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0 < y < x. x and y are odd integers.
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23 Feb 2022, 08:00
23 Feb 2022, 08:00
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Question Stats:
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37% (01:16) wrong based on 45 sessions
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\(0 < y < x\). x and y are odd integers.
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The remainder when xy divided by 2 |
\(\frac{ x}{y}\) |
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: 0 < y < x. x and y are odd integers.
[#permalink]
23 Feb 2022, 10:14
23 Feb 2022, 10:14
1
First for side A:
We are given that x and y are both odd numbers. Any odd number multiplied by another odd number will result in an odd number (it might be helpful to memorize/ familiarize yourself with these types of relationships because GRE tends to incorporate them into the test.) If you can't remember whether the product of a mixture of odd and even numbers is odd or even, just pick easy numbers to remember the rule, e.g., x=3 and y=1, 1x3=3 (odd, therefore we know that xy will always be odd (I'm not sure its worth going into depth about how the rules are determined, just try to remember them and then use easy examples to remember the rule if you forget)). So now that we know xy is odd, what happens when when we divide it by two? There will always be a remainder of 1 (which is kind of the definition of an odd number I think). So the value for side A is 1.
Side B:
Well, let's plug in numbers. If x is 3 and y is 1, x/y= 3. That's greater than one, but we need to determine that that is always the case by trying to minimize the ratio.
If we continue to increase x and y the ratio will diminish towards 1, but it will never reach it (99/97= ~1.02) (I'm using consecutive numbers here because that further minimizes the ratio and we are trying to look for a minimum answer). But honestly, plugging in numbers here is kind of a waste of time. We are given that x>y so the ratio x/y will always be greater than 1.
Therefore we have A=1 and B>1, so the answer should be B.
(I could be wrong, but hopefully that helps
)
We are given that x and y are both odd numbers. Any odd number multiplied by another odd number will result in an odd number (it might be helpful to memorize/ familiarize yourself with these types of relationships because GRE tends to incorporate them into the test.) If you can't remember whether the product of a mixture of odd and even numbers is odd or even, just pick easy numbers to remember the rule, e.g., x=3 and y=1, 1x3=3 (odd, therefore we know that xy will always be odd (I'm not sure its worth going into depth about how the rules are determined, just try to remember them and then use easy examples to remember the rule if you forget)). So now that we know xy is odd, what happens when when we divide it by two? There will always be a remainder of 1 (which is kind of the definition of an odd number I think). So the value for side A is 1.
Side B:
Well, let's plug in numbers. If x is 3 and y is 1, x/y= 3. That's greater than one, but we need to determine that that is always the case by trying to minimize the ratio.
If we continue to increase x and y the ratio will diminish towards 1, but it will never reach it (99/97= ~1.02) (I'm using consecutive numbers here because that further minimizes the ratio and we are trying to look for a minimum answer). But honestly, plugging in numbers here is kind of a waste of time. We are given that x>y so the ratio x/y will always be greater than 1.
Therefore we have A=1 and B>1, so the answer should be B.
(I could be wrong, but hopefully that helps
)
Re: 0 < y < x. x and y are odd integers.
[#permalink]
23 Feb 2022, 10:39
23 Feb 2022, 10:39
1
Carcass wrote:
\(0 < y < x\). x and y are odd integers.
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The remainder when xy divided by 2 |
\(\frac{ x}{y}\) |
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Let x = 3 and y = 1 ----> A = 3/2 reminder will be 1 & B = 3/1 ----------> B is greater
Let x = 15 and y = 3 ----> A = 45/2 reminder will be 1 & B = 15/3 ----------> B is greater
You can take any combination of x & y and divide it by 2, remainder will always be 1.
Ans is B
