# Math Help - anybody solve

1. ## anybody solve

A box of apples costs 2 Euros, a box of pears costs 3 Euros and a box of prunes costs 4 Euros.
Jeremy buys 8 boxes altogether and he has to pay 23 Euros. What is the biggest number of boxes of
prunes he could have bought?

2. Originally Posted by sri340
A box of apples costs 2 Euros, a box of pears costs 3 Euros and a box of prunes costs 4 Euros.
Jeremy buys 8 boxes altogether and he has to pay 23 Euros. What is the biggest number of boxes of
prunes he could have bought?
apples - 2
pears - 3
prunes - 4

You want to max out prunes, so this means that Jeremy would buy as many prunes as he could until he did not have enough money left to buy any more prunes.

So there are a number of ways to approach this.

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Subtraction Method - subtract 4 from the total for every box you want him to buy until you cannot subtract any more:
1 box: 23 - 4 = 19
2 boxes: 19 - 4 = 15
3 boxes: 15 - 4 = 11
4 boxes: 11 - 4 = 7
5 boxes: 7 - 4 = 3

And since he only has 3 leftover after buying 5 boxes, he cannot afford any more prunes. The remaining 3 could be spent buying pears.

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1 box: 0 + 4 = 4
2 boxes: 4+4 = 8
3 boxes: 8 + 4 = 12
4 boxes: 12 + 4 = 16
5 boxes: 16 + 4 = 20

And since 20 + 4 = 24 which is more than the 23 he spent, he cannot buy any more.

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Multiplication Method - know your multiplication tables, and see what is the greatest coefficient of 4 that leaves the total under 23:

let x be the number of prunes he bought
4x < 23

you can see that 4*5=20 and 4*6 = 24. since 20 < 23 and 24 > 23, you know that he must have spent 20, so x = 5, so he bought 5 boxes.

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Division Method - divide the money spent buy the price per item, discard any leftover since he cannot buy a partial box:

4/23 = 5 remainder 3

I personally used multiplication, to solve this one, I think it is the best option by far, if you become familiar enough with it, you should be able to do these in your head.

Here is another to try out the technique on: If Jeremy spent 35 Euros, what is the most prunes he could have bought? What is the most pears he could have bought? What is the most apples he could have bought?

3. Originally Posted by sri340
A box of apples costs 2 Euros, a box of pears costs 3 Euros and a box of prunes costs 4 Euros.
Jeremy buys 8 boxes altogether and he has to pay 23 Euros. What is the biggest number of boxes of
prunes he could have bought?
Box of prunes cost 4 Euros, but Jeremy pays 23 Euros for 8 boxes of fruit. Five boxes of prunes would leave three euros left to pay for only one box of pears for a total of six boxes. Four boxes of prunes costs 16 Euros leaving 7 euros to pay for four boxes of other fruit. Yet, another impossible task.

Three boxes of prunes cost 12 euros and leaves 11 euros to pay for five boxes of other fruit.

Four boxes of apples cost 8 Euros, and a box of pears make it 11 Euros exactly for five boxes of other fruit.

Answer is three boxes of prunes.

Why was there no math here? This has to be done trial and error because otherwise we have two equations with three unknowns (a - boxes of apples, b - pear boxes, c - prune boxes)

$A + B + C = 8$
$2A + 3B + 4C = 23$
We want: $max(C)$

This is actually a linear programming problem, but that algebra is gone from the recesses of my head!

4. Angel, I believe you missed the constraint of buying eight boxes.

5. Originally Posted by colby2152
...

$A + B + C = 8$
$2A + 3B + 4C = 23$
We want: $max(C)$

This is actually a linear programming problem, but that algebra is gone from the recesses of my head!
This is not necessary here:

From the 1rt equation you get: $A=8-B-C$ and therefore:

$2(8-B-C) + 3B + 4C = 23~\iff~2C+B=7$

That means:

$C=\frac72-\frac B2$

Since A, B, C must be natural numbers and C has a mximum value if B is minimized there is only one solution:

$B = 1~\implies~c = 3~\implies~A=4$

6. Originally Posted by earboth
This is not necessary here:

From the 1rt equation you get: $A=8-B-C$ and therefore:

$2(8-B-C) + 3B + 4C = 23~\iff~2C+B=7$

That means:

$C=\frac72-\frac B2$

Since A, B, C must be natural numbers and C has a mximum value if B is minimized there is only one solution:

$B = 1~\implies~c = 3~\implies~A=4$
Awesome finish earboth! The algebra was much simpler than I thought!

7. Originally Posted by colby2152
Angel, I believe you missed the constraint of buying eight boxes.
Good call >.<