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Re: What is the value of y?
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06 May 2016, 21:47
Bunuel wrote: What is the value of y?
(1) \(3x^24=y2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als cannot be negative. Not sufficient.
(2) \(3  y = 11\):
\(y<3\) > \(3y=11\) > \(y=8\); \(y\geq{3}\) > \(3+y=11\) > \(y=14\).
Two values for \(y\). Not sufficient.
(1)+(2) Since from (1) \(y\geq{2}\), then from (2) \(y=14\). Sufficient.
Answer: C.
Hope it's clear. Why is Y>=0? Can't we write 'Y>0'? If LHS is positive and RHS must also be positive then, RHS has to be greater than 0. As 0 is neither positive nor negative, how can we write Y>=0? some value= some value. Since LHS is an absolute value, RHS has to be positive as well. Why is RHS>=0 and not RHS>0?



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Re: What is the value of y?
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06 May 2016, 22:12
shekhar4847 wrote: Bunuel wrote: What is the value of y?
(1) \(3x^24=y2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als cannot be negative. Not sufficient.
(2) \(3  y = 11\):
\(y<3\) > \(3y=11\) > \(y=8\); \(y\geq{3}\) > \(3+y=11\) > \(y=14\).
Two values for \(y\). Not sufficient.
(1)+(2) Since from (1) \(y\geq{2}\), then from (2) \(y=14\). Sufficient.
Answer: C.
Hope it's clear. what if y is between 2 and 3? In that case we will have two values of y. Could you please clarify? Hi shekhar4847, I gives you \(y\geq{2}\)... II gives you two cases.. a) \(y<3\) > \(3y=11\) > \(y=8\); this means if y<3, it can have ONLY 8 as the value, so when we combine this with \(y\geq{2}\),which does not fit between 2 and 3.. hence WRONG b)\(y\geq{3}\) > \(3+y=11\) > \(y=14\) this means if \(y\geq{3}\), it can have ONLY 14 as the value, so when we combine this with \(y\geq{2}\), 14 fits in both the ranges.. hence CORRECT
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Re: What is the value of y?
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06 May 2016, 22:15
bimalr9 wrote: Bunuel wrote: What is the value of y?
(1) \(3x^24=y2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als cannot be negative. Not sufficient.
(2) \(3  y = 11\):
\(y<3\) > \(3y=11\) > \(y=8\); \(y\geq{3}\) > \(3+y=11\) > \(y=14\).
Two values for \(y\). Not sufficient.
(1)+(2) Since from (1) \(y\geq{2}\), then from (2) \(y=14\). Sufficient.
Answer: C.
Hope it's clear. Why is Y>=0? Can't we write 'Y>0'? If LHS is positive and RHS must also be positive then, RHS has to be greater than 0. As 0 is neither positive nor negative, how can we write Y>=0? some value= some value. Since LHS is an absolute value, RHS has to be positive as well. Why is RHS>=0 and not RHS>0? Hi, an ABSOLUTE value is NOT necessarily positive, it is NONNEGATIVE, so it can include both 0 and +ive value.. if in statement I x is 2 or 2, LHS will be 0 and hence y=2..
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Re: What is the value of y?
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07 May 2016, 01:49
dvinoth86 wrote: What is the value of y? (1) 3x^2 – 4 = y – 2 (2) 3 – y = 11 Y = ? stmt:1  it has two variables X AND Y. we need value of Y. insuff. stmt:2  Y=8 or Y=11 insuff. combined:  frm stmt1 3x^2 – 4 = y – 2 X^2 is positive. value of mod is positive. so left hand side is POSITIVE. frm stmt2 Y=8 or Y=11 so we can only pick 11 as the value to keep right hand side POSITIVE. Option C is right.
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What is the value of y?
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14 Aug 2016, 01:44
Hi Bunuel, Please correct me if i am wrong. i). 3x^2  4 =y2 3x^2 = y+ 10 OR 3x^2 = y +14 equating the 2 values of 3x^2 Cant we equate these 2 values?y=10 = y +14 y=2 ii). gives 2 values of y 8 & 14 therefore, answer is A



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Re: What is the value of y?
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14 Aug 2016, 02:39
shrashtisinghal wrote: Hi Bunuel, Please correct me if i am wrong. i). 3x^2  4 =y2 3x^2 = y+ 10 OR 3x^2 = y +14 equating the 2 values of 3x^2 Cant we equate these 2 values?y=10 = y +14 y=2 ii). gives 2 values of y 8 & 14 therefore, answer is A How do you know that we have 3x^2  12 =y2 or 12  3x^2 =y2 Since, we donot know which of the above equations is valid. We cannot deduce a single value. As per Modulus property, Mod(x) = x if x>0 = x if x<0. Hence A is not sufficient. In B also, we are getting two values of y, so not sufficient. But, when you combine the statements, you will see the equation is satisfied only for y =14. hence, C.
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Re: What is the value of y?
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18 Oct 2016, 00:52
dvinoth86 wrote: What is the value of y?
(1) 3x^2 – 4 = y – 2 (2) 3 – y = 11 FROM 1 (y2)/3 >= 0 , thus y>= 2 ... ... insuff from 2 3y = 11 or 3y = 11 , thus Y could be either 8 or 14.. insuff both together since from 1 y>= 2 and from 2 y = 8 or 14 then y = 4 ( common domain) ...C



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Re: What is the value of y?
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14 Aug 2017, 21:32
Bunuel wrote: What is the value of y?
(1) \(3x^24=y2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als cannot be negative. Not sufficient.
(2) \(3  y = 11\):
\(y<3\) > \(3y=11\) > \(y=8\); \(y\geq{3}\) > \(3+y=11\) > \(y=14\).
Two values for \(y\). Not sufficient.
(1)+(2) Since from (1) \(y\geq{2}\), then from (2) \(y=14\). Sufficient.
Answer: C.
Hope it's clear. Brilliant approach! I'd never thought it could be this easy.



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Re: What is the value of y?
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01 Mar 2019, 14:08
Bunuel wrote: What is the value of y?
(1) \(3x^24=y2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als cannot be negative. Not sufficient.
(2) \(3  y = 11\):
\(y<3\) > \(3y=11\) > \(y=8\); \(y\geq{3}\) > \(3+y=11\) > \(y=14\).
Two values for \(y\). Not sufficient.
(1)+(2) Since from (1) \(y\geq{2}\), then from (2) \(y=14\). Sufficient.
Answer: C.
Hope it's clear. Since we have the absolute value of (x^24), couldn't it be considered as the absolute value of (x+2)(x2)??? This way, you'd have two possible roots: 2 and 2. Then, since it's an absolute value, the LHS of the equation would give you always 6, making y=8. What did I miss?




Re: What is the value of y?
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