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If x, y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?
(A) (x+z)/z
(B) (y+z)/x
(C) (x+y)/z
(D) (xy)/z
(E) (yz)/x
(A) (x+z)/z
(B) (y+z)/x
(C) (x+y)/z
(D) (xy)/z
(E) (yz)/x
02 Jun 2015, 02:22
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Given
We are given 3 positive integers \(x\), \(y\) and \(z\) such that \(x\) is a factor of \(y\) and \(x\) is a multiple of \(z\). We are asked to find that among the options given which of them is not necessarily an integer.
Approach
Since \(x\) is a factor of \(y\), we can say that \(\frac{y}{x}\) is an integer...........(1)
Also, as \(x\) is a multiple of \(z\) i.e. \(z\) is a factor of \(x\), we can say that \(\frac{x}{z}\) is an integer............(2)
From the above two deductions, we can say that \(\frac{y}{z}\) will also be an integer as \(x\) divides \(y\) completely and \(z\) divides \(x\) completely..............(3)
Our endeavor would be to reduce the expressions in the options to one or more of the above 3 forms.
Working Out
(A) \(\frac{x +z}{z} = \frac{x}{z} + 1\) . Since \(\frac{x}{z}\) is an integer, the expression will be an integer
(B) \(\frac{y+z}{x} = \frac{y}{x} + \frac{z}{x}\) . \(\frac{y}{x}\) is an integer but we can't say if \(\frac{z}{x}\) is also an integer. So the expression need not necessarily be an integer.
Although we have got our answer, I am reducing the other expressions for solution purpose.
(C) \(\frac{x+y}{z} = \frac{x}{z} + \frac{y}{z}\). Both \(\frac{x}{z}\) and \(\frac{y}{z}\) are integers, hence the expression will also be an integer
(D) \(\frac{xy}{z}\). Since \(\frac{x}{z}\) is an integer, the expression will also be an integer.
(E) \(\frac{yz}{x}\). Since \(\frac{y}{x}\) is an integer, the expression will also be an integer.
Hence, answer is Option B
Hope this helps
Regards
Harsh
We are given 3 positive integers \(x\), \(y\) and \(z\) such that \(x\) is a factor of \(y\) and \(x\) is a multiple of \(z\). We are asked to find that among the options given which of them is not necessarily an integer.
Approach
Since \(x\) is a factor of \(y\), we can say that \(\frac{y}{x}\) is an integer...........(1)
Also, as \(x\) is a multiple of \(z\) i.e. \(z\) is a factor of \(x\), we can say that \(\frac{x}{z}\) is an integer............(2)
From the above two deductions, we can say that \(\frac{y}{z}\) will also be an integer as \(x\) divides \(y\) completely and \(z\) divides \(x\) completely..............(3)
Our endeavor would be to reduce the expressions in the options to one or more of the above 3 forms.
Working Out
(A) \(\frac{x +z}{z} = \frac{x}{z} + 1\) . Since \(\frac{x}{z}\) is an integer, the expression will be an integer
(B) \(\frac{y+z}{x} = \frac{y}{x} + \frac{z}{x}\) . \(\frac{y}{x}\) is an integer but we can't say if \(\frac{z}{x}\) is also an integer. So the expression need not necessarily be an integer.
Although we have got our answer, I am reducing the other expressions for solution purpose.
(C) \(\frac{x+y}{z} = \frac{x}{z} + \frac{y}{z}\). Both \(\frac{x}{z}\) and \(\frac{y}{z}\) are integers, hence the expression will also be an integer
(D) \(\frac{xy}{z}\). Since \(\frac{x}{z}\) is an integer, the expression will also be an integer.
(E) \(\frac{yz}{x}\). Since \(\frac{y}{x}\) is an integer, the expression will also be an integer.
Hence, answer is Option B
Hope this helps
Regards
Harsh
18 Mar 2014, 01:36
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SOLUTION
If x, y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?
(A) (x+z)/z
(B) (y+z)/x
(C) (x+y)/z
(D) (xy)/z
(E) (yz)/x
Given: \(z\) goes into \(x\) and \(x\) goes into \(y\). Note that it's not necessarily means that \(z<x<y\), it means that \(z\leq{x}\leq{y}\) (for example all three can be equal x=y=z=1);
Now, in all options but B we can factor out the denominator from the nominator and reduce it. For example in A: \(\frac{x+z}{z}\) as \(z\) goes into \(x\) we can factor out it and reduce to get an integer result (or algebraically as \(x=zk\) for some positive integer \(k\) then \(\frac{x+z}{z}=\frac{zk+z}{z}=\frac{z(k+1)}{z}=k+1=integer\)).
But in B. \(\frac{y+z}{x}\) we can not be sure that we'll be able factor out \(x\) from \(z\) thus this option might not be an integer (for example x=y=4 and z=2).
Answer: B.
Alternately you could juts plug some smart numbers and the first option which would give a non-integer result would be the correct choice.
If x, y, and z are positive integers such that x is a factor of y, and x is a multiple of z, which of the following is NOT necessarily an integer?
(A) (x+z)/z
(B) (y+z)/x
(C) (x+y)/z
(D) (xy)/z
(E) (yz)/x
Given: \(z\) goes into \(x\) and \(x\) goes into \(y\). Note that it's not necessarily means that \(z<x<y\), it means that \(z\leq{x}\leq{y}\) (for example all three can be equal x=y=z=1);
Now, in all options but B we can factor out the denominator from the nominator and reduce it. For example in A: \(\frac{x+z}{z}\) as \(z\) goes into \(x\) we can factor out it and reduce to get an integer result (or algebraically as \(x=zk\) for some positive integer \(k\) then \(\frac{x+z}{z}=\frac{zk+z}{z}=\frac{z(k+1)}{z}=k+1=integer\)).
But in B. \(\frac{y+z}{x}\) we can not be sure that we'll be able factor out \(x\) from \(z\) thus this option might not be an integer (for example x=y=4 and z=2).
Answer: B.
Alternately you could juts plug some smart numbers and the first option which would give a non-integer result would be the correct choice.
