Originally Posted by snookersfun
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Puzzles with numbers and things
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Congratulations to snookersfun and abextra who both solved rounds 106 and 107. They've also both solved round 108 (by Private Message), so anyone else who completes it, post the answer on the thread please..."If anybody can knock these three balls in, this man can."
David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.
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Originally Posted by abextra... and please ask Charlie to describe the progress of the break."If anybody can knock these three balls in, this man can."
David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.
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Round 109
something fast for between tournaments:
2 people start traveling at sunrise, one from A to B, the second from B to A. They travel at constant speed throughout the day without taking breaks.
After meeting at noon one person arrives at B at 4pm the second at A at 9pm. At which time was sunrise on that day?
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I assume that their longitudinal positions at sunrise are sufficiently close that we should assume that sunrise occurs simultaneously for each of them?"If anybody can knock these three balls in, this man can."
David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.
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Originally Posted by Royston636h 42m 30s
welcome to this thread. So, you found us, I thought you might like numbers...
About the result, I got a slightly different one, so maybe you want to recheck meanwhile (and I will maybe recheck mine, as I am known to throw in mistakes as well)!
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Originally Posted by snookersfunHi Royston63,
welcome to this thread. So, you found us, I thought you might like numbers...
About the result, I got a slightly different one, so maybe you want to recheck meanwhile (and I will maybe recheck mine, as I am known to throw in mistakes as well)!
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Originally Posted by Royston636h 42m 30s
Don't forget that round 108 is still open too!
http://www.thesnookerforum.com/showp...ount_1300.html"If anybody can knock these three balls in, this man can."
David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.
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Originally Posted by snookersfunsomething fast for between tournaments:
2 people start traveling at sunrise, one from A to B, the second from B to A. They travel at constant speed throughout the day without taking breaks.
After meeting at noon one person arrives at B at 4pm the second at A at 9pm. At which time was sunrise on that day? (Assuming it happened at the same time for both)
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Originally Posted by snookersfunNobody else having a try? There is a simple answer to it."If anybody can knock these three balls in, this man can."
David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.
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OK. Sunrise occurs at 6a.m.
Suppose
- sunrise occurs s hours before noon
- Person 1 travelling to A travels at speed a (units of distance per hour)
- Person 2 travelling to B travels at speed b (units of distance per hour)
Then distance between A and B can be expressed in 3 different ways:
(1) a(s+4) [total distance travelled by Person 1]
(2) b(s+9) [total distance travelled by Person 2]
(3) as + bs [sum of distances travelled by them both until noon]
Since (1) = (3), we get 4a = bs …..(4)
Since (2) = (3), we get 9b = as …..(5)
Multiply equations (4) and (5) and get 36ab = ab s^2
So s^2 = 36,
s=6
sunrise occurs 6 hours before noon."If anybody can knock these three balls in, this man can."
David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.
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... and that of course is the right answer! More fame and all for you d_g!
I had done it slightly different and just put that way up as well (as very elegant):
I divided the distance traveled into x and y (before/after noon).
Therefore speed of one person x/4 and second y/9 unit distances/h.
Now the times for each person to travel the respective parts until noon are
x/4y and y/9x respectively, which of course have to be equal.
x/4y=y/9x
or x²=4y²/9, or x=⅔ y
so distance x is ⅔ of y and as the faster person needs 4 hours to travel distance y, he needs 6 hours for distance x...
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