Calsoft3i InfotechABBAccentureAdobeAgilentAkamaiAmazonAmdocsAricent More 

Sample Aptitude Questions of Calsoft

A quick review on sample aptitude question
Views:20905
Instant Access to Free Material
  1. 7 cannibals of XYZ island, decide to throw a party. As you may be aware, cannibals are guys who eat human beings. The senior among them – Father Cannibal decides that any 6 of them will eat up one cannibal, then out of the remaining six – five of them will eat up one cannibal and so on till one is left. What is the time until one cannibal is left, if it takes one cannibal 3 hours to eat up one cannibal independently?
    1. 7 hrs 11 min
    2. 6 hrs 12 min
    3. 7 hrs 21 min
    4. 18 hrs 16 min
    Answer: Option C
    At the beginning 6 cannibals will eat one, so time required will be 180/6 = 30 min.
    Then out of the remaining six – five will devour one, so time required will be 180/5 = 36 min.
    Thus the time until one cannibal is left will be = (180/6 + 180/5 + 180/4 + 180/3 + 180/2 + 180/1) min
    = (30 + 36 + 45 + 60 + 90 + 180) min
    = 441 min
    = 7 hrs 21 min
  1. Three articles are purchased for Rs. 1050, each with a different cost. The first article was sold at a loss of 20%, the second at 1/3rd gain and the third at 60% gain. Later he found that their SPs were same. What was his net gain/loss?
    1. 14.28% gain
    2. 13% loss
    3. 12% loss
    4. 11.11% gain
    Answer: Option A
    Let us assume that their CPs are x, y & z respectively.
    According to the given condition 0.8x = 1.33y = 1.6z
    ⇒ (80/100)x = 400y/(3 × 100) = (160/100)z
    ⇒ x : y = 5 : 3 & y : z = 6 : 5
    Thus x : y : z = 10 : 6 : 5
    Hence CPs of the articles are x = (10/21) × 1050 = 500,
    y = (6/21) × 1050 = 300 &
    z = (5/21) × 1050 = 250.
    SP of the article with CP Rs. x is 0.8x = 0.8 × 500 = 400.
    Since SPs are same, the total SP will be 400 × 3 = 1200.
    Hence the gain % = (SP – CP)/CP × 100 = (1200 – 1050)/1050 × 100 = 14.28%.
  2. In the figure below, line DE is parallel to line AB. If CD = 3 and AD = 6, which of the following must be true?
    [image]
    1. ΔCDE ≈ ΔCAB
    2. (area of ΔCDE)/(area of ΔCAB) = ( CD/CA )2
    3. If AB = 4, then DE = 2
    1. I and II only
    2. I and III only
    3. II and III only
    4. I, II and III
    Answer: Option A
    Since the lines are parallel, ΔCDE ≈ ΔCAB and hence statements I and II are true. However, the ratio CD : CA = 3:9 ⇒ If AB = 4, then DE = 4/3. So statement III is false
  3. In a game of tennis, A gives B 21 points and gives C 25 points. B gives C 10 points. How many points make the game?
    1. 50 points
    2. 45 points
    3. 35 points
    4. 30 points
    Answer: Option C
    Suppose p points make the game. Hence the points played by A, B and C will be,
    A B C
    p points (p - 21)points (p - 25)points
      p points (p - 10)points
    Therefore, (p – 21) × (p – 10) = p × (p – 25)
    ⇒ p2 – 31p + 210 = p2 – 25p
    ⇒ 6p = 210
    ⇒ p = 35.
    Hence 35 points make the game.
  4. What is the value of a if x3 + 3x2 + ax + b leaves the same remainder when divided by (x – 2) and (x + 1)?
    1. 18
    2. 3
    3. -6
    4. Cannot be determined
    Answer: Option C
    Suppose the remainder is R.
    Substituting x = 2 and x = –1, we get R = 8 + 12 + 2a + b = –1 + 3 – a + b
    ⇒ 3a = –18
    ⇒ a = –6
Calsoft Preparation Links
  1. What is the range of values of k if (1 + 2k)x2 – 10x + k – 2 = 0 has real roots?
    1. –3 ≤ k ≤ 4.5
    2. –1.5 ≤ k ≤ 9
    3. k ≥ 4.5, k ≤ –3
    4. k ≥ 3, k ≤ –9
    Answer: Option A
    Since the given expression has real roots, we know that (–10)2 – 4(1 + 2k)(k – 2) ≥ 0
    100 – 8k2 + 12k + 8 ≥ 0
    8k2 – 12k – 108 ≤ 0
    2k2 – 3k – 27 ≤ 0
    –3 ≤ k ≤ 9/2
  2. A square, S1, circumscribes the circum circle of an equilateral triangle of side 10 cm. A square, S2, is inscribed in the in circle of the triangle. What is the ratio of the area of S1 to the area of S2?
    1. 4:1
    2. 32:1
    3. 8:1
    4. 2:1
    Answer: Option C
    [image]
    The height of the equilateral triangle is 5√3 cm.
    Since the height is also the median, we know that the circum-radius is 2/3 × 5√3 = 103/3 and the in-radius is 1/3 × 5√3 = 5√3/3.
    The diameter of the circumcircle is the side of square S1. So the area of S1 is (2 × 103/3)2 = 1200/9.
    The diameter of the in-circle is the diagonal of square S2. So the area of S2 is ½ × (2 × 5√3/3)2 = 300/18.
    Thus the ratio of areas S1 : S2 is 1200/9 : 300/18 = 8 : 1.
  3. The sum of the first n terms of an AP is Sn = 4n2 – 2n. Three terms of this series, T2, Tm and T32 are consecutive terms in GP. Find m
    1. 7
    2. 10
    3. 16
    4. 5
    Answer: Option A
    From the given information, S1 = 4 × 12 – 2 × 1 = 2 => T1 = 2. Now, S2 = 4 × 22 – 2 × 2 = 12.
    Since S2 = T1 + T2 and T1 = 2, we get T2 = 10. So, the 1st term of the AP is 2 and the common difference is 8. From this, we get T32 = 2 + 31 × 8 = 250.
    Since T2, Tm and T32 are consecutive terms in GP, we know that Tm / T2 = T32 / Tm ⇒ Tm2 = T2 × T32 = 2500 ⇒ Tm = 50.
    Since Tm is a term in the AP, we have 50 = 2 + 8(m – 1) ⇒ 48 = 8(m – 1) ⇒ m = 7.
  4. Three casks of equal capacities contain three liquids A, B & C in the ratio 1 : 2 : 3, 3 : 4 : 5 & 5 : 6 : 7 respectively. The mixtures from these casks are taken in the ratio 1 : 2 : 3 and poured into a 4th cask with the same capacity as that of the three casks and the cask is completely filled. What is the ratio of the liquids A, B and C in the resulting mixture?
    1. 25:36:47
    2. 16:21:26
    3. 3:4:5
    4. 1:2:3
    Answer: Option C
    ⇒(1 + 2 + 3) = 6, (3 + 4 + 5) = 12 & (5 + 6 + 7) = 18. Common multiple of (6, 12, 18) = 36. So let us fix the capacities of the four casks as 36 litres each.
      Liquid A Liquid B Liquid C
    Cask 1 (36 liters) 6 12 18
    Cask 2 (36 liters) 9 12 15
    Cask 3 (36 liters) 10 12 14
    Since the mixtures are taken in the ratio 1:2:3, 6litres, 12 litres and 18 litres mixture are drawn from the three casks respectively.
      Liquid A Liquid B Liquid C
    Cask 1 (6 liters) 1 2 3
    Cask 2 (12 liters) 3 4 5
    Cask 3 (18 liters) 5 6 7
    Cask 3 (36 liters) 9 12 15
    Hence the ratio of the liquids in the resulting mixture is 9 : 12 : 15 = 3 : 4 : 5
  5. In the figure below, the radii of the circles with centres at A and B are 5 cm and 3cm respectively. If ∠FAG = ∠DBC = 90° what is the perimeter of CEG? [image]
    1. 38.63cm
    2. 30.14cm
    3. 42.18cm
    4. 27.31cm
    Answer: Option E
    In ΔFAG, since AF = AG and ∠FAG = 90°, we know that ∠FGA = 45°. Similarly, ∠DCB = 45°. So, we can conclude that ΔCEG is an isosceles right triangle with hypotenuse CG = 16 cm. From this, we can calculate EG = CE = 8√2 cm. Thus the perimeter of ΔCEG = 8√2 + 8√2 + 16 = 16√2 +16 ≈ 38.63cm
Views:20905

Most Popular Articles

Placement Papers of all major companies

List of IT Companies

Aptitude Questions of all Major Companies

Test Pattern of all major companies

Technical Questions of all major companies

List of Core Companies

Interview Questions of all major companies

All About GATE Entrance Exam

How to prepare for Aptitude Tests?

GATE Previous Year Papers