Perplexing Inequality Issue: Is x less than 20?

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Question: Is x less than 20?
(St 1): The sum of x and y is less than 20.
(St 2): y is less than 20.

OA is E. My problem is NOT seeing why E is correct. The thing that has me baffled is why a straightforward algebraic approach doesn't work (see my next paragraph). Eliminating 1 & eliminating 2 is easy. Also, IF YOU PICK NUMBERS, it's relatively straightforward to see why E is correct and C is not. HERE's my question: I always thought you could add inequalities as long as the ineq symbol was pointed in the same direction.

When I saw this question, after eliminating S1 & S2, I quickly wrote statement 1 as x+y < 20 and I wrote statement 2 as y< 20. Then, I just subtracted S2 from S1 and I got x < 0. If x<0, then x is less than 20. So I picked C and moved on, thinking I'd cracked it. Is the fundamental issue that while you can ADD ineqs (as long as ineq symbol pointed in the same way), that you CAN't subtract them? I thought I'd shredded this one quickly and was surprised to see the result. Many thanks.

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by Jkight3107 » Fri Jan 15, 2010 5:15 pm
Question: Is x less than 20?
(St 1): The sum of x and y is less than 20.
(St 2): y is less than 20.

St 1 - The sum of x and y is less than 20 or x+y<20

Taking a logic approach -
x could be 10 and y could be 9 = 19 OR
x could be 25 and y could be -10 = 15

You have two solutions - thus statement 1 is insuff

St 2

St 2 doesn't even mention x so insuff

St 1 & St 2
If we know that y<20 and x+y<20 then

Using logic Y dosen't have to = a postive number.

You can still have multiple solutions as exampled in ST 1.

Thus E is the answer

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by beat2009 » Mon Jan 18, 2010 2:43 pm
Per my original post, I see how you can crack this by substituting numbers. That's not the question. I'm trying to figure out why a straightforward algebraic approach doesn't work. I had processed it using basic algebra and it came up with the right answer. Any ideas out there? Thanks much.

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by valleeny » Mon Jan 18, 2010 7:57 pm
If I'm not wrong, you can't do the normal algebraic approach when there's inequalities as when there is an inequality, a range of values are present.

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by beat2009 » Thu Jan 21, 2010 10:48 am
Thanks. I've used the algebraic approach with success in numerous other problems. And per some of the GMAT prep books, as long as you have the inequality symbol pointed in the same direction, you may add the inequalities. Instructors, anyone else: does ANYONE have a definitive answer on this? Again, I know how to solve the problem by picking numbers. My issue is I thought I was being quick and efficient by translating the phrases into inequalities and solving: turns out there's a hidden explosive in there somewhere. I'm trying to determine why the straight "translate the words into an inequality" approach fails in this case. Anyone?

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by Osirus@VeritasPrep » Thu Jan 21, 2010 12:53 pm
beat2009 wrote:Thanks. I've used the algebraic approach with success in numerous other problems. And per some of the GMAT prep books, as long as you have the inequality symbol pointed in the same direction, you may add the inequalities. Instructors, anyone else: does ANYONE have a definitive answer on this? Again, I know how to solve the problem by picking numbers. My issue is I thought I was being quick and efficient by translating the phrases into inequalities and solving: turns out there's a hidden explosive in there somewhere. I'm trying to determine why the straight "translate the words into an inequality" approach fails in this case. Anyone?
According to Manhattan GMAT you can ADD two inequalities pointing in the same direction, but you can never SUBTRACT them. You would never know whether or not to change the sign if you subtract the inequality

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by valleeny » Thu Jan 21, 2010 4:24 pm
Hi beat

can u give an example where you said you could add two inequalities pointing in the same direction?

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by beat2009 » Fri Jan 22, 2010 11:12 am
Valleeny,

If x < 10 and y <5, then x+y < 15. Take a look at the other post I made with a question to Stuart Kovinsky -> even the addition rule, while a decent rule of thumb, is not foolproof.

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by prinit » Fri Jan 22, 2010 12:44 pm
Question: Is x less than 20?
(St 1): The sum of x and y is less than 20.
(St 2): y is less than 20.
(St 1) x+y<20, Here x and y can be +ve or -ve.

consider the boundary conditions:
case1: x is +19 y is 0 or -ve.
case2: x is 21 and y is -ve.

in both cases st(1) is getting satisfied. but we can't answer whether x is less than 20 or not. so (St 1) is not sufficient.

(St 2): y is less than 20. Here y can't be greater than 20 but it does not mean that x is also less than 20. so (St 2) is not sufficient.

Now combine both. we still can't answer , because x+y<20 and y<20 does not reveal whether x is less than 20 or not. (see case 1 & case 2 closely, where y<20 is already getting applied.)

well.. If x < 10 and y <5, then x+y < 15 , this works fine and will always work fine. But if x+y < 15 then it does not mean that x < 10 and y <5. Hope it clarifies all.

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by shashank.ism » Fri Feb 19, 2010 6:25 am
beat2009 wrote:Question: Is x less than 20?
(St 1): The sum of x and y is less than 20.
(St 2): y is less than 20.

OA is E. My problem is NOT seeing why E is correct. The thing that has me baffled is why a straightforward algebraic approach doesn't work (see my next paragraph). Eliminating 1 & eliminating 2 is easy. Also, IF YOU PICK NUMBERS, it's relatively straightforward to see why E is correct and C is not. HERE's my question: I always thought you could add inequalities as long as the ineq symbol was pointed in the same direction.

When I saw this question, after eliminating S1 & S2, I quickly wrote statement 1 as x+y < 20 and I wrote statement 2 as y< 20. Then, I just subtracted S2 from S1 and I got x < 0. If x<0, then x is less than 20. So I picked C and moved on, thinking I'd cracked it. Is the fundamental issue that while you can ADD ineqs (as long as ineq symbol pointed in the same way), that you CAN't subtract them? I thought I'd shredded this one quickly and was surprised to see the result. Many thanks.
st.1 : x+y< 20 but x could be > 20 and y a -ve no. so insufficient
st.2 : y is less than 20 may be -ve and hence x could be > 20 insufficient
combined also y can be -ve and x could be > 20 insufficient
Ans E
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