Bank PO Questions

50.0 mL 0.0500 M calcium hydroxide, Ca(OH)2 (aq) has the highest pH value.

Sodium hydroxide and Calcium hydroxide are strong bases so they completely ionize in solution. Because they ionize at the same rate, and Calcium hydroxide has the higher pH calculated, then Calcium hydroxide, in fact, has the highest pH.

Nitric acid is a strong acid, so it completely ionizes in water. Whilst nitrous acid is a weak acid. Therefore, the nitric acid solution has the LOWEST ph.

A + B = 15 .... (1)

B x C = 84 .... (2)

C ÷ 4 = 3  .... (3)

B + D = 18 .... (4)

From (3), we get C = 12

From (2), we get B = 84/12 = 7

From (1), we get A = 15 - 7 = 8

From (4), we get D = 18 - 7 = 11

Hence, D = 11.

ANBC Full Form is Adjusted Net Bank Credit.

ANBC refers to Adjusted Net Bank Credit, is Net Bank Credit added to investments made by banks in non-SLR bonds (in held-to-maturity) or it is the credit equivalent of off-balance - sheet exposures, whichever is higher.

Net Bank Credit = Outstanding bank credits in India (-) Bills rediscounted with RBI/approved financial institutions.

When we divide 1234 by 8, remainder is 2

When we divide 1235 by 8, remainder is 3

When we divide 1237 by 8, remainder is 5

---> 2 x 3 x 5 = 30 

As 30 will not be the remainder because it is greater than 8,

when 30 divided by 8, remainder = 6.

16200 = 23×34×52

A perfect cube has a property of having the indices of all its prime factors divisible by 3.

Required number = 32×5 = 9x5 = 45.

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.

Hence, number of ways in which we can select the black balls

= 4C1 + 4C2 + 4C3 + 4C4
= 24-1 ........(A)

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red balls

Hence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
=23-1........(B)

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.

Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + … + 5C5
= 25..............(C)

From (A), (B) and (C), required number of ways
=  2524-123-1

Clearly, we may mark such sets of 3 numbers, in which the middle number is 8 and each of the two numbers on both sides of it is a factor of 8, as shown below :

2 8 3 8 2 4 8 2 4 8 6 8 2 8 2 4 8 3 8 2 8 6

So, there are two such 8's. Hence the answer is C.