Bank PO Questions

The comparison operator that is used in SQL is '='. Comparison operators test whether two expressions are the same. Comparison operators can be used on all expressions except expressions of the text, ntext, or image data types.

let sum =x . s.i.=x

rate =(100*x)/(x*6)=50/3%

now ,sum =x, s.i= 3x, rate =50/3%

time =(100*3x)/(x*50/3) = 18 years

A can write in 1hour = 32/6 pages

Similarly, B in 1 hour = 40/5 pages

Together (A+B) in 1 hour = 32/6 + 40/5 = 40/3

So, A+B write 40/3 pages in 1 hour

A+B write 110 pages in (3/40) x 110 Hours = 8 hours 15 min.

One day work of 6 boys and 8 girls is given as 6b + 8g = 1/10 -------->(I)

One day work of 26 boys and 48 women is given as 26b + 48w = 1/2 -------->(II)

Divide both sides by 2 in (I) and then multiply both sides by 5

Now we get, 15b + 20g = 1/4.

Therefore, 15 boys and 20 girls can do the same work in 4 days.

Given that,

Members in football team = n(f) = 10

Members in science club = n(s) = 14

Members in both football team and science club = n(f ^ s) = 5

Then, Members in only football team = n(F) = 10 - 5 = 5

Members in only science team = n(S) = 14 - 5 = 9

Hence, members in only football or science team = n(FUS) = n(F) + n(S) - n(f^s) 

= 9 + 5 - 5

= 9.

Hence, many students belong to only science club team or football team = 9.

only (c) is correct since it is divisible by 4.

Let the original number of element  be x then the new no.of elements will be 

            4x5 = K                               

So K must be divisible by 4

Since,      x =  Kx5/4

Let the present ages of A and B be 'x' and 'y' respectively

From the given data, 

[(x-2) + (y-2)]/2 = 26

=> x+y = 56

But given the age of A, 5 years hence is 40 yrs => present age of A = 40 - 5 = 35 yrs

=> x = 35 => y = 56 - 35 = 21

=> Age of B = 21 yrs

Given B is 5 years younger to C,

=> Age of C = 21 + 5 = 26 yrs

=> Required Difference between ages of A and C = 35 - 26 = 9 yrs.

Suppose there are 9 balls

Let us give name to each ball B1 B2 B3 B4 B5 B6 B7 B8 B9

Now we will divide all the balls into 3 groups.

Group1 - B1 B2 B3

Group2 - B4 B5 B6

Group3 - B7 B8 B9

Step1 - Now weigh any two groups. Let's assume we choose Group1 on left side of the scale and Group2 on the right side.

So now when we weigh these two groups we can get 3 outcomes.

Weighing scale tilts on left - Group1 has a heavy ball.
Weighing scale tilts on right - Group2 has a heavy ball.
Weighing scale remains balanced - Group3 has a heavy ball.
Lets assume we got the outcome as 3. i.e Group 3 has a heavy ball.

Step2 - Now weigh any two balls from Group3. Lets assume we keep B7 on left side of the scale and B8 on right side.

So now when we weigh these two balls we can get 3 outcomes.

Weighing scale tilts on left - B7 is the heavy ball.
Weighing scale tilts on right - B8 is the heavy ball.
Weighing scale remains balanced - B9 is the heavy ball.
The conclusion we get from this Problem is that each time weigh. We element 2/3 of the balls.

As we came to conclusion that Group3 has the heavy ball from Step1, we remove 6 balls from the equation i.e (2/3) of 9.

Simillarly we do the ame thing for the Step2.

Now going with this conclusion. We have 6561 balls.

Step - 1

Divided into 3 groups

Group1 - 2187Balls

Group2 - 2187Balls

Group3 - 2187Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 2

Divided into 3 groups

Group1 - 729Balls

Group2 - 729Balls

Group3 - 729Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 3

Divided into 3 groups

Group1 - 243Balls

Group2 - 243Balls

Group3 - 243Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 4

Divided into 3 groups

Group1 - 81Balls

Group2 - 81Balls

Group3 - 81Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 5

Divided into 3 groups

Group1 - 27Balls

Group2 - 27Balls

Group3 - 27Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 6

Divided into 3 groups

Group1 - 9Balls

Group2 - 9Balls

Group3 - 9Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 7

Divided into 3 groups

Group1 - 3Balls

Group2 - 3Balls

Group3 - 3Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 8

So now when we weigh 2 balls out of 3 we can get 3 outcomes.

Weighing scale tilts on left - left side placed is the heavy ball.
Weighing scale tilts on right - right side placed is the heavy ball.
Weighing scale remains balanced - remaining ball is the heavy ball.
So the general answer to this question is, it is always multiple of 3 steps.

For 9 balls  32= 9. therefore 2 steps

For 6561 balls 38 = 6561 therefore 8 steps