Question

How would you solve a problem like this, mathematically?

"A basket of oranges costs 20 dollars, a basket of pears costs 30 dollars and a basket of kiwi fruits costs 40 dollars.

Eight baskets of these fruits were bought for 230 dollars.

What is the largest possible number of baskets of kiwi fruits that were bought?"

The answer on the site is given by just putting in different values and testing if they work but I'd like to be able to solve it mathematically.

Answer 1

O=20 P=30 K=40

you could divide 230 by 40... which gives you 5.75, since you can't buy .75 of a basket, it's 5.

Answer 2

That's not a good general way of doing it, I think. If 30 were instead some different number your method would not work I think.

Answer 3

Oh, sorry. I missed that sentence.

Answer 4

I think putting up a system of equations
20x+30y+40z=230
x+y+z=8
And find out z's maximum integer value, would be a good start.

But x and y have to be integers too, so I don't know

Answer 5

20O+30P+40K=230

8x=230

O+P+K=8

Yes, you would want to create a list of equations, then find the maximum value of z, or in my case, K.

Answer 6

You would want to find a third equation, I think.

Answer 7

then you could plug in and cancel out O and P.

Answer 8

I don't think it's so easy to generalize :p
x, y, z all have to be whole numbers

Answer 9

Correct.

Answer 10

But it doesn't say you have to buy one of each either. So, you would want to find the maximum value of K that's 8 or less. Which is what I was initially trying to do.

Answer 11

No. But it does say that everything has to add up to 230. In this case, your earlier method works, but I don't think it's generalizeable

Answer 12

5 is the maximum whole number of Kiwi baskets you could buy for 230 that is less than the total number of baskets, 8.

Answer 13

Correct, but that's not my question :p

Answer 14

Haha. I'm sorry. I'm just trying to help. Your question is, can you do it mathematically? You have 3 variables and can only draw 2 equations from them. One has a limit of only being whole numbers. Instead of plugging in different random values, I chose the value we were looking for and derived it that way.

Answer 15

An equation that meets those limitations would be 5K+P=230

Answer 16

Where K=5 and P=1.

Answer 17

Sorry, no.

Answer 18

K=40 and P=30

Answer 19

40K+30P = 230 :p

Answer 20

You are correct but I think there is an additional way to solve for it mathematically, since there is a third and fourth constraint (x,y,z have to be all whole numbers), and (z has to be as large as possible). If only I would know how. Thanks for trying to help though...

Answer 21

Lol. Ok. Sorry!