A man sold one sort of oranges at 5 for $2, and another sort at 16 for $12. If he had sold them all at $0.5 each, he would have received $2 less. If he had sold all at 3 for $2, he would have received $18 more. How many of each sort did he sell?

1 Answer
Oct 15, 2015

Answer:

Let's call the amounts #xandy# and the money he got #z#

Explanation:

What he did can be translated into:
Equation (1): #z=2/5x+12/16y=2/5x+3/4y#

The second option can be translated into:
Equation (2): #z-2=0.5*(x+y)=1/2(x+y)#

The last option would be:
Equation (3): #z+18=2/3(x+y)#

If we look at the last two equations, we see that the difference between selling 2 for $1 or 3 for $2 is $20, or:

#2/3(x+y)-1/2(x+y)=20->#

#(2/3-1/2)(x+y)=(4/6-3/6)(x+y)=1/6(x+y)=20#

#->x+y=120->z-2=1/2*120=60->z=62#

We now have the total number of oranges and the price he got according to the original scheme. We can write #y=120-x#

Putting these into the first equation, we get:

#62=2/5x+3/4(120-x)=2/5x+90-3/4x->#

#62=8/20x+90-15/20x=90-7/20x->7/20x=28->#

#x=560/7=80->y=120-80=40#

Answer : 80 of the first sort, 40 of the other.
Note : Not quite satisfactory, as he could not have sold the second sort by the 16's, but 4 for $3 would do.