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12. A faulty watch gains 3 minutes per hour. If the watch is set to the correct time at 8:00 a.m., what is the correct time when the faulty watch shows 10:0 p.m. on the same day?
(A) 9:32 p.m.
(B) 9:26 p.m.
(C) 9:24 p.m.
(D) 9:20 p.m.
(E) 9:18 p.m.
Acing the New SAT I Math
p. 30
17 comments:
My answer is different from the book's.
When I tried to check the book's answer by making a list, I didn't get the answer they got.
My list:
bad watch: 8:00 am
good watch: 8:00 am
bad watch: 9:00 am
good watch: 8:57 am
bad watch: 10:00 am
good watch: 9:54 am
and so on, down to:
bad watch: 10:00 pm
good watch: 9:18 pm
thus:
time elapsed on good watch = 13 hrs & 18 minutes
what am I doing wrong?
There is some ambiguity in the question. I assume that the three minutes per hour is in a REAL hour, not a bad-watch hour. So try
bad watch: 9:03 am
good watch 9 am
bad watch: 9:27 pm
good watch: 9 pm
I'll leave you to work out the algebra, now that you can see the difference between gaining 3 minutes every hour and 3 minutes every 57 minutes.
Subtle difference. When the actual time is 9:00, the time on the faulty watch is 9:03.
Since 3 minutes is 1/20 of an hour, you could say that F = 1.05H where F is the hours since 8:00 on the faulty watch and H is the actual hours since 8:00.
Solving for H, and substituting 14 hours on the faulty watch (8am until 10pm), and H is 13 1/3 so 9:20.
I don't have the book to see their answer, but I came up with 9:20. I just mentally counted up:
8:00 a.m. -- watch reads 8:00 a.m.
It gains 3 minutes for every 60 minutes, so at 9:00 a.m. real time, the watch will read 9:03
proceeding this way, we get to real time:noon, watch reads 12:12
at 3:00 p.m., watch reads 3:21
at 6:00 p.m., watch reads 6:30
at 9:00 p.m., watch reads 9:39
at that point, there are 21 minutes left before the watch will read 10:00, but the watch is gaining 3 minutes per hour, or 1 minute every 20 minutes. So in 20 minutes it will read 10:00 but the actual time will be 9:20.
Or so it seemed to me, maybe I missed some key element. It seemed like a ratio problem to me.
Or another way: since the faulty watch gains 3 minutes every 60 minutes, it gains 1 minute every 20 minutes. So 20 actual minutes is 21 minutes on the faulty watch.
The 14 hours on the faulty watch is 840 minutes. Divide by 21 to get 40. Multiply by 20 to get 800 minutes. That's 40 minutes less, so 9:20.
oops, cancel the last time in my previous post. Typing faster than I was thinking. palisadesk managed to do it without the typo.
I find the problem's wording extremely confusing ---
THANK YOU!!
I solved as follows:
The faulty watch advances 63 minutes in 60 actual minutes. When the faulty watch advances from 0800 to 2200, it has advanced 14 hours or 14 * 60 = 840 minutes. This is 840 * 60/63 = 800 actual minutes, or 13 hours 20 minutes. Thus the correct time is 0800 + 1320 = 2120 or 9:20pm.
You know...I absolutely could not understand the language in this word problem. I kept stumbling over the "gain 3 minutes in an hour" idea: how do you gain 3 minutes in 60 minutes?
60 minutes is 60 minutes.
From there I got all balled up ... which is interesting, too.
My initial confusion seemed to infect my subsequent efforts to solve the problem ---
I'm curious what you all think of the wording in this problem?
As a nonfiction writer, I can tell you that it's not clear - but what do you think of it as 'math writing?'
I ask because of the various discussions people have had about whether questions are or are not properly worded ----
I don't think it is clear, and I guarantee that many ELL students would be completely confused. When I read it at first, I wasn't sure what "gain 3 minutes in an hour" meant, and when I first read the solutions, I thought people were adding the minutes in the wrong way. It would help if they gave an example like "when it is actually 10 am, the watch reads 10:03," but I guess they figure that makes it too easy.
The wording was fine with me because I knew clock terminology from growing up with pendulum clocks and from reading novels. A timepeice gaining 3 minutes is reading 3 minutes ahead of reality. Losing 3 minutes means that it reads 3 minutes earlier than it should.
I hazily recall that acccurate time was discussed in C.S. Forester's H. Hornblower books, as well as numerous scifi novels..Heinlein's Time for the Stars among others. Also I heard a brain teaser in grade school about a man who went around setting all the clocks in his house to the correct time, but when he came back to the first one, it was off...
The wording was fine with me, too, but only because I was familiar with the old ways of describing the accuracy of clocks, where a clock would "gain" a certain amount of time or "lose" a certain amount of time in a certain amount of time. This requires *knowledge* of how particular phrases used to be used when timepieces often had this level of inaccuracy. Since timepieces today rarely have such a problem, the old expressions used to describe the accuracy of timepieces have fallen into disuse, leaving many readers in the position of trying to parse the words instead of recognizing the expression. I've seen many similar problems in Civil War-era algebra texts.
This is an excellent illustration of E.D. Hirsch's notion that background *knowledge* is fundamental to comprehension. I interpreted this question (correctly) only by using knowledge I picked up from old literature over the years to recognize it, not from dictionary definitions of the individual words used to decode it.
I am **still** confused by the wording and I know about fast clocks because we always deliberately set our clocks fast on the farm as a means of combating chronic tardiness.
Actually, maybe that's the problem. Our kitchen clock didn't run fast; it ran correctly. We deliberately set the clock ahead. (Which is why the ploy didn't work. We'd look at the clock and discount the extra 10 minutes.)
The wording on this problem, and the interference from YEARS of 'fast clocks,' makes me think that a fast clock should read 3 minutes 'ahead' of the actual time.
At a real time of 9:00, the clock should read 9:03.
After years of practice, I have an ingrained habit of subtracting time from "fast" clocks. The dashboard clock in my car is 'fast,' too.
A timepeice gaining 3 minutes is reading 3 minutes ahead of reality.
I get completely turned around, reading this. It's hopeless.
It would help if they gave an example like "when it is actually 10 am, the watch reads 10:03,"
I desperately needed EXACTLY this!
I find that the gain/loss wording is far more clear than the "ahead / behind" wording.
I find the "ahead" formulation much more confusing.
"I arrived for an appt ahead of time" means I arrived EARLY. If you told me that a clock is ahead by 3 minutes an hour, I would wonder if you meant it was EARLY, and then I would have done the 9:57 thing: it reported 10:00 when the proper time was 9:57.
But if you said a bad clock that gains 3 minutes an hour is adding three minutes to its registered time compared to the proper time. So if the proper time is 10:00, the bad clock is 10:03.
>>It would help if they gave an example like "when it is actually 10 am, the watch reads 10:03,"
>>I desperately needed EXACTLY this!
But that's one of the points the problem is testing for - that the solver understands what 'per hour' refers to - the standard, not the faulty. Or the theoretical, not the experimental. It could have been posed as a percent error problem to tease out whether the solver understands the convention.
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