Here is how i solved these -
1)If d is a positive integer and f is the product of the 1st 30 positive integers ,what is the value of d?
a) 10^d is a factor of f
b) d>6

my ans:
From a)
The product of first 10 integers would atleast contain 3 zeros for sure towards the end due to the presence of 10, 20 and 30.
Using this info, d can be 1, 2 or 3. So A is insufficient.

From b)
d>6 says nothing, so insufficient.

Combining, in addition to 10,20 and 30, we can get zeros towards the end from, 2,5; 12,15, 25,4; which gives us 7 zeros at the end of the product. and since d>6, d has to be 7. So both together are sufficient.

2)if x<0,then SQRT(-X|X|) is
a)-x
b)-1
c)1
d)x
e)sqrt(x)
Consider x=-3,
Sqrt(--3|-3|) = sqrt(3.3)=3, which is -x.

3)A list consists of 5 different integers. Is the Average of the 2 greatest integers in the list greater than 70?
a) median of the list is 70
b) mean of the list is 70

Ans: a) median is 70, so the greatest 2 numbers have to be greater than 70 or will be both equal to 70.
So a is sufficient.

b) mean is 70. which again means that the greatest 2 numbers should be greater than 70 or both equal to 70.
hence sufficient.

4)In a class of 30 students,2 students did not borrow any books from the lib,12 borrowed 1 book,10 borrowed 2 books and the rest of the studets borrowed atleast 3 books.If the avg number of books borrowed per student was 2 ,what is the maximum number of books that any single student coupld have borrowed?
a)3
b)5
c)8
d)13
e)15
Given that avg books borrowed per student is 2. which means totally 30x2=60 books were borrowed.
its also given that 12 borrowed 1 book - 12x1 = 12
similarly, 10x2=20.
which adds up to 32. The rest 28 were borrowed by
30-2-12-10=6 students.
these 6 students have borrowed atleast 3 books. To maximize the no. of books for one person, it would be 3x5=15, and hence max of 13 books for one person.

hope this helps. Folks do correct in case you find any thing missing here.

Here is how i solved these -
1)If d is a positive integer and f is the product of the 1st 30 positive integers ,what is the value of d?
a) 10^d is a factor of f
b) d>6

my ans:
From a)
The product of first 10 integers would atleast contain 3 zeros for sure towards the end due to the presence of 10, 20 and 30.
Using this info, d can be 1, 2 or 3. So A is insufficient.

From b)
d>6 says nothing, so insufficient.

Combining, in addition to 10,20 and 30, we can get zeros towards the end from, 2,5; 12,15, 25,4; which gives us 7 zeros at the end of the product. and since d>6, d has to be 7. So both together are sufficient.

2)if x<0,then SQRT(-X|X|) is
a)-x
b)-1
c)1
d)x
e)sqrt(x)
Consider x=-3,
Sqrt(--3|-3|) = sqrt(3.3)=3, which is -x.

3)A list consists of 5 different integers. Is the Average of the 2 greatest integers in the list greater than 70?
a) median of the list is 70
b) mean of the list is 70

Ans: a) median is 70, so the greatest 2 numbers have to be greater than 70 or will be both equal to 70.
So a is sufficient.

b) mean is 70. which again means that the greatest 2 numbers should be greater than 70 or both equal to 70.
hence sufficient.

4)In a class of 30 students,2 students did not borrow any books from the lib,12 borrowed 1 book,10 borrowed 2 books and the rest of the studets borrowed atleast 3 books.If the avg number of books borrowed per student was 2 ,what is the maximum number of books that any single student coupld have borrowed?
a)3
b)5
c)8
d)13
e)15
Given that avg books borrowed per student is 2. which means totally 30x2=60 books were borrowed.
its also given that 12 borrowed 1 book - 12x1 = 12
similarly, 10x2=20.
which adds up to 32. The rest 28 were borrowed by
30-2-12-10=6 students.
these 6 students have borrowed atleast 3 books. To maximize the no. of books for one person, it would be 3x5=15, and hence max of 13 books for one person.

hope this helps. Folks do correct in case you find any thing missing here.

1.Statement 1 gives you a clue that 10^d means that the product of 1*2*...*30 will end in a certain number of 0's. How many, you ask? Well, lets see:

You can write out the product entirely and see for yourself: 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20*21*22*23*24*25*26*27*28*29*30

Automatically, pull out 10,20, and 30, since they end in 0. 10*20*30 = 6000, which gives you 3 zeros. Now, you have to remember that the only other number when multiplied that can yield a zero is 5. So you have 5, 15, and 25 left. Now find how many zeros you get when you multiply any of those numbers ending in 5:

25 *4 = 100, which gives you 2 zeros.
15 * 6 = 90, which gives you 1 zero
5 * 2 = 10, which gives you 1 zero.

You're total is 3 + 2 + 1 + 1 = 7. Choice C is your answer.
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2.Not sure about notation. Sorry
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3. Use numbers:
Make 2 sets is as follows {1,2,70,127,130}, {69,69,70,71,71} which satisfies the conditions

In both cases, the average of the 2 greatest integers are > median and average of the set. Always better to use numbers as examples here.
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4. Average # of books per student = 2, there are 30 students total, so total # of books = 60.

12 students borrowed 1 book, and 10 borrowed 2 books. That means that 12*1 + 10*2 = 32 books were used out of the 60, so there are 60 - 32 = 28 books remaining.

Those 28 books remain left for 6 other students (since there are 30 students, 2 which borrow nothing, leaving 28 students to borrow entirely, 12 borrow 1 book and 10 borrow 2 books, which leaves 6 students left who borrow at least 3).

So if we had 5 students who borrowed only 3 (maximizing the number of books the 6th person could take), then that means you would have 28 - (5 * 3) = 13 books left for the 6th student to take.

Oops, didn't see Ben's posts. I agree fully with him.

Thanks for your help guys!
But I still havent got the solution to the question 1 about the factors.

Any easier way to understand the problem?