Abigail and Bella have some money. How much did Abigail have at the beginning?

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[GMAT math practice question]

Abigail and Bella have some money. How much did Abigail have at the beginning?

1) The ratio of the amount of Abigail’s money to Bella’s is 5:3.
2) Abigail gives $300 to Bella, and then Bella gives Abigail 3/5 of the amount that Bella has. After that Abigail has 5 times as much money as Bella.
Source: — Data Sufficiency |

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Assume that a and b are the amounts of money Abigail and Bella have, respectively.

Since we have 2 variables (a and b) and 0 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2) together give us:
We have a:b = 5:3 or b = (3/5)a from condition 1).
From condition 2) we have (a - 300) + (b + 300)·(3/5) = (b + 300)(2/5)·5. Substituting b = (3/5)a gives us (a - 300) + ((3/5)a + 300)·(3/5) = 2((3/5)a + 300).
Then we have a – 300 + (9/25)a + 180 = (6/5)a + 600 or a + (9/25)a – (6/5)a = 720.
Thus, we have a = 4500.

The answer is unique, and conditions 1) and 2) together are sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.

Both conditions 1) & 2) together are sufficient.

Therefore, C is the correct answer.
Answer: C

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in Common Mistake Types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.