mstone wrote:A certain fruit stand sold apples for .70 each and bananas for .50 each. If a customer purchased both apples and bananas from the stand for a total of 6.30, what total number of apples and bananas did the customer purchase?

a) 10
b) 11
c) 12
d) 13
e) 14

the only equation i can set up for this is:

.70 a + .50 b = 6.30

how can i solve this without a second equation??? probably seems simple, but i'm confused.

You're 100% correct that if the only information we have is one equation and we have two variables, it is impossible to solve for those two variables.

However, in this question we have two other key pieces of information:

1) a and b > 0; and

2) a and b are integers.

It's the second piece of information which really allows us to solve, since if a and b didn't have to be integers there would be an infinite number of solutions.

When you're faced with this type of problem in general, always ask yourself "does the nature of the items require that they be integers?"

Some examples:

- people = always non-negative integers
- non-divisible objects = always non-negative integers
- divisible objects (e.g. litres of water) = non-negative numbers
- abstract variables that don't represent real things = no limitations

So, if this question had simply asked:

if .7a + .5b = 6.3, what's the value of a + b?

there would be no unique solution.

we will be having a equation like :

7X+ 5Y = 63 ( X: apples)

values of 7X will be 63-5Y that will be divisible by 7 leaving remainder zero.

Plug in the values of 5Y ranging from 5 to 60( multiples of 5 )

Now when 5Y =35,--> 7X =28:--------->X:4

SO X: 4 & Y: 7--- X+Y: 11

I think this is the easiest method..Anything more easy is welcome!!