fruit maths

francoisph wrote:hi,

please any ideas?

a certain fruit stand sold apples for 0.70$ each and bananas for 0.50$ 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?


thks

Hi,

at first it may seem like this problem is unsolvable since we have 2 variables and only 1 equation. However, we have 2 other key pieces of information:

1) the customer bought at least 1 of each; and
2) apples and bananas are indivisible, therefore our solution must contain integers.

Now, this question is a perfect example of why you should always post the answer choices, since backsolving (working backwards from the choices) is a perfect way to solve this problem.

However, we can also front-solve if forced to do so.

Turning our word problem into an equation:

70b + 50a = 630

Now let's brute force:

1 banana means that we need 580 worth of apples; 580 isn't divisible by 50.

2 bananas means that we need 490 worth of apples; 490 isn't divisible by 50.

3 bananas means that we need 420 worth of apples; 420 isn't divisible by 50.

4 bananas means that we need 350 worth of apples; 350 IS divisible by 50 and gives us 7 apples.

So, we have 4 bananas and 7 apples: choose "11".

you are the star of gmat

thks

the first line is 560



the explanation is official guide of gmat is not good at all
12th edition

0.50y=6.30-0.70x ok
then
0.50y=0.70(9-x) without any explanation ?
and
y=7/5(9-x)

francoisph wrote: the explanation is official guide of gmat is not good at all
12th edition

As I often say to my students:

"The Official Guide is a great source of questions and a horrible source of explanations."

the first line is 560

yes.

________

Although this was not a data sufficiency question, it is very helpful to think of the rule in terms of data sufficiency. When you have "n" linear distinct equations and "n" unknowns, you can solve for any of the unknowns unless:

(1) one or more equations isn't linear

or

(2) 2 equations aren't distinct (x+y = 10 is the same as 2x + 2y = 20)

The above are two situations in which you don't have sufficiency even though you have "n" equations for "n" unknowns. The reason is that some of the equations are either not linear or else not distinct.

But there are also exceptions to the rule; situations where you have sufficiency even though direct application of the rule may suggest otherwise:

(1) If you are asked for a relationship (ie, addition, subtraction, division, multiplication, etc.) between unknowns, typically you will only need "n-1" equations. For instance, if we are asked for the value of "x/y", then the equation "x/y = 10" is sufficient even though we don't know what "x" and "y" are. The reason is that "x/y" is actually just one unknown (you can call it "z"), and so one equation is sufficient. So at the end of the day, you can argue that this isn't really an exception.

(2) If you have a word problem dealing with indivisible objects (as we do in this question), you know that "n" objects must be a positive integer. This information makes sufficiency more likely. For example, if x and y must be positive integers, then the equation "x*y=5" only has two solutions sets: either x is 1 and y is 5 or vice-versa. But if there isn't a restriction on the kinds of numbers that x and y can be--if x and y can be nonintegers--then there are an infinite number of solutions.

(3) Other kind of information that limits the properties of the numbers involved.