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Following are some DS questions with typical traps, good for [#permalink]
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11 Oct 2009, 11:21
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This topic is locked. If you want to discuss this question please repost it in the respective forum. Following are some DS questions with typical traps, good for practising though. 1) If Z powered n equals 1 then what is the value of Z? {Z^n = 1} Stmt 1  n is a nonzero integer Stmt 2  Z > 0 2) Martha bought an armchair and coffee table at an auction and sold them at her store. Her gross profit from the purchase and sale of armchair was what percentage greater than the gross profit of purchase and sale of coffee table? Stmt 1  Martha paid 10 percent more for the armchair than for the coffee table Stmt 2  Martha sold the armchair for 20 percent more than she sold the coffee table. 3) If x, y, z are integers greater than 1, what is the value of x + y + z? Stmt 1  xyz = 70 Stmt 2  x/yz = 7/10 4) Is z equal to the median of the three positive integers x, y and z Stmt 1  x < y + z Stmt 2  y = z. I know, the questions were simple and silly and in my eagerness to be more careful, double checked the questions and got them wrong.
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Re: Some tough questions from GMATPrep [#permalink]
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11 Oct 2009, 12:58
SensibleGuy wrote: 1) If Z^n = 1, what is the value of Z? {Z^n = 1} Stmt 1. n is a nonzero integer Stmt 2. Z > 0 1. If n is a nonzero integer, 1^1, 1^(1), (1)^1, and (1)^(1) all are equal to 1. So z could be 1 or 1 as does n. NSF.. 2. If Z > 0, 1^1, 1^(1), 2^0, or (11)^0 all ends in 1. So Z could be 1, 2, ................infinite. NSF.. From 1 and 2: Z and n must be 1. SUFF... Thats C.
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Re: Some tough questions from GMATPrep [#permalink]
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13 Oct 2009, 06:45
SensibleGuy wrote: Following are some DS questions with typical traps, good for practising though. 3) If x, y, z are integers greater than 1, what is the value of x + y + z? Stmt 1  xyz = 70 Stmt 2  x/yz = 7/10 I know, the questions were simple and silly and in my eagerness to be more careful, double checked the questions and got them wrong. x,y,z, are >1 so we can get 2,5 and 7 as the only values of either x, y or z. so x+y+z = 14 A is suff x/yz = 7/10 =14/20 so we can have x = 14 and y or z can be 2 and 10. But we can also have x=7 and y or z as 2 and 5. so B is insuff



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Re: Some tough questions from GMATPrep [#permalink]
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13 Oct 2009, 06:49
SensibleGuy wrote: Following are some DS questions with typical traps, good for practising though.
4) Is z equal to the median of the three positive integers x, y and z Stmt 1  x < y + z Stmt 2  y = z.
I know, the questions were simple and silly and in my eagerness to be more careful, double checked the questions and got them wrong. A  x< y+z. Lets assume x =1 and y = 3 and z =5 then 1< 8 and y is the median ( 1,3,5) but if we consider z as 3 and y as 5 then z becomes the median. hence insuf B if y=z then irrespective the value of x z will be the median. if y=z=3 and x = 1 or 5 then z is median. Hence suff



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Re: Some tough questions from GMATPrep [#permalink]
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13 Oct 2009, 10:40
SensibleGuy wrote: 1) If Z^n = 1, what is the value of Z? {Z^n = 1} Stmt 1. n is a nonzero integer Stmt 2. Z > 0
1. If n is a nonzero integer, 1^1, 1^(1), (1)^1, and (1)^(1) all are equal to 1. So z could be 1 or 1 as does n. NSF.. 2. If Z > 0, 1^1, 1^(1), 2^0, or (11)^0 all ends in 1. So Z could be 1, 2, ................infinite. NSF..
From 1 and 2: Z and n must be 1. SUFF... Thats C.
In reference to the post above ....
i dont understand why the solution cannot be A. anything raised to 0 is 1, and 1 raised to any power is 1. stmt A says, N is not =0. hence Z has to be 1. so shouldn't A suffice? (i dont understand how there can be any value other than 1 for Z.)



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Re: Some tough questions from GMATPrep [#permalink]
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13 Oct 2009, 11:09
nitronori41 wrote: SensibleGuy wrote: 1) If Z^n = 1, what is the value of Z? {Z^n = 1} Stmt 1. n is a nonzero integer Stmt 2. Z > 0
1. If n is a nonzero integer, 1^1, 1^(1), (1)^1, and (1)^(1) all are equal to 1. So z could be 1 or 1 as does n. NSF.. 2. If Z > 0, 1^1, 1^(1), 2^0, or (11)^0 all ends in 1. So Z could be 1, 2, ................infinite. NSF..
From 1 and 2: Z and n must be 1. SUFF... Thats C.
In reference to the post above ....
i dont understand why the solution cannot be A. anything raised to 0 is 1, and 1 raised to any power is 1. stmt A says, N is not =0. hence Z has to be 1. so shouldn't A suffice? (i dont understand how there can be any value other than 1 for Z.) what if we consider n as 2 and Z = 1 ? we get Z^n = (1)^2 as 1 thereby we are getting multiple values of Z i.e both 1 and 1 hope this helps



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Re: Some tough questions from GMATPrep [#permalink]
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13 Oct 2009, 21:04
ah yes.. forgot to consider that possibility thanks




Re: Some tough questions from GMATPrep
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13 Oct 2009, 21:04






