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