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Re: The Discreet Charm of the DS
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10 Jun 2017, 12:47
KARISHMA315 wrote: Bunuel wrote: 5. What is the value of integer x?
(1) 2x^2+9<9x > factor qudratics: \((x\frac{3}{2})(x3)<0\) > roots are \(\frac{3}{2}\) and 3 > "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) > since there only integer in this range is 2 then \(x=2\). Sufficient.
(2) x+10=2x+8 > LHS is an absolute value, which is always non negative, hence RHS must also be nonnegative: \(2x+8\geq{0}\) > \(x\geq{4}\), for this range \(x+10\) is positive hence \(x+10=x+10\) > \(x+10=2x+8\) > \(x=2\). Sufficient.
Answer: D.
Hi Bunuel, a small doubt instead of \(2x+8\geq{0}\) > \(x\geq{4}\) this logic if we take x+10 to be both positive and negative we get 2 vale by solving equation x=2 and x=6. however by putting values back in eqtn we can see that only for x=2 equation is satisfying. Is this correct approach? Yes, that's also a correct way of solving. Good thing you did is that you did not forget to test both values after you got them.
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Re: The Discreet Charm of the DS
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06 Sep 2017, 00:43
Bunuel wrote: 11. If x and y are integers, is x a positive integer?
(1) x*y is a prime number > since only positive numbers can be primes, then: x*y=positive > x=positive. Sufficient
(2) x*y is nonnegative integer. Notice that we are told that x*y is nonnegative, not that it's positive, so x can be positive as well as zero. Not sufficient.
Answer: A. Hi Bunuel, I understood that x*y=positive; but if y<=0 then y=y and in this case x can be negative. Am I missing something? Thanks.



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Re: The Discreet Charm of the DS
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06 Sep 2017, 00:49
Yashodhan123 wrote: Bunuel wrote: 11. If x and y are integers, is x a positive integer?
(1) x*y is a prime number > since only positive numbers can be primes, then: x*y=positive > x=positive. Sufficient
(2) x*y is nonnegative integer. Notice that we are told that x*y is nonnegative, not that it's positive, so x can be positive as well as zero. Not sufficient.
Answer: A. Hi Bunuel, I understood that x*y=positive; but if y<=0 then y=y and in this case x can be negative. Am I missing something? Thanks. Yes. An absolute value of a number cannot be negative: a is positive or 0, no matter whether a itself is negative or not. If y <= 0, then y = y, yes, but even in this case y = negative = positive.
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Re: The Discreet Charm of the DS
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08 Jun 2018, 15:04
12. If 6a=3b=7c, what is the value of a+b+c? (1) ac=6b (2) 5b=8a+4c
Let 6a=3b=7c=K (where K is a constant) Thus, a=K/6 b= K/3 c=K/7
Therefore, we need to find the value of K/6 + K/3 + K/7 = 9K/14. OR if we find the value of just K it will give us the answer.
Stmt 1: ac=6b \(K/6*K/7 = 6*K/3;\)
\(K^2/422K=0\)
\(K(K/422)=0\)
So, K = 0 or K= 84. NOT Sufficient.
Stmt 2: 5b=8a+4c 5K/3=8K/6 +4K/7 40K/215K/3=0 5K/3=0 K=0 SUFFICIENT
Ans. B



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Re: The Discreet Charm of the DS
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09 Jun 2018, 01:36
Bunuel wrote: 2. Is xy<=1/2?
(1) x^2+y^2=1. Recall that \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) > \(x^22xy+y^2\geq{0}\) > since \(x^2+y^2=1\) then: \(12xy\geq{0}\) > \(xy\leq{\frac{1}{2}}\). Sufficient.
(2) x^2y^2=0 > \(x=y\). Clearly insufficient.
Answer: A. Dear sir, Why are not you taking (x+y)^2 >=0 i.e. x^2+y^2+2xy>=0?



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Re: The Discreet Charm of the DS
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10 Jun 2018, 00:16
BelalHossain046 wrote: Bunuel wrote: 2. Is xy<=1/2?
(1) x^2+y^2=1. Recall that \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) > \(x^22xy+y^2\geq{0}\) > since \(x^2+y^2=1\) then: \(12xy\geq{0}\) > \(xy\leq{\frac{1}{2}}\). Sufficient.
(2) x^2y^2=0 > \(x=y\). Clearly insufficient.
Answer: A. Dear sir, Why are not you taking (x+y)^2 >=0 i.e. x^2+y^2+2xy>=0? The point is that if you consider \((x+y)^2 \ge 0\), you'll get \(xy\geq{\frac{1}{2}}\), which is not helpful at all. Actually since both \(xy\geq{\frac{1}{2}}\) (from \((x+y)^2 \ge 0\)) and \(xy \le \frac{1}{2}\) (from \((xy)^2 \ge 0\)) are true, then we get that \(\frac{1}{2} \leq{xy} \leq{\frac{1}{2}}\). But only the approach given in the solution above gives you the answer we are looking for, while another one gives you an inequality which is not helpful. Hope it's clear.
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Re: The Discreet Charm of the DS
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18 Jul 2018, 05:26
Bunuel wrote: 11. If x and y are integers, is x a positive integer?
(1) x*y is a prime number > since only positive numbers can be primes, then: x*y=positive > x=positive. Sufficient
(2) x*y is nonnegative integer. Notice that we are told that x*y is nonnegative, not that it's positive, so x can be positive as well as zero. Not sufficient.
Answer: A. Shouldn't we consider zero as positive ? 2 is not sufficient as y can be zero so x can be negative or positive



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Re: The Discreet Charm of the DS
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18 Jul 2018, 05:29
teaserbae wrote: Bunuel wrote: 11. If x and y are integers, is x a positive integer?
(1) x*y is a prime number > since only positive numbers can be primes, then: x*y=positive > x=positive. Sufficient
(2) x*y is nonnegative integer. Notice that we are told that x*y is nonnegative, not that it's positive, so x can be positive as well as zero. Not sufficient.
Answer: A. Shouldn't we consider zero as positive ? 2 is not sufficient as y can be zero so x can be negative or positive ZERO. 1. 0 is an integer. 2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even. 3. 0 is neither positive nor negative integer (the only one of this kind).4. 0 is divisible by EVERY integer except 0 itself.
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Re: The Discreet Charm of the DS
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03 Oct 2018, 07:33
Bunuel wrote: 5. What is the value of integer x?
(2) x+10=2x+8 > LHS is an absolute value, which is always non negative, hence RHS must also be nonnegative: \(2x+8\geq{0}\) > \(x\geq{4}\), for this range \(x+10\) is positive hence \(x+10=x+10\) > \(x+10=2x+8\) > \(x=2\). Sufficient.
Hi, until now whenever I saw modulus, I was told to begin by removing the mod and considering both signs. Therefore, I approached it as x+10=2x+8 and (x+10)=2x+8. Just wanted to understand when to use this method and when to use the take it as an absolute value method? Thanks!



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Re: The Discreet Charm of the DS
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10 Apr 2019, 01:20
With respect to question no 8 above, (If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?), why would the answer not just be A? Since we can determine the least possible value of both a and b, and thereby deduce that x would be greater than 0.77.



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Re: The Discreet Charm of the DS
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10 Apr 2019, 01:28
tejasvis7 wrote: With respect to question no 8 above, (If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?), why would the answer not just be A? Since we can determine the least possible value of both a and b, and thereby deduce that x would be greater than 0.77. Solution is here: https://gmatclub.com/forum/thediscreet ... l#p1039662Here is what it says about (1): a+b>14 > the least value of a is 6 (6+9=15>14), so in this case x=0.69d<0.77(7) but a=7 and b=9 is also possible, and in this case x=0.79d>0.77(7). Not sufficient.
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The Discreet Charm of the DS
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13 Apr 2019, 06:41
Bunuel wrote: 12. If 6a=3b=7c, what is the value of a+b+c?
Given: \(6a=3b=7c\) > least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) > \(a=7x\), \(b=14x\) and \(c=6x\).
(1) ac=6b > \(7x*6x=6*14x\) > \(x^2=2x\) > \(x=0\) or \(x=2\). Not sufficient.
(2) 5b=8a+4c > \(5*14x=8*7x+4*14x\) > \(70x=80x\) > \(10x=0\) > \(x=0\) > \(a=b=c=0\) > \(a+b+c=0\). Sufficient.
Answer: B. Hi BunuelCould you please change 4*14x (as highlighted) to 4*6x because c=6x. (small typo)
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Re: The Discreet Charm of the DS
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13 Apr 2019, 10:02
dhritidutta wrote: Bunuel wrote: 12. If 6a=3b=7c, what is the value of a+b+c?
Given: \(6a=3b=7c\) > least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) > \(a=7x\), \(b=14x\) and \(c=6x\).
(1) ac=6b > \(7x*6x=6*14x\) > \(x^2=2x\) > \(x=0\) or \(x=2\). Not sufficient.
(2) 5b=8a+4c > \(5*14x=8*7x+4*14x\) > \(70x=80x\) > \(10x=0\) > \(x=0\) > \(a=b=c=0\) > \(a+b+c=0\). Sufficient.
Answer: B. Hi BunuelCould you please change 4*14x (as highlighted) to 4*6x because c=6x. (small typo) _______________________ Edited. Thank you.
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The Discreet Charm of the DS
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01 May 2019, 09:02
Bunuel wrote: piyushksharma wrote: Bunuel wrote: 5. What is the value of integer x?
(1) 2x^2+9<9x > factor qudratics: \((x\frac{3}{2})(x3)<0\) > roots are \(\frac{3}{2}\) and 3 > "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) > since there only integer in this range is 2 then \(x=2\). Sufficient.
(2) x+10=2x+8 > LHS is an absolute value, which is always non negative, hence RHS must also be nonnegative: \(2x+8\geq{0}\) > \(x\geq{4}\), for this range \(x+10\) is positive hence \(x+10=x+10\) > \(x+10=2x+8\) > \(x=2\). Sufficient.
Answer: D.
Hope it helps. Hi bunuel, Isn't x+10=2x+8 be written as Either x+10=2x+8 or x+10=(2x+8) ? and then this should be solved? Please help on this one. We goth that x is more than or equal to 4. Now, for this range x+10>0 so x+10 expands only as x+10 (x+10=x+10). Hi Bunuel, A doubt on the above note. You have mentioned that absolute values are nonnegative and hence you have considered 2x+8>=0 But in some questions we solve by taking 2 cases when modulus is present in the question. Using this casebased approach, in the above given equation, case 1 would be x+10=2x+8 and case 2 would be x10=(2x+8) So, following are my questions: 1. Why have we ignored 2x+8<0 in your solution? 2. When do we use the 2 case approach and when do we use the 'absolute values are nonnegative approach? This is a bit confusing. Please assist on this.




The Discreet Charm of the DS
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