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06 Oct 2021, 13:26
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Question Stats:
49% (01:36) correct
51%
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If x and y are different positive integers, what is the value of x?
(1) 3x + 5y = 40
(2) 80 – 6x = 10y
(1) 3x + 5y = 40
(2) 80 – 6x = 10y
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06 Oct 2021, 15:51
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If x and y are different positive integers, what is the value of x?
(1) 3x + 5y = 40
(2) 80 – 6x = 10y
Evaluate question stem first. Not much to analyze, looks like we are just interested in finding x value.
(1) First thought may be that this is insufficient, because we are given only one linear equation with 2 variables. However, this one equation may be enough since we know that both x and y are different positive integers. There are only so many solutions to this problem that will result in this optimal scenario, so plug and see what result we get.
y = 1 gives 3x = 35, x will not be an integer. This solution is out.
y = 2 gives 3x = 30, x = 10, this is an integer! But we still have to check if there are other solutions that result in integer values.
Keep plugging until you get to y = 5. Then we see that 3x = 15, x = 5. This is an integer solution! But hold on - x = 5, and y = 5, and the question says they are different values. So this is out. We still need to check for more values.
y = 6 gives 3x = 10, not an integer.
y = 7 gives 3x = 5, not an integer.
y = 8 gives 3x = 0, x = 0. Although this is an integer, it is not positive. So this is not a possible solution.
We can stop here since as y > 8 gives negative solution for x which is not allowed.
So the only possible solution is x = 10, y = 2. Hence, sufficient.
(2) 80 – 6x = 10y
This looks different, but let us divide the equation by 2: 40 - 3x = 5y
And rearrange: 3x + 5y = 40. Exact same equation as (1), so we already know that only 1 solution exists. Hence, sufficient.
Answer is (D)
(1) 3x + 5y = 40
(2) 80 – 6x = 10y
Evaluate question stem first. Not much to analyze, looks like we are just interested in finding x value.
(1) First thought may be that this is insufficient, because we are given only one linear equation with 2 variables. However, this one equation may be enough since we know that both x and y are different positive integers. There are only so many solutions to this problem that will result in this optimal scenario, so plug and see what result we get.
y = 1 gives 3x = 35, x will not be an integer. This solution is out.
y = 2 gives 3x = 30, x = 10, this is an integer! But we still have to check if there are other solutions that result in integer values.
Keep plugging until you get to y = 5. Then we see that 3x = 15, x = 5. This is an integer solution! But hold on - x = 5, and y = 5, and the question says they are different values. So this is out. We still need to check for more values.
y = 6 gives 3x = 10, not an integer.
y = 7 gives 3x = 5, not an integer.
y = 8 gives 3x = 0, x = 0. Although this is an integer, it is not positive. So this is not a possible solution.
We can stop here since as y > 8 gives negative solution for x which is not allowed.
So the only possible solution is x = 10, y = 2. Hence, sufficient.
(2) 80 – 6x = 10y
This looks different, but let us divide the equation by 2: 40 - 3x = 5y
And rearrange: 3x + 5y = 40. Exact same equation as (1), so we already know that only 1 solution exists. Hence, sufficient.
Answer is (D)
06 Oct 2021, 16:10
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BrentGMATPrepNow
Note: I created this question to expose a common myth that says a single equation with two variables cannot be solved. In many cases, the myth is true and the equation can't be solved, BUT all of that changes when we restrict the values of x and y (as I've done here)
Given: x and y are different positive integers
Target question: What is the value of x?
Statement 1: 3x + 5y = 40
Since the values of x and y are restricted to POSITIVE INTEGERS, it's quite possible that there's only one possible solution to the equation. Let's find out!
Since 5y and 40 are both divisible by 5, we know that 3x must also be divisible by 5 (in order for the equation to be true)
So, we need only test two cases: x = 5 and x = 10
If x = 5, then y = 5. So, that's one solution to the equation.
Similarly, if x = 10, then y = 2. That's another solution to the equation.
Since we're told x and y are DIFFERENT, it must be the case that x = 10, then y = 2, which means the answer to the target question is x = 10
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: 80 – 6x = 10y
You may have noticed that the three terms in this equation, are all TWICE the terms given in statement 1's equation.
So, if we take 80 – 6x = 10y, and divide both sides by 2, we get the equivalent equation: 40 - 3x = 5y
From here, if we add 3x to both sides of the equation we get 40 = 3x + 5y, which is the same equation we had and statement 1.
Since we already know that statement 1 is SUFFICIENT, we can be certain that statement 2 is also SUFFICIENT
Answer: D
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