Free GMAT Practice Problems with Solutions
Sample 1

Detailed answer to GMAT problems in sample 1.

  1. ABC is a triangle whose sides AB and BC have length of 3 and 5 respectively. Which of the following could be the length of side AC?

    (I) 8
    (II) 12
    (III) 0.5

    A) (I) only
    B) (II) only
    C) (III) only
    D) (I) and (II) only
    E) None

    Solution

    The length of the third side of a triangle is always less than the sum and greater than the difference of the lengths of other two sides.

    The sum S and the difference D of the given sides is.

    S = 3 + 5 = 8

    D = 5 - 3 = 2

    None of the values suggested in (I), (II) or (III) satisfy the above condition. Answer E

  2. Given the following

    If √ x = 3, then x4 =

    Solution

    Square both sides of the given equation

    (√ x)2 = 32

    Simplify to find x

    x = 9

    Elevate both sides to the power 4

    x4 = 94

    Simplify

    x4 = 6561

  3. If n is an odd integer, then which of the following is even?

    (I) n2
    (II) n2 + 1
    (III) 3n2 - 1

    A) (I) only
    B) (II) only
    C) (III) only
    D) (I) and (II) only
    E) (II) and (III) only

    Solution

    Since n is odd integer, it can be written as

    n = 2 k + 1 , where k is an integer

    Let us express n2 in terms of k as follows

    n2 = (2 k + 1)2 = 4 k2 + 4 k + 1

    Let rewrite n2 as follows

    n2 = 2(2 k2 + 2 k) + 1

    Hence n2 is odd.

    We now express n2 + 1 in terms of k.

    n2 + 1 = 2(2 k2 + 2 k) + 1 + 1 = 2(2 k2 + 2 k + 1)

    Hence n2 + 1 is even.

    We now express 3 n2 - 1 in terms of k.

    3 n2 - 1 = 3 [2(2 k2 + 2 k) + 1 ] - 1

    = 3 [2(2 k2 + 2 k) ] + 3 - 1

    = 6(2 k2 + 2 k) + 2

    Hence 3 n2 - 1 is even.

    Answer E.

  4. 82100 + 42101 =

    Solution

    Use the fact that 8 = 23 and 4 = 22 to rewrite the given expression as follows

    82100 + 42101 = 232100 + 22 2101

    Use rules of exponent to simplify

    = 2103 + 2103

    = 22103 = 2104

  5. If 3x + 5y = 5 and x + 3y = 20, then 2x + 4y =

    Solution

    Add the left and right hand sides of the given equations to obtain a new equation

    (3x + 5y) + (x + 3y) = (5) + (20)

    Simplify

    4x + 8y = 25

    Divide all taerms of the above equation by 2

    2x + 4y = 25 / 2

  6. If n is positive and less than 1, then which of the following is true?

    (I) n2 - n < 0
    (II) n3 < n
    (III) n + 1 < 1

    A) (I) only
    B) (II) only
    C) (III) only
    D) (I) and (II) only
    E) (II) and (III) only

    Solution

    n is positive and less than 1 is translated as follows

    0 < n < 1

    Multiply all terms of the above inequality by n to obtain

    0 < n2 < n

    which also gives

    n2 - n < 0

    Hence statement (I) is true

    Multiply all terms of the above inequality by n to obtain

    0 < n3 < n2

    Since n2 < n, we have

    0 < n3 < n

    Hence statement (II) is true

    For n = 0.75, statement (III) is not true

    Hence statement (I) and (II) only are true.

  7. Which of the following has the greatest value?

    A) 250%
    B) 2 + 1/2
    C) 5 0.5
    D) 1 / 0.1
    E) 4

    Solution

    Use decimal numbers to rewrite the above expressions

    A) 250% = 250/100 = 2.5
    B) 2 + 1/2 = 2 + 0.5 = 2.5
    C) 5 0.5 = 2.5
    D) 1 / 0.1 = 10
    E) 4 = 4

    The expression 1 / 0.1 has the greatest value.

  8. (4x2 - 4) / (- 3x + 3) =

    Solution

    Factor numerator as follows

    (4x2 - 4) = 4(x2 - 1) = 4(x - 1)(x + 1)

    Factor denominator as follows

    (- 3x + 3) = - 3(x - 1)

    Substitute in the given expression and simplify

    (4x2 - 4) / (- 3x + 3) = [ 4(x - 1)(x + 1) ] / [ - 3(x - 1)]

    = (- 4/3)(x + 1)

  9. For what values of b and c will the equation x2 + bx = c have the solutions 2 and -3?

    Solution

    Since x = 2 is a solution of x2 + bx = c, then

    (2)2 + b(2) = c or 4 + 2b = c

    Since x = -3 is a solution of x2 + bx = c, then

    (-3)2 + b(-3) = c or 9 - 3b = c

    We now have a system of simultaneous equations in b and c to solve. Combining the above equation, we obtain

    4 + 2b = 9 - 3b

    Solve for b

    5b = 5 , b = 1

    Substitute b by 1 in the equations 4 + 2b = c and solve for c

    4 + 2(1) = c or c = 6

    Values for b and c

    b = 1 , c = 6

  10. (√12 - √3)(-√12 + √3)

    Solution

    Rewrite as

    (√12 - √3)(-√12 + √3) = - (√12 - √3) (√12 - √3)

    and simplify

    = - (12 - 3) = -9



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