Question

\(15^{th}\) August of a year falls on Wednesday. Then what day is \(2^{nd}\) October of that year?

• A. Wednesday
• B. Tuesday
• C. Monday
• D. Sunday

Hint:

For calendar questions, count total days and take the remainder after division by \(7\).

Answer 1

The Correct Option is B

Concept:
To find the day of a later date, count the number of days between the two dates and divide by \(7\). The remainder gives the day shift.

Step 1: Count days from \(15^{th}\) August to \(2^{nd}\) October. From \(15^{th}\) August to \(31^{st}\) August: \[ 16\text{ days} \] September has: \[ 30\text{ days} \] From \(1^{st}\) October to \(2^{nd}\) October: \[ 2\text{ days} \] Total difference: \[ 16+30+2=48 \]

Step 2: Find odd days. \[ 48\div 7=6\text{ remainder }6 \] So the day shifts by \(6\) days.

Step 3: Count from Wednesday. Starting from Wednesday: \[ 1\rightarrow Thursday \] \[ 2\rightarrow Friday \] \[ 3\rightarrow Saturday \] \[ 4\rightarrow Sunday \] \[ 5\rightarrow Monday \] \[ 6\rightarrow Tuesday \]

Step 4: Final answer. \[ \boxed{\text{Tuesday}} \]