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

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