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The number of touchscreen personal organizers currently in stock is p times what it was at the beginning of the year. If the number of touchscreen organizers currently in stock is t and the number in stock at the beginning of the year is x (where t, x and p do not equal zero), are there more touchscreen personal organizers in stock now than there were at the beginning of the year?

Re: The number of touchscreen personal organizers currently in stock is p [#permalink]

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20 Jan 2015, 08:53

Bunuel wrote:

The number of touchscreen personal organizers currently in stock is p times what it was at the beginning of the year. If the number of touchscreen organizers currently in stock is t and the number in stock at the beginning of the year is x (where t, x and p do not equal zero), are there more touchscreen personal organizers in stock now than there were at the beginning of the year?

(1) p > x

(2) tp > t

Kudos for a correct solution.

question is t > x? assume that x=int; t=int; p: case 1 p=int; case 2 0<p<1. t=px

1st statement; p > x Since x is an integer case 2 of p won't happen, meaning that p is an integer --> t = x(int) --> t > x.

2nd statement: tp>t --> t(p-1)>0 --> t > 0 we already know that; p>1. Case 2 of p won't happen. Least value of p = 2 --> t=2x --> t > x.

Answer is D.
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The number of touchscreen personal organizers currently in stock is p times what it was at the beginning of the year. If the number of touchscreen organizers currently in stock is t and the number in stock at the beginning of the year is x (where t, x and p do not equal zero), are there more touchscreen personal organizers in stock now than there were at the beginning of the year?

Re: The number of touchscreen personal organizers currently in stock is p [#permalink]

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12 May 2015, 12:44

1

This post was BOOKMARKED

Stock at the beginning of the year is x

and , t = p*x where t,p,and x are not equal to Zero

1) If p>x , then t > x ......... So this statement is sufficient

2) If tp > t , then, p > 1 which again gives t > x ............ sufficient

Answer D

Bunuel wrote:

The number of touchscreen personal organizers currently in stock is p times what it was at the beginning of the year. If the number of touchscreen organizers currently in stock is t and the number in stock at the beginning of the year is x (where t, x and p do not equal zero), are there more touchscreen personal organizers in stock now than there were at the beginning of the year?

(1) p > x

(2) tp > t

Kudos for a correct solution.

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Now, since x denotes the number of items, x must be a non-negative integer (because you cannot have -1 devices or 0.5 devices) . We are already given that x is not equal to zero. This means, x is a positive integer for sure (the minimum possible value of x is 1)

So, we can safely divide both sides of the above inequality with x, without impacting the sign of inequality.

And thus, the question simplifies to: Is p > 1?

St. 1 says: p > x

This means, p > (positive integer x, whose minimum possible value is 1)

=> p > 1 for sure

Sufficient.

St. 2 says: tp > t

As explained above for x, t also is a positive integer for sure. So, we can safely divide both sides of the above inequality with t, without impacting the sign of inequality.

Re: The number of touchscreen personal organizers currently in stock is p [#permalink]

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13 May 2015, 23:42

Bunuel wrote:

The number of touchscreen personal organizers currently in stock is p times what it was at the beginning of the year. If the number of touchscreen organizers currently in stock is t and the number in stock at the beginning of the year is x (where t, x and p do not equal zero), are there more touchscreen personal organizers in stock now than there were at the beginning of the year?

(1) p > x

(2) tp > t

Ans: D Solution: we know numbers x=number of TPO at the begining, t=TPO at the end of the year, p=multiplier we also know that these numbers are integers and only p can be any value except zero. [t=p*x given] 1) p>x ; if p>x then t>x, because x is integer and non negative. as you cannot have -1 TPO. (I think) [Sufficient] 2) tp>t ; t(p-1)>0 ; so now for this to be true, t>0 & (p-1)>0 OR t<0 & (p-1)<0 as we know t>0 second condition does not follow and with first condition we know that P>1 so this statement alone is sufficient to ans the question. [Sufficient]

Ans. D [Both statement alone are sufficient to answer the question]
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Re: The number of touchscreen personal organizers currently in stock is p [#permalink]

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25 Feb 2016, 20:22

Bunuel wrote:

The number of touchscreen personal organizers currently in stock is p times what it was at the beginning of the year. If the number of touchscreen organizers currently in stock is t and the number in stock at the beginning of the year is x (where t, x and p do not equal zero), are there more touchscreen personal organizers in stock now than there were at the beginning of the year?

(1) p > x

(2) tp > t

Kudos for a correct solution.

ok, so we are given: t=px. where t, p, and x - are not zero.

1. p>x. since x can't be zero, suppose p=x. px=x^2. x^2 is always greater than x when x is greater than 0. note that it can't be a non-integer, since we are speaking about touchscreen personal organizers, and these cannot be half/third/etc.

so 1 is sufficient.

2. tp>t. ok, we know t=px p=t/x t*t/x>t t^2/x >t knowing that t is not 0, and can't be negative - touchscreen personal organizers negative? :D we can divide by t: t/x>1. so t>x. sufficient.

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25 Sep 2017, 11:57

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