Bank PO Questions

Starting From his house, Sumanth walks 200 feet straight towards North; we can write it as N200. Similarly, the other distances and directions can be considered and written as follows : (Notice that he covers 300 yards towards East; one yard = 3 feet and hence 300 yards = 900 feet)

 N200S100E900 Adding all south and all North distances and finding the net result, and also adding all East and all West distances and finding the net result, this can further be summarized as : N100E900

 Thus, the direction from his original position is North-East.

Here the most stable carbocation is the option (2).

Carbocations are always hungry for electrons. The stability of carbocations increases from primary to secondary to tertiary. Here, option (c) shows the most stable carbocation because it can easily withdraw electrons from two other carbon atoms and the more electronegative oxygen atom.

Carbocation stability depends on hyperconjugation. The tertiary carbocation is stabilized by electron releasing inductive effect.

Number of runs scored more to increse the ratio by 1 is 26 - 14 = 12
To raise the average by one (from 14 to 15), he scored 12 more than the existing average.
Therefore, to raise the average by five (from 14 to 19), he should score 12 x 5 = 60 more than the existing average. Thus he should score 14 + 60 = 74.

Total number of female members = 115

Let the number of males = x

Number of females = y

=> x + y = 115  ......(1)

According to the given data,

x + y(0.50) = 69 ......(2)

From (1) & (2), we get

0.5y = 46

=> y = 46 x 2 = 92

=> x = 115 - 92 = 23

Hence, required number of gentlemen in the family = x = 23.

Assuming Sripad has scored the least marks in subject other than science,
Then the marks he could secure in other two are 58 each.
Since the average mark of all the 3 subject is 65.
i.e (58+58+x)/3 = 65
116 + x = 195
x = 79 marks.

Therefore, the maximum marks he can score in maths is 79.

LCM of 4, 6, 8 and 10 = 120
120) 1000 (8
960
------
40

The least number of four digits which is divisible by 4, 6, 8 and 10 => 1000 + 120 - 40 = 1080.

Let us assume S as number of shirts and T as number of trousers
Given that each trouser cost = Rs.70 and that of shirt = Rs.30
Therefore, 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 T
Simplifying by taking 3 as common factor i.e, 3(27-S) / 7
In 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 get
S = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.
The ratio of T:S = 9:6 = 3:2.