Number System


  1. The product of two numbers is 24 times the difference of these two numbers. If the sum of these numbers is 14, the larger number is









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    Let the first number be x.
    ∴  Second number = 14–x
    ∴  x (14 – x ) = 24 (x – 14 + x)
    ⇒  x (14 – x ) = 24 (2x – 14)
    ⇒  14x – x2 = 48x – 336
    ⇒  x2 + 34x – 336 = 0
    ⇒  x2 + 42x – 8x – 336 = 0
    ⇒  x (x + 42) – 8 (x + 42) = 0
    ⇒  (x + 42) (x – 8) = 0
    ∴  x = 8 as x ≠ – 42
    ∴  Second number = 14 – 8 = 6
    ∴  Larger number = 8
    Note : : It is preferable to solve it
    by oral calculation with the help
    of given alternatives.

    Correct Option: B

    Let the first number be x.
    ∴  Second number = 14–x
    ∴  x (14 – x ) = 24 (x – 14 + x)
    ⇒  x (14 – x ) = 24 (2x – 14)
    ⇒  14x – x2 = 48x – 336
    ⇒  x2 + 34x – 336 = 0
    ⇒  x2 + 42x – 8x – 336 = 0
    ⇒  x (x + 42) – 8 (x + 42) = 0
    ⇒  (x + 42) (x – 8) = 0
    ∴  x = 8 as x ≠ – 42
    ∴  Second number = 14 – 8 = 6
    ∴  Larger number = 8
    Note : : It is preferable to solve it
    by oral calculation with the help
    of given alternatives.


  1. Five times of a positive integer is equal to 3 less than twice the square of that number. The number is









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    Let the positive integer be x.
    ∴  2x2 – 5x = 3
    ⇒  2x2 – 3 = 0
    ⇒  2x2 – 6x + x – 3 = 0
    ⇒  2x (x – 3) + 1 (x – 3) = 0
    ⇒  (x – 3) (2x + 1) = 0

    ∴  x = 3 and x = –
    1
    2

    is not admissible.

    Correct Option: A

    Let the positive integer be x.
    ∴  2x2 – 5x = 3
    ⇒  2x2 – 3 = 0
    ⇒  2x2 – 6x + x – 3 = 0
    ⇒  2x (x – 3) + 1 (x – 3) = 0
    ⇒  (x – 3) (2x + 1) = 0

    ∴  x = 3 and x = –
    1
    2

    is not admissible.



  1. The sum of the digits of a two digit number is 10. The number formed by reversing the digits is 18 less than the original number. Find the original number.









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    Let the two digit number be 10x + y where x > y.
    Here, x + y = 10    ...(i)
    and, 10x + y – 10y – x = 18
    ⇒  9x – 9y = 18
    ⇒  9 (x – y) = 18
    ⇒  x – y = 2   ...(ii)
    Solving equations (i) and (ii),
    x = 6 and y = 4
    ∴  Number = 10 × 6 + 4 = 64

    Correct Option: C

    Let the two digit number be 10x + y where x > y.
    Here, x + y = 10    ...(i)
    and, 10x + y – 10y – x = 18
    ⇒  9x – 9y = 18
    ⇒  9 (x – y) = 18
    ⇒  x – y = 2   ...(ii)
    Solving equations (i) and (ii),
    x = 6 and y = 4
    ∴  Number = 10 × 6 + 4 = 64


  1. Of the three numbers, the second is twice the first and is also thrice the third. If the average of these three numbers is 44, the largest number is









  1. View Hint View Answer Discuss in Forum

    Let the third number be x.
    ∴  Second number = 3x

    First number =
    3
    x
    2

    According to the question,
    3x
    + 3x + x = 44 × 3
    2

    ⇒ 
    3x + 2x + 6x
    = 44 × 3
    2

    ⇒  11x = 88 × 3
    ⇒  x =
    88 × 3
    = 24
    11

    ∴  The largest number
    = 3x = 3 × 24 = 72

    Correct Option: C

    Let the third number be x.
    ∴  Second number = 3x

    First number =
    3
    x
    2

    According to the question,
    3x
    + 3x + x = 44 × 3
    2

    ⇒ 
    3x + 2x + 6x
    = 44 × 3
    2

    ⇒  11x = 88 × 3
    ⇒  x =
    88 × 3
    = 24
    11

    ∴  The largest number
    = 3x = 3 × 24 = 72



  1. The sum and product of two numbers are 12 and 35 respectively. The sum of their reciprocals will be









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    Let the numbers be x and y respectively.
    ∴  x + y = 12 and xy = 35

    ∴ 
    1
    +
    1
    =
    x + y
    =
    12
    xyxy35

    Correct Option: A

    Let the numbers be x and y respectively.
    ∴  x + y = 12 and xy = 35

    ∴ 
    1
    +
    1
    =
    x + y
    =
    12
    xyxy35