is it a Gmat math style?

So based on the information given we can set up this formula:
x: 23-cent pencil quantity
y: 21-cent pencil quantity
We have: Total Cost = x.23 + y.21

(1) We know x + y = 6 but we dont know the total cost so x can be 1 and y can be 5 or x = 2 and y = 4
=> can't get x value => Not sufficient => Eliminate answers A D

(2) 23x + 21y = 130
Based on the formula, just assume that the formula is 20x + 20y = 130 <=> 20(x+y)=130 => x+ y ~ 6
Try couple variations:
x=1,y=5: 23 + 105 < 130
x=2,y=4: 46 + 84 = 130 => get x
x=3,y=3: 69 + 63 > 130
The larger the value x, the sum will much larger than 130
=> Based on this info alone, we can get x
=> B is sufficient

Ans is B

connit, it looks like something hard for my understanding...

I thing...

1. 23x+21y=130;

2. x+y=6;

3. x=6-y;

4. 23x+21(6-x)=130;

5. 23x+126-21x=130;

6. 2x=4

7. x=2

So, we need both 1 and 2.

IMO C.

OA is B!
my answer is C. I'm confused. what is the indicator that tells us to replace the numbers into x and y? it takes a lot of time to do those calculations since we don't have much time in the real test, for example:
x+y = 15 !!! we can not do: (1,14); (2,13); (3,12)...!!!

kiennguyen-

The question and prompts should be a type of red flag to you. If they aren't yet, they will be after you review additional problems.

The stem sets up a basic problem (that the GMAT will expect any test taker to recognize) in:
x=#pencils, y=#pens, C=total cost; .23x + .21y = C

You see that you have three variables (x, y, C).

Prompt 1 tells you x + y.
Prompt 2 tells you C.

You should instantly be cautious when the question is very easy to interpret, and you are given exactly what you need (3 equations with 3 variables). The instinct is the say "C".

However, you haven't accurately validated Prompt 2 by itself.

A red flag should be that 21 and 23 are not "easy" numbers (like 3 and 4, 2 and 5, etc.) that could have multiple combinations to form a total (in this case 130). There may be only 1 value for each x and y that works for our stem equation (.23x + .21y = C).

So, let's test it:
.23x + .21y = C

Let's quickly find the highest possible value for x or y. Since the price of y < price of x, y has the potential to be greater that x.

So, let's divide 21 into 130. At most, this will be 6R4. So, y must be <=5.
If y=5, x=(130-(5*21))/23 = 25/23 Fail (must divide evenly)
If y=4, x=(130-(4*21))/23 = 46/23 = 2 This is possible
If y=3, x=(130-(3*21))/23 = 67/23 Fail
If y=2, x=(130-(2*21))/23 = 88/23 Fail
If y=1, x=(130-(1*21))/23 = 109/23 Fail

So, y=4 is the only solution. And in that solution, x=2.

So, the answer is B.

You don't need all the algebra to solve this. You just need to recognize the initial "tricky" premise of the question. And, you need to do a bit of validation.

Hope that helps clear it up.