hard algebra question!!!!plz help!!!!!?

Question

someone buys three kinds of fruit for 40 dollars, 10 dollars, and 1 dollar each. he pays 259 dollars for 100 pieces of fruit. how many pieces of the cheapest kind did he buy?

Answer 1

Let's call the 40 dollar fruit X, the 10 dollar fruit Y, and the 1 dollar fruit Z.

(X)(40) + (Y)(10) + (Z)(1) = $259

You can set up an equation, but for such a simple question, it would probably be easier to guess and check. Let's try:

(2)(40) + (5)(10) + (93)(1) = ?

80 + 50 + 93 = 223

Close for a first try! Now remember, wherever you add another fruit, you have to remove one from somewhere else so that the total number of fruits does not exceed 100.

(2)(40) + (7)(10) + (91)(1) = ?

80 + 70 + 91 = 241

Good, we're closer. Let's try again:


(2)(40) + (9)(10) + (89)(1) = ?

80 + 90 + 89 = 259

Yay!!! There you go. There are 2 of the $40 fruit, 9 of the $10 fruit, and 89 of the $1 fruit. A better way would have been to start out knowing that the number of $1 fruits had to end in 9 since it's the only way to get a total of $259.

Hope this helps! Good luck!

Answer 2

89
2 @40
9 @10
89+2+ 9= 100
89*1+2*40+9*10 = 259

Answer 3

x + y + z = 100
x + 10y + 40z = 259

you know that there are 9, 19, 29, etc of the 1 dollar fruit.

the only number that fits is 89 of the 1 dollar fruits, 9 of the 10 fruit and 1 of the 40 dollar fruits.

Answer 4

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Answer 5

first you have to set up three equations:
a=one kind of fruit
b=second kind of fruit
c=third kind of fruit

dollars equation:
40a+10b+1c = 259

pieces equation:
a+b+c = 100

so now you have to find a,b and c

40a+10b+1c = 259
c = 259 - 40a - 10b now you can use this equation in the pieces equation

a + b + (259 - 40a - 10b) = 259
a + b + 259 - 40a - 10b = 259 subtract 259 from both sides
a + b - 40a - 10b = 0
-39a - 9b = 0
-9b = 39a
-9b/39 = a

so now using the equation of a + b +c = 100
(-9b/39 ) + b + (259 - 40a - 10b) = 100 and you can substitute the a so you get:

(-9b/39 ) + b + (259 + (360/39 ) - 10b) = 100 and solve for b

then put b into -9b/39 = a
then using b and a put those into a+b+c = 100 or 40a+10b+1c = 259

hope thats helpful

Answer 6

This is one of a class of problems called Diophantine. There is no algebraic or other neat way to solve it; just start with the maximum number of the expensive ones, and see if you can make the numbers fit. If that does not work, try one less of those, and so on.