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If xyz ≠ 0, is 3x/2 + y + 2z = 7x/2 + y? [#permalink]
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13 May 2016, 06:44
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If xyz ≠ 0, is 3x/2 + y + 2z = 7x/2 + y? (1) y = 3 and x = 2 (2) z = x Source: Manhattan Prep "Fractions, Decimals, Percents" First they grouped all the "like" terns together and ended up with the equation z = x? Answer: Only statement 2 is sufficient by itself. BWhy? I would have answered A because I plugged in the numbers given in statement 1 and solved for z. Then I came up with z = 2. Then I plugged in the numbers for all 3 variables in the original equation (including z = 2) with the result of left side of equation = right side of equation so I can clearly answer the question with statement 1?
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Last edited by Bunuel on 13 May 2016, 07:12, edited 1 time in total.
Renamed the topic and edited the question.



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Re: If xyz ≠ 0, is 3x/2 + y + 2z = 7x/2 + y? [#permalink]
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13 May 2016, 07:09
Silviax wrote: Source: Manhattan Prep "Fractions, Decimals, Percents" If xyz not equal 0, is 3x/2 + y + 2z = 7x/2 + y? (1) y=3 and x = 2 (2) z = x First they grouped all the "like" terns together and ended up with the equation z = x? Answer: Only statement 2 is sufficient by itself. BWhy? I would have answered A because I plugged in the numbers given in statement 1 and solved for z. Then I came up with z = 2. Then I plugged in the numbers for all 3 variables in the original equation (including z = 2) with the result of left side of equation = right side of equation so I can clearly answer the question with statement 1? Hi, you have to find if \(\frac{3x}{2}+ y + 2z = \frac{7x}{2}+ y?\)... But there are three variables and 'z' does not cancel out but remains till end.. so we cannot sove the equation. we donot have to find z=2, we should be given z=2 for equation to be correct.. remember it is asking you IS A=B.. so you require both A and B
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If xyz ≠ 0, is 3x/2 + y + 2z = 7x/2 + y? [#permalink]
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13 May 2016, 07:18
Silviax wrote: If xyz ≠ 0, is 3x/2 + y + 2z = 7x/2 + y? (1) y = 3 and x = 2 (2) z = x Source: Manhattan Prep "Fractions, Decimals, Percents" First they grouped all the "like" terns together and ended up with the equation z = x? Answer: Only statement 2 is sufficient by itself. BWhy? I would have answered A because I plugged in the numbers given in statement 1 and solved for z. Then I came up with z = 2. Then I plugged in the numbers for all 3 variables in the original equation (including z = 2) with the result of left side of equation = right side of equation so I can clearly answer the question with statement 1? Notice that the question asks is 3x/2 + y + 2z = 7x/2 + y? Which can be simplified to is z = x? We are NOT given that z = x, we are asked to find whether it's true. (1) says that x = 2. So the question becomes is x = 2. We don't know that, thus the statement is not sufficient. (2) says that z = x. Now, if z = x, then z = x can only be true when z = x = 0. Since we are told that xyz ≠ 0, then none of the variables is 0, therefore we can say that z ≠ x. Sufficient. Answer: B. Hope it's clear.
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Re: If xyz ≠ 0, is 3x/2 + y + 2z = 7x/2 + y? [#permalink]
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13 May 2016, 09:15
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Information from the question: xyz is not equal to 0 Question asked: 3x/2+y+2z= 7x/2+y? 3x/2+2z= 7x/2? 2z=4x/2? 2z=2x? z=x? So basically, the question is weather z=x? Only statement 2 talks about both z and x. B is the answer
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If xyz ≠ 0, is 3x/2 + y + 2z = 7x/2 + y? [#permalink]
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13 May 2016, 09:24
No I still don't get it because statement 1 gives us two values for two variables. So if plug in these values in my equation I get: 3(2)/2 + 3 + 2z = 7(2)/2 + 3 6/2 + 3 + 2z = 14/2 + 3 3 + 3 + 2z = 7 + 3 6 + 2z = 10 2z = 4 z = 2 So z = 2 results from the equation with the provided values for x and y and if I plug in all 3 values now I can see that both equation are the same so z = 2 and x = 2 thus z = x Where is my thinking error here? chetan2u wrote: Silviax wrote: Source: Manhattan Prep "Fractions, Decimals, Percents" If xyz not equal 0, is 3x/2 + y + 2z = 7x/2 + y? (1) y=3 and x = 2 (2) z = x First they grouped all the "like" terns together and ended up with the equation z = x? Answer: Only statement 2 is sufficient by itself. BWhy? I would have answered A because I plugged in the numbers given in statement 1 and solved for z. Then I came up with z = 2. Then I plugged in the numbers for all 3 variables in the original equation (including z = 2) with the result of left side of equation = right side of equation so I can clearly answer the question with statement 1? Hi, you have to find if \(\frac{3x}{2}+ y + 2z = \frac{7x}{2}+ y?\)... But there are three variables and 'z' does not cancel out but remains till end.. so we cannot sove the equation. we donot have to find z=2, we should be given z=2 for equation to be correct.. remember it is asking you IS A=B.. so you require both A and B



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Re: If xyz ≠ 0, is 3x/2 + y + 2z = 7x/2 + y? [#permalink]
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13 May 2016, 09:29
Silviax wrote: No I still don't get it because statement 1 gives us two values for two variables. So if plug in these values in my equation I get: 3(2)/2 + 3 + 2z = 7(2)/2 + 3 6/2 + 3 + 2z = 14/2 + 3 3 + 3 + 2z = 7 + 3 6 + 2z = 10 2z = 4 z = 2 So z = 2 results from the equation with the provided values for x and y and if I plug in all 3 values now I can see that both equation are the same so z = 2 and x = 2 thus z = x Where is my thinking error here? chetan2u wrote: Silviax wrote: Source: Manhattan Prep "Fractions, Decimals, Percents" If xyz not equal 0, is 3x/2 + y + 2z = 7x/2 + y? (1) y=3 and x = 2 (2) z = x First they grouped all the "like" terns together and ended up with the equation z = x? Answer: Only statement 2 is sufficient by itself. BWhy? I would have answered A because I plugged in the numbers given in statement 1 and solved for z. Then I came up with z = 2. Then I plugged in the numbers for all 3 variables in the original equation (including z = 2) with the result of left side of equation = right side of equation so I can clearly answer the question with statement 1? Hi, you have to find if \(\frac{3x}{2}+ y + 2z = \frac{7x}{2}+ y?\)... But there are three variables and 'z' does not cancel out but remains till end.. so we cannot sove the equation. we donot have to find z=2, we should be given z=2 for equation to be correct.. remember it is asking you IS A=B.. so you require both A and B We don't know whether 3x/2 + y + 2z = 7x/2 + y. We are asked to find out whether 3x/2 + y + 2z = 7x/2 + y. Please read my reply above.
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Re: If xyz ≠ 0, is 3x/2 + y + 2z = 7x/2 + y? [#permalink]
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13 May 2016, 09:52
The question asked is effectively, as Divya pointed out is if x=y which is the simplified version of the equations in questions. only statement 2 answers this.



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Re: If xyz ≠ 0, is 3x/2 + y + 2z = 7x/2 + y? [#permalink]
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08 Sep 2017, 20:26
Silviax wrote: If xyz ≠ 0, is 3x/2 + y + 2z = 7x/2 + y? (1) y = 3 and x = 2 (2) z = x Source: Manhattan Prep "Fractions, Decimals, Percents" First they grouped all the "like" terns together and ended up with the equation z = x? Answer: Only statement 2 is sufficient by itself. BWhy? I would have answered A because I plugged in the numbers given in statement 1 and solved for z. Then I came up with z = 2. Then I plugged in the numbers for all 3 variables in the original equation (including z = 2) with the result of left side of equation = right side of equation so I can clearly answer the question with statement 1? Rewrite and see the pattern here Essentially 3x/2 +2x + y = 7x/2 + Y This would mean that X would have to be equal to Y, x=y beccause 3x/2 +4x/2 + y = 7x/2 + Y St 1 Clearly Insuff because we know nothing about Z St 2 x/2 + y 7x/2 +y = this could only be true if X were 0 but neither x nor y nor z can be 0 Suff




Re: If xyz ≠ 0, is 3x/2 + y + 2z = 7x/2 + y?
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