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Is rst <= 1?

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Is rst <= 1?  [#permalink]

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New post 30 Dec 2010, 06:23
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Difficulty:

  95% (hard)

Question Stats:

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Is rst ≤ 1?

(1) rs + rt = 5
(2) r + st = 2

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Re: Algebra DS  [#permalink]

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New post 30 Dec 2010, 06:48
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gmatpapa wrote:
Is rst ≤ 1?

(1) rs + rt = 5
(2) r + st = 2


Is rst ≤ 1?

(1) rs + rt = 5 --> \(r=\frac{5}{s+t}\) --> question becomes: is \(\frac{5st}{s+t}\leq{1}\)? --> is \(\frac{st}{s+t}\leq{\frac{1}{5}}\)? Now, if \(s\) and \(t\) are large enough positive numbers (for example 2 and 3) then the asnwer will be NO but if one of them equals to zero then the answer will be YES. Not sufficient.

(2) r + st = 2 --> \(st=2-r\) --> question becomes: is \(r(2-r)\leq{1}\)? --> is \(2r-r^2\leq{1}\)? --> is \(r^2-2r+1\geq{0}\)? --> is \((r-1)^2\geq{0}\)? As square of some expression is always more than or equal to zero then the answer to this question is always YES. Sufficient.

Answer: B.

Hope it's clear.
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Re: Algebra DS  [#permalink]

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New post 30 Dec 2010, 20:49
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gmatpapa wrote:
Is rst ≤ 1?

(1) rs + rt = 5
(2) r + st = 2
(A) Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.


The question is testing you on the following concept:
If the sum of two non negative numbers is constant, their product is greatest when the numbers are equal. e.g. if a + b = 10, then ab is maximum when a = b = 5. Maximum value of ab = 5*5 = 25


Stmnt 1: Given rs + rt is constant, the maximum value of rs*rt will be when rs = rt = 5/2. Maximum value of \(r^2st= (5/2)*(5/2) = 25/4.\)
But we get no information about maximum value of rst so not sufficient.

Stmnt 2: Given r + st = 2, then maximum value of r*st will be when r = st = 1.
Maximum value of rst = 1 i.e. rst <= 1. Sufficient.

Answer (B).

Note here that though they haven't said that the numbers are non-negative, we can easily see that the product can be less than 1. We only need to worry about the product greater than 1. In that case, since the sum is positive and product we need is positive, we only need to worry about positive numbers.
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Re: Algebra DS  [#permalink]

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New post 30 Dec 2010, 09:03
Yes it is clear. With your explanation I also learnt a situation where equation can be manipulated to check sufficiency. very useful.. Kudos!!
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Re: Algebra DS  [#permalink]

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New post 19 Jan 2011, 22:12
Bunuel wrote:
gmatpapa wrote:
Is rst ≤ 1?

(1) rs + rt = 5
(2) r + st = 2


Is rst ≤ 1?

(1) rs + rt = 5 --> \(r=\frac{5}{s+t}\) --> question becomes: is \(\frac{5st}{s+t}\leq{1}\)? --> is \(\frac{st}{s+t}\leq{\frac{1}{5}}\)? Now, if \(s\) and \(t\) are large enough positive numbers (for example 2 and 3) then the asnwer will be NO but if one of them equals to zero then the answer will be YES. Not sufficient.

(2) r + st = 2 --> \(st=2-r\) --> question becomes: is \(r(2-r)\leq{1}\)? --> is \(2r-r^2\leq{1}\)? --> is \(r^2-2r+1\geq{0}\)? --> is \((r-1)^2\geq{0}\)? As square of some expression is always more than or equal to zero then the answer to this question is always YES. Sufficient.

Answer: B.

Hope it's clear.


Hello Bunnel,

In the second statement you have concluded that rst > = 0 which does not state that it has to be >=1. which the Q has asked to find... since it can have values from 0 to 1.
Please correct me if I am wrong..
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Re: Algebra DS  [#permalink]

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New post 19 Jan 2011, 22:41
jullysabat wrote:
Bunuel wrote:
gmatpapa wrote:
Is rst ≤ 1?

(1) rs + rt = 5
(2) r + st = 2


Is rst ≤ 1?

(1) rs + rt = 5 --> \(r=\frac{5}{s+t}\) --> question becomes: is \(\frac{5st}{s+t}\leq{1}\)? --> is \(\frac{st}{s+t}\leq{\frac{1}{5}}\)? Now, if \(s\) and \(t\) are large enough positive numbers (for example 2 and 3) then the asnwer will be NO but if one of them equals to zero then the answer will be YES. Not sufficient.

(2) r + st = 2 --> \(st=2-r\) --> question becomes: is \(r(2-r)\leq{1}\)? --> is \(2r-r^2\leq{1}\)? --> is \(r^2-2r+1\geq{0}\)? --> is \((r-1)^2\geq{0}\)? As square of some expression is always more than or equal to zero then the answer to this question is always YES. Sufficient.

Answer: B.

Hope it's clear.


Hello Bunnel,

In the second statement you have concluded that rst > = 0 which does not state that it has to be >=1. which the Q has asked to find... since it can have values from 0 to 1.
Please correct me if I am wrong..


Hi,

Let me answer that before Bunuel the Boss comes. If you see, What has been concluded is not \(rst >= 0\) but \((r-1)^2>=0\). Follow the procedure Bunuel has demonstrated. The statement has been manipulated, leading to mean that the question: Is \(rst< = 1\) is the same as Is \((r-1)^2 > =0\).

Hope its clear!
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Re: Is rst <= 1?  [#permalink]

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New post 23 Jan 2018, 18:49
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Re: Is rst <= 1?   [#permalink] 23 Jan 2018, 18:49
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