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For real numbers, A, B, C, and D on the number line

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For real numbers, A, B, C, and D on the number line [#permalink] New post   04 Jan 2014, 23:57
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For real numbers, A, B, C, and D on the number line above, is |A| + |B| < |C| + |D| ?

(1) |B| < |C|
(2) The product of any two of the numbers A, B, C and D is between 0 and 1.

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Re: For real numbers, A, B, C, and D on the number line [#permalink] New post   05 Jan 2014, 06:27
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For real numbers, A, B, C, and D on the number line above, is |A| + |B| < |C| + |D| ?

Notice that from the diagram we have that A<B<C<D.

(1) |B| < |C|. This one is clearly insufficient. Consider: A=1, B=2, C=3, and D=4 for an YES answer and A=-10, B=2, C=3, and D=4 for a NO answer. Not sufficient.

(2) The product of any two of the numbers A, B, C and D is between 0 and 1. So, we have that the product of ANY two of the numbers is positive. This implies either that all the numbers are negative or that all the numbers are positive. In the first case the answer is NO and in the second case the answer is YES. Not sufficient.

(1)+(2) If all the numbers were negative, then because we also have that B < C, then we would have that |B| > |C|, which contradicts the first statement, thus we have the second case: all the numbers are positive. Therefore since A<B<C<D, then |A| + |B| < |C| + |D|. Sufficient.

Answer: C.

Hope it's clear.
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Re: For real numbers, A, B, C, and D on the number line [#permalink] New post   13 Jan 2015, 10:26
Bunuel wrote:

(2) The product of any two of the numbers A, B, C and D is between 0 and 1. So, we have that the product of ANY two of the numbers is positive. This implies either that all the numbers are negative or that all the numbers are positive. In the first case the answer is NO and in the second case the answer is YES. Not sufficient.



Hi Bunuel,

I have been reading your posts quite regularly, and they are very helpful. I am really grateful to you for providing assistance to people like us.

As far as this question is concerned i have a doubt on s(2). As you have mentioned above "that all the numbers are negative or that all the numbers are positive."
I have been working on one or two examples and came across this one
Suppose B and C are -1/2 and -1/3. Their product is 1/6, which is between 0 and 1. Now the possible values can be
A<B<C<D :- -1 < -1/2 < -1/3 <1 . In this case |A| + |B| IS NOT LESS THAN |C| + |D| insufficient

Suppose A and B are -1/2 and -1/3. Again their product is 1/6, which is between 0 and 1. Now the possible values can be
A<B<C<D :- -1/2 < -1/3 < 1 < 2. In this case |A| + |B| < |C| + |D| sufficient.

So, how can we say "that all the numbers are negative or that all the numbers are positive." ? and it doesn't say that The product of every two of the numbers A, B, C and D is between 0 and 1.

Thanks
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Re: For real numbers, A, B, C, and D on the number line [#permalink] New post   13 Jan 2015, 14:06
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Hi pahujanavdeep,

Unfortunately, your example does NOT fit the "restriction" that is described in Fact 2:

Fact 2: The product of any two of the numbers A, B, C and D is between 0 and 1.

This tells us that if we take ANY two of the 4 values and multiply them....then the product will be a POSITIVE fraction.

Your example:
A = -1
B = -1/2
C = -1/3
D = 1

If one of the two numbers chosen is D, then the product will be a NEGATIVE fraction. This is NOT a viable option given what Fact 2 tells us. The ONLY ways for the product of ANY two of those values to be a POSITIVE fraction is if ALL 4 values are negative OR ALL 4 are positive.

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Re: For real numbers, A, B, C, and D on the number line [#permalink] New post   14 Jan 2015, 06:34
Thank you Rich :)

I got confused with the wording. But it make sense now. Appreciate it :-D

Regards
Navdeep
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Re: For real numbers, A, B, C, and D on the number line [#permalink] New post   28 Aug 2016, 13:29
Bunuel wrote:


Notice that from the diagram we have that A<B<C<D.



where does the diagram tell us that?
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Re: For real numbers, A, B, C, and D on the number line [#permalink] New post   28 Aug 2016, 21:29
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Hi nycgirl212,

The prompt specifically refers to the number line that's presented. On a number line, numbers increase from left to right. Thus, based on the given number line, A < B < C < D.

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Re: For real numbers, A, B, C, and D on the number line [#permalink] New post   29 Aug 2016, 08:31
EMPOWERgmatRichC wrote:
Hi nycgirl212,

The prompt specifically refers to the number line that's presented. On a number line, numbers increase from left to right. Thus, based on the given number line, A < B < C < D.

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so even though it doesn't explicitly state that the numbers increase from left to right, we are automatically supposed to assume that they do when we see a number line?
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Re: For real numbers, A, B, C, and D on the number line [#permalink] New post   29 Aug 2016, 09:15
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nycgirl212 wrote:
EMPOWERgmatRichC wrote:
Hi nycgirl212,

The prompt specifically refers to the number line that's presented. On a number line, numbers increase from left to right. Thus, based on the given number line, A < B < C < D.

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so even though it doesn't explicitly state that the numbers increase from left to right, we are automatically supposed to assume that they do when we see a number line?


Please check the highlighted part below.

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Re: For real numbers, A, B, C, and D on the number line [#permalink] New post   15 Mar 2021, 07:09
idinuv wrote:


For real numbers, A, B, C, and D on the number line above, is |A| + |B| < |C| + |D| ?

(1) |B| < |C|
(2) The product of any two of the numbers A, B, C and D is between 0 and 1.


Hi Bunuel,

I am thinking of thr following numbers. lets go straight to test choice C.

so let the numbers be: A=-100000, B=0.00000000002, C=0.000000000003, D=0.000000000004.
statement A and B are satisfied here. B<C and product of any two numbers is betweem 0 and one. but A+B is greater than B+C bcs of module symbol.

another case:
A=0.0000000001, B=0.00000000002, C=0.0000000003, D=10000000000

In this case all conditions are also met, but A+B is less than C+D.

Am I missing anything here?

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Re: For real numbers, A, B, C, and D on the number line [#permalink] New post   15 Mar 2021, 07:55
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Foreheadson wrote:
idinuv wrote:


For real numbers, A, B, C, and D on the number line above, is |A| + |B| < |C| + |D| ?

(1) |B| < |C|
(2) The product of any two of the numbers A, B, C and D is between 0 and 1.


Hi Bunuel,

I am thinking of thr following numbers. lets go straight to test choice C.

so let the numbers be: A=-100000, B=0.00000000002, C=0.000000000003, D=0.000000000004.
statement A and B are satisfied here. B<C and product of any two numbers is betweem 0 and one. but A+B is greater than B+C bcs of module symbol.

another case:
A=0.0000000001, B=0.00000000002, C=0.0000000003, D=10000000000

In this case all conditions are also met, but A+B is less than C+D.

Am I missing anything here?

Posted from my mobile device


I think your reading of the second statement is not correct.

"The product of any two of the numbers A, B, C and D is between 0 and 1" means that the product of ANY two of the four numbers is between 0 and 1. So,

0 < AB < 1
0 < AC < 1
0 < AD < 1
0 < BC < 1
0 < BD < 1
0 < CD < 1

Now, if A = -100000, B=0.00000000002, C=0.000000000003, D=0.000000000004, then AB, AC and AD are negative not between 0 and 1.
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Re: For real numbers, A, B, C, and D on the number line [#permalink] New post   15 Mar 2021, 15:57
Bunuel wrote:
I think your reading of the second statement is not correct.

"The product of any two of the numbers A, B, C and D is between 0 and 1" means that the product of ANY two of the four numbers is between 0 and 1. So,

0 < AB < 1
0 < AC < 1
0 < AD < 1
0 < BC < 1
0 < BD < 1
0 < CD < 1

Now, if A = -100000, B=0.00000000002, C=0.000000000003, D=0.000000000004, then AB, AC and AD are negative not between 0 and 1.


Oh, I see the mistake now. I was confused with |x| symbols and assumed the same here for no reason.

Thanks a lot for a quick and comprehensive reply. I really appreciate it.

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Re: For real numbers, A, B, C, and D on the number line [#permalink]
15 Mar 2021, 15:57
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