UPSC CAPF

Comparing the volumes: $V_A = 2$, $V_B = 5$, $V_C = 6$.
Block C has the largest volume (6 cubic units) and is therefore the heaviest.

Let’s look for a relationship between the columns within each row.
Observe $R_1$: $1, 7, 9$. $1^2 + 7 + 1 = 1 + 7 + 1 = 9$. So $C_1^2 + C_2 + C_1 = C_3$.
Observe $R_3$: $3, 105, 117$. Let’s test the same relationship: $3^2 + 105 + 3 = 9 + 105 + 3 = 117$. This relationship holds for R3 as well.
Let’s apply this relationship to $R_2$: $C_1=2, C_2=14$.
The missing number ($C_3$) should be $C_1^2 + C_2 + C_1 = 2^2 + 14 + 2 = 4 + 14 + 2 = 20$.

From “SERVANT is coded as 6789450”:
S -> 6
E -> 7
R -> 8
V -> 9
A -> 4 (Matches the code from WOMAN)
N -> 5 (Matches the code from WOMAN)
T -> 0

We need to find the code for “VOTERS”. We use the established codes for each letter:
V -> 9
O -> 2
T -> 0
E -> 7
R -> 8
S -> 6
Combining these, VOTERS is coded as 920786.

The final position is (3, 4) and the starting position is (0, 0).
The distance from the starting point is the straight-line distance between (0, 0) and (3, 4).
Using the distance formula: Distance = $\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$
Distance = $\sqrt{(3 – 0)^2 + (4 – 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ km.

We need to find the day of the week for 1st December of the previous year.
Let the current year be Y. We are finding the day of 1st December, Y-1.
The period is from 1st December, Y-1 to 6th January, Y.
Number of days in December Y-1 = 31 days.
Number of days in January Y up to 6th = 6 days.
Total number of days = 31 + 6 = 37 days.

To find the day of the week 37 days before Tuesday, we find the number of odd days in 37.
Number of odd days = $37 \pmod 7$.
$37 = 5 \times 7 + 2$. The remainder is 2.
So, 37 days is equal to 5 weeks and 2 days.
The day of 1st December, Y-1 is 2 days before Tuesday.
Tuesday – 1 day = Monday.
Monday – 1 day = Sunday.

The times in the interval [4:00 p.m., 10:00 p.m.] are:
4:05, 4:38, 5:10, 5:43, 6:16, 6:49, 7:21, 7:54, 8:27, 9:00, 9:32.
All these 11 times are within the specified range.

However, the option C is 10. This suggests a different interpretation. A common interpretation in such problems is to count the number of times the hands form a right angle strictly *between* the listed hour marks.
Times within (4:00, 5:00): 4:05, 4:38 (2)
Times within (5:00, 6:00): 5:10, 5:43 (2)
Times within (6:00, 7:00): 6:16, 6:49 (2)
Times within (7:00, 8:00): 7:21, 7:54 (2)
Times within (8:00, 9:00): 8:27 (1) – 9:00 is a boundary
Times within (9:00, 10:00): 9:32 (1) – 9:00 and 10:00 are boundaries
Summing these gives 2 + 2 + 2 + 2 + 1 + 1 = 10.
This interpretation excludes the instance at 9:00 p.m., which falls exactly on an hour boundary.