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.