# SAT Math Questions under a Preparatory Series

The questions give you a glimpse of what the mathematical section reveals as part of the SAT paper. A variety of questions are available using the same kind of logic and algebraic usage.

**1) ****What will the remainder be if 16 ^{3} + 17^{3} + 18^{3} + 19^{3} is divided by 70?**

**A. 0
B. 69
C. 1
D. 34
E. 47**

How to answer: The correct answer for the above question is A, O. The tactic behind answering this math problem is by applying the following – 16^{3} + 19^{3} + 17^{3} + 18^{3}

= (16 + 19)(16^{2} – 16 * 19 + 19^{2}) + (17 + 18)(17^{2} – 17*18 + 18^{2})

= 35 (16^{2} – 16 * 19 + 19^{2} + 17^{2} – 17*18 + 18^{2})

= 35 (even number).

When 35 is multiplied by an even number, the product is then divisible by 70 therefore leaving the answer as 0.

**2) ****If 3 students have to sit in a row, there are 9 students who do not have a seat. But if 9 students were made to sit in a row instead, then 5 rows are left empty. How many students need to be seated in a row to leave an equal number of students sitting in all the rows?**

A. 5

B. 4

C. 8

D. 7

E. 6

How to answer: Assume that the number of rows in the classroom is ‘n’. Three students are made to sit in a row leaving 3n students unseated. Therefore 9 students in total do not have a place to sit, leaving the equation at (3n+9) students within the classroom. If 9 students are made to sit in a row, it leaves 5 rows empty, deriving n-5 rows, therefore 9*(n-5) students in the classroom.

The students in the class remain at a fixed number no matter how they are rearranged leaving the equation now at 3n+9=9* (n-5) which amounts to 3n+9 = 9n-45 or 6n = 54 or n = 9. There are nine rows in total within the classroom, where 9*(n-5) equals 9*(9-5) = 36 students. If these 36 students need to be seated in 9 rows, then an equal number of them will seat themselves in each row, leaving us with 36/9= 4 students per row, where option B is the answer.

**3) ****How many ways can 4 boys occupy 3 different rooms where no room is left empty?**

A. ^{3}C_{1}* 2!

B. ^{4}C_{2}* 3C_{1}

C. 4!

**D. ****4!/2!**

**E. ^{4}C_{2} * 3!**

How to answer: There needs to be 2 boys assigned to a single room, where one room will hold exactly a pair. This has to done by selecting two boys out of the four which are going to share the rooms. This is done using ^{4}C_{2 }where we then assign which two boys get which room. The assigning requires 3 rooms to contain 3 sets of boys, that is, 1, 1, and 2 within each set. This is done using 3! The final answer is then ^{4}C_{2 }*3!