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Mark owns four low quality watches. Watch1 loses 15 minutes every hour. Watch2 gains 15 minutes every hour relative to

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by BTGmoderatorDC » Sun Apr 18, 2021 4:30 pm
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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



OA A

Source: Veritas Prep
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Source: — Problem Solving |

BTGmoderatorDC wrote: ↑
Sun Apr 18, 2021 4:30 pm
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



OA A

Solution:

We are given that all four watches are reset to the correct time at 12 noon. Since watch 1 loses 15 minutes every hour, let’s pit the correct time against watch 1’s time (i.e., the time shown on watch 1):

Correct time ----- Watch 1’s time
1:00 pm ----- 12:45 pm
2:00 pm ----- 1:30 pm
3:00 pm ----- 2:15 pm
4:00 pm ----- 3:00 pm


We see that when the correct time is 4:00 pm, watch 1’s time is 3:00 pm. Now let’s see watch 2’s time. Since watch 2 gains 15 minutes every hour relative to watch 1, then in 3 hours, it will gain 15 x 3 = 45 minutes on watch 1 when watch 1’s time is 3:00 pm. Thus watch 2’s time is 3:45 pm.

Now let’s see watch 3’s time. Since watch 3 loses 20 minutes every hour relative to watch 2, then in 3¾ hours, it will lose 20 x 3¾ = 20 x 15/4 = 75 minutes or 1 hour 15 minutes to watch 2 when watch 2’s time is 3:45 pm. Thus, watch 3’s time is 2:30 pm.

Finally, let’s see watch 4’s time. Since watch 4 gains 20 minutes every hour relative to watch 3, then in 2½ hours, it will gain 20 x 2½ = 20 x 5/2 = 50 minutes on watch 3 when watch 3’s time is 2:30 pm. Thus watch 4’s time is 3:20 pm.

Now if we pit the correct time against watch 4’s time, we see that when the correct time is 4 pm, watch 4’s time is 3:20 pm. Thus, watch 4 loses 40 minutes in 4 hours, or 10 minutes per hour against the correct time. Thus, in 12 hours (from 12 noon to 12 midnight), watch 4 will lose 10 x 12 = 120 minutes, or 2 hours against the correct time. Therefore, when the correct time is 12 midnight, watch 4’s time is 10 pm.

Answer: A

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BTGmoderatorDC wrote: ↑
Sun Apr 18, 2021 4:30 pm
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



OA A

Source: Veritas Prep
Let the speed of a correct minute hand be \(S\) (1 revolution in 60mins).

Speed of watch-1\(=\dfrac{3}{4}\cdot S\)

Speed of watch-2 relative to watch-1\(=\dfrac{5}{4}\cdot \dfrac{3}{4} \cdot S\)

Speed of watch-3 relative to watch-2\(=\dfrac{2}{3}\cdot \dfrac{5}{4} \cdot \dfrac{3}{4} \cdot S=\dfrac{5}{8}\cdot S\)

Speed of Watch-4 relative to watch-3\(=\dfrac{5}{8} \cdot \dfrac{4}{3} \cdot S=\dfrac{5}{6}\cdot S\)

i.e. watch-4 will lose \(\dfrac{1}{6}\) of 1 hour, every hour. Total loss in 12 hours \(= \dfrac{1}{6} \cdot 12\) \(= 2\).

So Watch-4 will show \(10:00\) PM.
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