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goldenfox [79]
3 years ago
6

Jackie's watch loses two minutes every hour. Adam's watch gains one minute every hour. They both set their watches at the radio

at 6:00 am, then start their journey to the airport. When they arrive (at the same time) their watches are twelve minutes apart. What is the real time they arrive at the airport?
Mathematics
2 answers:
Troyanec [42]3 years ago
7 0
<span>10:00 am If you look at the problem, you'll realize that Jackie's and Adam's watches diverge from each other by 3 minutes every hour of real time. So when they arrive, their watches differ by 12 minutes, so they've actually spend 12/3 = 4 hours travel time. Since they both started traveling at exactly 6:00 am using a common authoritative time standard (the radio). The current time is now 6:00 am + 4 hours = 10:00 am.</span>
JulijaS [17]3 years ago
5 0
Ratio of loss of tinme of Jackies warch to the gain of time of Adam's watch is 2 : 1

Let the number of hour ellapsed at the time they are twelve minutes apart be x, then

2x + x = 12

⇒ 3x = 12

⇒ x = 12 / 3 = 4

Thus, after 4 hours Jackie's watch has lost 8 minutes and Adam's watch has gained 4 minutes.

Therefore, the real time at the time they arrive at the airport is 10:00 am.
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DE is 24, so GH = 12.

JH is half of DF.  Since G is the midpoint of DF, DG is also half of DF.  So JH = DG = 8.

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Step-by-step explanation:

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