Problem solving question number 2! - more advanced

Bugz

Fledgling Freddie
Joined
May 18, 2004
Messages
7,297
Sally is helping her dad dig worms for fishing.

He uses 4 worms on the hook and reckons on having to replace an average of one of these each time he casts. He casts 12 times an hour. Today he will fish 5 ponds, spending an hour at the first four and an hour and a half at the fifth. He always discards worms on the hook when walking between ponds.

How many worms are in the tin for there to be 'exactly enough' ?


Here's my analysis:

First 4 ponds: 4 starting worms for each, 12 casts therefore 12 replacement worms for each

Thus (4+12) x 4

Last pond: 4 starting worms, 12 casts/hour therefore in 90 mins 18 casts => 18 replacement worms

Thus 4 + 18

Therefore total answer is 64 + 22 = 86.

Who agrees?
 

kirennia

Part of the furniture
Joined
Dec 26, 2003
Messages
3,857
Sally is helping her dad dig worms for fishing.

He uses 4 worms on the hook and reckons on having to replace an average of one of these each time he casts. He casts 12 times an hour. Today he will fish 5 ponds, spending an hour at the first four and an hour and a half at the fifth. He always discards worms on the hook when walking between ponds.

How many worms are in the tin for there to be 'exactly enough' ?


Here's my analysis:

First 4 ponds: 4 starting worms for each, 12 casts therefore 12 replacement worms for each

Thus (4+12) x 4

Last pond: 4 starting worms, 12 casts/hour therefore in 90 mins 18 casts => 18 replacement worms

Thus 4 + 18

Therefore total answer is 64 + 22 = 86.

Who agrees?

Don't forget he doesn't have to replace the worm for the first cast at each pond so minus 5 from your total ;)
 

Bugz

Fledgling Freddie
Joined
May 18, 2004
Messages
7,297
Ah fuck - that's where I'm going wrong.

Ta!
 

Aoami

I am a FH squatter
Joined
Dec 22, 2003
Messages
11,223
((4 + 11) x 4) + (4 + 17) + (4 x 4)

I got 97, but i'm probably wrong cos i'm shit at maths. I don't even know why i tried to do it tbh.
 

old.Tohtori

FH is my second home
Joined
Jan 23, 2004
Messages
45,210
Silly wording again, to me atleast: "spending an hour at the first four and an hour and a half at the fifth. "

shouldn't that be "spending and hour at each of the first four" :D

Or does she spedn 15 minutes at each pond of the first four.

But if she spends 1 hour on each first lake;

12/hour, is 16 for the last lake (1.5h).
12 for the first four.

There are no excess worms.

Starts off with 4 worms, throws 4 times, takes 20 minutes.
Reloads instantly(magic)
Throws another 4, 20 minutes.
Reload.
Throws another for, 20 minutes, hour full, moves to next lake.

So, 12+12+12+12+16= 66 worms.

In basis, 12 worms/hour.


Now, if she uses 15minutes/first lake, second lake, etc;

15 minutes/lake, 3 worms(12/4), 1 discarded. Meaning, 4/lake.

Last lake, still 16 worms.

So, 4+4+4+4+16= 32

NOTE: Only one worm is lost/cast. 12 casts/hour, 12 worms/hour.
 

Bugz

Fledgling Freddie
Joined
May 18, 2004
Messages
7,297
I do agree Tohtori that the wording needs clarification in areas but then I'm not the one who sets the questions so the best I can do is suck it up and have a go! :)
 

Aoami

I am a FH squatter
Joined
Dec 22, 2003
Messages
11,223
((4 + 11) x 4) + (4 + 17) + (4 x 4)

I got 97, but i'm probably wrong cos i'm shit at maths. I don't even know why i tried to do it tbh.

It's either 97 or 81, i can't decide if that 4 x 4 needs to be at the end or not, it's there to account for the discarding of the worms between ponds, so the addition of 4 at the beginning of each new pond, but i might have accounted for it earlier.
 

old.Tohtori

FH is my second home
Joined
Jan 23, 2004
Messages
45,210
I do agree Tohtori that the wording needs clarification in areas but then I'm not the one who sets the questions so the best I can do is suck it up and have a go! :)

Aye, but in any case, one worm lost/cast, that's the whole needed amount if there's no discard worms(1 hour at each fist 4 lakes) or 1 discard worm/lake if it's 15 minutes/first four.
 

gohan

I am a FH squatter
Joined
Jul 24, 2004
Messages
6,338
its 81



15 worms per pond per hour (4 + 11 replacments)

so 15x4 = 60

then 15 for the first hour + 6 for the extra 30 mins (12/2) = 21

60+21 = 81


i think the wording on this one is very good no room for error, tho i didnt see that the wording on the bat/ball was too bad people looking for faults where they don't lie


#edit# i do see the dodgey wording says an hour at the first four not at EACH, but as it doesnt give distances/times to move between i think its a safe bet thats its an hour at each
 

gohan

I am a FH squatter
Joined
Jul 24, 2004
Messages
6,338
It's either 97 or 81, i can't decide if that 4 x 4 needs to be at the end or not, it's there to account for the discarding of the worms between ponds, so the addition of 4 at the beginning of each new pond, but i might have accounted for it earlier.

no u dont need the 4x4 :p

cos you already equated that with the (4+11) x4 other wise it would be:

(11x4) + (4+17) + (4x4) = 81
or 4(11+4) +(4+17) = 81
 

old.Tohtori

FH is my second home
Joined
Jan 23, 2004
Messages
45,210
No but...it's quite clear:

He uses 1: 4 worms on the hook and reckons on having to replace an average of 2:one of these each time he casts. He casts 3:12 times an hour.

1: One hook. The 4 are loaded before first throw.
2: One is lost at each cast. This means, 4 casts before replacement worms.
3: 12 casts, 12 worms lost.

Unless it's "loads 4, replaces to full 4 each cast", in which case she's a silly girl, but it goes:

12 lost worms and 3 after last cast(each lake discard) discarded on the way to last lake.

Which would be 15+15+15+15+16=76

She discard only between lakes and only way to lose worms is casting, which is 12/hour.

If you count the exact amount needed worms in tin for all casts, you only count 1 worm lost/cast +discard.

Also, he(said she before, who cares, he's a trannyfisher like OD) wouldn't replace the lost worm if he waas going to discard them after last cast.
 

Huntingtons

Resident Freddy
Joined
Jan 19, 2004
Messages
10,770
what a shitty fisherman. he casts and looses 1 worm each time but hes not catching any fish?!

and toh, it has nothing to do with Sally and all to do with his dad!
 

Lamp

Gold Star Holder!!
Joined
Jan 16, 2005
Messages
23,342
Whats the worms point of view? No ones asking the worms how they feel. Worms have rights too, you know ! Equal rights for worms !
 

gohan

I am a FH squatter
Joined
Jul 24, 2004
Messages
6,338
No but...it's quite clear:

He uses 1: 4 worms on the hook and reckons on having to replace an average of 2:one of these each time he casts. He casts 3:12 times an hour.

1: One hook. The 4 are loaded before first throw.
2: One is lost at each cast. This means, 4 casts before replacement worms.
3: 12 casts, 12 worms lost.

Unless it's "loads 4, replaces to full 4 each cast", in which case she's a silly girl, but it goes:

12 lost worms and 3 after last cast(each lake discard) discarded on the way to last lake.

Which would be 15+15+15+15+16=76

She discard only between lakes and only way to lose worms is casting, which is 12/hour.

If you count the exact amount needed worms in tin for all casts, you only count 1 worm lost/cast +discard.

Also, he(said she before, who cares, he's a trannyfisher like OD) wouldn't replace the lost worm if he waas going to discard them after last cast.



15 +15 +15 +15 + 21 =86


last time hes thier for an hour and a half

if he does 12 on 1 hour he will do 18 in 1:30 + the 3 that keep it at 4 worms per cast =21



dont argue with me you wont win
 

gohan

I am a FH squatter
Joined
Jul 24, 2004
Messages
6,338
Sally is helping her dad dig worms for fishing.

He uses 4 worms on the hook and reckons on having to replace an average of one of these each time he casts.

which keeps it at 4

4 worms go in 3 come out 1 gets added on
4 in 3 out 1 added ect

He casts 12 times an hour.

so the first 11 times he will replace the worm the 12th he wont bother he will through the remaining 3 away

so 4 original + 11 replacements = 15

Today he will fish 5 ponds, spending an hour at the first four and an hour and a half at the fifth. He always discards worms on the hook when walking between ponds.


so the first 4 ponds are identicle

15 x 4 = 60


the last he spends and extra 30 mins or and extra 50%

so still starting with 4 worms he does his 11 replacements THEN needs a futher 6 replacements at a rate of 1 every 5 mins 6x5 =30


so 4 original + 11 replacements + 6 additional replacements =21 on the last pond

How many worms are in the tin for there to be 'exactly enough' ?


60+21 = 81



cannot make it any clearer
 

old.Tohtori

FH is my second home
Joined
Jan 23, 2004
Messages
45,210
15 +15 +15 +15 + 21 =86


last time hes thier for an hour and a half

if he does 12 on 1 hour he will do 18 in 1:30 + the 3 that keep it at 4 worms per cast =21

dont argue with me you wont win

But...15+15+15+15+21 is 81 :p

And by the way, the question doesn't say if he replenishes to four after each cast, just that he loses one at each cast.

These questions need to be challenged! Bugz go in there nd give 'em what's what! :D
 

gohan

I am a FH squatter
Joined
Jul 24, 2004
Messages
6,338
But...15+15+15+15+21 is 81 :p

And by the way, the question doesn't say if he replenishes to four after each cast, just that he loses one at each cast.

These questions need to be challenged! Bugz go in there nd give 'em what's what! :D


hahahah cant belive i put 86 :<

it isnt worded beatifully but i does say he uses 4 worms per cast and loses one each time implying that he need to add 1 to replenish to 4
 

old.Tohtori

FH is my second home
Joined
Jan 23, 2004
Messages
45,210
hahahah cant belive i put 86 :<

it isnt worded beatifully but i does say he uses 4 worms per cast and loses one each time implying that he need to add 1 to replenish to 4

We would so give hell about it in math class though :D

Oh and yes, know where i f*cked up as i counted 16 for the last lake. Don't know how :(

Oh yeah! The question only says he discard worms while moving to next pond, there' no pon after the last, so no discard! :D

So it's 15+15+15+15+18!= 78!

Oh yeah but nevermind....

The question is flawed! There can't be exact amount in the tin as the silly fisherman has 3 leftovers!
 

gohan

I am a FH squatter
Joined
Jul 24, 2004
Messages
6,338
We would so give hell about it in math class though :D

Oh and yes, know where i f*cked up as i counted 16 for the last lake. Don't know how :(

Oh yeah! The question only says he discard worms while moving to next pond, there' no pon after the last, so no discard! :D

So it's 15+15+15+15+18!= 78!

Oh yeah but nevermind....

The question is flawed! There can't be exact amount in the tin as the silly fisherman has 3 leftovers!

also can't have an exact about as he ESTIMATE he lose 1 worm on AVERAGE per cast

so exact goes right out the window lol
 

Ezteq

Queen of OT
Joined
Jan 4, 2004
Messages
13,457
74


bearing in mind my mathematical abilities are probably worse than the worms. anyhoo here's how i did it lol:

so he has 4 worms per hook and recasts 12 times each time he casts he only has to replace 1 worm so thats the 4 on the hook plus the replacement worms for each cast = 12 (afaik 4 + 12 = 16...i may be wrong) so he does this 4 times (worms so far = 64) but only half an hour (which would be 6 casts, 6 casts plus the 4 original worms = 10) meaning he used 10 worms

which i believe would be 74... however, like i said i'm not good at sums :(

so Bugs whats the right answer?
 

Bugz

Fledgling Freddie
Joined
May 18, 2004
Messages
7,297
The question is flawed in two ways from what I can see:

a) as Tohtori said: an hour at the first four indicates either 1 hour at each or 1 hour for all 4.

b) Statistically 12 an hour. But unless we can confirm each one is exactly divided up into the hour i.e. one in the first 1/12th, one in the 2/2th etc. then we do not know how many will be used in the hour and a half. It could be anything from 12 to 24.

That said - the answer is in fact 81.

First 4 ponds: 4 starting worms for each, 12 casts therefore 12 replacement worms for each.

Thus (4+12) x 4

Last pond: 4 starting worms, 12 casts/hour therefore in 90 mins 18 casts => 18 replacement worms

Thus 4 + 18

Therefore total answer is 64 + 22 = 86.

However the FIRST CAST is the one with ALL four starting worms on. So the first cast for each pond will be 0 replacement worms.

Thus 86 - 5 = 81.

Scarily enough - that one was an official practice question and even the wording on that was sloppy!
 

Users who are viewing this thread

Top Bottom