Vtyjkyuk

The question goes as:

A wall clock and a Table clock are set to correct time today on 10 pm. The wall clock loses 3 minute in 1st hour, 6 minutes in the second hour and 9 minutes in the third hour and so on. The table clock loses 5 minutes in the 1st hour, 10 minutes in the second hour and 15 minutes in the third hour and so on. When will they show the same time?

My approach:

In the first hour, the difference between the two clocks would be $2$ (obtained from $5-3$) minutes.

In the second hour, it'll be four minutes and so on. This would form an arithmetic progression with $a$ = 2 and $d = 2$. I, then, formulated the problem as:

$$2 + 4 + 6+ 8 + dots + n = 720 $$

The RHS is $720$ because I assumed they'll meet after 12 hours.

With this, I got the root as $23.337$ hours, so I arrived at the answer as $10 , textPM + 23.337$ hours i.e $9:20 , textPM $.

Is this correct?

EDIT: I realised this equation won't give an integral answer, and we need one as $n$ on the LHS represents the number of terms. So instead of that, I wrote it as:

$$2 + 4 + 6 + dots + n = 720 times k$$ where $k in (1,2,3,4, dots)$.

Using this method, for $k = 9$, I get the value of $n$ $textas$ $80 , texthours$.

Does this seem correct?

It is ok. Let $f(t)$, and $g(t)$ be the functions in hours that indicate the hours of the two clocks, we consider the time starts at that 10:00. Then we have:
$$
beginaligned
f(t) &=tleft( 1-frac 3120(t+1)right) ,\
g(t) &=tleft( 1-frac 5120(t+1)right) .
endaligned
$$
This is so because we expect interpolation polynomials of degree two (as in physics a uniformly accelerated movement), and the values in $0,1,2$ shoud (only) fit. In our case we have $f(0)=0$, $f(1)=1-6/120=1-3/60$, and $f(2)=2(1-9/120)=2-9/60$. Similarly for $g$. Then we solve the equation $f(t)-g(t)=12$. (The clock period is $12$ hours. We will soon see why this multiple of $12$ gets first hit.) This equation has the solution:

sage: var('t');
sage: f(t) = -3/120*t^2 + (1-3/120)*t
sage: g(t) = -5/120*t^2 + (1-5/120)*t
sage: (f(t)-g(t)).factor()
1/60*(t + 1)*t
sage: solve( f(t)-g(t) == 12, t )
[t == -1/2*sqrt(2881) - 1/2, t == 1/2*sqrt(2881) - 1/2]
sage: ( 1/2*sqrt(2881) - 1/2 ).n()
26.3374738006393

So the exact solution is $frac 12(sqrt2881-1)$. We solve the same equation as $displaystyleunderbrace2+4+dots+2n_=n(n+1)=720$ in the OP. (Stated in terms of the $n$, which also counts the hours.)

Here is the statistic of true hours, the times of the two clocks, and the difference.

sage: for t in [0..28]:
....: print ( "t=%2s wall=%5.2f table=%5.2f :: diff = %8.5f min"
....: % ( t, f(t), g(t), 60*(f(t)-g(t)) ) )
....: 
t= 0 wall= 0.00 table= 0.00 :: diff = 0.00000 min
t= 1 wall= 0.95 table= 0.92 :: diff = 2.00000 min
t= 2 wall= 1.85 table= 1.75 :: diff = 6.00000 min
t= 3 wall= 2.70 table= 2.50 :: diff = 12.00000 min
t= 4 wall= 3.50 table= 3.17 :: diff = 20.00000 min
t= 5 wall= 4.25 table= 3.75 :: diff = 30.00000 min
t= 6 wall= 4.95 table= 4.25 :: diff = 42.00000 min
t= 7 wall= 5.60 table= 4.67 :: diff = 56.00000 min
t= 8 wall= 6.20 table= 5.00 :: diff = 72.00000 min
t= 9 wall= 6.75 table= 5.25 :: diff = 90.00000 min
t=10 wall= 7.25 table= 5.42 :: diff = 110.00000 min
t=11 wall= 7.70 table= 5.50 :: diff = 132.00000 min
t=12 wall= 8.10 table= 5.50 :: diff = 156.00000 min
t=13 wall= 8.45 table= 5.42 :: diff = 182.00000 min
t=14 wall= 8.75 table= 5.25 :: diff = 210.00000 min
t=15 wall= 9.00 table= 5.00 :: diff = 240.00000 min
t=16 wall= 9.20 table= 4.67 :: diff = 272.00000 min
t=17 wall= 9.35 table= 4.25 :: diff = 306.00000 min
t=18 wall= 9.45 table= 3.75 :: diff = 342.00000 min
t=19 wall= 9.50 table= 3.17 :: diff = 380.00000 min
t=20 wall= 9.50 table= 2.50 :: diff = 420.00000 min
t=21 wall= 9.45 table= 1.75 :: diff = 462.00000 min
t=22 wall= 9.35 table= 0.92 :: diff = 506.00000 min
t=23 wall= 9.20 table= 0.00 :: diff = 552.00000 min
t=24 wall= 9.00 table=-1.00 :: diff = 600.00000 min
t=25 wall= 8.75 table=-2.08 :: diff = 650.00000 min
t=26 wall= 8.45 table=-3.25 :: diff = 702.00000 min
t=27 wall= 8.10 table=-4.50 :: diff = 756.00000 min
t=28 wall= 7.70 table=-5.83 :: diff = 812.00000 min

This is showing also how "realistic" the table clock makes time a better time. I really need this clock!
There is a point between $t=26$ and $t=27$ where we hit the $12$ hours, i.e. $720$ minutes. The solution is indeed located in this interval.

With this, I got the root as 23.337 hours, so I arrived at the answer as 10PM+23.337 hours i.e 12:20AM.

23.33333.. is 2/3 of an hour less than a day, or 40 min less than a day. So 10PM+23.3333... is 10PM+1day-40 min, or 9:20PM.

There are several different interpretations of this problem. For one thing, "lose a minute" is ambiguous. It could mean "be 1 minute behind" or "be 1 more minute behind". Under the first interpretation, the difference increases by 2 minutes each hour, so it will take 30 hours to differ by an hour, and 360 hours, or 15 days, to differ by 12 hours. There is now a further ambiguity as to what type of clocks they are; if they are 24-hour clocks, then it will take 720 hours, or 30 days, to show the same time.

If "lose a minute" means "be 1 more minute behind", then we need to make further assumptions. We can fit a quadratic equation to the data points given: after one hour, the wall clock shows 57 minutes. After 2 hours, it shows 111 minutes, etc. That gives the points (1,57), (2,11), (3,168). If x is the actual time in hours, and y is the shown time in minutes, then $y = 60x-frac3x(x+1)2$ fits these data points, but if the problem expects us to therefore conclude that this is the right equation, it is asking us to make something of a leap. Similarly, the equation $y = 60x-frac5x(x+1)2$ fits the table clock. The difference is then $frac2x(x+1)2$, or just $x(x+1)$. Solving for $x(x+1) = 720$, we get:

$x(x+1) = 720$

$x^2+x-720=0$

This gives the solutions 26.3375 and −27.3375. The negative solution corresponds to some past time when they showed he same time, so we should take the positive one. This corresponds to 24+2.3375 hours, or 1 day, 2 hours, and .3375 hours. .3375 hours is 20.25 minutes. .25 minutes is 15 seconds. So this corresponds to 1 day, 2 hours, 20 minutes, and 15 seconds. Adding this to the initial time of 10PM gets to 12:20:15 AM on the second day (the day after the day after the clocks are set).

This matches the time you got, so perhaps your number 23.337 was a typo.