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Sample Aptitude Questions of Adobe

Practice important Aptitude questions asked in Adobe
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  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.
    Hence option 3.
  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 ΔCDE / area 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, Triangle CDE is similar to TriangleCAB and hence statements I and II are true.
    However, the ratio CD : CA = 3:9
    So,If AB = 4, then DE = 4/3. So statement III is false Hence option A
  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
    2. 45
    3. 35
    4. 30
    Answer: Option C
    When B scored p -10 then C scored p - 25.
    When B scores 1 then C scores (p-25)/(p-10)
    So when B scores p points then C will score (p-25)/(p-10) × p
    As per question (p-25)/(p-10) × p = p -10 . Solving this we get p = 35
    A B C
    p points (p-21) points (p-25) points
      p points (p-10) points
  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
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  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
    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 = 10√3/3 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 × 10√3/3)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.
    [image]
    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
    Explanation: 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 , T3 and T32 are consecutive terms in GP, we know that Tm / T2 = T32 / Tm
    ⇒ (Tm )2 = T2 × T32 = 2500
    ⇒ Tm = 50.
  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 four casks as 36 liters 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, 6 liters, 12 liters and 18 liters 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 4(36 liters) 9 12 15
    Hence the ratio of the liquids in the resulting mixture is 9 : 12 : 15 = 3 : 4 : 5
    Hence option C
  5. A trader sells two bullocks for Rs. 8,400 each, neither losing nor gaining in total. If he sold one of the bullocks at a gain of 20%, the other is sold at a loss of
    1. 20 %
    2. 18%
    3. 14%
    4. 21 %
    5. None of these
    Answer: Option A
    A trader sells two bullocks for Rs. 8,400 each, neither losing nor gaining in total.
    As he sold one bullocks at a gain of 20%, it means 120% of C.P = 8400
    We get C.P of one bullock = 7000
    So gain on one bullock = Rs 1400
    Other bullocks is sold at lose and there is neither losing nor gaining in total
    So loss on 2nd bullock = 1400/7000×100 = 20%
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