GRE
Practice Test 37
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X is the sum of prime factors of 57 and Y is the sum of prime factors of 91
value of X-Y 5
  • Solution

    Prime factors of 57: 19 and 3. ∴ Sum = 22

    Prime factors of 91: 13 and 7. ∴ Sum = 20

    ∴ X - Y = 2.

    Col. A = 2

Distance between 2 stations P and Q is 90 miles.

A train starts from P towards Q at 60 miles per hour. A bird starts at the same time at a constant speed ox 90 miles per hour from Q towards moving train.

On reaching the train, it turns back instantaneously and returns to Q. The bird makes its journey from Q to train and back continuously till train reaches station Q.

Total distance in miles travelled by the bird. 120
  • Solution

    There is no need to work out the distance at which train and the bird meet.

    Distance to be covered by train = 90 miles.

    Speed of train = 60 miles per hour.

    ∴ Time = \(\frac{90}{60}\)= 1.5 hours.

    ∴ The bird also travels for 1.5 hours at 90 miles per hour.

    ∴ Distance covered by bird = 90 × 1.5 = 135 miles = Col. A

A 27″ × 27″ square metal plate needs to be fixed by a carpenter on to a wooden board. The carpenter uses nails all along the edges of the square such that there are 28 nails each side of the square. Each nail is at a constant distance from neighbouring nails.
Total number of nails used by the carpenter 108
  • Solution

    The figure shows the metal plate 27° × 27°

    Consider side AB and CD.

    Total number of nails on these 2 sides = 28 × 2

    Consider side AD and BC.

    Total number nails on these 2 sides = 2(28 - 2) = 52

    Nails A and D should not counted for side AD.

    Same is the case with side BC.

    ∴ Col. A = 56 + 52 = 108

Suzan and Kate have a collection of billiard balls. If Kate gives Suzan 2 balls from her collection, then they would have an equal number of balls. However, if Suzan happens to give Kate 2 balls from her collection, then Kate would have twice as many balls as Suzan.
Number of balls with Kate 14
  • Solution

    Let Suzan and Kate have S and K balls respectively.

    According to first condition,

    S + 2 = K - 2 ∴K - S = 4....(I)

    According to second condition,

    2(S - 2) = K + 2

    ∴ 2S - 4 = K + 2 ∴ K - 2S = -6....(II)

    Subtracting (II) from (I)

    S = 10 ∴ K = 14

    ∴ Col.A = 14 = Col.B

A set of football matches is to be played in a “round-robin” fashion i.e. every participating team play a match against every other team once and only once. 36 matches are totally played.
Number of teams participated 12
  • Solution

    Let there be a teams in all. One has to choose any 2 teams out of n, to have a match.

    This can be done in nc2, number of ways.

    ∴ nc2 = 36 ∴ \(\frac{n(n+1)}{2}\) = 36

    ∴ n(n -1) = 72 ∴ 9 × 8 = 72 ∴ n = 9

Mark has 91 marbles. He arranges them in n rows, such that the nth row contains n marbles, (n – 1)th row contains (n – 1) marbles and so on.
The number of marbles in the nth row 14
  • Solution

    1st row contains 1 marble, 2nd row → 2,

    3rd row → 3.........and so on.

    ∴ Total number of marbles is the add» of first n natural numbers.

    ∴ Sum =\(\frac{n(n+1)}{2}\)= 91 ∴ n(n+ 1)= 182

    ∴ 13 × 14 = 182 ∴ n = 13 = Col.A

A three digit number is to be formed in the following manner. Third digit (units digit) is the square root of first digit. 2nd digit is the sum of first and third digits.
The number of such 3 digit numbers possible 2
  • Solution

    Let the digit in unifs place be 1.

    ∴ The number is 121.

    Let the digit in unit's place be 2.

    ∴ The number is 462.

    If the number in the unit's place is 3 or more, then the given conditions do not give a 3 digit number.

    ∴ Col. A = 2 = Col. B

I participated in a race. (1⁄5)th of those who were running ahead of me was exactly equal to (5⁄6)th of those running behind me. A total of 32 individuals ran in the race and nobody was running with me.
The number of individuals running behind me 7
  • Solution

    Total number of contestants excluding myself = 31.

    Let x individuals run ahead of me.

    ∴ 31 - x run behind me.

    ∴ x⁄5 = 5⁄6(31 - x)

    ∴ 6x = (25 × 31)- 25x

    ∴ 31x = 25 × 31 ∴ x = 25

    ∴ 6 persons run behind me.

There is a wrestling competition amongst 10 wrestlers every wrestler will fight against every other wrestler.
total no. of fights 40
  • Solution

    umber of fights can be calculated with the theory of combinations, 10 wrestlers at a time.

    i.e. 10C2 = \(\frac{10\times 9}{2}=45\)

    Col. A = 45.

Ralph claims that if the digits of his present age are reversed, then the age of his son can be obtained. He further states that one year ago this age was twice as that of his son and that the sum of his son’s present age and his age equals 110.
Ralph’s age his son’s 40 years hence.
  • Solution

    Total of their present age is 110.

    Possible ages can be 64 - 46,73 - 37,82 - 28,91- 19

    But 1 year ago his age was twice that of his son.

    ∴ His age was 72 and age of his son was 36.

    ∴ Today age of Ralph is 73 and age of his son is 37.

    ∴ Col. A = 73, Col. B = 77

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