Catch the Football
A father and his son are tossing a football. Each time the father receives the ball, he throws it and then steps back 3 feet. When the total distance the ball has been thrown reaches 195 feet, the boy fails to catch the ball.
(amendment: the Dad throws first)
How many times is the ball thrown?
What is the distance of the last throw?
(amendment: the Dad throws first)
How many times is the ball thrown?
What is the distance of the last throw?
posted by Zaux at
7:31 AM
36 Comments:
Well, if the initial distance is "x,"
The number of throws is equal to (195-x)/3
And the distance is equal to (195-x)/3 + x
1st throw to father = 96'
father steps back 3'
throws back to son who fails to catch = 99'
2 throws
last throw 99'
regards, Curtis
no correct answers yet
Okay, so with initial distance x, the number of throws is expressed by:
2x+2(x+3)+2(x+6)+2(x+9)...=195, or
sum(2(x+(3(i-1))))=195
and the number of throws is equal to the highest value of i.
There were 6 throws made, the father started at 6 feet away, increasing by three feet until being 21 feet away when the total throws add to 195 feet.
OK ...
how many throws back and forth?
and the length of the last one?
there is a unique solution.
still no correct solution
You don't say who threw the ball first. But the answer is about 15 throws and 45 feet. I'll let someone else sharpen that up now I've done the English comprehension part and as I've got to go out.
Austin's got the right idea. His post hadn't appeared when I started writing my comment.
guys ... I slightly changed the wording ... take a look
Who threw first?
Curtis's answer looks good to me.
OK. Start 64 feet apart. Dad throws to son, son throws back, dad steps back 3 feet and throws to son, who misses. That's 195 feet and 3 throws, last throw was 67 - 0 feet.
There are an infinite number of valid answers, most of which only differ by infinitesimal quantities.
There isn't enough information to determine a unique answer.
My solution (Curtis) doesn't work now the wording of the puzzle has changed
regards, Curtis
Chris ... there is a published solution. I am going to provide it, so you can see if you can works backwards and determine how it was accomplished.
Solution:
9 throws
1st - 15 ft.
last - 27 ft.
However, if they are 97 1/2 ft apart, son throws to father, father back, then steps backward and son drops....it holds true.
If only using whole integers,Dad must throw 1st as if son throws 1st then the total distance thrown will always total an even number.
regards, Curtis
so every 2 throws (every time the father receives the ball), the distance increases by 3 ft.
I am tired (again). When I read the amendment, I still read it as the father stepped back 3 feet before throwing. Perhaps I'd have take a different view if I read it right.
But there is nothing in the problem to say the answers must be in the integers. I gues that I'll have to accept that it must be in order that there is a unique solution. I'm pretty sure that it is necessary to assume that they are equally good catchers and throwers.
I'm done in as I had to cycle to the shops as my car battery has died. I'm not as used to physical excercise as Knightmare is ;)
ok .... I removed part of a sentence for clarity ... does not change the meaning.
Hi Zaux. WHat I meant is, that I interpreted it as dad catches the ball, stepped back three feet and then throws. You changed it to say that he caught it, threw it, the steps back 3 feet. I still managed to think it was as my first interpretation. The blood rushing through my brain has made me quite fuzzy.
I just tried to solve this thing without considering the published solution, and I go the first toss as 96 ft. ... as Curtis did ... but I think I am missing something
If d is the initial distance apart, then
195 = n*d+3(n^2-1)/4
where n is #throws, is odd and =< 15
(n^2 -1) = (n-1)(n+1). But n is odd so both n-1 and n+1 are even and so (n^2 -1)/4 is an integer.
So, looking at the 195 = ..., in mod 3 => n, d or both must be divisible by 3.
195 = nd + 3(n^2 -1)/4
Choose n = 3 => 195 = 3d + 6 => d = 63.
n = 9 => 195 = 9d + 60 => d = 15
n = 5 => 195 = 5d + 18 => d = 35.4
n = 7 => 195 = 7d + 36 => d = 22.71..
So have two good answers:
n = 3, d = 63 and n = 9, d = 15.
But the other answers don't break any rules (unless you arbitrarily say d must be integral).
n = 11 => 195 = 11d + 90 => d = 9.54..
n = 13 => 195 = 13d + 126 => d = 5.3...
n = 15 => 195 = 15d + 168 => d = 1.8
I've not shown derivation of the starting equation, but you can easily verify it with a few test cases.
Last throws.
n = 3 => D = 66
n = 5 => D = 41.4..
n = 7 => D = 29.71..
n = 9 => D = 27
n = 11 => D = 24.54..
n = 13 => D = 23.3...
n = 15 => D = 22.8
Hi Chris ... this problem made me mad ... problems should be worded such that there is clarity
I lost faith in the publisher's wording of the problem ... so I explored the possibilities.
The published solution is:
9 throws
First throw - 15 ft.
Last throw - 27 ft.
Knowing the intended outcome, I will explore possiblities to make sure the problem statement is worded correctly (as to who throws first, and whether the Dad throws and then steps back, or steps back and then throws)
4 possiblities:
1. Dad throws first.
Once he receives, he throws, then steps back
2. Dad throws first.
Once he receives, he steps back, then throws.
3. Son throws first.
Once Dad receives, he thows, then steps back.
4. Son throws first.
Once Dad receives, he steps back, then throws.
let x = distance of throw
Situation 1.
Throw 1 - Dad throws- distance x
Throw 2 - Son throws- distance x+3
Throw 3 - Dad throws- distance x+3
Throw 4 - Son throws- distance x+6
Throw 5 - Dad throws- distance x+6
Throw 6 - Son throws- distance x+9
Throw 7 - Dad throws- distance x+9
Throw 8 - Son throws- distance x+12
Throw 9 - Dad throws- distance x+12
Son misses
Total distance thrown:
9x + 60 = 195 ft.
Solving x = 15 ft.
So, the situation where Dad throws first and then steps back is the correct wording. I have added an ammendment to the problem statement to reflect this info.
Just for "grins and giggles", I went the same gyrations for each of the other situations. The outcome is the same for each one:
9x + 48 = 195
Solving x = 16 1/3 ft.
Knowing that this is not the intended solution negates its validity.
I went through these gyrations so that others who may want to attempt this problem will at least have accurate beginning info.
Hi Zaux. I worked out the formula as it saves a lot of labour in the end. If d is the initial distance, n is #throws (and odd) the final throw is D = d + 3(n-1)2. Foolishly I lost track, and did most the work with d. But I did end up with two integral solutions:
n = 3 => d = 63 and D = 66.
n = 9 => d = 15 and D = 27.
I've also done a fairly nice solution for the "How much fabric?" problem as well as finding an on-line virtual Pickett N4-ES slide rule (their flagship one); I am sure it's the one I used to own (I might still have it in my attic, along with the Otis King model L). I wonder how many 13 year olds today could have mastered one - I learnt the lot, after my dad gave me a crash course in logarithms.
Ooops, D = d + 3(n-1)/2
The original problem was set by someone who deoesn't really understand the rules. Some problems beg that they should be confined to the integers; this one doesn't. e.g. the son could fail to catch by a hairs breadth or by almost 3 feet?
At Zaux's request I post my solution.......
Catch the Football
A father and his son are tossing a football. Each time the father receives the ball, he throws it and then steps back 3 feet. When the total distance the ball has been thrown reaches 195 feet, the boy fails to catch the ball.
(amendment: the Dad throws first)
How many times is the ball thrown?
What is the distance of the last throw?
1f) x
1s)x
2f)x+3
2s)x+3
3f)x+6
3s)x+6
...
Nf)x+(N-1)*3
Ns)x+(N-1)*3
N=1,2,3,4.....
S=a+2a+3a+4a+5a....+na
S/a=1+2+3+4+5+...n
S/a=(n+1)/2*n
S=a/2*(n+1)*n
for 0,6,12,18,24..(N-1)*6
in this case a=6 n=N-1
S=3*N*(N-1)=3N^2-3N
must subtract (N-1)*3 for sons missing pass back
S=3N^2-3N-(3*(N-1))=3N^2-6N+3
Now to sum the xs (S=(2*N-1)*x)
Total sum= 3N^2-6N+3+(2N-1)*x
195=3N^2-6N+3+(2N-1)*x
(3)N^2+(2x-6)*N+(-192-x)=0
x=(192-3N^2+6N)/(2N-1)
and distance of last throw D= x+(N-1)*3
and to translate N into # of throws is N*2-1
for
N=1, 1 throw, x=195,D=195
N=2, 3 throws,x=64,D=67
N=3, 5 throws,x=183/5=36.6,D=42.6
N=4, 7 throws, x=24,D=33
N=5, 9 throws, x=147/9=16.333.....,D=28.33333....
N=6, 11 throws, x=120/11,D=25.90909....
N=7, 13 throws, x=87/13,D=24.69231
N=8, 15 throws, x=48/15,D=24.2
N=9, 17 throws, x=3/17,D=24.17647
beyond N=9 x must be -ve so we discard further possibilities
Answer:
Multiple answers:
1 throw,D=195
3 throws,D=67
5 throws,D=42.6
7 throws,D=33
9 throws,D=28.33333....
11 throws, D=25.90909....
13 throws,D=24.69231
15 throws, D=24.2
17 throws, D=24.17647
Cam
Hi Cam ...
thanks for the solution you posted ... but, am I missing something? The first throw, by Dad is 15ft. And the last throw which the boy misses is 27 ft ... a total of 9 throws. Did I mis-interpret your solution, or are you not in agreement with the 15ft. - 27ft. - 9 throw scenario?
Cam didn't make the father step back after his first throw, that's why he gets a different answer. In fairnes, the rule was that the father catches, the ball, throws it, then steps back. As the father started with the ball, he hadn't caught it, and so hadn't had the full sequence that triggered him to step back.
Cam also considered the degenerate case, where the boy missed on the first throw, at 195 feet.
9 throw scenario
D-15 S-15
D-18 S-18
D-21 S-21
D-24 S-24
D-27
Total:183
183 != 195
So I disagree with the 15ft. - 27ft. - 9 throw scenario
My 9 throw scenario requires
9 throws, x=147/9=16.333.....,D=28.33333....
Cam
9 throw scenario
D-16 1/3 S-16 1/3
D-19 1/3 S-19 1/3
D-22 1/3 S-22 1/3
D-25 1/3 S-25 1/3
D-28 1/3
Total:195
Cam
Catch the Football
Revised answer:
Previous treatment had father step back then throw as opposed to throw then step back. Now corrected. As stated, this makes first 3 throws same distance.
A father and his son are tossing a football. Each time the father receives the ball, he throws it and then steps back 3 feet. When the total distance the ball has been thrown reaches 195 feet, the boy fails to catch the ball.
(amendment: the Dad throws first)
How many times is the ball thrown?
What is the distance of the last throw?
1f)x
1s)x
2f)x
2s)x+3
3f)x+3
3s)x+6
4f)x+6
4s)x+9
...
Nf)x+(N-2)*3
Ns)x+(N-1)*3
N=2,3,4.....
S=a+2a+3a+4a+5a....+na
S/a=1+2+3+4+5+...n
S/a=(n+1)/2*n
S=a/2*(n+1)*n
for 0,3,9,15,21,27..(2N-3)*3)
Add 0,0,6,12,18,24…(N-2)*6
To 0,3,3, 3, 3, 3
in this case a=6 n=N-2
S=3*(N-1)*(N-2)=3*(N^2-3N+2)=3N^2-9N+6
add sum of 0,3,3,3,…
Add 3*(N-1)=3*N^2-9N+6+3*(N-1)
S=3*N^2-6N+3
must subtract (N-1)*3 for sons missing pass back
S=3N^2-6N+3-3*(N-1)=3N^2-9N+6
Now to sum the xs (S=(2*N-1)*x)
Total sum= 3N^2-9N+6+(2N-1)*x
195=3N^2-9N+6+(2N-1)*x
(3)N^2+(2x-9)*N+(-189-x)=0
x=(189-3N^2+9N)/(2N-1)
and distance of last throw D= x+(N-2)*3 (for N>=2)
and to translate N into # of throws is N*2-1
for
N=1, 1 throw, x=195,D=195
N=2, 3 throws,x=65,D=65
N=3, 5 throws,x=37.8,D=40.8
N=4, 7 throws, x=25.28571,D=31.28571
N=5, 9 throws, x=17.666.....,D=26.666....
N=6, 11 throws, x=12.27272,D=24.27272....
N=7, 13 throws, x=8.076923,D=23.076923
N=8, 15 throws, x=4.6,D=22.6
N=9, 17 throws, x=1.588235,D=22.588235
beyond N=9 x must be -ve so we discard further possibilities
Answer:
1 throw, D=195
3 throws, D=65
5 throws, D=40.8
7 throws, D=31.28571
9 throws, D=26.666....
11 throws, D=24.27272....
13 throws, D=23.076923
15 throws, D=22.6
17 throws, D=22.588235
Example for 9 throws:
F=17.66666667 S=17.66666667
F=17.66666667 S=20.66666667
F=20.66666667 S=23.66666667
F=23.66666667 S=26.66666667
F=26.66666667
TOTAL:195
Cam
Post a Comment
Links to this post:
Create a Link
<< Home