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505-555 Level|   Arithmetic|   Number Properties|                     
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AbdurRakib
Is the product of two positive integers x and y divisible by the sum of x and of y ?

(1) x=y

(2) x=2

We can write the question:

Is xy/(x + y) = integer?

Statement One Alone:

x = y

Since x = y, we have:

Is xy/(x + y) = integer?

Is (x)(x)/(x + x) = integer?

Is x^2/2x = integer?

Is x/2 = integer?

Since we don’t have a value of x, we cannot determine whether x/2 is an integer. For example, if x = 2, then x/2 is an integer; however, if x = 3, then x/2 is not an integer. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

x = 2

Only knowing that x = 2 is not enough to determine whether xy/(x + y) = integer. Statement two alone is not sufficient to answer the question.

Statements One and Two Together:

Using the statements together, we see that since x is 2, we can say that 2/2 = 1, which is an integer.

Answer: C
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As x and y are positive integers,
our requirement is whether xy/x+y is an integer.

Statement 1: x=y
we can assume x=2 and y=3, then xy/x+y =6/5= Non Integer
again if, x=2 and y=2, then xy/x+y =1= Integer
Statement ! Not Sufficient

Statement 2: x=2
That means y can take any integer value
If y=3, then xy/x+y =6/5= Non Integer assumed x is 2
and if y=2, then xy/x+y =1= Integer assumed x is 2
Statement 2 not sufficient

Combining Both:
x=2 and y=2, xy/x+y=4/4=1=Integer
Clearly sufficient

Answer C
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Is the product of two positive integers x and y divisible by the sum of x and of y ?

(1) x=y

(2) x=2

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.

Since we have 2 variables (x and y) 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)
Since we have x = y = 2 from both conditions, we have xy = 4 and x + y = 4.
xy is divisible by x + y and we have a unique answer 'yes'.
Since both conditions together yield a unique solution, they are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If x = y = 2, then we have xy = 4, x + y = 4 and xy is divisible by x+y.
If x = y = 1, then we have xy = 1, x + y = 2 and xy is not divisible by x+y.
Since condition 1) does not yield a unique solution, it is not sufficient

Condition 2)
Since we have have any information on the variable y, this is not sufficient obviously.

Since condition 2) does not yield a unique solution, it is not sufficient

Therefore, C is the answer.

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.
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Is the product of two positive integers x and y divisible by the sum of x and of y ?

(1) x=y

(2) x=2
 
 
Answer: Option C

Video solution by GMATinsight

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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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