Maths question

[united4ever]

Anyone able to answer and tell me the easiest way to arrive at the answer?

Okay, this looks fun :)

If you've got T starfish, C octopuses and L lobsters, then you've got

Eyes = T + 2C + 2L
Legs = 5T + 8C + 10L

But we can simplify the problem by spotting that if we multiply the Eyes equation by 5, we can subtract the Legs equation from it to eliminate both T and L, leaving only C:

5*Eyes = 5T + 10C + 10L
Legs = 5T + 8C + 10L

Subtract one from the other to get:

5*Eyes - Legs = 2C

We used both the Eyes and the Legs equation to derive this one, so we can now throw one of the original ones away. Throw away the Legs equation because it's the messier one - so now we're working with:

5*Eyes - Legs = 2C
Eyes = T + 2C + 2L

We want to find T, C and L in three different scenarios, where all are non-negative integers, and (T + C + L) is prime. Work through them...

Problem 1: The Botany Bay Tank contains 175 legs and 41 eyes, so

2C = 5*Eyes - Legs = 205 - 175 = 30, so C = 15.
T + 2C + 2L = T + 30 + 2L = 41, so T + 2L = 11.

T must be odd (because 2L is even, and 11 is odd), so T could only be 1, 3, 5, 7, 9 or 11. Try them all, checking each time to see if (C+T+L) is prime.

If T = 1 then L = 5. (C+T+L) = 21 which is not prime.
If T = 3 then L = 4. (C+T+L) = 22 which is not prime.
If T = 5 then L=3. (C+T+L) = 23 which is prime.
If T = 7 then L=2. (C+T+L) = 24 which is not prime.
If T = 9 then L=1. (C+T+L) = 25 which is not prime.
If T = 11 then L=0. (C+T+L) = 26 which is not prime.

So the only feasible solution is C=15, T=5, L=3, i.e. 15 octopuses, 5 starfish and 3 lobsters.

Problem 2: The Sargasso Sea Tank contains 92 legs and 22 eyes, so

2C = 5*Eyes - Legs = 110 - 92 = 18, so C = 9.
T + 2C + 2L = Eyes = 22, so T + 2L = 4.

T must be even (because 2L is even and 4 is even), so T could only be 0, 2 or 4. Try them all, checking each time to see if (C+T+L) is prime.

If T = 0 then L = 2. (C+T+L) = 11 which is prime.
If T = 2 then L = 1. (C+T+L) = 12 which is not prime.
If T = 4 then L = 0. (C+T+L) = 13 which is prime.

So there are two feasible solutions:
C=9, T=0, L=2, i.e. 9 octopuses, no starfish and 2 lobsters
C=9, T=4, L=0, i.e. 9 octopuses, 4 starfish and no lobsters.

Problem 3: The La Manche Tank contains 46 legs and 10 eyes, so

2C = 5*Eyes - Legs = 50 - 46 = 4, so C = 2.
T + 2C + 2L = 10, so T + 2L = 6.

T must be even (because 2L is even and 6 is even), so T could only be 0, 2, 4 or 6. Try them all, checking each time to see if (C+T+L) is prime.

If T = 0 then L = 3. (C+T+L) = 5 which is prime.
If T = 2 then L = 2. (C+T+L) = 4 which is not prime.
If T = 4 then L = 1. (C+T+L) = 3 which is prime.
If T = 6 then L = 0. (C+T+L) = 2 which is not prime.

So there are two feasible solutions:

C=2, T=0, L=3, i.e. 2 octopuses, no starfish and 3 lobsters
C=2, T=4, L=1, i.e. 2 octopuses, 4 starfish and 1 lobster.

OP - I don't think I've made any mistakes, but do go through it and check my working. The whole thing took far longer than it should have done because I made a couple of false starts before spotting that simplification at the beginning which made the problem a lot easier to solve. (There might well be quicker ways to do it, but it's the weekend and I've got stuff I need to get on with :) )

Oops - I forgot this is supposed to be somebody's homework (although the original post is several days old). My apologies: if a moderator would like to delete the second and third scenarios and replace them with "Left as an exercise for the reader", I shan't be offended.

Many thanks, this is so helpful. I shall go through it and try to explain it to my son.

There were 2 more questions we couldn't get (attached). If you, or anyone, is willing and able then any help appreciated.

Once again, Thanks so much!

On the assumption that somebody else might be working on the mice right now (because it's the first one), I'll try comparing the combine harvesters :)

First thing I'll say is that I think this is a seriously tough question, not least because as I see it, there are two whopping ambiguities:

(a) Are we talking about a single crop that has to be harvested using one of the three scenarios, or an unspecified number of crops? If it's the former, the Kate machine will have finished the job in just over a day, whereas the Abi will take a week to do a rather better job. If it's the latter, we're going to end up with a formula for the profit that depends on the number of crops (but which won't be a simple multiple of the number of crops because the harvester itself only has to be bought once).

(b) Does the fact that one of the harvesters can harvest 85% of a crop in one day mean that it can harvest 100% of the crop in just over one day (about 1.18 days at a constant rate, in which case Keith would need to be paid for two days instead of one), or does it mean that the harvester is only 85% efficient (in which case it can only ever harvest 85% of a crop anyway, and it will take a day to do it)?

I'm going to assume we're talking about a single crop that has to be harvested, and that the three percentages relate to the efficiencies of the harvesters - but those assumptions should be stated explicitly at the top of the answer, because if whoever marks it reads the question a different way, I'd lose marks for not explaining my reasoning.

I'm also going to assume that Margaret's objective is to make as much money as possible from this one harvest, even though the harvester itself is obviously a long-term asset with far lower cost implications over subsequent years (e.g. storage, maintenance and fuel) - and that assumption should also be stated explicitly, because for all we know, Margaret's objective might be to maximize her profit over several years.

Okay, so here we go. The total profit should be the profit from selling whatever percentage of the crop is harvested at £300 per percentage point, minus the cost of employing the driver for as many days as he's needed, minus the price of the harvester.

Kate's Kwick Kombine: can harvest 85% of the crop in just one day at £300 per percentage point, so the total profit should be (85 x £300) - (1 x £150) - £15,000 = £10,350.

I think I'd better leave out the calculations for the other two harvesters because it's a homework question, and I wouldn't want to get shouted at for doing the whole thing, but I'll note that by my calculations it's a pretty close-run thing as to which harvester is the best buy.

(For the record, I suspect that the assumptions I've made probably are correct, simply because under those assumptions the actual calculations are fairly straightforward. If we were supposed to factor in extra fractions of a day needed to harvest 100% of the crop, everything would be a lot harder - but that's just applying the principle that if the mechanics of the problem look suspiciously messy, you've probably misunderstood it!)

Well, it looks like nobody's brave enough to attempt to solve the mice infestation problem, so I'll see what I can do. The first thing to say however is that I can't see a smart way to solve it that would be any quicker or less prone to error than by setting it up in a spreadsheet and just slogging through it. I don't know if you're allowed to use a spreadsheet for this, but if you're not, I'd say the chances of actually getting it right are pretty small - just because there are so many steps where a tiny slip would result in an error being introduced. It's therefore essential to explain your working as you go, just to make sure you get credit for all the steps you do get right.

I started by just working out how many mice there should be on the morning of 1st August, which I think ought to be 256 (8 on 1st March, 16 on 1st April, 32 on 1st May, 64 on 1st June, 128 on 1st July, 256 on 1st August). Later that morning, both farmers buy a cat - and now I'll set up a spreadsheet in which I can track the mouse population on a daily basis until it dies out. Most days it's a simple subtraction of the three mice (or two mice after the first two months) that Linda's cat catches at noon, but there's a different calculation when the month changes. When that happens, (i) Richard's cat reduces the population by one-third (to the nearest mouse) late on the last day of the month, then (ii) the population doubles in the early hours of the next morning, then (iii) Linda's cat catches another three (or two) mice later that same day. So we need to model that sequence of events in the spreadsheet.

Set up a column of dates, starting at 1st August (it doesn't matter what year, provided the date you're trying to find occurs before 28th February, which I believe it does). Now you've got to be careful to get the sequence of events right - and make sure you are clear on at what time during the day the number of mice in the spreadsheet relates. I'm going to report the number of mice in the morning of each day - so against 1st August I'm going to assign 256 mice to that date. Then subtract three mice each day until you get to 31st August, when there will be 256 - (3x30) = 166 mice in the morning.

Now we've got to be careful. At noon that day, Linda's cat catches another three mice, leaving 163. At sundown, Richard's cat catches one-third of those 163 mice, leaving 109 mice (to the nearest mouse), but in the early hours of the morning the population doubles to 218, so on the morning of 1st September I reckon there should be 218 mice.

Now we carry on through September just as in August, losing three mice each day. On 1st October we repeat the calculation we did on 1st September. But now Linda's cat starts catching only two mice each day instead of three - so throughout October, only two mice are caught by Linda's cat each day. On 1st November we repeat the calculation we did on 1st October, remembering to reduce the number caught by Linda's cat from 3 to 2 before Richard's cat wipes out one-third of the remainder.

This goes on until early in the New Year, around which time I reckon the final two mice will get caught by Linda's cat. As before, I shan't do the whole thing because its a homework question - but I hope there's enough information above to enable you to tackle it as I think it should be addressed. Good luck :)

You are a star @DodoApplet. Thanks so much. I agree their are some ambiguities. I will look at it with my son together and your explanation will help massively.

My son is only year 7. He is good at Maths and I think his teacher wanted to challenge him but perhaps this was a step too far for now. I am not good at Maths at all so I can see my ability to help him is fading as he grows up. Got to encourage him to speak with his teacher when he has questions and also become an independent learner.

Once again, thank you very much.

You're very welcome . I'll cheerfully admit that when I was your son's age, the fish tank and harvester problems would have given me panic attacks, and the mice problem would probably have needed several hours of checking and rechecking the arithmetic, as we didn't have spreadsheets (or even calculators) in those days. If the teacher reckons they make good material for presenting your son with a challenge, I'd say that's quite a compliment!