Joe bought only twenty cent stamps and thirty cent stamps. How many

Agreed--the x>8 is the key.

C for me.

Let no.of 0. 20 cent stamps "x", and 0.30 cent stamps "y"
stmnt (1) is not sufficient because:
x(0.20)+(x+8)(0.30)=?. we do not know x and total amount. Eliminate it.

stmnt (2) is not sufficient because:
x(0.20) +y(0.30)= 2.50. two unknowns in one equation. eliminate it.

(1)+(2)
x(0.20)+(x+8)(0.30)= 2.50.
we can easily solve "x" and"y" values. so answer is C

Bunnel, thanks for the explanation.

For detailed explanation on how to solve for integral solutions, check out the following post:

https://gmatquant.blogspot.com/2010/11/integral-solutions-of-ax-by-c.html

1--> no suff.
2--> .20x+ .30y= 2.5 muliple by 5 to remove clumsy.

1x +1.5y= 12.5 ---> now if x= 11 then y=1 or if x= 6 then y= 5 and similar way you could get different count for 20 cent stamps.

now 1+2 --> from one we got more than 8 hence it must be suff cuz, if x= 9 or 10 , you wont get proper value for y as y should be an integer.

Hence 11 - 20 cents stamps

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Assume a and b are number of stamps with twenty cents and thirty cents, respectively.

Since we have 2 variables (a and b) and 0 equations, C is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2):

a > 8
0.2a + 0.3b = 2.5
2a + 3b = 25

Case 1: a = 9
We have 18 + 3b = 25 or 3b = 7. It doesn't have an integer solution.

Case 2: a = 10
We have 20 + 3b = 25 or 3b = 7. It doesn't have an integer solution.

Case 3: a = 11
We have 22 + 3b = 25 or 3b = 3. Thus the pair of a = 11, b = 1 is a solution.

Case 4: a = 12
We have 24 + 3b = 25 or 3b = 1. It doesn't have an integer solution.

Thus we have a unique solution.

Both conditions together are sufficient.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT(Common Mistake Type) 4(A).

Condition 1) a > 8.
We can't specify the value of a from the condition 1).
The condition 1) is not sufficient.

Condition 2) 2a + 3b = 25
We have solutions a = 2, b = 7 and a = 5, b = 5.
Since it doesn't have a unique solution, the condition 2) is not sufficient.

Therefore, the answer is C.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

B-trap answer for this type of question.
I figure out that:
+The answer could be B or C depends on which info you have from the 1st statement
If:
+It's a condition as this question, the answer is C
+It gives extra info such as the total number of objects, the answer is B
Correct me if I'm wrong, in case of not having enough time to list all possible pairs, guessing would be great to save time and move on.

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