<p>Hi guys,</p>
<p>I just wanted to make this post for hard questions in SAT math test before 19th november,so please post any question you have with its answer.</p>
<p>Those are questions from real test I solved today.</p>
<p>For sequence A, the nth term is 10n + 58.
For sequence B, the nth term is 2^n;</p>
<p>15)The nth terms of sequence. A and B are defined above above for all positive integers n. What is the least value of n for which the nth term of sequence B is greater than nth term of sequence A?</p>
<p>2^n > 10n + 58;</p>
<p>We can just input a value we know it works and then decrease n by 1 each time,instead of doing complicated algebra because this would require log() to solve.</p>
<p>2^10 > 10(10) + 58 = 1024 > 10(100) + 58; We know this work lets decrease by one.
2^9 > 10(9) + 58; 512 > 10(9) + 58; works too;
2^8 > 10(8) + 58 ; 252 > 10(8) + 58; works too;
2^7 > 10(8) + 58; 128 > 10(7) + 58; doesn’t work;</p>
<p>So answer here is 8;</p>
<p>Let the function g be defined for all values of x by g(x) = x(x - 1). If m is a positive number and g(m + 2) = 12, what is the value of m?</p>
<p>(m + 2)(m + 2 -1) = 12 - > (m + 2)(m + 1) =</p>
<p>m^2 + m + 2m + 2 = 12;
m^2 + 3m - 12 = 0;</p>
<p>We do here quadratic equation then we get the values:</p>
<p>M = -5 OR m = 2;</p>
<p>Please post your question you think is hard and you answer aswell.</p>
<p>18) </p>
<p>Grade 9:
Hall A = 155
Hall B = ?
Hall C = 25
TotalOfgrade 9 = ?</p>
<p>Grade 10:
Hall A = 85;
Hall B = 100;
Hall C = 20;
TotalOfGrade 10 = 205;</p>
<p>Grade 11:
Hall A = 20;
Hall B = 70;
Hall C = 115;
TotalOfGrade 11 = 180</p>
<p>Grade 12:
Hall A = 0;
Hall B = 75;
Hall C = 115;
Total Grade 12 = 190;</p>
<p>Sum of Hall A in all grades = 260;
Sum of Hall B in all grades = ?;
Sum of Hall C in all grades = 260;</p>
<p>18)
At Eastern High School, each student is assigned one locker. The locker may be either A,B, or C the partially completed table above shows the number of students from each grade assigned a locker in each of the three halls. If 20 percent of the students in grade 9 were assigned a locker in hall B. What is the total number of students who were assigned a locker in hall B?</p>
<p>We know that Grade 9 has Hall A,B,C and he said .2 percent student in 9 were assigned a locker in B,so we know that A + C = .8X;</p>
<p>.8x = 155 + 25;
x = 225;</p>
<p>Grade 9 Hall B = 225 - 155 - 25 = 45; for</p>
<p>then total students from Hall B are :</p>
<p>45 + 100 + 70 + 75 = 290;</p>
<p>Yeah, I remember these questions. These are from some recent exams.</p>
<p>Yes its october 2010 exam.</p>
<p>Another hard question plus how to solve it:</p>
<p>16) In a survey, 78 people were asked about two television programs, X and Y. Of people Surveyed 56 watch programs X, and 42 watch program Y, and 7 watch neither programs. How many of the people surveyed watch both programs ?</p>
<p>A)15
B)20
C)27
D)29
E)36;</p>
<p>This can be either solved by logic of by Algebra :</p>
<p>Lets see algebra way:
X = 42 - B;
Y = 65 - B;</p>
<p>X + Y = 78 - 7 = 71;</p>
<p>71 = 42 + - B + 56 - B;
71 - 98 = -2B;
-27 = -2B;
2B = 27;</p>
<p>Or we can do this by logic Of those 78 people 7 people didn’t watch,so 71 and we know that 56 + 42 = All watched plus the overlap.</p>
<p>so 56 + 42 - 71 = 21 now we remove how many acctually watched and get the overlap only.</p>
<p>^ So the answer is 27?</p>
<p>Yes I wrote 21 wrong my mind must have lagged or thought of number 21 while I acctually solved it as 27;</p>
<p>Redit:</p>
<p>so 56 + 42 - 71 = 27 = 98 - 71 = 27 now we remove how many acctually watched and get the overlap only.</p>
<p>I hate how your minds laggs and thinks of answer completely different what you got it as :@ I lost 2 question in my the exam I solved yesterday because of this.</p>
<p>It’s okay. You’re usually sharper in the real thing, due to anxiety, so you probably won’t be making too many stupid mistakes.</p>