GRE
Practice Test 31
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‘P’ is a set of integers that are both positive factors of 72 and also multiples of 12. ‘K’ is a set of integers that are positive factors of the number 56. ‘m’ is a member of set ‘P’ and ‘n’ is a member of set ‘K’.
minimum possible value of m – n -16
  • Solution

    P = {12, 24,36, 72}

    K = {I, 2,4,7,8, 14,28, 56}

    Minimum value of 'm' = 12

    Maximum value of 'n' = 56

    ∴ Col. A = m - n = 12 = 56 = -44, Col. B = -16

    Col. A < Col. B

 

Total numbers between 280 and 440 that leaves a remainder 7 when divided b 6, 8, 9 3
  • Solution

    LCM of 6, 8 and 9 is 72.

    Multiples of 72 between 280 and 440 are 288, 360, 432.

    ∴ The actual numbers which yield a remainder 7 when divided 72 are 288 + 7 = 295, 360 + 7 = 367 and 432 + 7 = 439.

    ∴ Col. A = 3, Col. B = 3

ABCD is a rhombus.

AC = 12 cm, BD = 16 cm

Height of the rhombus in cm. 10 cm
  • Solution

    Area of rhombus =\(\frac{d_{1}d_{2}}{2}=\frac{12 \times 16}{2}= 96cm^{2}\)

    Also, are of rhombus = base × height Consider t.DOC

    LDOC = 90°,DO = 8cm,OC = 6 cm

    ∴ By Pythagoras theorem, DC = 10 cm

    ∴ 10 × h = 96 ∴h = 9.6cm

    ∴ Col. A = 9.6 cm, Col. B = 10 cm

    (Height will always be less than Hypotenuse)

Malcolm is the majority owner of a departmental store and therefore receives 40% of the profit. Hillary, Susan, Murphy and Brain are minority owners and each receives 2S% of the remaining profit. Malcolm, Hillary and Murphy together receive $49000.
total profit made in the business in dollars 64000
  • Solution

    Malcolm receives 40% of total profit.

    Hillary, Susan, Murphy and Brian each receive = 25% of 60% of total profit = 15% of total profit.

    ∴ Malcolm, Hillary and Murphy together receive 40% + 15% + 15% = 70% of total profit.

    ∴ 70% of profit = $49000

    ∴ Total profit = $70000

    ∴ Col. A = $70000, Col. B = $64000

The figure shows a trapezium ABCD
BE = 8, AE = 2 (ED), AD = 2 (Be)

Area of trapezium ABCD 54
  • Solution

    BC || AD and they are cut by a transversal BE.

    ∴ Alternate angles are congruent

    ∴ BEA = 60 ∴ABE = equilateral triangle.

    ∴ AE = 8, ED = 4, BC = 6

    The altitude of triangle ABE = length CD

    ∴ Height of trapezium

    (CD)= √3⁄2 × 8 = 4√3

    ∴ Col.A = 1⁄2(Sum of II sides) × (height)

    = 1⁄2 × (8 + 4 + 6) × 4√3

    = \(\frac{72\sqrt{3}}{2}\)= 36√3 = 36 × 1.732

    Col. A = 54 = 36 × 1.5 ∴ Col. A > Col.B

\(a^{2}b^{2}+\frac{74}{2}=107,a=\sqrt{17.5}\)
Value of 3b + 3 10
  • Solution

    a = \(\sqrt{17.5}\) ∴ a2 = 17.5

    a2b2 + 37 = 107 ∴ a2b2 = 70

    ∴ 17.5b2 = 70 ∴ b2 = 4 ∴ b = 2

    ∴ Col.A = 3b + 3 = 6 + 3 = 9

\(\sqrt[3]{\sqrt{8}}=2^{p}\)
3p 1.5
  • Solution

    L.H.S. = \(\sqrt[3]{8^{\frac{1}{2}}}\)

    = 81⁄6 = 23⁄6 = 21⁄2 = 2P

    ∴ p = 1⁄2

    ∴ Col.A = 3P = 3⁄2 = 1.5 = Col.B

Mathew and Johnny each are required to polish 440 metal components. Mathew takes 10 hours longer to polish 440 components than does Johnny. Johnny polishes 10% more components per hour than does Mathew.
Number of components produced by mathew per hour 4.4
  • Solution

    Let Johnny require x hours to finish the job.

    ∴ Mathew requires x + 10 hours

    Let Mathew polish y components per hour

    Johnny polishes 1.1y components in one hour

    (x + 10)y = 440

    ∴ xy + 10y = 440

    ∴ 1.1xy = 440

    ∴ xy = 400 ∴ 400 + 10y = 440

    ∴ y = 4 and x = 100

    ∴ (Mathew polishes 4 components per hour and therefore requires 110 hours.

    Johnny polishes 4.4 components; per hour and requires 100 hours.)
    ∴ Col. A = 4, Col. B = 4.4

ma × mb × mc = m – 17

m > 0, a, b, c are each different negative integers.

smallest possible value of c -15
  • Solution

    According to laws of indices

    ma × mb × mc = ma+b+c = m-17

    ∴ a + b + c = -17

    For c to have the smallest possible value, 'a' and 'b' need to be maximum

    The maximum values possible for 'a' and 'b' are -1 and-2

    ∴ c + (-1) + (-2) = -17

    ∴ c = -14 ∴ Col.A > Col.B

In the subtraction problem shown below, x and y represent different single digit numbers.
x + y 8
  • Solution

    y - x = 6...........subtracting in units place.

    Also there is carry-over of 1.

    ∴ x > y

    So possible values of (y, x) are (1, 5), (2,6), (3, 7), (4,8), (5, 9).

    Since there is a carryover for hundred's place, the value of x is 6. ∴ y = 2.

    ∴ The subtraction is as follows

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