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Aptitude miscellaneous
  1. 999
    98
    		× 99
    99

    is equal to :








  1. View Hint View Answer Discuss in Forum

    999 +
    98
    		 × 99
    99

    = 999 × 99 + 98
    = (1000 – 1) 99 + 98
    = 99000 – 99 + 98 = 98999

    Correct Option: A

    999 +
    98
    		 × 99
    99

    = 999 × 99 + 98
    = (1000 – 1) 99 + 98
    = 99000 – 99 + 98 = 98999

  1. The length of a road is one kilometre. The number of plants required for plantation at a gap of 20 metres in both sides of the road is








  1. View Hint View Answer Discuss in Forum

    Length of the road = 1000 metre
    Number of plants on one side of the road

    =
    1000
    		 + 1 = 51
    20

    ∴  Total number of plants
    = 2 × 51 = 102

    Correct Option: B

    Length of the road = 1000 metre
    Number of plants on one side of the road

    =
    1000
    		 + 1 = 51
    20

    ∴  Total number of plants
    = 2 × 51 = 102

  1. A man has some hens and cows. If the number of heads : number of feet = 12 : 35, find out the number of hens, if the number of heads alone is 48.








  1. View Hint View Answer Discuss in Forum

    Number of hens = x
    ∴  Number of cows = 48 – x
    ∴  2x + (48 – x ) × 4 = 35 × 4
    ⇒  2x + 192 – 4x = 140
    ⇒  2x = 192 – 140 = 52
    ⇒  x = 26

    Correct Option: B

    Number of hens = x
    ∴  Number of cows = 48 – x
    ∴  2x + (48 – x ) × 4 = 35 × 4
    ⇒  2x + 192 – 4x = 140
    ⇒  2x = 192 – 140 = 52
    ⇒  x = 26

  1. The sum of a natural number and its square equals the product of the first three prime numbers. The number is








  1. View Hint View Answer Discuss in Forum

    Let the required number be x.
    ∴  x2 + x = 2 × 3 × 5
    ⇒  x2 + x – 30 = 0
    ⇒  x2 + 6x – 5x – 30 = 0
    ⇒  x (x + 6) – 5 (x + 6) = 0
    ⇒  (x – 5) (x + 6) = 0
    ⇒  x = 5

    Correct Option: C

    Let the required number be x.
    ∴  x2 + x = 2 × 3 × 5
    ⇒  x2 + x – 30 = 0
    ⇒  x2 + 6x – 5x – 30 = 0
    ⇒  x (x + 6) – 5 (x + 6) = 0
    ⇒  (x – 5) (x + 6) = 0
    ⇒  x = 5

  1. If the digits in the unit and the ten’s places of a three digit number are interchanged, a new number is formed, which is greater than the original number by 63. Suppose the digit in the unit place of the original number be x. Then, all the possible values of x are








  1. View Hint View Answer Discuss in Forum

    Let the two digit number be
    10y + x where x > y
    ∴  10x + y – 10y – x = 63
    ⇒  9x – 9y = 63
    ⇒  x – y = 7
    ∴  x = 7, 8, 9 and y = 0, 1, 2

    Correct Option: B

    Let the two digit number be
    10y + x where x > y
    ∴  10x + y – 10y – x = 63
    ⇒  9x – 9y = 63
    ⇒  x – y = 7
    ∴  x = 7, 8, 9 and y = 0, 1, 2