Bunuel wrote:
Two hands of a watch are pointing at 12 o'clock. One hand spins at a rate of 80 revolutions per minute, while the other spins at a rate of \(67 \frac{1}{2}\) revolutions per minute. After how many seconds will hands be pointing at 12 o'clock again for the first time?
A. 6
B. 12
C. 18
D. 24
E. 48
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