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21 Jul 2014, 21:57
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
79% (01:11) correct
21%
(01:06)
wrong
based on 144
sessions
History
Date
Time
Result
Not Attempted Yet
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.
(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.
21 Jul 2014, 22:43
Kudos
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Gnpth
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.
22 Jul 2014, 03:09
Kudos
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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 ?
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.
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.
