Bank PO Questions

In the given number puzzle,

Sum of Row 1 = 14

Sum of Row 2 = 23

Sum of Row 3 = 13 + ?

Sum of Column 1 = 16 

Sum of Column 2 = 23 

Sum of Column 3 = 11 + ?

Diagonal 1 sum = 17

Diagonal 2 sum = 14 + ?

From the above analysis,

Sum of Row 2 = Sum of Column 2 = 23

Sum of Row 1 = Sum of Column 3 = 14 = 11 + ?    .............(1)

Sum of Row 3 = Sum of Column 1 = 16 = 13 + ?    .............(2)

Diagonal 1 sum = Diagonal 2 sum = 17 = 14 + ?    .............(3)

Hence,

From (1), (2) & (3)

? = 3

Let S be the sample space.
Here n(S)= 23 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.

Let the required height of the building be x meter

More shadow length, More height(direct proportion)

Hence we can write as

(shadow length) 40.25 : 28.75 :: 17.5 : x
⇒ 40.25 × x = 28.75 × 17.5
⇒ x = 28.75 × 17.5/40.25
= 2875 × 175/40250
= 2875 × 7/1610
= 2875/230
= 575/46
= 12.5 cm

Let 'M' be the maximum marks in the examination.

Therefore, Madhu got 32% of M = 32M/100 = 0.32M
And Kumar got 42% of M = 42M/100 = 0.42M.
In terms of the maximum marks Kumar got 0.42M - 0.32M = 0.1M more than Madhu. ---- (1)

The problem however, states that Kumar got 16 marks more than the cut-off mark and Madhu got 8 marks less than the cut-off mark. Therefore, Kumar has got 16 + 8 = 24 marks more than Madhu. ---- (2)

Now, Equating (1) and (2), we get
0.1M = 24 => M = 24/0.1 = 240
'M' is the maximum mark and is equal to 240 marks.
We know that Madhu got 32% of the maximum marks.
Therefore, Madhu got 32 x 240/100 = 76.8 marks.
We also know that Madhu got 8 marks less than the cut-off marks.

Therefore, the cut-off marks will be 8 marks more than what Madhu got
= 76.8 + 8 = 84.8.

The intersection of a column and a row in a spread sheet is called as a cell or a bon.