•   We know that January 01, 2001 is Monday and December 31, 2000 is a Sunday (It is a fact).

•   Total number of odd days between December 31, 2000  to June 1 2099:-

•   There are 98 complete years + 5 months (Jan 2099 to May 2099) + One day of June (i.e. June 1)

•   98 complete years have 24 leap years having 2 odd days each and 74 ordinary years having one odd day each.

•   So total number of odd days in 98 years = 24×2 + 74×1 = 48 + 74 = 122 (i.e. 122/7 = 3 odd days).

•   In year 2099 (Ordinary Year, having Feb of 28 days) Jan to May we will get number of days as follows 3 (Jan) + 0 (Feb) + 3 (March) + 2 (April) + 3 (May) + 1 (June) = 3 + 3 + 2 + 3 + 1 = 12 = 5 odd days (12/7)

•   So, total odd days = 3 + 5 = 8 = 1 (remainder when we divide 8 by 7)

•   Thus June 1, 2099 will be Monday. (considering 31 dec 2000 as base)

•   So, first Sunday of June will be on 7th.

•   1 day = 24 hr, 1hr = 60 minutes, 1minute = 60 seconds.

•   So,  x minutes = 60 x seconds;

•   x hours = 60 x minutes = 3600 x seconds;

•   x days = 24 x hours = 24×60 x minutes = 24 × 3600 x seconds = 86400 x seconds;

•   x weeks = 7 x days = 7 × 24 x hours = 7 × 24 × 60 x minutes = 7 × 24 × 3600 x seconds = 6,04,800 x seconds.

•   Thus in x weeks, x days, x hours, x minutes and x seconds there will be

•   6,04,800 x + 86400 x + 3600 x + 60 x + x = 694861 x seconds.

•   Formula – Angle between minute hand and hour hand θ = 11/2 M~30H.

•   Where M is number of Minutes and H is number of Hours.

•   Put θ = 0, and H = 3 to find M between 3 and 4, when both hands coincide.

•   0 = 11/2 M – 30 × 3

•   M = 16:36

•   So the minute and hand will coincide at 3:16:36 which is between 3 and 4.Thus, statement, 1 is true.

•   Hour hand and second hand coincide in once in every minute. So, statement, 2 will also be true.

•   Let's calculate maximum value of petrol price = 80 + 0.1 × 100 = 90

•   Now, equate it with the price of diesel

•   ∴ 90 = 69 + 0.15 × n

•   ∴ n= (90-69)/0.15)= 140.

•   2021 is non-leap year.

•   ∴ 31 + 28 + 31 + 30 + 20 = 140 days.

•   Thus, 20th May is the required date.

Option D: Between 7:05 hours and 7:10 hours
To determine when the hour and minute hands of a clock make an angle of 180°, we need to calculate the positions of the hands at different times.
     ● Step 1: Calculate the angle moved by the hour hand.  
           ○ The hour hand moves 360° in 12 hours, which is 30° per hour.
           ○ At 7:00, the hour hand is at 7 × 30° = 210°.
     ● Step 2: Calculate the angle moved by the minute hand.  
           ○ The minute hand moves 360° in 60 minutes, which is 6° per minute.
           ○ At 7:00, the minute hand is at 0°.
     ● Step 3: Determine the time when the angle between the hour and minute hands is 180°.  
           ○ Let the time be 7:x minutes.
           ○ The angle of the hour hand at 7:x is 210° + 0.5x° (since it moves 0.5° per minute).
           ○ The angle of the minute hand at 7:x is 6x°.
           ○ We need the absolute difference between these angles to be 180°:
         $$|(210 + 0.5x) - 6x| = 180$$
         $$|210 - 5.5x| = 180$$
     ● Step 4: Solve the equation.  
           ○ Case 1: $210 - 5.5x = 180$
         $$210 - 5.5x = 180 \implies 5.5x = 30 \implies x = \frac{30}{5.5} \approx 5.45$$
           ○ Case 2: $5.5x - 210 = 180$ (not possible as it would imply a negative x)
     ● Conclusion: The hands make an angle of 180° at approximately 7:05.45, which is between 7:05 and 7:10.  
     Therefore, the correct answer is Option D: Between 7:05 hours and 7:10 hours.

No. of days in February is 29 if it is a leap year.

No. of days left in January = 31 - 12 = 19

Total no. of days till 12th July = (19 + 29 + 31 + 30 + 31 + 30 + 12) = 182

No. of odd days = 182 ÷ 7 = 26 weeks and 0 odd days.

Directions for the following 7 (Seven) items:

Read the following five passages and answer the items that follow. Your answers to these items should be based on the passages only.

•  In 24 hours, the correct clock moves 24 × 60 = 1440 minutes, but the incorrect clock will move 1440 + 10 = 1450 min. So now we have basic relation between the correct and incorrect clock.

•  Now by the condition given in question the incorrect clock has moved 24 + 24 + 10 = 58 hours (i.e. 58×60 minutes).

•  If the incorrect clocks moves 1450 minutes, correct clock moves 1440 minutes.

•  If the incorrect clock moves 1 minute, correct clock moves 1440/1450 minutes.

•  But in our case incorrect clock has moved 58 hrs × 60 min /hr = 3480 minutes.

•  So, if the incorrect clock moved 3480 minutes, the correct clock will have moved (1440×3480)/1450 = 3456 minutes.

•  Converting 3456 min into hours we have 3456/60 = 573/5 or 57 hours 36 minutes.

•  To have the same calendar two things should be matched. First – the first day of the year and second – the year type i.e., ordinary or leap.

•  Let the first day of year 2009 is X day. It means first day of 2010 will be X+1 day, as there is one odd day in one ordinary year.

•  First day of 2011 will be X+2 day. (As there is one odd day in one ordinary year)

•  First day of 2012 will be X+3 day. (As there is one odd day in one ordinary year)

•  First day of 2013 will be X+5 day. (As there are TWO odd days in one LEAP year)

•  First day of 2014 will be X+6 day. (As there is one odd day in one ordinary year); and

•  First day of 2015 will be X+7 day. (As there is one odd day in one ordinary year)

•  After every 7 days, the same day appears. So, the first day of 2015 will be same as 2009.

•  Since both are ordinary years. We can use the calendar of 2009 in 2015.

•   A watch loses 2 minutes in every 24 hours, while another watch gains 2 minutes in every 24 hours.

•   Hence, every day the time difference between the two watches will keep on increasing by 4 minutes.

•   They will again show the same identical time when this difference increases to 24 hours, i.e. 24 × 60 minutes or 1440 minutes.

•   If 4 minutes are gained per day, then 1440 minutes will be gained in 1440 / 4 = 360 days.

•   At 5 o’clock the clock will strike 5 times. It’s given that it takes 12 seconds to do so.

•   Now, at 10 o’clock the clock will strike 10 times. Hence, the time taken by it to strike = 12 × 2 = 24 seconds

•   Class starts at 11:00 am and lasts till 2:27 pm.

•   Total time = 207 minutes

•   Total rest time = 3x5 = 15 minutes

•   Effective time to be divided = 207 - 15 = 192 minutes

•   Duration of one period = 194/4 = 48 minutes

We know that, exactly at 6 pm, angle between minute hand is 180°.

Now, to cover this angle, minute hand takes 180°/ 6° = 30 min.

Since, the minute hand is 3 min ahead of the hour hand.

Difference in angles = 3 x 5.5= 16.5°

⸫  180° + 16. 5° = 6 x x - 0.5 (x - 3)

=  196.5 = 6x- 0.5x+ 1.5

=  195 = 5.5x = x = 195 / 5.5 = 35. 45 = 36 min

⸫  Required time = 6 : 36 pm

Option C: 65
To solve this problem, we need to determine when the hour and minute hands of the clock will be aligned one above the other after initially being aligned between 8 o'clock and 9 o'clock.
  ● Step 1: Calculate the initial position of the hour and minute hands at 8 o'clock.  
        ○ At 8:00, the hour hand is at 240 degrees (since each hour represents 30 degrees, and 8 hours is \(8 \times 30 = 240\) degrees).
        ○ The minute hand is at 0 degrees.
  ● Step 2: Determine the rate of movement for each hand.  
        ○ The minute hand moves at 6 degrees per minute (since it completes 360 degrees in 60 minutes).
        ○ The hour hand moves at 0.5 degrees per minute (since it completes 30 degrees in 60 minutes).
  ● Step 3: Set up the equation for when the hands are aligned.  
        ○ Let \( t \) be the time in minutes after 8:00 when the hands are aligned.
        ○ The position of the minute hand after \( t \) minutes is \( 6t \) degrees.
        ○ The position of the hour hand after \( t \) minutes is \( 240 + 0.5t \) degrees.
  ● Step 4: Solve for \( t \) when the hands are aligned.  
        ○ The hands are aligned when their positions are equal:
      \[
      6t = 240 + 0.5t
      \]
        ○ Simplify the equation:
      \[
      6t - 0.5t = 240
      \]
      \[
      5.5t = 240
      \]
        ○ Solve for \( t \):
      \[
      t = \frac{240}{5.5} \approx 43.64
      \]
  ● Step 5: Calculate the next alignment after the initial one.  
        ○ The hands align every 360 degrees relative to each other.
        ○ The relative speed of the minute hand compared to the hour hand is \( 6 - 0.5 = 5.5 \) degrees per minute.
        ○ Time for the next alignment after the initial one:
      \[
      \frac{360}{5.5} \approx 65.45
      \]
  ● Step 6: Round to the nearest integer.  
        ○ The nearest integer to 65.45 is 65.
 Therefore, the two hands will be aligned again approximately 65 minutes after the initial alignment.