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 163 + 173 + 183 + 193 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 – 163 + 193 + 173 + 183
= (16 + 19)(162 – 16 * 19 + 192) + (17 + 18)(172 – 17*18 + 182)
= 35 (162 – 16 * 19 + 192 + 172 – 17*18 + 182)
= 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. 3C1* 2!

B. 4C2* 3C1

C. 4!

D. 4!/2!

E. 4C2 * 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 4C2 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 4C2 *3!