Bank PO Questions

SP of first article = Rs.1000

Profit = 20%

 CP = SP x 100100+P% = Rs. 25003

SP of Second Article = Rs.1000

 Loss = 20%

 CP = SP x 100100-L% = Rs. 1250

So, Total SP = Rs.2000; Total CP = Rs.6250/3

As the CP is more than SP, he makes a loss.

Loss = CP - SP = (6250/3) - 2000 = Rs.(250/3)

So, Loss Percent = Lossx100/CP = 4%

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The given number series is 3, 5, 13, 43, 167

3

3 x 1 + 2 = 5

5 x 2 + 3 = 13

13 x 3 + 4 = 43

43 x 4 + 5 = 177 (not equal to 167)

Hence, the odd number in the given number sequence is 167. 

Here the given series 1, 6,  ?, 15, 45, 66, 91 follows a

Pattern in the series is, +5, +9, +13,...,.. +21, +25

So +4 is getting increased at every term addition.

Missing Number in the series will be 15+(9+4) = 15 + 13 = 28.

1 x 1 + 1 x 1 = (1 + 1) = 2

2 x 2 + 2 x 2 = (4 + 4) = 8

8 x 3 + 3 x 3 = (24 + 9) = 33

33 x 4 + 4 x 4 = (132 + 16) = 148

148 x 5 + 5 x 5 = (740 + 25) = 765.

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 %

Given number of boxes = 14

Number of workers = 4

Now, number of whole boxes per worker = 14/4 = 3.5 

Hence, number of whole boxes per each coworker = 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.