SAT questions from real exam

<p>Hi guys,</p>

<p>Was solving some real exams got these 4 questions wrong.</p>

<p>Family - Number of Consecutive Nights</p>

<p>Jackson 10
Callan 5
Epstein 8
Liu 6
Benton 8</p>

<p>8) The table above shows the number of consectuvie nights that each of five families stayed at a certain hotel during a 14-night period. If the Lie family's stay did not overlap with the Benton Family's stay, which of the following could be a night on which only one of the five families stayed at the hotel?</p>

<p>1)The 3rd
2)The 5th
3)The 6th
4)The 8th
5)The 10th</p>

<p>14)How many different ordered pairs (x,y) are there such that x is an even integer, where 4<= x <= 10, and y is an integer , where 4 < y < 10
1)8
2)12
3)20
4)30
5)36</p>

<p>A sphere of radius r inside a cube touches each one of the six sides of the cube. what is the volume of the cube, in terms of r?</p>

<p>1) r^3
2) 2r^3
3) 4r^3
4)4/3PIr^3
5)8r^3</p>

<p>In the figure above, SQUARE RSTU is inscribed in the circle. What is the degree measure of ar ST?</p>

<p>1)45
2)60
3)90
4)120
5)180.</p>

<p>No.8
I’m not very good with these kind of problems but I think it’s the tenth</p>

<p>No.14
answer is 20</p>

<p>one with the sphere is 8r^3</p>

<p>one with the square is 90</p>

<p>tell me which ones I got correct so i can tell you how i solved it.</p>

<p>Both are correct can you please tell me how you solved it.</p>

<p>I can’t really explain the first one, it’s confusing.</p>

<p>14) 4<= x <= 10, and y is an integer , where 4 < y < 10</p>

<p>X= 4 possibilities and y= 5 possibilities</p>

<p>4*5 = 20 which is the right answer</p>

<p>the one with the sphere) you have to realize the the side of the cube equals 2r and that the volume of the cube equals the side^3 bottom line, 2r<em>2r</em>2r= 8r^3</p>

<p>the one with the square) this is a fairly easy one, the sides of the squares are equal and each side intersects with two points on the circle dividing it into 4 equal arcs which means that the length of one arc is 360/4 = 90</p>

<p>tell me if your still confused and i will give a more elaborate expalanation</p>

<p>for no. 14) first you have 4<=x<=10 , 4<y<10 & the imp one is that( x is an even integer )
just write the numbers between 4 &10 ,but take care he said less than (or equal) it will be
4<=x<=10 (4,5,6,7,8,9,&10) because x is an even integer we gonna delete the odd numbers (5,7,&9)
4<y<10 (5,6,7,8,&9)<br>
finally we have (4,6,8,&10) for x & (5,6,&9) for y .to get the answer we we gonna write (x,y) (4,5),(4,6),(4,7),(4,8),(4,9) …till (10,9) and u will get 2o just by counting them .</p>

<p>for 14 I really got bad error I could have solved it,but I counted all numbers even/odd I made 10 - 4 + 1 = 7; and 10 - 4 - 1 = 5;</p>

<p>so I got my result as 7 * 5 = 35;</p>

<p>Thanks alot guys I hope someone come and explain number 8 later,because I don’t really see a way to solve it he doesn’t give any information!</p>

<p>if Liu stayed the first 6 nights and each family came on the 4th night then it makes sense because Jackson the one with the most amount of nights if he comes on the fourth night then he’ll leave on the 14th night</p>

<p>For number 1.
Jackson 10
Callan 5
Epstein 8
Liu 6
Benton 8</p>

<p>Make a table. We’re gonna try and push all the families to the bottom, let me demonstrate. Push the Jackson family from the 3rd to the 13th nights. That’s 10 nights. Then push all the other families under the Jackson family, which shouldn’t be a problem, since they all stayed less than 10 nights. Take a look at the table, and figure out on which night was the hotel occupied by only 1 family. Do you get it now?</p>

<p>Jacksons stayed from 3rd - 13th (10 nights)
Callan from 5th - 10th (5 nights)
Epstein from the 4th - 12th (8 nights)
Benton, same as Epstein. (8 nights)
Liu from the 4th till the 10th. (6 nights)</p>

<p>As you can see, they’ve all been pushed below the Jackson family, which was present since the 3rd night. The answer is A) 3rd night.</p>

<p>I did the same as Sir…sort of</p>

<p>You really only need to worry about 3 families…
10days for Jackson
not overlapping Lui and Benton (together equal 14 days)</p>

<p>So you know that you need either benton or lui family to be the one by themselves. </p>

<p>If you put the 10 night jackson stay on nights 1-10, then they it will not work with any answer available. </p>

<p>So… put in the Jacksons from the 5-14th night. Then put the Liu family in nights 1-6, and Benton nights 7-14. All the rest could stay anytime from 4-14. </p>

<p>So, the only family present on days 1-3 are the Liu family. </p>

<p>your answer is still night 3.</p>

<p>Lots of ways, but no matter how hard you try, you’ll still end up with night 3 being the only night in which only one family staying was possible.</p>

<p>Except, sir, you can not have bentons and Epstines staying the same nights, as that would over lap with Lui. Benton and Lui can not overlap.</p>

<p>I really don’t understand your method, but at the same time it doesn’t matter, what matters is that the answer is the same. OP can choose whichever method he sees fit.</p>

<p>Thanks guys for number 8 I see its logic now.</p>

<p>@SirWanksalot</p>

<p>Your method is wrong. Lui and Benton can’t overlap; it’s in the question.</p>

<p>@SirWanksalot Jackson can’t come on the 3rd day cause his stay will overlap with Liu if he’s staying the first 6 days, Jackson has to come on the fourth day and every other family has to come after Jackson to make the 3rd day occupied by one family which is Liu</p>

<p>My bad, I’ve done this question before and chose A, but I forgot all the info and didn’t actually read it. I vividly remember my solution now.</p>

<p>Nights 1-6 : Liu
Nights 4-10: Jackson
Nights 7-14: Bentons, and Epsteins
Nights 9-14: Callans</p>

<p>There you have it.</p>

<p>How do you do the last two problems? </p>

<p>Also, on the hotel problem why isn’t it the tenth night?</p>

<p>@000ooo000ooo</p>

<p>It can’t be on the 10th night because if you put all the families except for the Benton’s in front like:
Nights #1-6: Liu</p>

<h1>1-8: Epstein</h1>

<h1>1-5: Callan**</h1>

<h1>1-10: Jackson**</h1>

<h1>7-14: Benton</h1>

<p>On the 10th night, the Jackson and Benton families are in the hotel. Therefore, there are two families that night so it can’t be the 10th night!</p>