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Question Stats:
33% (02:28) correct 67% (02:19) wrong based on 169 sessions
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If \(x\) and \(y\) are negative numbers, is \(x \lt y\)? (1) \(3x + 4 \lt 2y + 3\) (2) \(2x  3 \lt 3y  4\)
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Re M2718
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16 Sep 2014, 01:27
Official Solution: (1) \(3x + 4 \lt 2y + 3\). Rearrange: \(3x \lt 2y1\). \(x\) can be some very small number, for instance 100 and \(y\) some large enough number for instance 3 and the answer would be YES, \(x \lt y\) BUT if \(x=2\) and \(y=2.1\), then the answer would be NO, \(x \gt y\). Not sufficient. (2) \(2x  3 \lt 3y  4\). Rearrange: \(x \lt 1.5y\frac{1}{2}\). Rewrite as \(x \lt y+(0.5y\frac{1}{2}) = y + \text{negative}\). So, \(x \lt y\) (as \(y+\text{negative}\) is "more negative" than \(y\)). Sufficient. Answer: B
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Re: M2718
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14 Nov 2015, 23:53
Hello,
Is there any specific approach to solve this type of questions. I find to solve these questions very difficult.
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Re: M2718
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05 Mar 2016, 22:55
Bunuelwhy can we apply same logic, which you applied in option B, in option A. We have x<2/3 y 1/3. please tell me if I am missing anything. y is negative and we are subtracting and ve number.



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Re: M2718
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05 Mar 2016, 23:54
sudhirmadaan wrote: Bunuelwhy can we apply same logic, which you applied in option B, in option A. We have x<2/3 y 1/3. please tell me if I am missing anything. y is negative and we are subtracting and ve number. Hi, I'll try to answer this for you ..
We know x and y are negative numbers ..
and the answer to the Q lies in 1)knowing that if a ive number is multiplied with a bigger positive number/integer, The product will become even smaller.. meaning 3x<2x etc 2) role of < and > signs so if we want to check on the relative values of x and y, they have to brought in some way to same Coeff..
Can you spot the difference in two statements given keeping the above info in mind..
the statements are..
(1) 3x+4<2y+3 So, 3x+1<2y.. 3x<2y1 now x and y are negative integers.. so 3x< 2x.. but we cannot make a relation between 2x and 2y1.. Insuff
(2) 2x−3<3y−4 2x<3y1.. now 3x<2x.. so we can write 3x<2x<3y1.. or 3x<3y1 x<y1/3.. since both x and y are negative numbers and x is lesser than sum of y, a ive number, and another ive number.. x<y Suff
B
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Re: M2718
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13 Jun 2016, 08:35
(1) 3x+4<2y+3
3x+1<2y
1.5x+0.5<y
If 1.5 times x (i.e. a ve term) and 0.5 added to it is still less than y, then x should be less that y..isn't it?
what am I doing wrong?
(2) 2x−3<3y−4
2x3+4<3y
2x+1<3y
.66x+.33<y
I used the same logic as stated above....X being ve, .66 times x is also ve and 0.33 added to it is still less than y. Hence x should be less than y.
What is wrong with my approach in (1) ?



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GraphicallyOne easy way to do this is to see then the lines intersect by graphing them. In the attached, red is (1), blue is (2), and yellow is the prompt. You can see that (1) intersects with y>x when x, y negative, and thus you can pick options that allow the red line to be true, but can literally go on either side of y > x. For the blue line, that is also possible, but only when y and x are positive. Thus, we know that (2) is sufficient for all x,y negative. AlgebraicallyAnother easy way to do this is to see when the lines intersect, but by using algebra. (1) 2y > 3x+1 Pretend this is 2y=3x+1. Set this equal to what we are comparing it to, y=x, or, 2y=2x. 2y=3x+1=2x They intersect at x = 1. Either side of this, you will a Y or N answer when seeing if (1) is sufficient. (2) 3y > 2x+1 Pretend this is 3y=2x+1. Set this equal to what we are comparing it to, y=x, or, 3y=3x. 3y=2x+1=3x They intersect at x=1. If less than 1, you will always get the same answer when seeing if (2) is sufficient. Since x < 0 < 1, (2) is sufficient.
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Why are we using picking numbers for statement 1 and solving the second statement using inequalities and algebra? Both inequalities are of the same nature. Can someone please explain the sufficiency of the first statement using algebra?



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Bunuel wrote: Official Solution:
(1) \(3x + 4 \lt 2y + 3\). Rearrange: \(3x \lt 2y1\). \(x\) can be some very small number, for instance 100 and \(y\) some large enough number for instance 3 and the answer would be YES, \(x \lt y\) BUT if \(x=2\) and \(y=2.1\), then the answer would be NO, \(x \gt y\). Not sufficient. (2) \(2x  3 \lt 3y  4\). Rearrange: \(x \lt 1.5y\frac{1}{2}\). Rewrite as \(x \lt y+(0.5y\frac{1}{2}) = y + \text{negative}\). So, \(x \lt y\) (as \(y+\text{negative}\) is "more negative" than \(y\)). Sufficient.
Answer: B Question: Bunuelcase1) 3x < 2y 1 y=0 => 3x < 1 => x < 0.33 x=0 => y > 0.5 (but y < 0 given). Therefore, we can have both x < y (0.33 < 0.1) and x > y (0.33 > 0.4). Thus 1 NS case2) x < (3/2)y  1/2 y = 0 => x < 1/2 x = 0 => y > 1/3 (but y < 0 given). Therefore, we can have both x < y (0.5 < 0.1) and x > y (0.5 > 0.6). Thus 2 NS Case1+2) x < 0.33 & x < 0.5 => x < 0.33 & y < 0. Thus 1+2 NS => E What am i doing wrong? Can you please clarify Bunuel...



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Re: M2718
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03 Oct 2017, 12:57
manishtank1988 wrote: Bunuel wrote: Official Solution:
(1) \(3x + 4 \lt 2y + 3\). Rearrange: \(3x \lt 2y1\). \(x\) can be some very small number, for instance 100 and \(y\) some large enough number for instance 3 and the answer would be YES, \(x \lt y\) BUT if \(x=2\) and \(y=2.1\), then the answer would be NO, \(x \gt y\). Not sufficient. (2) \(2x  3 \lt 3y  4\). Rearrange: \(x \lt 1.5y\frac{1}{2}\). Rewrite as \(x \lt y+(0.5y\frac{1}{2}) = y + \text{negative}\). So, \(x \lt y\) (as \(y+\text{negative}\) is "more negative" than \(y\)). Sufficient.
Answer: B Question: Bunuelcase1) 3x < 2y 1 y=0 => 3x < 1 => x < 0.666 x=0 => y > 0.5 (but y < 0 given). Therefore, we can have both x < y (0.33 < 0.1) and x > y (0.33 > 0.4). Thus 1 NS case2) x < (3/2)y  1/2 y = 0 => x < 1/2 x = 0 => y > 1/3 (but y < 0 given). Therefore, we can have both x < y (0.33 < 0.1) and x > y (0.33 > 0.4). Thus 2 NS What am i doing wrong? Can you please clarify Bunuel... We are told that BOTH x and y are negative numbers. Can you give an example when that condition and \(2x  3 \lt 3y  4\) are satisfied and x is not less than y?
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Re: M2718
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04 Oct 2018, 00:07
[url]@chetan2u[/url] I noted that you mentioned that the 2nd witll equate to x<y1/3 but if I take x=2 and y=2 then x<y does not hold true
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Re: M2718
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04 Oct 2018, 00:59
Superg8 wrote: [url]@chetan2u[/url] I noted that you mentioned that the 2nd witll equate to x<y1/3 but if I take x=2 and y=2 then x<y does not hold true Hi.. If you take x=y,how can x be less than y.. X and y are NEGATIVE.. AND statement II tells us x<y1/3..( this is true we don't have to prove this) You can't take x=y=2.. If y =2, x<21/3=5/3 so x<5/3 it can be 3,10
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Re: M2718
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05 Oct 2018, 11:30
Bunuel wrote: Official Solution:
(1) \(3x + 4 \lt 2y + 3\). Rearrange: \(3x \lt 2y1\). \(x\) can be some very small number, for instance 100 and \(y\) some large enough number for instance 3 and the answer would be YES, \(x \lt y\) BUT if \(x=2\) and \(y=2.1\), then the answer would be NO, \(x \gt y\). Not sufficient. (2) \(2x  3 \lt 3y  4\). Rearrange: \(x \lt 1.5y\frac{1}{2}\). Rewrite as \(x \lt y+(0.5y\frac{1}{2}) = y + \text{negative}\). So, \(x \lt y\) (as \(y+\text{negative}\) is "more negative" than \(y\)). Sufficient.
Answer: B How did you conclude that (0.5y−0.5) is more negative. if y is positive then that may not be the case.



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Re: M2718
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05 Oct 2018, 11:33
sakuac wrote: Bunuel wrote: Official Solution:
(1) \(3x + 4 \lt 2y + 3\). Rearrange: \(3x \lt 2y1\). \(x\) can be some very small number, for instance 100 and \(y\) some large enough number for instance 3 and the answer would be YES, \(x \lt y\) BUT if \(x=2\) and \(y=2.1\), then the answer would be NO, \(x \gt y\). Not sufficient. (2) \(2x  3 \lt 3y  4\). Rearrange: \(x \lt 1.5y\frac{1}{2}\). Rewrite as \(x \lt y+(0.5y\frac{1}{2}) = y + \text{negative}\). So, \(x \lt y\) (as \(y+\text{negative}\) is "more negative" than \(y\)). Sufficient.
Answer: B How did you conclude that (0.5y−0.5) is more negative. if y is positive then that may not be the case. It is given that y is negative so 0.5y is negative and add another negative term to it 0.5, you get smaller negative number
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Keeping it SIMPLE  One trick to remember is that if u can prove \(x = y\) then u can answer YES or NO to a \(x<y\) or \(x>y\) question. Statement 1  \(3x + 4 < 2y + 3\) Rearrange : \(3x  2y < 1\) Simply try\(x = y = 2\) We see that above condition is satisfied. But answer to Is \(x < y?\) is NO. Try \(x = 3\) and \(y = 2\) We see that above condition is satisfied. Hence answer to Is\(x < y?\) is YES. Statement 2  \(2x  3 < 3y  4\) Rearrange : \(2x  3y < 1\) Before testing values. Just observe the equation. The (\(3y\)) part has will always be positive since y is negative. Also, we need to get a negative value so (\(2x)\) part has to be more than the \((3y)\) part. Is this possible if \(x\) and \(y\) are equal? No. Hence we have different values of \(x\) and \(y\). This should be sufficient. Bunuel chetan2u please confirm.



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Re: M2718
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15 May 2019, 01:50
I got this wrong for stupidly overlooking the "negative numbers" part in the stem. Took me a couple of minutes to realise why @bunuel's solution said y= negative haha
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Re: M2718
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05 Sep 2019, 04:49
What about this approach? Notice that whether our expressions for our inequalities sit to the left or to the right of zero, our relationship between the expressions and x an y should be preserved. so I am going to test negative values only. You should get a similar result if you test positive values only 3x+4<2y+ 3. First lets investigate what happens to x and y if we let these expressions be close together on the number line. Let 3x+4 = =2 and 2y+3=1. This gives x=2 and y=2 so x is not < y. now lets investigate when thet expressions are far apart: Let 3x+4=12 and 2y+3=2. Then x=16/3 and y = 5/2 and x is <y. Since we get a yes sometimes and a no sometimes NS.
For (2) we take the same approach
Let 2x3 = 2 and 3y4 = 1. then x= 1/2 and y = 8/3 and x is not <y Now let 2x3 = 12 and 3y4 = 2. Then x =9/2 and y = 2/3 and x is still not <y. Since we get a no answer when the gap between expressions is small and large we conclude suff
B










