Problem Solving

Source: Veritas Prep

Mark owns four low quality watches. Watch1 loses 15 minutes every hour. Watch2 gains 15 minutes every hour relative to watch1 (that is, as watch1 moves from 12:00 to 1:00, watch2 moves from 12:00 to 1:15). Watch3 loses 20 minutes every hour relative to watch2. Finally, watch4 gains 20 minutes every hour relative to watch3. If Mark resets all four watches to the correct time at 12 noon, what time will watch4 show at 12 midnight that day?

A. 10:00
B. 10:34
C. 11:02
D. 11:48
E. 12:20

The OA is A

$$Let\ speed\ of\ correct\ hour\ =H$$
If mark resets four watches to 12 noon, what time will watch 4 get to show 12 am.
After every 1 hour watch 1 lose 15 min
Speed of watch 1 $$=\left(\frac{1}{1}-\frac{1}{4}\right)\cdot S=\frac{3}{4}\cdot S$$
Speed of watch 2 relative to watch 1 when watch 2 gains 15 min every hour ;
$$\frac{75}{60}\cdot\frac{3}{4}\cdot S$$
$$=\frac{5}{4}\cdot\frac{3}{4}\cdot S$$
Speed of watch 3 relative to watch 2 when watch 3 loses 20 min every hour ;
$$\left(\frac{1}{1}-\frac{20}{60}\right)\cdot\frac{5}{4}\cdot\frac{3}{4}\cdot S$$
$$\frac{4}{3}\cdot\frac{5}{8}\cdot S=\frac{5}{8}\cdot S$$
Speed of watch 4 relative to watch 3 when watch 4 gains 20 minutes every hour ;
$$\frac{80}{60}\cdot\frac{5}{8}\cdot S$$
$$=\frac{4}{3}\cdot\frac{5}{8}\cdot S=\frac{20}{24}\cdot S=\frac{5}{6}\cdot S$$
i.e watch 4 will lose $$\left(\frac{1}{1}-\frac{5}{6}\right)=\frac{1}{6}of\ \ 1hour\ for\ every\ hour$$
Total loss in 12 hours $$=\frac{1}{6}\cdot\frac{12}{1}=\frac{12}{6}=\ 2\ hours$$
By 12:00 am (midnight) watch could have lost 2 hours, hence it will show (12-2 = 10 hours)
$$Answer\ is\ option\ A$$