<p>It'd be nice to have all of the hardest math problems from each month's SAT posted in one thread.</p>
<p>1) Open BlueBook
2) Go to test number #1.
3) Look at answer page for test #1.
4) Under the math sections find the ones that have numbers 4 or 5 next to them or the letter “H”.
5) Repeat process for test #2, #3, #4, #5, #6, #7, #8, #9, and #10.</p>
<p>here is one.What is the greatest possible radius of a sphere that can fit in a cube with side 2?
If one side of a triangle is 12(leg) and the other is 6(leg),What is the greatest number of triangles that can fit if each have a side length of 4?</p>
<p>Thanks for replying avidstudent, but I’m looking for new math problems that people commonly have trouble with on past SATs.</p>
<p>All three questions were given the difficulty level of 5.</p>
<p>May 2010:</p>
<p>Section 2
18. What is the greatest four-digit integer that meets the following three restrictions?
(1) All of the digits are different.
(2) The greatest digit is the sum of the other three digits.
(3) The product of the four digits is divisible by 10 and not equal to zero.</p>
<p>Section 6
17. The first term of a sequence is a. Each term after the first is p more than the preceding term. Which of the following expressions is the sum of the first 6 terms of the sequence?
a. 6a + 5p
b. 6a + 15p
c. 6a + 18p
d. 12a + 5p
e. 12a + 30p</p>
<p>Section 9
16. The sum of r and 5 is less than the square of the difference between r and 5 but greater than the square of one-half r.
Which of the following statement is equivalent to the statement above?
a. 1/4(r^2) < r + 5 < (r-5)^2
b. 1/4(r^2) < r + 5 < r^2 - 5
c. 1/2(r^2) < r + 5 < (r-5)^2
d. 1/2(r^2) < r + 5 < r^2 - 5
e. 1/2(r^2) < r^2 - 5 < r^2 + 5</p>
<p>AvidStudent: Where did you find a copy of the May 2010 SAT? Also, do you happen to have the answers as well? </p>
<p>Thanks for posting those questions to share :)</p>
<p>A friend of mine gave me some old released QAS booklets.
Yes, I do have the answers and when you or anyone else finishes the problem they can gladly PM to see if they got the questions correct.</p>
<p>8521, b, a?</p>
<p>^
Try 9531</p>
<p>^<a href=“3”>quote</a> The product of the four digits is divisible by 10 and not equal to zero.
[/quote]
</p>
<p>9 x 5 x 3 x 1 = 135</p>
<p>8521, b, a</p>
<p>From October 2010 SAT-Level 5</p>
<p>The integers 1-6 appear on the six faces of a cube, one on each face. If three such cubes are rolled, what is the probability that the sum of the numbers on the top faces is 17 or 18?</p>
<p>A) 1/108
B) 1/54
C) 1/27
D) 1/18
E) 1/16</p>
<p>^
There are only 4 ways that they sum to 17 or 18(5,6,6 6,5,6 6,6,5 and 6,6,6). There are 6^3=216 possible rolls. Thus, the probability is 4/216 or 1/54</p>
<p>
</p>
<p>How did you guys solve this?</p>
<p>^
Since it’s divisible by 10, one of the digits must be 5 and one must be even.</p>
<p>“Section 2
18. What is the greatest four-digit integer that meets the following three restrictions?
(1) All of the digits are different.
(2) The greatest digit is the sum of the other three digits.
(3) The product of the four digits is divisible by 10 and not equal to zero.”</p>
<p>As said above, one digit must be 5, and the other must be 8 or 2 or 4, for the product to be divisible by 10.
Because the sum must be equal to the largest number, 5 and 8 don’t work (sum would be 13, which is two digits) and 5 and 4 don’t work (the sum would be 9, but there must be another digit, which would increase the sum to a two-digit number).
So 5 and 2 are in the number.
The only numbers that 5 and 2 and some unknown number could sum to would be 9 or 8. Because all digits must be different the number of possibilities is reduced to only 1, as 2 has already been used.</p>
<p>So, 5,2,1, sum to 8. To arrange them so the number is the largest, the orderis highest first.</p>
<p>8,5,2,1
8521</p>
<p>"
Section 6
17. The first term of a sequence is a. Each term after the first is p more than the preceding term. Which of the following expressions is the sum of the first 6 terms of the sequence?
a. 6a + 5p
b. 6a + 15p
c. 6a + 18p
d. 12a + 5p
e. 12a + 30p</p>
<p>This is the easiest of the bunch, IMO.
Turn the expression into an arithmetic series:
a+(a+p)+(a+p+p)+(a+p+p+p)+(a+p+p+p+p)+(a+p+p+p+p+p)</p>
<p>Add up the a’s and p’s to get:
6a+15p
Alternatively, use a formula to get the summation but this is pretty simple.
Section 9
16. The sum of r and 5 is less than the square of the difference between r and 5 but greater than the square of one-half r.
Which of the following statement is equivalent to the statement above?
a. 1/4(r^2) < r + 5 < (r-5)^2
b. 1/4(r^2) < r + 5 < r^2 - 5
c. 1/2(r^2) < r + 5 < (r-5)^2
d. 1/2(r^2) < r + 5 < r^2 - 5
e. 1/2(r^2) < r^2 - 5 < r^2 + 5
"
This is converting the statement into inequalities.
(r/2)^2<r+5<(r-5)^2
Then rearrane a bit to get the answer.</p>
<p>
</p>
<p>(3) gives us 5 and 2.
9 as the greatest digit doesn’t work. It would give: 952_, where _ would have to be 2 for 5 + 2 + _ to total 9, which goes against (1).
8 as the greatest digit satisfies the requirements: 8521.</p>
<p>
</p>
<p>a.</p>
<p>bump…</p>
<p>I found this one tricky :</p>
<p>There are 27 students in Mr White’s homeroom. What is the probability that at least 3 of them have their birthdays in the same month?
A) 0
B) 3/27
C) 3/12
D) 1/2
E) 1</p>
<p>ANSWER </p>
<p>E - 1
If there were no month in which at least 3 students had a birthdays, then each month would have the birthday of a most 2. But that’s not possible; even if there were 2 birthday’s in January, 2 in February … only 24 students would be accounted for. It is guaranteed that, with more than 24 students, at least 1 month will have 3 or more birthdays. The probability is 1 [/spoiler]</p>