Bank PO Questions

Nehru trophy is related to the sports game football.

Java is a general-purpose computer-programming language that is concurrent, class-based, object-oriented, and specifically designed to have as few implementation dependencies as possible.

Java is used to develop mobile apps, web apps, desktop apps, games and much more...

The Bhindawas Bird Sanctuary (BBS) is located in Jhajjar district, Haryana. It is home to many birds such as Blue Peafowl, Gray Francolin, Black Francolin, Shikra, Black Kite, Greater Coucal, White-throated Kingfisher, Coppersmith Barbet and Indian Roller. It is in news because the sanctuary will soon be developed as a beautiful tourist resort and the additional facilities to be provided to the tourists coming to this place would include arrangements for camping for night stay, beautification of the lake, and opening of dispensary for birds and the wild life.

The given number series 8, 18, 56, 232, 1165 

8

8 x 2 + 2 = 18

18 x 3 + 3 = 57 (not 56)

57 x 4 + 4 = 232

232 x 5 + 5 = 1165

Hence, the odd man in the given series is 56.

The given number series is 120, 36, 6, 2, 1, 1

120

120/5 = 24 (not 36)

24/4 = 6

6/3 = 2

2/2 = 1

1/1 = 1

Hence, the odd man in the given number series is 24.

The given series 24.3, 23.1, 20.7, 17.1, 12.4 follows a pattern that

24.3

24.3 - 1.2 = 23.1

23.1 - 2.4 = 20.7

20.7 - 3.6 = 17.1

17.1 - 4.8 = 12.3 not 12.4

Hence, the odd man or odd number in the given series is 12.4

Interest for 1 year is the same whether it's simple interest or the compound interest.

Now interest of third year = 8575 - 7350 = 1225; means principal for this interest is 7350 if comound interest is takenIf 7350 is the principal interest = 1225

if 100 is the principal interest =(1225/7350) * 100 = 50/3 %

When a thing increases for two successive times the overall increase on initial amount = a + b + (a * b)/100

Therefore overall interest for two years = 50/3 + 50/3 + [(50/3)*(50/3)]/100 = 325/9 %

Therefore amount after 2 years = 100 + 325/9 = 1225/9

If 1225/9 is the amount principal =100

if 7350 is the amount principal =(900/1225)*7350 =5400

So sum = 5400; Rate = 50/3 %

1. ANSWER : C 

 Explanation -  Number of students who passed half-yearly exams in the school 

= (Number of students passed in half-yearly but failed in annual exams) + (Number of students passed in both exams)
= (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)

= 288.

Also, Number of students who passed annual exams in the school

= (Number of students failed in half-yearly but passed in annual exams) + (Number of students passed in both exams)

= (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)

= 288.

Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.

Thus Statements (a), (b) and (d) are false and Statement (c) is true. 

 2. ANSWER : D 

 Explanation -  Since the classification of the students on the basis of their results and sections form independent groups, so the total number of students in the class: 

 = (28 + 23 + 17 + 27 + 14 + 12 + 8 + 13 + 6 + 17 + 9 + 15 + 64 + 55 + 46 + 76)

= 430. 

 3. ANSWER : D 

 Explanation- Pass percentages in at least one of the two examinations for different sections are:

  For Section A = 14+6+6428+14+6+64×100 = 84112×100% = 75%

 For Section B =12+17+5523+12+17+55×100 % = 78.5%

 For Section C = 8+9+4617+8+9+46×100%= 78.75%

 For Section D = 13+15+7627+13+15+76×100%= 79.39%

  Clearly ,the pass percentage is maximum for Section D 

 4. ANSWER : A

 Explanation - Total number of students passed in annual exams in a section 

= [ (No. of students failed in half-yearly but passed in annual exams) + (No. of students passed in both exams) ] in that section

 Success rate in annual exams in Section A= 14+64112 × 100% = 69.64%

 Similarly, success rate in annual exams in:

 Section B = 12+55107×100% =  62.62% 

 Section C = 8+4680×100% = 67.5% 

 Section D = 89131×100% = 67.94% 

 Clearly, the success rate in annual examination is maximum for Section A. 

 5. ANSWER : D 

 Explanation - Total number of failures in half-yearly exams in a section

  = [ (Number of students failed in both exams) + (Number of students failed in half-yearly but passed in Annual exams) ] in that section

 Failure rate in half-yearly exams in Section A %= 37.5 %

 Similarly, failure rate in half-yearly exams in:

  Section B = 32.71%

  Section C = 31.25%

  Section D = 30.53%

  Clearly, the failure rate is minimum for Section D.

 = (Number of students passed in half-yearly but failed in annual exams) + (Number of students passed in both exams)

 = (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)

 = 288.

 Also, Number of students who passed annual exams in the school

 = (Number of students failed in half-yearly but passed in annual exams) + (Number of students passed in both exams)

 = (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)

 = 288.

Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.

Thus Statements (a), (b) and (d) are false and Statement (c) is true.