Sample Aptitude Questions of OpenText Cordys

  1. The fourth root of 24010000 is
    1. 7
    2. 49
    3. 490
    4. 70

    Answer : Option D.
    √24010000 = 4900
    Again, √4900 = 70
    4√√24010000 = 70
  2. The greatest 4 digit member which is a perfect square, is
    1. 9999
    2. 9909
    3. 9801
    4. 9081

    Answer : Option C.
    The greatest four -digit perfect square will be the square of the greatest two digit number, hence it will be 99 × 99 = 9801.
  3. A piece of work can be done by Ram and Shyam in 12 days, by Shyam and Hari in 15 days and by Hari and Ram in 20 days. Ram alone will complete the work in
    1. 30 days
    2. 32 days
    3. 36 days
    4. 42 days

    Answer : Option A.
    (Ram’s + Shyam’s) 1 day’s work = 1/12
    (Shyam’s + Hari’s) 1 day’s work = 1/15
    (Hari’s + Ram’s) 1 day’s work = 1/20
    Adding all three,
    2 (Ram’s + Shyam’s + Hari’s 1 day’s work = 1/12 + 1/15 + 1/20
    = (5+4+3)/60 = 1/5
    ∴ (Ram’s + Shyam’s + Hari’s) 1 day’s work = 1/10
    ∴Ram’s 1 day’s work
    = 1/10 - 1/15 = (3-2)/30 = 1/30
    ∴ Ram alone will do the work in 30 days.
  4. 3 men or 5 women can do a work in 12 days. How long will 6 men and 5 women take to finish the work?
    1. 4 days
    2. 5 days
    3. 6 days
    4. 7 days

    Answer : Option A.
    3 men = 5 women
    6 men + 5 women = 15 women
    ∴ By M1 D1 = M2 D2
    5 × 12 = 15 × D2
    D2 = (5 * 12)/15 = 4 days.
  5. Pipe A can fill a tank in 30 mins and Pipe B can fill it in 45 mins. If both are opened together, then after how much time must B be turned off so that the tank gets filled in 24 mins.
    1. 9 mins
    2. 12 mins
    3. 6 mins
    4. 18 mins

    Answer : Option A.
    Work done by pipe A is 24 mins = 24/30 = 4/5
    This means work left for B is 1/5. B can fill the entire tank in 45 mins. So now it has to fill only 1/5 of it which should take 45/5 = 9 mins.
  6. The difference of perimeter and diameter of a circle is X unit. The diameter of the circle is
    1. X/(π - 1) unit
    2. X/(π + 1) unit
    3. X/π unit
    4. (X/π - 1) unit

    Answer : Option A.
    If the diameter of the circle be d units, then
    π d – d = X
    ⇒ d(π - 1) = X
    => d = X/(π - 1) units
  7. The perimeter of the base of a right circular cylinder of volume V is ‘a’ unit, then the height of the cylinder is
    1. (4a2V)/π unit
    2. (4a2π)/V unit
    3. (πa2V)/4 unit
    4. (4πV)/a2 unit

    Answer : Option D.
    If the radius of base of cylinder be r units and its height be h units, then
    2 π r = a
    => r = a/2π units
    ∴ Volume of cylinder = πr2h
    => V = π * (a2)/(4π2) * h => h = (4πV)/a2 units
  8. A sphere of diameter 6 cm is dropped in a right circular cylindrical vessel party filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is just completely submerged in water, then the rise of water level in the cylindrical vessel is
    1. 2 cm
    2. 1 cm
    3. 3 cm
    4. 4 cm

    Answer : Option B.
    Volume of sphere = (4/3)πr3
    (4/3)π*3*3*3 = 36π cu. cm
    If the water level rises by h cm,
    Then
    πr2h = 36 π ⇒ 6 × 6 × h = 36
    ⇒ h = 1 cm
  9. A shopkeeper marks his goods 20% above his cost price and gives 15% discount on the marked price. His gain percent is
    1. 5%
    2. 7%
    3. 2%
    4. 1%

    Answer : Option C.
    If the CP of goods be Rs. 100, then
    Marked Price = Rs. 120
    Hence, S.P. = (120 x 85)/100 = Rs. 102
    Therefore, Gain Percent = (102 - 100)*100/100
    => Gain Percent = 2%
    Hence, option C is the answer.
  10. A shopkeeper earns a profit of 12% on selling a book at 10% discount on printed price. The ratio of the cost price to printed price of the book is
    1. 45 : 56
    2. 50 : 61
    3. 90 : 97
    4. 99 : 125

    Answer : Option A.
    CP of the book = Rs. X
    Printed price = Rs. Y
    Therefore, (y * 90)/100 = x * 112/100 => x/y = 90/112 = 45/56.
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