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23 Sep 2023, 11:32
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E
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If \(x\) is odd, and \(x = 4 + 3n\). Is \(n + y =\) even ?
1) \(y\) is odd
2) \(n*y\) = odd
1) \(y\) is odd
2) \(n*y\) = odd
23 Sep 2023, 12:10
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n=(x-4)/3.
1) If y is some odd number, then y is an integer. Lets say y is 3. If n is 1 y+n is even. If n is 4/3 however, it is not an integer and therefore not even.
2) yn is odd.
Lets say n=1/3 and y=3. In this case, y+n will not be an integer. But if y=5 and n=1, then y+n is even. Ins.
3) We know for sure that y is some odd integer. However, n can be a fraction or an integer. Since there is no way of knowing what n is, it can either be even (if n is some odd integer) or non-even (if n is some fraction)
E.
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1) If y is some odd number, then y is an integer. Lets say y is 3. If n is 1 y+n is even. If n is 4/3 however, it is not an integer and therefore not even.
2) yn is odd.
Lets say n=1/3 and y=3. In this case, y+n will not be an integer. But if y=5 and n=1, then y+n is even. Ins.
3) We know for sure that y is some odd integer. However, n can be a fraction or an integer. Since there is no way of knowing what n is, it can either be even (if n is some odd integer) or non-even (if n is some fraction)
E.
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23 Sep 2023, 12:27
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Ans - D
Given - x is odd, and 4+3n=x, so 3n= odd- even = odd hence 3n= odd and this is possible only when n is odd.
Now 1) y is odd.
n+y = odd+odd= even hence sufficient.
2) n*y= odd.
As n is odd, so y has to be odd for the product to be odd. Hence y is odd
Therefore n+y is even and sufficient.
So ans is D
Posted from my mobile device
Given - x is odd, and 4+3n=x, so 3n= odd- even = odd hence 3n= odd and this is possible only when n is odd.
Now 1) y is odd.
n+y = odd+odd= even hence sufficient.
2) n*y= odd.
As n is odd, so y has to be odd for the product to be odd. Hence y is odd
Therefore n+y is even and sufficient.
So ans is D
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