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At RW Press, the advertising rate, a, is inversely [#permalink]
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 Updated on: 24 Oct 2012, 05:20
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At RW Press, the advertising rate, a, is inversely proportional to the productivity measure, b, which, in turn, is inversely proportional to the labor cost, c. Is the labor cost at least 200 when the productivity measure is at least 100? (1) When \(C\geq{100}\), \(B\geq{50}\) (2) When \(C=100\), \(a=100\)
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Originally posted by LM on 24 Oct 2012, 04:46.
Last edited by Bunuel on 24 Oct 2012, 05:20, edited 1 time in total.
Renamed the topic and edited the question.
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Re: At RW Press, the advertising rate, a, is inversely [#permalink]
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31 Oct 2012, 06:22
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LM wrote: At RW Press, the advertising rate, a, is inversely proportional to the productivity measure, b, which, in turn, is inversely proportional to the labor cost, c. Is the labor cost at least 200 when the productivity measure is at least 100?
(1) When \(C\geq{100}\), \(B\geq{50}\)
(2) When \(C=100\), \(a=100\) Hi, Based on Q's we can say that a=k1/b where K1 is a constant and a -A.R and b is Productivity ----> Eq 1 b=K2/C where K2 is constant C is labour cost----> Eq 2 As per St 1 When C>/100, B>/50 ---> When C =100, B=50, Value of K2 is 5000 Putting B value as 100 in Eq 2, we can find value of C St 2 Solving in similar way, we end up only with ratio of K1/K2 and hence cannot find value of C against given value of B My ans choice is A
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Re: At RW Press, the advertising rate, a, is inversely [#permalink]
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08 May 2013, 01:13
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Until the OA is posted, this answer will be based on my own understanding, therefore susceptible to be incorrect.  The wording of this problem seems off to me. Anyway, let's solve this. First of all, we need to understand and rephrase the problem. We are told that the advetising rate "a" is inversely proportional to the productivity measure "b". Algebraically it means that there's a constant C such as : \(a=\frac{C}{b}\) (1) Likewise, we are told that the productivity measure "b" is inversely proportional to the labor cost "c". Algebraically it means that there's a constant D such as : \(b = \frac{D}{c}\) (2) So we're told that to see whether the labor cost is at least 200 when the productivity measure is at least 100. Meaning that we need to find the value of the constant D. Statement 1 : when\(c >=100\) then \(b >=50\) Using equation (2), we get : \(D = b*c\) which means that given the statement, \(D >= 5000\). If we consider the lowest possible value of D, i.e 5000, we get for \(b >= 100\), \(\frac{D}{c} >= 100\), meaning that \(\frac{5000}{c} >=100\), therefore \(c <= 50\). Which means that the labor cost will be at most 50 when the productivity measure is at least 100 given \(D = 5000\). In fact the more D increases, the more c decreases. So statement 1 is sufficient. Statement 2 : when \(c = 100\) then \(a = 100\) Since a and c are not directly linked, then using both equations 1 & 2, we can find the link between the two variables : We have : \(a=\frac{C}{b}\) and \(b = \frac{D}{c}\) Therefore, by putting b's expression in a, we get \(a = c*\frac{C}{D}\) Since a = c, then we get C = D. And since there's no further information to be gained from this statement (remember that we're looking for D's value), then this statement is insufficient. The final answer is A.Hope that helped 
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Re: At RW Press, the advertising rate, a, is inversely [#permalink]
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06 Feb 2014, 02:32
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My approach was very simple. If two variables are inversely proportional, it simply means if one goes up then other goes down and vice versa. Stmt 1 then gives straight forward NO answer to our question. when C>=100 B>=50. If B increases to 100 then C decreases further and will be <100. Hence C cannot be >=200 when B>=100. Sufficient.Stmt 2 Is a bit tricky. Nothing is mentioned about B and we just know the relation between a & b. b & c. given a=100 and c=100. Now if a goes up b comes down. thus c increases. if b were 900, increasing a & c to >=200 may have b at 400 or 100. we don't know. hence insufficient.A is the answer.
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At RW Press, the advertising rate, a, is inversely proportional to the [#permalink]
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15 Apr 2015, 05:27
At RW Press, the advertising rate, a, is inversely proportional to the productivity measure, b, which, in turn, is inversely proportional to the labor cost, c. Is the labor cost at least 200 when the productivity measure is at least 100?
(1) When c=>100, b=>50.
(2) When c=100, a=100.
---what I struggled with the question---
I think the correct choice should have been E because,
let's take c= 1000 and b=50, then if b= 100; c=500 that is greater than 200
let's take c= 100 and b=50, then if b= 100; c=50 that is less than 200, this inconsistency make the first stament insufficient.
Also, both together are not sufficient.
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Re: At RW Press, the advertising rate, a, is inversely [#permalink]
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15 Apr 2015, 05:51
Takoz89 wrote: At RW Press, the advertising rate, a, is inversely proportional to the productivity measure, b, which, in turn, is inversely proportional to the labor cost, c. Is the labor cost at least 200 when the productivity measure is at least 100?
(1) When c=>100, b=>50.
(2) When c=100, a=100.
---what I struggled with the question---
I think the correct choice should have been E because,
let's take c= 1000 and b=50, then if b= 100; c=500 that is greater than 200
let's take c= 100 and b=50, then if b= 100; c=50 that is less than 200, this inconsistency make the first stament insufficient.
Also, both together are not sufficient. Merging similar topics. Please refer to the discussion above.
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Re: At RW Press, the advertising rate, a, is inversely [#permalink]
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04 Feb 2016, 19:19
we are told that a=1/b, and b=1/c, or a=c.
B and D we can eliminate right away, because we are told the same information. C=100 while A=100.
Thus, it must be A, C, or E.
C is highly unlikely, as there is no much new information given
picked A...but confused with the terms used here...
i think it should be specified that big letters are constants..
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Re: At RW Press, the advertising rate, a, is inversely [#permalink]
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08 Jun 2017, 11:11
BunuelI could understand the above explanation....can you please help ? If in the relation between c and b, if the constant is 30000,can't b=150 and c=200?
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Manager
Joined: 02 Mar 2017
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At RW Press, the advertising rate, a, is inversely [#permalink]
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30 Jul 2017, 06:07
Bunuel Please help. Why is everyone plugging the value of C= 100 and b=50 for option A, we can try c=100000 and b=50. From this option A can be both true or false. NO?
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Senior Manager
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At RW Press, the advertising rate, a, is inversely [#permalink]
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09 Apr 2018, 05:17
Hi Bunuel, I have a question, Let the relation between b and c be b = k/c , where k is constant Statement 1: when c >= 100, b >= 50 Let c = 100, b = 50, then k = 5000 if b >= 100, then c <= 50 ( C is lesser than 200) but as per statement 1, if we take c = 1000 (which is greater than 100) and b = 50 then k = 50000 now if b>= 100, then c <= 500 now C could be greater than 200 Can you please tell me where i am going wrong? Thanks
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At RW Press, the advertising rate, a, is inversely
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09 Apr 2018, 05:17
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