# Free GRE Practice Questions with Solutions Sample 2

Solutions and detailed explanations to questions and problems similar to the questions in the GRE test. sample 2

## Solution to Question 1

w is the mean of a, b, c and d is written as
w = (a + b + c + d) / 4
w is the average W of m(a + k), m(b + k), m(c + k) and m(d + k) is given by
W = [ m(a + k) + m(b + k) + m(c + k) + m(d + k) ] / 4
W = m [ a + b + c + d + 4 k] / 4 = m [a + b + c + d ] / 4 + m k
= m (w + k)
If w is the average of a, b, c and d, then the average W of m(a + k), m(b + k), m(c + k) and m(d + k) is given by
W = m (w + k)

## Solution to Question 2

Let r and R be the radii of the smaller and larger circles respectively. The radius of the larger circle is three times the radius of the smaller circle leads to
R = 3r
Areas A1 of smaller and A2 of larger circles are given by
A1 = Pi r2
A2 = Pi R2 = Pi (3r)2 = 9 Pi r2
ratio R of areas larger / smaller is equal to
R = 9 Pi r2 / Pi r2 = 9

## Solution to Question 3

Let S1 be the total salary of the group of 20 employess. Hence
35,000 = S1 / 20
S1 = 20 * 35,000 = $700,000 Let S2 be the total salary of the group of 30 employess. Hence 40,000 = S2 / 30 S2 =$1,200,000
The average of all 50 employers is given by
(700,000 + 1,200,000) / 50 = \$38,000

## Solution to Question 4

Rewrite the given expression using the fact that 48 = 3 � 16
√48 = √(3 � 16)
Use the formula √(a � b) = √a � √a to rewite √(3 � 16) as
√48 = √(3 � 16) = √3 � √16
= 4 √3

## Solution to Question 5

Since the three angles are in the ration 2:4:3, their sizes they may be written in the form
Size of A = 2 k , Size of B = 4 k and size of C = 3 k , where k is a constant.
The sum of the angles of a ny triangle is equal to 180�; hence
2 k + 4 k + 3 k = 180
Solve for k
9 k = 180 , k = 20
The smallest angle is A and its size is equal to 2 k
2 k = 2 � 20 = 40�

## Solution to Question 6

If n is even, it can be written as follows
n = 2 k , where k is an integer
If m is odd, it can be written as follows
m = 2 K + 1 , where K is an integer
We now express n + m in terms of k and K
n + m = 2 k + 2 K + 1 = 2(k + K) + 1
n + m is odd
We now express n - m in terms of k and K
n - m = 2 k - (2 K + 1) = 2 k - 2 K - 1
n - m = 2 (k - K) - 1
n - m is odd
We now express n * m in terms of k and K
n * m = (2 k)(2 K + 1) = 2( k(2K + 1) )
n * m is even
We now express n2 + m2 + 1 in terms of k and K
n2 + m2 + 1 = (2 k)2 + (2 K + 1)2 + 1 = 4 k2 + 4 K2 + 4 K + 1 + 1
= 2 ( 2 k2 + 2 K2 + 2 K + 1)
n2 + m2 + 1 is even
Statement D is true.

## Solution to Question 7

Use the facts that 25 = 52 and 125 = 53 to rewrite the given expression as follows
5100 + 2550 + 3(12534 / 25) = 5100 + (52)50 + 3( (53)34 / (52))
Use formula for exponents to simplify
= 5100 + 5100 + 3( 5102 / 52)
= 5100 + 5100 + 3( 5100)
= 5 * 5100
= 5101

## Solution to Question 8

Factor numerator as follows
6x10 - 2x9 = 2x9 (3x - 1)
Factor denominator as follows
9x2 - 1 = (3x - 1)(3x + 1)
Substitute numerator and denominator by their factored forms and simplify the given expression
[ 6x10 - 2x9 ] / (9x2 - 1) = [2x9 (3x - 1) ] / [(3x - 1)(3x + 1)]
= 2x9 / (3x + 1)

## Solution to Question 9

Expand by multiplication or using the identity (x + y)2 = x2 + 2 x y + y2.
(- 2x + 6)2 = (-2x)2 + 2 (-2x)(6) + 62
= 4 x2 - 24 x + 36

## Solution to Question 10

The sum of all interior angles of a polygon of n sides is given by.
(n - 2) * 180
and is equal to 1800�. Hence
(n - 2) * 180 = 1800
Solve for n
(n - 2) = 10
n = 12

## More References and Links to More Maths Practice Tests

Free GRE Quantitative for Practice
Free Practice for GAMT Maths tests
Free Compass Maths tests Practice
Free Practice for SAT, ACT Maths tests
Free AP Calculus Questions (AB and BC) with Answers