# 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

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