Quantitative Aptitude प्रश्न और उत्तर का अभ्यास करें

प्र: The calendar for the year 1988 is same as which upcoming year ? 1545 0

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    2012
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  • 2
    2014
    सही
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  • 3
    2016
    सही
    गलत
  • 4
    2010
    सही
    गलत
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उत्तर : 3. "2016"
व्याख्या :

Answer: C) 2016 Explanation: We already know that the calendar after a leap year repeats again after 28 years. Here year 1988 is a Leap year, then the same calendar will be in the year = 1988 + 28 = 2016.

प्र: A bus covers its journey at the speed of 80km/hr in 10hours. If the same distance is to be covered in 4 hours, by how much the speed of bus will have to increase ? 1606 0

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    85 km/hr
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  • 2
    95 km/hr
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    105 km/hr
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  • 4
    120 km/hr
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उत्तर : 4. "120 km/hr"
व्याख्या :

Answer: D) 120 km/hr Explanation: Initial speed = 80km/hrTotal distance = 80 x 10 = 800kmNew speed = 800/4 =200km/hrIncrease in speed = 200 - 80 = 120km/hr

प्र: In a slip test, K got the 15th rank and he was 44th from the bottom of the list of passed students. 4 students did not take up the slip test and 3 students were failed. What is the total strenght of the class ? 1257 0

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    63
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  • 2
    62
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  • 3
    64
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  • 4
    65
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उत्तर : 4. "65"
व्याख्या :

Answer: D) 65 Explanation: Total strength of the class is given by 15 + 44 + 4 + 3 - 1 = 65

प्र: Find the odd man out from the given alternatives ? 2688 0

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    97
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  • 2
    233
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  • 3
    437
    सही
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  • 4
    599
    सही
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उत्तर : 3. "437"
व्याख्या :

Answer: C) 437 Explanation: Except 437 all others are prime numbers.

प्र: Karthik could cover a distance of 200 km in 22 days while resting for 2 hrs. per day. In how many days (approx) he will cover a distance of 250 km while resting for 2 hrs. per day and moving with 2/3rd of the previous speed ? 1791 0

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    41.25 days
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    37.5 days
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    39.75 days
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  • 4
    40 days
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उत्तर : 1. "41.25 days"
व्याख्या :

Answer: A) 41.25 days Explanation: Given Karthik can cover the distance of 200 kms resting 2 hrs per day in 22 days.   Let the initial speed be '1'   Hence, Time = 22(24hrs - 2)   Noe new speed = 2/3   Let the number of days he take be 'D'   Therefore, 2224-2x3200=D24-2x2250   D = 11 x 3 x 5/4 = 41.25 days. = 41 1/4 days.

प्र: A can build a wall in 16 days while B can destroy it in 8 days. A worked for 5 days. Then B joined with A for the next 2 days. Find in how many days could A build the remaining wall ? 2354 0

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    12 days
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  • 2
    14 days
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  • 3
    13 days
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  • 4
    16 days
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उत्तर : 3. "13 days"
व्याख्या :

Answer: C) 13 days Explanation: Now, Total work = LCM(16, 8) = 48 A's one day work = + 48/16 = + 3 B's one day work = - 48/8 = -6 Given A worked for 5 days to build the wall => 5 days work = 5 x 3 = + 15 2days B joined with A in working = 2(3 - 6) = - 6 Remaining Work of building wall = 48 - (15 - 6) = 39 Now this remaining work will be done by A in = 39/3 = 13 days.

प्र: In how many ways the letters of the word NUMERICAL can be arranged so that the consonants always occupy the even places ? 2314 0

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    5!
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  • 2
    6!
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  • 3
    4!
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  • 4
    Can't determine
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उत्तर : 4. "Can't determine"
व्याख्या :

Answer: D) Can't determine Explanation: NUMERICAL has 9 positions in which 2, 4, 6, 8 are even positions. And it contains 5 consonents i.e, N, M, R, C & L. Hence this cannot be done as 5 letters cannot be placed in 4 positions. Therefore, Can't be determined.

प्र: Find the total numbers greater than 4000 that can be formed with digits 2, 3, 4, 5, 6 no digit being repeated in any number ? 1614 0

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    120
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  • 2
    256
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  • 3
    192
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  • 4
    244
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उत्तर : 3. "192"
व्याख्या :

Answer: C) 192 Explanation: We are having with digits 2, 3, 4, 5 & 6 and numbers greater than 4000 are to be formed, no digit is repeated.   The number can be 4 digited but greater than 4000 or 5 digited.   Number of 4 digited numbers greater than 4000 are4 or 5 or 6 can occupy thousand place => 3 x P34 = 3 x 24 = 72.    5 digited numbers = P55 = 5! = 120  So the total numbers greater than 4000 = 72 + 120 = 192

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