Bank Exams प्रश्न और उत्तर का अभ्यास करें

प्र: How many 6-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that the digits should not repeat and the second last digit is even ? 2540 0

  • 1
    521
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  • 2
    720
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  • 3
    420
    सही
    गलत
  • 4
    225
    सही
    गलत
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उत्तर : 2. "720"
व्याख्या :

Answer: B) 720 Explanation: Let last digit is 2when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.so for last digit = 2, total numbers=240   Similarly for 4 and 6When last digit = 4, total no. of ways =240and last digit = 6, total no. of ways =240so total of 720 even numbers are possible.

प्र: Which memory management technique involves dividing the memory into fixed sized blocks ? 1743 2

  • 1
    Paging
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  • 2
    Scaling
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  • 3
    Whiping
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  • 4
    Tracking
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उत्तर : 1. "Paging"
व्याख्या :

Answer: A) Paging Explanation: Paging is a memory management technique in which process address space is broken into blocks of the same size called pages.

प्र: A man spend Rs. 810 in buying trouser at Rs. 70 each and shirt at Rs. 30 each. What will be the ratio of trouser and shirt when the maximum number of trouser is purchased ? 3430 0

  • 1
    1:3
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  • 2
    2:1
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  • 3
    3:2
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  • 4
    2:3
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उत्तर : 3. "3:2"
व्याख्या :

Answer: C) 3:2 Explanation: Let us assume S as number of shirts and T as number of trousersGiven that each trouser cost = Rs.70 and that of shirt = Rs.30Therefore, 70 T + 30 S = 810=> 7T + 3S = 81......(1)T = ( 81 - 3S )/7 We need to find the least value of S which will make (81 - 3S) divisible by 7 to get maximum value of TSimplifying by taking 3 as common factor i.e, 3(27-S) / 7In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7 Hence T = (81-3xS)/7 = (81-3x6)/7 = 63/7 = 9 Hence for S, put T in eq(1), we getS = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.The ratio of T:S = 9:6 = 3:2.

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उत्तर : 4. "a,e,c"
व्याख्या :

Answer: D) a,e,c Explanation: The statements  a,e,c  are logically sequenced.

प्र: A wholesale dealer allows a discount of 20 % on the marked price to the retailer. The retailer sells at 5% below the marked price. If the customer pays Rs.19 for an article, what profit is made by the retailer on it ? 1678 0

  • 1
    Rs. 4
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  • 2
    Rs. 3
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  • 3
    Rs. 5
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  • 4
    Rs. 1
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उत्तर : 2. "Rs. 3"
व्याख्या :

Answer: B) Rs. 3 Explanation: Let wholesaler dealers marked price = 100%, Retailer's C.P = 80% And the retailer sells at 5% less than the marked price => S.P = 95%If S.P of 95% of the retailer costs Rs.19 to customer,so its C.P of 80% will cost 80 x 19/95 = 16 Profit made by the retailer = 19-16 =  Rs.3

प्र: Total number of students in 3 classes of a school is 333 . The number of students in class 1 and 2 are in 3:5 ratio and 2 and 3 class are 7:11 ratio . What is the strength of class that has highest number of students ? 2631 0

  • 1
    125
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  • 2
    155
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    गलत
  • 3
    135
    सही
    गलत
  • 4
    165
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उत्तर : 4. "165"
व्याख्या :

Answer: D) 165 Explanation: Ratio C1:C2 = 3:5 and C2:C3 = 7:11So C1:C2:C3 = 21 : 35 : 55Let the strength of three classes are 21x, 35x and 55x respectively, then Given that 21x + 35x + 55x = 333=> 111x = 333 or x=3So strength of the class with highest number of students =55x = 55x3 = 165.

प्र: The top and bottom of a tower were seen to be at angles of depression 30° and 60° from the top of a hill of height 100 m. Find the height of the tower ? 2300 1

  • 1
    42.2 mts
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  • 2
    33.45 mts
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  • 3
    66.6 mts
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  • 4
    58.78 mts
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उत्तर : 3. "66.6 mts"
व्याख्या :

Answer: C) 66.6 mts Explanation: From above diagramAC represents the hill and DE represents the tower Given that AC = 100 m angleXAD = angleADB = 30° (∵ AX || BD ) angleXAE = angleAEC = 60° (∵ AX || CE) Let DE = h Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1) tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)∵ BD = CE and Substitute the value of CE from equation 1 100/√3 = 100−h(√3) => h = 66.66 mts The height of the tower = 66.66 mts.

प्र: 5/9 of the part of the population in a village are females. If 30 % of the females are married. The percentage of unmarried males in the total males is ? 1894 0

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    62.5 %
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  • 2
    125 %
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  • 3
    84.32 %
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  • 4
    46.87 %
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उत्तर : 1. "62.5 %"
व्याख्या :

Answer: A) 62.5 % Explanation: Let total population = p number of females = 5p/9number of males =(p-5p/9) = 4p/9married females = 30% of 5p/9 = 30x5p/100x9 = p/6married males = p/6unmarried males =(4p/9-p/6) = 5p/18Percentage of unmarried males in the total males = {(5p/18)/4p/9}x100 = (5p/18)x(9/4p) = 125/2 % = 62.5%

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