GRE Math Problem 60

x>0

Column A	Column B

Solution for GRE Math Problem 59:

The perimeter of the square is
 Answer:(C) 3x + 4

GRE Math Problem 42

If 3^a=b and 3^c=d, then bd=
(A) 3^ac
(B) 3^a+c
(C) 6^a+c
(D) 9^ac
(E) 9^a+c


Solution for GRE Math Problem 41:
Let’s build this sequence: 1, 1x3-1=2, 2x3-1=5, 5x3-1=14, 14x3-1=41, 41x3-1=122 – this is the smallest term greater than 100. It is greater than 120
Answer:A

GRE Math Problem 31

Compare two values:

Column A	Column B
The remainder when 10100 is divided by 575	The remainder when 10100 is divided by 755

Solution for GRE Math Problem 30:
The average of three digits is 2 means, that their sum is 6. There are 7 tripples of digits with the sum of 6. they are: 6,0,0; 5,1,0; 4,2,0; 4,1,1; 3,3,0; 3,2,1; 2,2,2.
6,0,0 gives only one 3-digit number: 600.
For 5,1,0 we have 510, 501, 150, 105.
For 4,2,0: 420, 402, 240, 204.
For 4,1,1: 411, 141, 114.
For 3,3,0: 330, 303.
For 3,2,1: 321, 312, 231, 213, 123, 132.
For 2,2,2: 222.
Total 21 numbers.
Answer: Column A is greater.

GRE Math Problem 29

What is the average (arithmetic mean) of 3³⁰, 3⁶⁰, and 3⁹⁰?
(A) 3⁶⁰
(B)3¹⁷⁷
(C)3¹⁰ + 3²⁰ + 3³⁰
(D) 3²⁷ + 3⁵⁷ + 3⁸⁷
(E)3²⁹+3⁵⁹+3⁸⁹

Solution for GRE Math Problem 28:

Is the class average before the adjustment was A, then the average of missed points was 100-A. After the adjustment, the average of missed points became . So, the average of points gained became
Answer:(A)

GRE Math Problem 22

Compare two values:

Column A	Column B
The units digit of 1414	The units digit of 1616

Solution for GRE Math Problem 21:

The area of the circle with the radius of R is R². To cet the are of the ring, we should substract the area of the inner circle from the area of the outer circle. So, the are of the smallest circle is . The area of the whole figure is x4²=16. The area aof the yellow resion is x3²-x1²=8. It is exactly the half of the whole figure's area. It means, that the area of the shaded resion equals the area of the yellow region.
Answer:C