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Bob and his brother Jim live 95 miles apart and like to meet and play golf together,
the course is 15 miles further for Bob as it is from Jim but Bob averages 10 mph faster
than Jim so if they want to meet at the course at 9:00 and Jim leaves at 8:00
what time would Bob need to leave?
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Let me make some/several notes as I read your post.
First, the assignment is written unprofessionally.
From the post, it is difficult to understand the meaning of the assignment:
it needs to be re-formulated to make it clear.
So, it is not suprised, that the other tutor @mananth reads, understands and
interprets it incorrectly; so, his solution is incorrect, too.
Second, in English, problem's writers and all other writers NEVER use so long
statements as in your post. They usually use punctuation signs like dots and
commas to subdivide their long thoughts into short understandable sentenses.
Also, professional writers tend to place their sentences in logic order,
because their goal is not to fool readers; in opposite, their goal is to teach
the students to present their own thoughts in logic form.
It is a necessary condition in order for the students love Math;
otherwise, they will hate it.
Therefore, below I place my formulation in the form as it should be.
Bob and Jim live 95 miles apart along a highway and like to meet and play golf together.
The golf course is located somewhere near the highway between the Bob's and Jim's homes
15 miles further from Bob's home as it is from Jim's home.
They want to meet at the course at 9:00 am. Jim leaves at 8:00 am.
Bob averages 10 mph faster than Jim. At what time would Bob need to leave?
Solution
First, from the condition it is clear that the distance from Bob's home to the golf course
is 55 miles, while that from the Jim's home to the golf course is 40 miles.
If it is not obvious for you, then you can write these equations for distances
B + J = 95 miles (1)
B - J = 15 miles (2)
and add them. You will get
2B = 95 + 15 = 110 miles ---> B = 110/2 = 55; then from equation (1) J = 95-55 = 40 miles.
+--------------------------------------------------------+
| OK. Half of the problem is just solved. |
| Starting from this point, I will work on other half. |
+--------------------------------------------------------+
Jim leaves at 8:00 am and arrives at 9:00 am - so, Jim spends precisely 1 hour for his trip.
It tells us that the Jim's average speed is 40 mph.
From it, we conclude that Bob's average speed is 40+10 = 50 mph.
So, Bob's travel time is = 1 hours = 1 hours = 1 hour and 6 minutes.
It means that Bob should leave 6 minutes before 8:00 am, i.e. at 7:54 am.
ANSWER. Bob should leave at 7:54 am, or 6 minutes before Jim.
Solved, with explanations.
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As I see, tutor @math_tutor2020 successfully retold my solution in his own words.
I don't know, for what reason he did it. The solution did not become better of it.
In any case, thanks for popularizing my ideas.
As I worked in Science and Engineering for many years, I know that it is not welcome
(is not a good style) to re-told the published works of other people without a reference.