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Aptitude miscellaneous
  1. The sum of squares of three positive integers is 323. If the sum of squares of two numbers is twice the third, their product is








  1. View Hint View Answer Discuss in Forum

    Let the positive numbers be x, y and z (respectively).
    ∴  x2 + y2 + z2 = 323       ... (i)
    and   x2 + y2 = 2z       ... (ii)
    ∴  z2 + 2z = 323
    ⇒  z2 2z – 323 = 0
    ⇒  z2 + 19z – 17z – 323 = 0
    ⇒  z (z + 19) – 17 (z + 19) = 0
    ⇒  (z – 17) (z + 19) = 0
    ⇒  z = 17 because z ≠ –19
    ∴  x2 + y2 = 2 × 17 = 34
    = 32 + 52
    ∴  xyz = 3 × 5 × 17 = 255

    Correct Option: A

    Let the positive numbers be x, y and z (respectively).
    ∴  x2 + y2 + z2 = 323       ... (i)
    and   x2 + y2 = 2z       ... (ii)
    ∴  z2 + 2z = 323
    ⇒  z2 2z – 323 = 0
    ⇒  z2 + 19z – 17z – 323 = 0
    ⇒  z (z + 19) – 17 (z + 19) = 0
    ⇒  (z – 17) (z + 19) = 0
    ⇒  z = 17 because z ≠ –19
    ∴  x2 + y2 = 2 × 17 = 34
    = 32 + 52
    ∴  xyz = 3 × 5 × 17 = 255

  1. Each member of a club contributes as much rupees and as much paise as the number of members of the club. If the total contribution is Rs. 2525, then the number of members of the club is








  1. View Hint View Answer Discuss in Forum

    Number of members in the club = x (let)
    According to the question,

    x2 +
    x2
    		 = 2525
    100

    ⇒ 
    100x2 + x2
    		 = 2525
    100

    ⇒  101x2 = 252500
    ⇒  x2 =
    252500
    		 = 2500
    101

    ⇒  x = √2500 = 50

    Correct Option: D

    Number of members in the club = x (let)
    According to the question,

    x2 +
    x2
    		 = 2525
    100

    ⇒ 
    100x2 + x2
    		 = 2525
    100

    ⇒  101x2 = 252500
    ⇒  x2 =
    252500
    		 = 2500
    101

    ⇒  x = √2500 = 50

  1. A number when divided by 6 leaves remainder 3. When the square of the same number is divided by 6, the remainder is:








  1. View Hint View Answer Discuss in Forum

    Let the number (n) be 6m + 3
    where m = quotient.
    On squaring both sides,
    n2 = (6m + 3)2
    = 36 m2 + 36 m + 32
    ∴  Required remainder = 32
    ∴  Remainder on dividing 9 by 6 = 3

    Correct Option: D

    Let the number (n) be 6m + 3
    where m = quotient.
    On squaring both sides,
    n2 = (6m + 3)2
    = 36 m2 + 36 m + 32
    ∴  Required remainder = 32
    ∴  Remainder on dividing 9 by 6 = 3

  1. The product of two positive integers is 2048 and one of them is twice the other. The smaller number is








  1. View Hint View Answer Discuss in Forum

    Let the smaller number be x.
    ∴  Larger number = 2x
    According to the question,
    2x2 = 2048

    ⇒  x2 =
    2048
    		 = 1024
    2

    ∴  x = √1024 = 32

    Correct Option: A

    Let the smaller number be x.
    ∴  Larger number = 2x
    According to the question,
    2x2 = 2048

    ⇒  x2 =
    2048
    		 = 1024
    2

    ∴  x = √1024 = 32

  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