Quantitative Aptitude Practice Question and Answer

Q: A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most  ? 1527 0

  • 1
    215
    Correct
    Wrong
  • 2
    268
    Correct
    Wrong
  • 3
    254
    Correct
    Wrong
  • 4
    216
    Correct
    Wrong
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Answer : 1. "215"
Explanation :

Answer: A) 215 Explanation: Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts, one of them is a successful attempt.   Maximum number of unsuccessful attempts = 216 - 1 = 215.

Q: Rs.1200 divided among P, Q and R. P gets half of the total amount received by Q and R. Q gets one-third of the total amount received by P and R. Find the amount received by R ? 4694 0

  • 1
    Rs. 1100
    Correct
    Wrong
  • 2
    Rs. 500
    Correct
    Wrong
  • 3
    Rs. 1200
    Correct
    Wrong
  • 4
    Rs. 700
    Correct
    Wrong
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Answer : 2. "Rs. 500"
Explanation :

Answer: B) Rs. 500 Explanation: Let the amounts to be received by P, Q and R be p, q and r.p + q + r = 1200p = 1/2 (q + r) => 2p = q + r Adding 'p' both sides, 3p = p + q + r = 1200=> p = Rs.400 q = 1/3 (p + r) => 3q = p + r Adding 'q' both sides, 4q = p + q + r = 1200=> q = Rs.300 r = 1200 - (p + q) => r = Rs.500.

Q: Find the odd one out of the group ? 1298 0

  • 1
    708
    Correct
    Wrong
  • 2
    392
    Correct
    Wrong
  • 3
    618
    Correct
    Wrong
  • 4
    816
    Correct
    Wrong
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Answer : 2. "392"
Explanation :

Answer: B) 392 Explanation: The sum of the digits in 708, 618 and 816 is 15, but not in 392.

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Answer : 3. "37"
Explanation :

Answer: C) 37 Explanation: The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.Number of cans of Maaza = 368/16 = 23Number of cans of Pepsi = 80/16 = 5Number of cans of Sprite = 144/16 = 9The total number of cans required = 23 + 5 + 9 = 37 cans.

Q: Five men and nine women can do a piece of work in 10 days. Six men and twelve women can do the same work in 8 days. In how many days can three men and three women do the work  ? 3754 0

  • 1
    18 days
    Correct
    Wrong
  • 2
    20 days
    Correct
    Wrong
  • 3
    16 days
    Correct
    Wrong
  • 4
    14 days
    Correct
    Wrong
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Answer : 2. "20 days"
Explanation :

Answer: B) 20 days Explanation: (5m + 9w)10 = (6m + 12w)8  => 50m + 90w = 48w + 96 w    => 1m = 3w 5m + 9w = 5m + 3m = 8m   8 men can do the work in 10 days. 3m +3w = 3m + 1w = 4m So, 4 men can do the work in (10 x 8)/4 = 20 days.

Q: A work which could be finished in 11 days was finished 4 days earlier after 4 more men joined. The number of men employed was ? 1394 0

  • 1
    7
    Correct
    Wrong
  • 2
    8
    Correct
    Wrong
  • 3
    9
    Correct
    Wrong
  • 4
    10
    Correct
    Wrong
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Answer : 1. "7"
Explanation :

Answer: A) 7 Explanation: As the work is sameM1D1 = M2D2 Here Let the number of men employed was M=> 11M = 7(M+4)=> 11M - 7M = 28=> M = 7

Q: A, B, C and D enter into partnership. A subscribes 1/3 of the capital B 1/4, C 1/5 and D the rest. How much share did A get in a profit of Rs.2490 ? 2577 0

  • 1
    Rs. 820
    Correct
    Wrong
  • 2
    Rs. 830
    Correct
    Wrong
  • 3
    Rs. 840
    Correct
    Wrong
  • 4
    Rs. 850
    Correct
    Wrong
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Answer : 2. "Rs. 830"
Explanation :

Answer: B) Rs. 830 Explanation: Let the total amount in the partnership be 'x'.Then A's share = x/3B's share = x/4C's share = x/5D's share = x - (x/3 + x/4 +x/5) = 13x/60 A : B : C : D = x/3 : x/4 : x/5 : 13x/60 = 20 : 15 : 12 : 13 A's share in the profit of Rs. 2490 = 20 (2490/60) = Rs. 830.

Q: The number of sequences in which 4 players can sing a song, so that the youngest player may not be the last is ? 1202 0

  • 1
    2580
    Correct
    Wrong
  • 2
    3687
    Correct
    Wrong
  • 3
    4320
    Correct
    Wrong
  • 4
    5460
    Correct
    Wrong
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Answer : 3. "4320"
Explanation :

Answer: C) 4320 Explanation: Let 'Y' be the youngest player. The last song can be sung by any of the remaining 3 players. The first 3 players can sing the song in (3!) ways. The required number of ways = 3(3!) = 4320.

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