Quantitative Aptitude प्रश्न और उत्तर का अभ्यास करें

प्र: 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  ? 1700 0

  • 1
    215
    सही
    गलत
  • 2
    268
    सही
    गलत
  • 3
    254
    सही
    गलत
  • 4
    216
    सही
    गलत
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उत्तर : 1. "215"
व्याख्या :

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.

प्र: 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 ? 4875 0

  • 1
    Rs. 1100
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    गलत
  • 2
    Rs. 500
    सही
    गलत
  • 3
    Rs. 1200
    सही
    गलत
  • 4
    Rs. 700
    सही
    गलत
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उत्तर : 2. "Rs. 500"
व्याख्या :

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.

प्र: Find the odd one out of the group ? 1445 0

  • 1
    708
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  • 2
    392
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  • 3
    618
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    गलत
  • 4
    816
    सही
    गलत
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उत्तर : 2. "392"
व्याख्या :

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

प्र: A drink vendor has 368 liters of Maaza, 80 liters of Pepsi and 144 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required ? 1469 0

  • 1
    47
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  • 2
    46
    सही
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  • 3
    37
    सही
    गलत
  • 4
    35
    सही
    गलत
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उत्तर : 3. "37"
व्याख्या :

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.

प्र: 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  ? 3946 0

  • 1
    18 days
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  • 2
    20 days
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  • 3
    16 days
    सही
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  • 4
    14 days
    सही
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उत्तर : 2. "20 days"
व्याख्या :

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.

प्र: 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 ? 1546 0

  • 1
    7
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  • 2
    8
    सही
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  • 3
    9
    सही
    गलत
  • 4
    10
    सही
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उत्तर : 1. "7"
व्याख्या :

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

प्र: 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 ? 2760 0

  • 1
    Rs. 820
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    गलत
  • 2
    Rs. 830
    सही
    गलत
  • 3
    Rs. 840
    सही
    गलत
  • 4
    Rs. 850
    सही
    गलत
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उत्तर : 2. "Rs. 830"
व्याख्या :

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.

प्र: The number of sequences in which 4 players can sing a song, so that the youngest player may not be the last is ? 1312 0

  • 1
    2580
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  • 2
    3687
    सही
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  • 3
    4320
    सही
    गलत
  • 4
    5460
    सही
    गलत
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उत्तर : 3. "4320"
व्याख्या :

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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