Hand 1 takes 1 min for 80 revolutions i.e. it takes 1/80 min for 1 revolution.
Similarly Hand 2 takes 135/2 revolutions in 1 min, i.e. 2/135 min for 1 revolution
Time when both of them will be together at 12 is LCM(1/80, 2/135) = 2/5 min = 24 seconds

Hi Sharad, could you please explain in detail how you arrived at 2/5 min ?

One hand spins at a rate of 80 revolutions per minute.
80 revolutions in 60 seconds, so 1 revolution in \(\frac{60}{80} = \frac{3}{4}\) seconds.

while the other spins at a rate of \(67 \frac{1}{2}\) revolutions per minute.
That is 135/2 in 60seconds or 1 revolution in \(\frac{60}{\frac{135}{2}}\) or \(\frac{120}{135} = \frac{8}{9}\) seconds

Thus, first hand will be at the same location after every 3/4 seconds, so at each multiple of 3/4 => 3/4, 6/4, 9/4....
Also, the second hand will be at the same location after every 8/9 seconds, so at each multiple of 8/9 => 8/9, 16/9....
They will be together whenever the multiple of 3/4 coincides with multiple of 8/9, which means nothing but LCM of the two.

LCM of a fraction = \(\frac{LCM \ of \ numerator}{HCF \ of \ denominator}\)
Thus LCM(\(\frac{3}{4},\frac{8}{9}\))=\(\frac{LCM(3,8)}{HCF(4,9)}=\frac{24}{1}=24\)


D

Here, you can also solve it by using options.

One hand spins at a rate of 80 revolutions per minute.
80 revolutions in 60 seconds or 8 revolutions in 6 seconds
Now all the options are multiple of 6, so surely the first hand WILL be at 12 o'clock at each given option.

So, any choice that tells us that the second hand is also at 12 o'clock will be our answer, given it is the smallest in such possible choices.

while the other spins at a rate of \(67 \frac{1}{2}\) revolutions per minute.
That is 135/2 in 60 seconds or 135 in 120 seconds or 9 revolutions in 8 seconds.
So any choice that multiplied by 9/8 gives an integer is our answer.
D and E are possible but D is the least of two as 24*9/8=27.

D

Hi Chetan, why did you took hcf of denominator ? is there any logic behind taking hcf of denominator and lcm of numerator . please explain