GRE
Practice Test 21
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(3x 19)+ (3x 7)+ (104 x 0.125) (26 × 1⁄2) + (20 × 3)+ (6 × 3)
  • Solution

    Col. A: (3 × 19) + (3 × 7)+ (104 × 1⁄8)

    = 3(19 + 7) + ( 26 × 1⁄2)

    = 3 × 26 + 26 × 1⁄2 = 26 × 31⁄2

    Col. B: 26 × 1⁄2 + 3(20 + 6) = 26 × 31⁄2 = Col. A

12x + 12x+1 = 3x + 3x+1 + 3x+2
x 1
  • Solution

    12X(1 + 12)= 3X(1 + 3 + 9) ∴ 12X × 12 = 3X × 12

    ∴ 12X = 3X

    This equality holds true only for x = 0

be : ac: ab = 1 : 2 : 3
Ratio of \(\frac{a}{bc}:\frac{b}{ac}\) 4: 1
  • Solution

    bc : ac : ab = 1 : 2 : 3

    ∴ \(\frac{abc}{a}:\frac{abc}{b}:\frac{abc}{c}= 1 : 2 : 3\)

    ∴ 1⁄a : 1⁄b : 1⁄c = 1 : 2 : 3 ∴ a : b : c = 1:1⁄2:1⁄3

    ∴ \(\frac{a^{2}}{b^{2}}=\frac{1}{1/4}=\frac{4}{1}\) ∴ Required ratio = 4 : 1

The price of a commodity is increased by 30% and then decreased by 30%.
The % denoting the new price compared to the original price 100 %
  • Solution

    Col. A: Let the initial price be $100.

    ∴ Increased price after 30% increase = $130

    Decreased price after 30% decrease = 130 × 0.7 = $91

    ∴ The price of commodity has changed by $9.

    Let the required number be x

    ∴ 9 = x × 100 ∴ x = \(\frac{9}{100}\)

AB = 1 meter

Perimeter of square PQRS = 4 times the perimeter of square ABCD

Perimeter of SDCR \(\frac{19}{2}\) meters
  • Solution

    Perimeter of ABCD = 4x

    ∴ Perimeter of PQRS = 16x

    Also DS = CR = BQ = AP

    DS + BQ = ℓ(SQ)- ℓ(DB)

    ∴ DS + BQ = 3x√2. i.e. DS + CR = 3x√2.

    Perimeter of SDCR = x + 4x + 3x√2.

    = x(5 + 3√2) = 9.2x = \(\frac{18.4x}{2}\)

    Col. B: \(\frac{19x}{2}\)

Out of a group of ducks, 4 times the square root of all the ducks were seen swimming in the water and five ducks remained out of the water.
Number of ducks in group 36
  • Solution

    Let the number of ducks be x

    ∴ 4-√x + S = x ∴ 4-√x = x - S

    Squaring both sides

    16x = x2 - 10x + 2S

    ∴ x2 -26x + 2S = 0

    ∴ (x - 25)(x -1) = 0

    ∴ x = 25 or x = 1

    x = 1 is not possible ∴ x = 25

HCF of 1⁄2,3⁄4,5⁄6,7⁄8\(\frac{9}{10}\) HCF of 1⁄4,3⁄2,5⁄8,7⁄6\(\frac{9}{10}\)
  • Solution

    HFC of fractions = \(\frac{HCF\, of\, numerators}{LCM\, of\, denominators}\)

    Note that in both the columns, fractions have numerators 1,3, S, 7, 9 and denominators 2, 4, 6, 8, 10

    ∴ Both the sets of fractions have the same HCF.

Digit in the units place of (213)13 9
  • Solution

    It is essentially, digit in unit place of 313. Power cycle for number 3 is 3, 9, 7, 1 and frequency is 4 .

    ∴ \(\frac{13}{4}\) = 3(1⁄3)

    Three cycles are completed at 312 and a new cycle beings at 313.

    ∴ (3)13 ends in 3 ∴ (213)13 ends in 3.

Of the two tasks P and Q, P is completed by 11 men in 4 days and Q is completed by 4 men in 22 days.
Number of da s in which all 15 men will finish both tasks. 12
  • Solution

    Task P is completed by 11 men in 4 days.

    ∴ 1 man completes the task in \(\frac{11\times 4}{1}\)= 44 days.

    Similarly, task Q is completed by 4 men in 22 days ⇒ the same task can be completed by 1 man in 88 days.

    ∴ Both tasks can be completed by 1 man in 88 + 44 = 132 days.

    ∴ 15 men can do it = \(\frac{132}{15}\) = 8(4⁄5) days.

Jack sells 25 figs for 96 dollars and suffers a loss of 4%.
Number of fi s he should sell for 15 dollars to incur a loss of 25% 5
  • Solution

    Selling price of 25 figs = $96.

    Loss is 4% of cost.

    ∴ Cost should be $100

    ∴ Cost for 1 fig = $4

    If cost is 100%, loss is 25% then selling price is 75%.

    ∴ 75% of cost = $15

    ∴ 100% cost = $20

    Cost per fig is $4

    ∴ $20 will be the cost of 5 figs

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