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

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by AAPL » Tue Jul 31, 2018 6:15 am

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Answers

A

B

C

D

E

Stats

Difficulty—

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 has twice as many suits in the year 2000 as she did in the year 1995.
(2) Maria had two more suits in the year 2000 than she did in the year 1995.

The OA is C.

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.

Has anyone another strategic approach to solving this DS question? Regards!
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Source: — Data Sufficiency |
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by deloitte247 » Sat Aug 18, 2018 1:23 pm
Let the number of suits in 1990 = x
Number of suits in 1995 = 2x
Question : How many suits did maria have in the year 1990?
We need to find x.
Statement 1 : Maria had twice as many suits in the year 2000 as she did in the year 1995.
Thus the number of suits in 2000 = twice of 1995
= 2 * 2x = 4x.
The information in statement 1 is however not enough to find the value of x, so statement 1 is NOT SUFFICIENT.

Statement 2 : Maria had 2 more suits in the year 2000 than she did in the year 1995.
Thus, number of suits in 2000 = 2x + 2
The information in statement 2 is however not enough to find the value of x, so statement 2 is NOT SUFFICIENT.
Statement 1 and 2 together:
4x = 2x + 2
4x - 2x = 2
$$\frac{2x}{2}=\ \frac{2}{2}$$
x = 1

Statement 1 and 2 is SUFFICIENT together.
Option C is CORRECT.
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