rite2deepti wrote:

Joe bought only twenty cent stamps and thirty cent stamps. How many twenty cent stamps did he buy?

(1) Joe bought more than 8 twenty cent stamps.

(2) Joe bought a total of $2.50 worth of stamps.

The OA says C but I am getting B ..guys can you please explain

I applied my reasoning that .2x+.3y=2.50

so .2*5=1.0

.3*5=1.5

Hence we add and we can get 2.50 ..so B should be sufficient.I tried but there are no other integer choices for this combination.

Please help ....

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.

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