Trick Question apple/orange/watermelon

Don't know whether this is the right place to ask. Somebody ask me this and i can't figure it out.

anyone?

1 apple= 10 cent
1 orange = 30 cent
1 watermelon =$7

How do i buy 100 items for $100

Hi,

You can write that as a linear diophantine equation, then google for solutions.

Alternatively, ask whether you can buy the melon in slices.

Alan

Personnaly, I could not eat 12 watermelons.

At those prices though, if you don't get those 12 watermelons you'll end up with a lot of apples & oranges, could be even more difficult to get through. Even 52 apples is a lot for one person, I think a fruit salad is the best plan.

Kind of frustrating when this question is lingering on mind.

In terms of programming or in terms of maths,i can't really figure it out.

Solve it like any other "word problem" in algebra ...

Code:

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let x = number of apples
let y = number of oranges
let z = number of watermellons

Equations:

(eq 1) :      x +      y +      z = 100
(eq 2) : 0.10*x + 0.30*y + 7.00*z = 100

Other restrictions:

(r 1) : 0 <= x <= 100
(r 2) : 0 <= y <= 100
(r 3) : 0 <= z <= 100
(r 4) : x , y, and z must be integers


For ease : multiply both sides of (eq 2) by 10

(eq 1) :      x +      y +    z =  100
(eq 2) :      x +    3*y + 70*z = 1000

2 equations ... 3 unknowns

solve for x in (eq 1) and substitute into (eq 2)

(eq 1) : x = 100 - y - z
(eq 2) : (100-y-z) + 3*y + 70*z = 1000

simplify (eq 2)

(eq 2) : 2*y + 69*z + 100 = 1000

solve for y in equation 2

(eq 2) : y = (900 - 69*z) / 2

brute force :

set z =  0 ... yields y = 450 (too big)
set z =  1 ... yields y = 415.5 (too big and not an integer)
set z =  2 ... yields y = 381 (too big)
set z =  4 ... yields y = 312 (too big)
set z =  6 ... yields y = 243 (too big)
set z =  8 ... yields y = 174 (too big)
set z = 10 ... yields y = 105 (too big)
set z = 12 ... yields y =  36

check :

z = 12 , y = 36 ---> x = 52  (from eq 1)

eq 2 : 52 + 3*36 + 12*70 must equal 1000

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-U r a genius-