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If x, y, and z are integers greater than 0 and x = y + z, what is the

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If x, y, and z are integers greater than 0 and x = y + z, what is the [#permalink]

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If x, y, and z are integers greater than 0 and x = y + z, what is the value of (y - z)/y?

(1) (x - y)/y = 4/5
(2) z/y = 4/5
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Re: If x, y, and z are integers greater than 0 and x = y + z, what is the [#permalink]

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New post 11 Nov 2010, 02:13
amirdubai1982 wrote:
Question: If x, y, and z are integers greater than 0 and x = y + z, what is the value of ?

(1) (x-y)/y=4/5
(2) z/y=4/5


Question should be: If x, y, and z are integers greater than 0 and x=y+z, what is the value of (y-z)/y?

\(\frac{y-z}{y}=1-\frac{z}{y}=?\) So, basically the question asks about the value of \(\frac{z}{y}\).

(1) \(\frac{x-y}{y}=\frac{4}{5}\) --> cross multiply --> \(5x-5y=4y\) --> \(5x=9y\) --> as \(x=y+z\) (or as \(5x=5y+5z\)) then \(9y=5y+5z\) --> \(4y=5z\) --> \(\frac{z}{y}=\frac{4}{5}\). Sufficient.

(2) \(\frac{z}{y}=\frac{4}{5}\), directly gives the value of the ratio we need. Sufficient.

Answer: D.
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Re: If x, y, and z are integers greater than 0 and x = y + z, what is the [#permalink]

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New post 11 Nov 2010, 05:01
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amirdubai1982 wrote:
Question: If x, y, and z are integers greater than 0 and x = y + z, what is the value of (y-z)/y?

(1) (x-y)/y=4/5
(2) z/y=4/5

A)Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
B)Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
C)BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D)EACH statement ALONE is sufficient.
E)Statements (1) and (2) TOGETHER are NOT sufficient.


Data from question stem: x = y + z. Since we need the value of \(1 - \frac{z}{y}\), let us divide this equation by y. We get \(\frac{x}{y} = 1 + \frac{z}{y}\)

Question: What is \(\frac{(y-z)}{y} = 1 - \frac{z}{y}\)
To get z/y, we should either have x/y (as seen above) or z/y

1.\(\frac{(x - y)}{y} = \frac{4}{5}\)
or \(\frac{x}{y} - 1 = \frac{4}{5}\)
From here, we get the value of x/y. Sufficient.

2. This directly gives us the value of z/y. Sufficient.
Answer (D)
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Re: If x, y, and z are integers greater than 0 and x = y + z, what is the [#permalink]

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New post 07 Sep 2013, 00:41
Bunuel wrote:
amirdubai1982 wrote:
Question: If x, y, and z are integers greater than 0 and x = y + z, what is the value of ?

(1) (x-y)/y=4/5
(2) z/y=4/5


Question should be: If x, y, and z are integers greater than 0 and x=y+z, what is the value of (y-z)/y?

\(\frac{y-z}{y}=1-\frac{z}{y}=?\) So, basically the question asks about the value of \(\frac{z}{y}\).

(1) \(\frac{x-y}{y}=\frac{4}{5}\) --> cross multiply --> \(5x-5y=4y\) --> \(5x=9y\) --> as \(x=y+z\) (or as \(5x=5y+5z\)) then \(9y=5y+5z\) --> \(4y=5z\) --> \(\frac{z}{y}=\frac{4}{5}\). Sufficient.

(2) \(\frac{z}{y}=\frac{4}{5}\), directly gives the value of the ratio we need. Sufficient.

Answer: D.


Do we always have to reduce ratio's to their lowest forms?

What I did was this, please tell me why this reasoning is wrong.

we need to find the value of \(\frac{y-z}{y}\) or \(1- \frac{y}{z}\) so we need to find the value of
\(\frac{y}{z}\)


(1)\(\frac{x-y}{y} = \frac{4}{5}\)

\(\frac{x}{y} -1 = \frac{4}{5}\) ,\(\frac{x}{y} = \frac{4}{5} + 1\), \(\frac{x}{y}= \frac{9}{5}\)

now given x= y+z so if \(\frac{x}{y} = \frac{9}{5}\) then we can take x= 9 and y=5 then z=4 here \(\frac{z}{y} =\frac{4}{5}\)

but if \(\frac{x}{y} = \frac{9}{5}\)then we can also take x= 18 and z=10 can't we? then z= 8 here \(\frac{z}{y} =\frac{8}{10}\)

So I thought statement 1 did not give us the actual values of x and y, so I thought there were many possibilities hence marked statement as insufficient.

so this was my confusion , I though since we got a ratio at the end of simplification of 1 we do not know the actual values of x and y , since we got \(\frac{x}{y} = \frac{9}{5}\), \(\frac{x}{y}\) could also be \(\frac{18}{10}\),or \(\frac{27}{15}\), if x and y differs so will z and hence will \(\frac{z}{y}\) unless we are always reducing ratios to their lowest form

please tell me how can we be sure of x and y from this ratio

thanks
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Re: If x, y, and z are integers greater than 0 and x = y + z, what is the [#permalink]

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New post 07 Sep 2013, 05:27
stne wrote:
Bunuel wrote:
amirdubai1982 wrote:
Question: If x, y, and z are integers greater than 0 and x = y + z, what is the value of ?

(1) (x-y)/y=4/5
(2) z/y=4/5


Question should be: If x, y, and z are integers greater than 0 and x=y+z, what is the value of (y-z)/y?

\(\frac{y-z}{y}=1-\frac{z}{y}=?\) So, basically the question asks about the value of \(\frac{z}{y}\).

(1) \(\frac{x-y}{y}=\frac{4}{5}\) --> cross multiply --> \(5x-5y=4y\) --> \(5x=9y\) --> as \(x=y+z\) (or as \(5x=5y+5z\)) then \(9y=5y+5z\) --> \(4y=5z\) --> \(\frac{z}{y}=\frac{4}{5}\). Sufficient.

(2) \(\frac{z}{y}=\frac{4}{5}\), directly gives the value of the ratio we need. Sufficient.

Answer: D.


Do we always have to reduce ratio's to their lowest forms?

What I did was this, please tell me why this reasoning is wrong.

we need to find the value of \(\frac{y-z}{y}\) or \(1- \frac{y}{z}\) so we need to find the value of
\(\frac{y}{z}\)


(1)\(\frac{x-y}{y} = \frac{4}{5}\)

\(\frac{x}{y} -1 = \frac{4}{5}\) ,\(\frac{x}{y} = \frac{4}{5} + 1\), \(\frac{x}{y}= \frac{9}{5}\)

now given x= y+z so if \(\frac{x}{y} = \frac{9}{5}\) then we can take x= 9 and y=5 then z=4 here \(\frac{z}{y} =\frac{4}{5}\)

but if \(\frac{x}{y} = \frac{9}{5}\)then we can also take x= 18 and z=10 can't we? then z= 8 here \(\frac{z}{y} =\frac{8}{10}\)

So I thought statement 1 did not give us the actual values of x and y, so I thought there were many possibilities hence marked statement as insufficient.

so this was my confusion , I though since we got a ratio at the end of simplification of 1 we do not know the actual values of x and y , since we got \(\frac{x}{y} = \frac{9}{5}\), \(\frac{x}{y}\) could also be \(\frac{18}{10}\),or \(\frac{27}{15}\), if x and y differs so will z and hence will \(\frac{z}{y}\) unless we are always reducing ratios to their lowest form

please tell me how can we be sure of x and y from this ratio

thanks


Notice that we need to find the ratio of z to y (z/y). Now, you are right, we CANNOT get the VALUES of x, y, and z, from x=y+z and (x-y)/y=4/5, but we CAN get the RATIO of z to y (as shown in my post).

Does this make sense?
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Collection of Questions:
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Re: If x, y, and z are integers greater than 0 and x = y + z, what is the [#permalink]

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New post 08 Sep 2013, 02:41
Bunuel wrote:
stne wrote:
Do we always have to reduce ratio's to their lowest forms?

What I did was this, please tell me why this reasoning is wrong.

we need to find the value of \(\frac{y-z}{y}\) or \(1- \frac{y}{z}\) so we need to find the value of
\(\frac{y}{z}\)


(1)\(\frac{x-y}{y} = \frac{4}{5}\)

\(\frac{x}{y} -1 = \frac{4}{5}\) ,\(\frac{x}{y} = \frac{4}{5} + 1\), \(\frac{x}{y}= \frac{9}{5}\)

now given x= y+z so if \(\frac{x}{y} = \frac{9}{5}\) then we can take x= 9 and y=5 then z=4 here \(\frac{z}{y} =\frac{4}{5}\)

but if \(\frac{x}{y} = \frac{9}{5}\)then we can also take x= 18 and z=10 can't we? then z= 8 here \(\frac{z}{y} =\frac{8}{10}\)

So I thought statement 1 did not give us the actual values of x and y, so I thought there were many possibilities hence marked statement as insufficient.

so this was my confusion , I though since we got a ratio at the end of simplification of 1 we do not know the actual values of x and y , since we got \(\frac{x}{y} = \frac{9}{5}\), \(\frac{x}{y}\) could also be \(\frac{18}{10}\),or \(\frac{27}{15}\), if x and y differs so will z and hence will \(\frac{z}{y}\) unless we are always reducing ratios to their lowest form

please tell me how can we be sure of x and y from this ratio

thanks


Notice that we need to find the ratio of z to y (z/y). Now, you are right, we CANNOT get the VALUES of x, y, and z, from x=y+z and (x-y)/y=4/5, but we CAN get the RATIO of z to y (as shown in my post).

Does this make sense?


So we are taking \(\frac{x}{y} = \frac{9}{5} =\frac{18}{10} =\frac{27}{15}\)

if we are taking all these to be equal, then ratio's are being reduced to their lowest forms.

That was my original question. " Do we need to reduce ratio's to the lowest forms "
The answer must be yes, because if the answer is no then \(\frac{9}{5}\) would not be equal \(\frac{18}{10}\) , and different values of x and y would give different values of z, and we would have different \(\frac{z}{y}\)

So yes , ratio's must be reduced to their lowest forms,Please can you reaffirm.
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Re: If x, y, and z are integers greater than 0 and x = y + z, what is the [#permalink]

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New post 08 Sep 2013, 04:45
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stne wrote:
Bunuel wrote:
stne wrote:
Do we always have to reduce ratio's to their lowest forms?

What I did was this, please tell me why this reasoning is wrong.

we need to find the value of \(\frac{y-z}{y}\) or \(1- \frac{y}{z}\) so we need to find the value of
\(\frac{y}{z}\)


(1)\(\frac{x-y}{y} = \frac{4}{5}\)

\(\frac{x}{y} -1 = \frac{4}{5}\) ,\(\frac{x}{y} = \frac{4}{5} + 1\), \(\frac{x}{y}= \frac{9}{5}\)

now given x= y+z so if \(\frac{x}{y} = \frac{9}{5}\) then we can take x= 9 and y=5 then z=4 here \(\frac{z}{y} =\frac{4}{5}\)

but if \(\frac{x}{y} = \frac{9}{5}\)then we can also take x= 18 and z=10 can't we? then z= 8 here \(\frac{z}{y} =\frac{8}{10}\)

So I thought statement 1 did not give us the actual values of x and y, so I thought there were many possibilities hence marked statement as insufficient.

so this was my confusion , I though since we got a ratio at the end of simplification of 1 we do not know the actual values of x and y , since we got \(\frac{x}{y} = \frac{9}{5}\), \(\frac{x}{y}\) could also be \(\frac{18}{10}\),or \(\frac{27}{15}\), if x and y differs so will z and hence will \(\frac{z}{y}\) unless we are always reducing ratios to their lowest form

please tell me how can we be sure of x and y from this ratio

thanks


Notice that we need to find the ratio of z to y (z/y). Now, you are right, we CANNOT get the VALUES of x, y, and z, from x=y+z and (x-y)/y=4/5, but we CAN get the RATIO of z to y (as shown in my post).

Does this make sense?


So we are taking \(\frac{x}{y} = \frac{9}{5} =\frac{18}{10} =\frac{27}{15}\)

if we are taking all these to be equal, then ratio's are being reduced to their lowest forms.

That was my original question. " Do we need to reduce ratio's to the lowest forms "
The answer must be yes, because if the answer is no then \(\frac{9}{5}\) would not be equal \(\frac{18}{10}\) , and different values of x and y would give different values of z, and we would have different \(\frac{z}{y}\)

So yes , ratio's must be reduced to their lowest forms,Please can you reaffirm.


No, that's not correct. You don't have to reduce. 9/5 is the same ratio as 18/10.

If you consider x/y=18/10, you'll get that z/y=8/10, which is the same as 4/5.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If x, y, and z are integers greater than 0 and x = y + z, what is the [#permalink]

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New post 08 Sep 2013, 05:24
Bunuel wrote:
No, that's not correct. You don't have to reduce. 9/5 is the same ratio as 18/10.

If you consider x/y=18/10, you'll get that z/y=8/10, which is the same as 4/5.


Well I guess we all have different ways to look at it
if x/y=9/5 = 18/10
The fact that their are equal is not easily visible to me, till i get both the numerator and denominator to have the same value as the other ratio's , and that I get when I reduce them

So I guess without reducing them if one can ascertain that the ratio's are same then no need to reduce them.

Main thing is to realize that 9/5 = 18/10 = 27/15

Let me know if I am still missing something, thank you +1
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Re: If x, y, and z are integers greater than 0 and x = y + z, what is the [#permalink]

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New post 18 Sep 2016, 19:58
D is correct.

(1) (x-y)/y = 4/5

Rewrite equation given at the beginning to z = x -y and substitute into (y-z)/y which gives you (2y-x)/y (a)

Now, looking at equation given to us from (1), cross multiply --> 5x-5y = 4y --> x = (9/5)y

Plug this into equation (a)

SUFFICIENT

(2) z/y = 4/5 --> Rewrite to z = (4/5)y

Plug into (y-z)/y

SUFFICIENT

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