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Maria had twice as many suits in the year 1995 as she did

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Maria had twice as many suits in the year 1995 as she did  [#permalink]

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New post 21 Jul 2014, 21:57
1
2
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

77% (00:52) correct 23% (00:59) wrong based on 75 sessions

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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.

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Re: Maria had twice as many suits in the year 1995 as she did  [#permalink]

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New post 21 Jul 2014, 22:43
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.
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Re: Maria had twice as many suits in the year 1995 as she did  [#permalink]

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New post 22 Jul 2014, 03:09
1
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 ?

We need to find the number of suits in 1990, while knowing that by 1995 that number doubled. So, basically we need to find the number of suits in 1995.

(1) Maria had twice as many suits in the year 2000 as she did in the year 1995. This gives an equation connecting 1995 and 2000. Not sufficient.

(2) Maria had 2 more suits in the year 2000 than she did in the year 1995. This also gives an equation connecting 1995 and 2000. Not sufficient.

(1)+(2) We have two different linear equations connecting 1995 and 2000, which means that we can solve for 1995. Sufficient.

Answer: C.

As you can see we don't even need to form any of the equations or solve anything.
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Re: Maria had twice as many suits in the year 1995 as she did  [#permalink]

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New post 17 May 2018, 18:12
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.


We can let the number of suits in 1990 = x, and the number of suits in 1995 = 2x.

We need to determine x.

Statement One Alone:

Maria had twice as many suits in the year 2000 as she did in the year 1995.

Thus the number of suits in 2000 = 4x; however, we cannot determine x. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

Maria had 2 more suits in the year 2000 than she did in the year 1995.

Thus the number of suits in 2000 = 2x + 2; however, we cannot determine x. Statement one alone is not sufficient to answer the question.

Statements One and Two Together:

Using the two statements together, we have:

4x = 2x + 2

2x = 2

x = 1

Answer: C
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Re: Maria had twice as many suits in the year 1995 as she did  [#permalink]

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New post 28 Sep 2018, 18:30
Let S1990, S1995, and S2000 denote the number of suits Maria had in 1990, 1995, and 2000, respectively. We know from the question stem that S1995 = 2(S1990), and we need to solve for S1990. We will have sufficiency if we can determine the value of S1995

Statement (1): insufficient. This tells us that S2000 = 2(S1995), but without knowing the value of S2000, we can't solve for S1995. We don't have the information necessary to find S1990, so statement (1) is insufficient. Eliminate (A) and (D).

Statement (2): insufficient. This tells us that S2000 = 2 + S1995, but we can't solve for S1995, and thus we don't have sufficiency. Eliminate (B).

Statements (1) and (2): sufficient. Together, then statements give us two linear equations with two variables (S2000 and S1995), so we could solve for S1995, which will give us sufficiency. Choice (C) is correct.
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Re: Maria had twice as many suits in the year 1995 as she did &nbs [#permalink] 28 Sep 2018, 18:30
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