DS Algebra

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DS Algebra

by prachi18oct » Tue Jun 16, 2015 7:24 am
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.


I got this one right. But the solution mentioned was not precise. So to validate my reasoning, I am putting this up here.

From (1), there can be many combinations which will lead to different answer.
r(s+t) = 5
r = 1 ; s = 3 ; t = 2 => rst > 1
r = -1 ; s = -3 ; t = -2 => rst < 1 INSUFFICIENT

From (2), r +st = 2
Picking various numbers, we can see that the product rst will never be > 1
r = 1 ; st = 1 ( s = 5; t = 1/5) rst =1 YES
r = -3 ; st = 5 ( s = 25 ; t = 1/5) rst < 1 YES
r = 5/2 ; st = -1/2 => rst < 0 YES
SUFFICIENT.

Please suggest alternaye solutions

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by GMATGuruNY » Tue Jun 16, 2015 7:48 am
prachi18oct wrote:Is rst ≤ 1?

(1) rs + rt = 5
(2) r + st = 2
Statement 1: rs + rt = 5
r(s+t) = 5.
If r=1, s=1 and t=4, then rst > 1.
If r=1, s=0 and t=5, then rst < 1.
INSUFFICIENT.

RULE:
If x+y = k, where k is a positive constant, then the greatest possible value of xy occurs when x=y.

EXAMPLE: x+y = 10
If x=y=5, then xy = 5*5 = 25.
If x=4 and y=6, then xy = 4*6 = 24.
If x=3 and y=7, then xy = 3*7 = 21.
As the case above illustrates, the greatest possible value of xy occurs when x=y.

Statement 2: r + st = 2
Since r + st is equal to a positive constant, the greatest possible value of rst occurs when r=st.
r=st when r=1 and st=1, implying that the greatest possible value of rst = 1*1 = 1.
Thus, rst ≤ 1.
SUFFICIENT.

The correct answer is B.
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by Ian Stewart » Tue Jun 16, 2015 10:25 am
prachi18oct wrote:Is rst ≤ 1?

(1) rs + rt = 5
(2) r + st = 2
You can also see why S2 is sufficient using algebra. We can see that when r=s=t=1 we can get a 'yes' answer to the question. We want to know if it is possible to get a 'no' answer. That is, we want to know when it will be true that rst > 1.

Using S2, we know that st = 2 - r. If we replace 'st' with '2-r' in the inequality "rst > 1", we'll find out for which values of r this inequality is true. That is, we'll discover for which values of r the answer to the question is 'no' :

rst > 1
r(2-r) > 1
2r - r^2 > 1
0 > r^2 - 2r + 1
0 > (r - 1)^2

But a square can never be negative, so there is no value of r which can make this inequality true. So the answer to the question can never be 'no', and Statement 2 is sufficient. Since S1 is not sufficient as we can see by testing values, the answer is B.
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