Spectator N0 2651 By Fieldfare
Crosswords1 min ago
Maths For 9 Year Olds
9 Answers
My grandaughter (aged 9) has been asked the following for her homework. We are baffled as to how to tackle this, and would appreciate any help.
A farmer buys 100 animals at a cost of £100.
Bulls cost £10, cows cost £5 and calves cost 50p.
How many of each animal does he buy?
A farmer buys 100 animals at a cost of £100.
Bulls cost £10, cows cost £5 and calves cost 50p.
How many of each animal does he buy?
Answers
1 to 9 of 9
Algebra....
13:24 Mon 16th Mar 2015
As far as I can tell there's going to be a certain amount of trial and error involved. But the logic goes something like this:
we know that you can't buy half a cow, or half a bull or something, so always there is a whole number of each type of animal bought. Moreover the high cost of bulls implies that the farmer can't have bought more than 10 of these (as 11*£10 = £110 is too much), and similarly he can't have bought any more than 20 cows (21*£5 = £105 is also too much). These give us useful constraints, the bull one being the most useful.
At this point I'm inclined to turn to algebra for the next step, although once I've solved the problem I'll see if there's another way around it.
Let B = number of bulls, C = number of cows and V = number of calves. Then we have the equations:
B 9B + 4C = V/2 or 18B + 8C = V
Hence in particular the number of calves is at least 18 times larger than the number of bulls; as 18*6 = 108, this also tells us that the farmer can have bought at most 6 bulls, and similarly 8*13 = 104 tells us that the farmer can have bought at most 13 cows. Also the number of calves bought has to be even (if it were odd then you would have spent some amount of money ending in 50p, ie not equal to 100).
Now I suppose you can write V= 18B + 8C and B+C+V=100 and put these together to get 19B + 9C = 100, or:
C= (100-19B)/9
Now it is a matter of trial and error, running through the possible numbers of B which are 0,1,2,3,4,5 until you find one that gives a whole number solution to this equation. In fact, and rather luckily, B=1 works, giving C=(100-19)/9=81/9 =9 and nothing else does (check this), so the farmer buys 1 bull, 9 cows and 90 calves.
To check: this equates to spending 1+£10 + 9*£5 +90*£0.50 = 10 + 45 + 45 = 100. So that works and is the only solution. Phew.
* * * * * *
That honestly seems too much work for a nine-year-old to do. So how's this for an alternative?
200 calves can be bought for £100, but obviously this is too many animals. So let us remove calves and replace them with cows, while fixing the total spend at 100. This can be done by removing ten calves at a time and replacing them with 1 cow (as a cow costs the same as ten calves, 50p*10 = £5), until we get as close to 100 animals as possible. If we do this eleven times, we end up with 11 cows and (200-110) = 90 calves, which is just one animal too many, but still have spent £100. So now we can try replacing cows with bulls: 2 cows to a bull. But taking away only two cows and replacing them with one bull fixes the spend at £100 while buying 100 animals, as we have bought 1 bull, (11-2) = 9 cows and still 90 calves.
Take your pick which solution you prefer!
we know that you can't buy half a cow, or half a bull or something, so always there is a whole number of each type of animal bought. Moreover the high cost of bulls implies that the farmer can't have bought more than 10 of these (as 11*£10 = £110 is too much), and similarly he can't have bought any more than 20 cows (21*£5 = £105 is also too much). These give us useful constraints, the bull one being the most useful.
At this point I'm inclined to turn to algebra for the next step, although once I've solved the problem I'll see if there's another way around it.
Let B = number of bulls, C = number of cows and V = number of calves. Then we have the equations:
B 9B + 4C = V/2 or 18B + 8C = V
Hence in particular the number of calves is at least 18 times larger than the number of bulls; as 18*6 = 108, this also tells us that the farmer can have bought at most 6 bulls, and similarly 8*13 = 104 tells us that the farmer can have bought at most 13 cows. Also the number of calves bought has to be even (if it were odd then you would have spent some amount of money ending in 50p, ie not equal to 100).
Now I suppose you can write V= 18B + 8C and B+C+V=100 and put these together to get 19B + 9C = 100, or:
C= (100-19B)/9
Now it is a matter of trial and error, running through the possible numbers of B which are 0,1,2,3,4,5 until you find one that gives a whole number solution to this equation. In fact, and rather luckily, B=1 works, giving C=(100-19)/9=81/9 =9 and nothing else does (check this), so the farmer buys 1 bull, 9 cows and 90 calves.
To check: this equates to spending 1+£10 + 9*£5 +90*£0.50 = 10 + 45 + 45 = 100. So that works and is the only solution. Phew.
* * * * * *
That honestly seems too much work for a nine-year-old to do. So how's this for an alternative?
200 calves can be bought for £100, but obviously this is too many animals. So let us remove calves and replace them with cows, while fixing the total spend at 100. This can be done by removing ten calves at a time and replacing them with 1 cow (as a cow costs the same as ten calves, 50p*10 = £5), until we get as close to 100 animals as possible. If we do this eleven times, we end up with 11 cows and (200-110) = 90 calves, which is just one animal too many, but still have spent £100. So now we can try replacing cows with bulls: 2 cows to a bull. But taking away only two cows and replacing them with one bull fixes the spend at £100 while buying 100 animals, as we have bought 1 bull, (11-2) = 9 cows and still 90 calves.
Take your pick which solution you prefer!
Hmm. All I'm saying is that if you have two equations and three unknowns some level of trickery/ guesswork is also required. It's not really enough, then, to write down just the two starting equations and leave it there -- I also added the manipulations needed, for example, and then presented a nice little method to avoid algebra altogether.
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