gmatusa2010 wrote:Apples cost $3 dollar , Oranges $2 dollar. $17 is spent. How much Apple and Oranges were bought?


1) 3 Times oranges is equal to 4 times Apples

2) Oranges = Apple + 1

Data Sufficiency


Fairly easy problem but I brought it up to bring up discussion regarding weighted averages. On more difficult weighted averages where the answer is not obvious how do you treat 2)? we already know 1 is sufficient.

D. Since you already know how to tackle A (which I thought was harder one to crack, btw), I'll tell you my approach on 2.

Per B, o = a + 1
Also, Given that: 3a + 2o = 17
Substituting, o = a + 1,
3a + 2(a+1) = 17
3a + 2a + 2 = 17
5a = 15
a = 15/3
a = 3
o = 3 +1 = 4

This type of question is best approached with a number of equations/ number of variables approach. In general, when the number of variables equals the number of equations the problem is solveable (except in the case of even exponents or equations that are multiples of each other.)

The statement provides the equation 3A + 2O = 17 - two variables

Statement 1 gives us 3O = 4A - this is a second equation, not a multiple therefore the problem is solveable. AD.

Statement 2 gives us the equation O = A+1 - another, different equation but again, 2 variables, 2 equations, you can solve it

Therefore the answer is D.

what if the 17 is larger? Isn't it possible for there to be multiple pairs of 3s and 2s that will generate the Total number? My actual question is not how you solve this particular problem but more how do you treat weight averages where the relationship between two items is ADDITIVE. I've actually ran across a problem (OG 10 maybe) where if you perform the 2 equation 2 variable analysis you would be wrong because the additive relationship yield multiple answers. Anybody who uses the weighted average shortcut please chime in.

gmatusa2010 wrote:Apples cost $3 dollar , Oranges $2 dollar. $17 is spent. How much Apple and Oranges were bought?


1) 3 Times oranges is equal to 4 times Apples

2) Oranges = Apple + 1

Data Sufficiency


Fairly easy problem but I brought it up to bring up discussion regarding weighted averages. On more difficult weighted averages where the answer is not obvious how do you treat 2)? we already know 1 is sufficient.

This is not correctly worded, sounds like written on memory basis, isn't it?

I take it as

Apples cost $3 each, and oranges cost $2 each. If $17 were spent in buying the two fruits, then how many apple(s) and orange(s) were bought?


(1) 3 times the number of oranges bought is equal to 4 times the number of apples bought.

(2) The number of oranges bought is 1 more than the number of apples bought.


Do not bother about the weighted average concept in (2), when it's so plane and simple as under

3 A + 2 O = 17 and O = A + 1

Hence, 3 A + 2 (A + 1) = 17, or A = 3, and so O = 4.

[spoiler]D[/spoiler]

You can take inkling from

https://www.beatthegmat.com/gmat-prep-pr ... 65012.html

just to catch why it's not so important here

ur right. well i just made up the problem to actually discuss weighted average. in your opinion, when is it a good time to think about these problems in the form of weighted averages? How am I sure there's only one combination of A and O that add up to 17?

Or Lets flip the problem. 3 Apples and 2 Oranges are purchased. How much does an apple and an orange cost EACH?

gmatusa2010 wrote:ur right. well i just made up the problem to actually discuss weighted average. in your opinion, when is it a good time to think about these problems in the form of weighted averages? How am I sure there's only one combination of A and O that add up to 17?

Or Lets flip the problem. 3 Apples and 2 Oranges are purchased. How much does an apple and an orange cost EACH?

Hope you have seen the link I sent, already.


When we know that A and O are whole numbers, we are infact supplied with an additional info that could possibly replace another linear equation in A and O in some peculiar cases frequently tested on GMAT.

As in (1), 3 O = 4 A with 3 A + 2 O = 17 as already given, could be a luxury here. It can be answered straight from the stem itself, had it been the peculiar case I was talking earlier about.

From stem itself

A = (17 - 2 O)/3

We know that A and O are whole numbers, hence, 17 minus an even number less than 17 must be a multiple of 3 in order to make A a possible whole number. Now, try such an even number, I found 2 working, 8 too working; this is not the peculiar case I was talking earlier about. (1) is no luxury here, and as we get a dissimilar linear equation in A and O, we are set free to shout SUFFICIENT, instead still weighing some average.

If you flip, things would slip, because then we would call you to supply two refreshed statements to follow your flip.

I still couldn't find a necessity of weighted average so far, please let the two refreshed statements make us feel a need of it, now.