How many minutes is it until six o’clock if fifty minutes ago it was four times as many minutes past three o’clock? 

Exactly 24 times.

We know that 1 year = 365.25 days

1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds

=> 1 Year = 365.25 x 24 x 60 x 60 Seconds

But 1 billion seconds = ?

1 Year = 365.25 x 24 x 60 x 60 Seconds

    ?     =  1000000000 Seconds

? = 1000000000365.25 x 24 x 60 x 60 years = 31. 68 years.

At 9 o’clock, the hour hand is at 9 and the minutes hand is at 12, i.e., the two hands are 15 min. spaces apart.
So, the minute hand should gain = (30 - 15) minutes = 15 minutes
55 minutes will be gained in 60 min.
15 minutes spaces will be gained in ((60/55) x 15) min. = 180/11 min.
The hands will be in the same straight line but not together i.e.,in 180 degrees at 180/11 min. past 9.

In this problem , it has considered that 65 mins = 1hr
So mins has increased by 5 mins so multiply 5 x 24 = 120 mins extra ,
That is now per day it adds 2hr extra, so divide 1440/26 = 59.384 days =~ 60 days.

Let the required number of minutes = M
Total minutes between 3o'clock and 6o' clock = 180 minutes
M + 50 + 4M = 180
5M + 50 = 180
5M = 130
M = 26 minutes.

Apply the formula for finding the angle:
Angle=(30h-(11/2)m( where m stands for minutes and h stands for hours)
Hence, 30x11 - 11/2x35 = 330 - 192.5 = 137.5 degrees
The reflex angle = 360 - 192.5 degree = 222.5 degrees.

For 1 hour, 20 minutes is slower, then for
06 miuntes - 2 minutes is slower
30 minutes - 10 minutes
Actual time | false time
12 12
01 PM 12:40 PM (20 minutes slower)
02 PM 01:20 PM
03 PM 02:00 PM

So, when actual time is 3pm, the false time is 2 pm.
So, it loses 60 minutes when it shows 2 pm.

Since, in one hour, two hands of a clock coincide only once, so, there will be value.
Required time T = 211H×30 + Ao minutes past H.
Here H - initial position of hour hand = 2 (since 2 O'clock)
A° = Required angle = 0° (Since it coincides)

T = 2112×30 + 0o minutes past 2 

=> 11011 minutes past 2