Solution:
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.
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Now we will solve this DS question using the Variable Approach.
Let’s apply the 3 steps suggested previously.
Follow the first step of the Variable Approach by modifying and rechecking the original condition and the question.
We have to find the value of \(\frac{x}{\left(x+y\right)}\) for positive integers x and y.
Follow the second and the third step: From the original condition, we have 2 variables (x and y). To match the number of variables with the number of equations, we need 2 equations. Since conditions (1) and (2) will provide 1 equation each, C would most likely be the answer.
Recall 3 Principles and choose C as the most likely answer.
Let’s look at both conditions (1) and (2) together.
Conditions (1) and (2) tell us that \(\frac{y}{x}=\frac{y-39}{x-21}\) and that the least common multiple of x and y is 1001. Then, we get y(x - 21) = x(y - 39), xy - 21y = xy - 39x.
=> If we subtract xy from both sides, we get, -21y = -39x, and dividing both sides by -1 gives us 21y = 39x.
=> If we divide both sides by 3, we get 7y = 13x or the ratio of x : y = 7 : 13.
=> We can have multiple combinations that will give us the ratio of 7:13 between x and y.
=> We know that 1001 is divisible by both x and y from condition (2). Since 1001 = 7*11*13, x should be 7*11 and y should be 13*11, so we get\(\frac{x}{x+y}=\frac{7\cdot11}{7\cdot11\ +\ 13\cdot11}=\frac{7}{7+\ 13}=\frac{7}{20}\) .
The answer is unique, so both conditions (1) and (2) combined are sufficient, according to Common Mistake Type 2, which states that the number of answers should be only one. So, C seems to be the answer.
Since this question is an integer question, which is also one of the key questions, we should apply CMT 4(A), which states that if an answer C is found too easily, either A or B should be considered as the answer.
Let’s look at each condition separately:
Condition (1) tells us that x : y = 7 : 13 as shown above. To form a ratio, we need a common number. Let the common number be k. So, we get x = 7k and y = 13k.
Then, \(\frac{x}{x+y}=\frac{7k}{7k\ +\ 13k}=\frac{7k}{20k}=\frac{7}{20}\).
The answer is unique, so the condition is sufficient, according to Common Mistake Type 2, which states that the number of answers should be only one.
Condition (2) tells us that the least common multiple of x and y is 1001. Then. since 1001 = 7*11*13, (x, y) can be (7*11, 13*11), (7*11, 7*13), or (7*13, 13*11).
The answer is not unique, so the condition is not sufficient, according to Common Mistake Type 2, which states that the number of answers should be only one.
If the question has both C and A as its answer, then A is the answer rather than C according to the definition of DS questions.
Here, we should know that condition (1) with a ratio wins over condition (2) with a number.
Also, we should remember the solving process of the relationship between the Variable Approach, and Common Mistake Type 3, and Common Mistake Type 4(A or B).
Also, we should know CMT 4(A), which states that if an answer C is found too easily, either A or B should be considered as the answer.
Condition (1) alone is sufficient.
Therefore, A is the correct answer.
Answer: A
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B, or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
