glass beads can be represented as 4x. There are 4 times the number of glass beads than wood beads. </p>
<p>The number of red beads can be represented as 3x, and the number of blue beads is x. There are 3 times as many red glass beads as blue glass beads, and remember, the number of glass beads must add up to 4x. </p>
<p>The problem then states that there are 12 red beads. Therefore, 3x = 12. x = 4.</p>
<p>Finally, the problem asks for how many beads there are in all. There are 5x beads in all, or 5(4), which equals 20.</p>
<p>3) The number of squares on the boundary increase in increments of 4. </p>
<p>2<em>2 square -> 4 squares on boundary
3</em>3 square -> 8 squares on boundary
4<em>4 square -> 12 squares on boundary
5</em>5 square -> 16 squares on boundary</p>
<p>K is 4 times n minus the four corner squares that cannot be counted twice. This means that k has to be a multiple of 4. Only 52 is. </p>
<p>Fwiw, n is (52+4)/4 or 14. Just in case you feel the urge to play with the numbers. </p>
<p>By the way, and before someone recommends to use plug numbers, be aware that this would be exactly why ETS expects one to do … and waste precious time. This problem could and should be solved by solely looking at the properties of the answers and the properties of a square board.</p>
<p>I agree and disagree with Xiggi. Certainly the solution he gave is the quickest and easiest. BUT, I think that the average student will be frightened just looking at this problem, and will not know where to begin. This is after all a Level 5 problem, and appeared as the last question of a math section on an actual SAT.</p>
<p>It does not take that long to try a few values of n, and recognize a pattern:</p>
<p>n=3</p>
<hr>
<hr>
<hr>
<p>So k=8</p>
<p>n=4</p>
<hr>
<hr>
<hr>
<hr>
<p>So k=12</p>
<p>n=5</p>
<hr>
<hr>
<hr>
<hr>
<hr>
<p>So, k=16</p>
<p>Thus we see that k must be divisible by 4, and the answer is choice (E).</p>
<p>As far as the argument that this method wastes time, I would argue that it doesn’t matter. This is the last question of the section. If you’re a really strong student, then you probably have all the time in the world. If you’re a weaker student, and you run out of time on this question it ultimately won’t hurt your score.</p>
<p>Thanks alot guys for your help I am studying alot these days to get 800,so I am covering everything I get wrong and understand concepts of solving them.</p>