Simplification

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Aptitude miscellaneous
  1. If the sum of squares of two real numbers is 41 and their sum is 9, then the sum of cubes of these two numbers is








  1. View Hint View Answer Discuss in Forum

    Let the two real numbers be x and y.
    According to the question,
    x2 + y2 = 41
    x + y = 9
    ∴  (x + y)2 = x2 + y2 + 2xy
    ⇒  81 = 41 + 2xy
    ⇒  2xy = 81 – 41 = 40

    ⇒  xy =
    40
    		 = 20
    2

    ∴  x3 + y3 = (x + y)3 – 3xy (x + y)
    = (9)3 – 3 × 20 (9)
    = 729 – 540 = 189

    Correct Option: C

    Let the two real numbers be x and y.
    According to the question,
    x2 + y2 = 41
    x + y = 9
    ∴  (x + y)2 = x2 + y2 + 2xy
    ⇒  81 = 41 + 2xy
    ⇒  2xy = 81 – 41 = 40

    ⇒  xy =
    40
    		 = 20
    2

    ∴  x3 + y3 = (x + y)3 – 3xy (x + y)
    = (9)3 – 3 × 20 (9)
    = 729 – 540 = 189

  1. Find the least number which must be subtracted from 18265 to make it a perfect square.








  1. View Hint View Answer Discuss in Forum


    ∴  Required answer = 40

    Correct Option: C


    ∴  Required answer = 40

  1. If a perfect square, not divisible by 6, be divided by 6, the remainder will be








  1. View Hint View Answer Discuss in Forum

    Remainder on dividing 32 = 9 by 6 = 3
    Remainder on dividing 42 = 16 by 6 = 4
    Remainder on dividing 52 = 25 by 6 = 1

    Correct Option: C

    Remainder on dividing 32 = 9 by 6 = 3
    Remainder on dividing 42 = 16 by 6 = 4
    Remainder on dividing 52 = 25 by 6 = 1

  1. The greatest perfect square number of 6 digits is








  1. View Hint View Answer Discuss in Forum

    Largest 6-digit number = 999999

    ∴  Required perfect square number = 999999 – 1998 = 998001

    Correct Option: B

    Largest 6-digit number = 999999

    ∴  Required perfect square number = 999999 – 1998 = 998001

  1. three numbers are such that their sum is 50, product is 3750 and the sum of their reciprocals is 31/50 .Find the sum of the squares of the three numbers.








  1. View Hint View Answer Discuss in Forum

    x + y + z = 50 ; xyz = 3750

    ∴ 
    1
    		 +
    1
    		 +
    1
    		 =
    yz + zx + xy
    xyzxyz

    =
    31
    150

    ⇒  xy + yz + zx =
    31
    		xyz
    150

    =
    31
    		× 3750 = 775
    150

    ∴  (x + y + z )2 = x2 + y2 + z2 + 2(xy + yz + zx)
    ⇒  (50)2 + x2 + y2 + z2 + 2 × 775
    ⇒  2500 = x2 + y2 + z2 + 1550
    ⇒  x2 + y2 + z2 = 2500 – 1550
    = 950

    Correct Option: C

    x + y + z = 50 ; xyz = 3750

    ∴ 
    1
    		 +
    1
    		 +
    1
    		 =
    yz + zx + xy
    xyzxyz

    =
    31
    150

    ⇒  xy + yz + zx =
    31
    		xyz
    150

    =
    31
    		× 3750 = 775
    150

    ∴  (x + y + z )2 = x2 + y2 + z2 + 2(xy + yz + zx)
    ⇒  (50)2 + x2 + y2 + z2 + 2 × 775
    ⇒  2500 = x2 + y2 + z2 + 1550
    ⇒  x2 + y2 + z2 = 2500 – 1550
    = 950