Bank Exams Questions

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Three persons A, B, C can be arranged in a queue in six different ways, ie ABC, CBA, BAC, CAB, BCA, ACB. But since there are only 3 persons ahead of C, so C should be in front of the queue. Thus, there are only two possible arrangements, ie CBA and CAB.

We may consider the two cases as under:

Case I:  ←3C↔8B↔5A→21

Clearly, number of persons in the queue = (3+1+8+1+5+1+21=) 40

Case II:  ←3C A↔5B

Number of persons between A and C                  

= (8 - 6) = 2   

∵[C↔8B A→21B]

Clearly number of persons in the queue = (3+1+2+1+21) = 28

Now, 28 < 40. So, 28 is the minimum number of persons in the queue.                                                     

A can do the work = 18 days
B can do the work = 18/2 = 9 days
(A + B)'s 1 day work = 1/18 + 1/9 = 1/6
=> In 3 days = 3x1/6 = 1/2 work is completed.

difference in C.I and S.I in 2years =Rs.32

S.I for 1year =Rs.400

S.I for Rs.400 for one year =Rs.32

rate=[100*32)/(400*1)%=8%

difference between in C.I and S.I for 3rd year

=S.I on Rs.832= Rs.(832*8*1)/100=Rs.66.56

Quantity1-
Rakhi’s marks= 115
Meenakshi’s marks= 115 - 20 = 95
Puneeta’s marks= 95 + 65= 160
Ankita’s marks=160 - 35= 125
Simpy’s marks= 125+ 67= 192
Total maximum marks= 192 + 108= 300
Required percentage marks of Simpy
= 192300×100=64%
Quantity2-
Let length and breadth be 100.

After increase in length it become 160, then
reduction in breadth be ‘x’
Now, 160*x= 100*100
Hence, x = 10000160=62.5

100-62.5=37.5%