Gnpth wrote:
Maria had twice as many suits in the year 1995 as she did in the year 1990. How many suits did Maria have in the year 1990 ?
1) Maria had twice as many suits in the year 2000 as she did in the year 1995.
2) Maria had 2 more suits in the year 2000 than she did in the year 1995.
Here's how I solve this one. Again I use plugging in because I find it to be simplest during the time pressure of the test.
Let's start by evaluating statement 1 by itself:
if x = number of suits in YR90, then
YR95 = 2x
YR00 = 2 (2x) = 4 x
That tells us nothing about what x is equal to. Therefore we can eliminate A and D.
Now let's evaluate statement 2 by itself.
if x = number of suits in YR90, then
YR95 = 2x
YR00 = (2x) + 2
If you plug in numbers, x could equal 1, 100, 1000 or any other number. Therefore this doesn't give us the answer either. We can eliminate B.
Now we need to choose between C and E.
Let's try both statements together:
if x = number of suits in YR90, then
YR95 = 2x
YR00 = 2 (2x) = 4 x
YR00 = (2x) + 2
4 x = (2x) + 2
2x = 2
x = 1
We've got an answer! Therefore, the question can be answered with both statements and the correct answer choice is C.