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The Discreet Charm of the DS 02 Aug 2023, 12:11
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3,4,5,6,7
they are 5 consecutive nos with median and mean both same as 5 and I think the only pair with the given condition,
Am I wrong in this case ?



Bunuel
4. How many numbers of 5 consecutive positive integers is divisible by 4?

(1) The median of these numbers is odd --> the median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.

(2) The average (arithmetic mean) of these numbers is a prime number --> in any evenly spaced set the arithmetic mean (average) is equal to the median --> mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.

Answer: D.
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The Discreet Charm of the DS 02 Aug 2023, 13:59
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3,4,5,6,7
they are 5 consecutive nos with median and mean both same as 5 and I think the only pair with the given condition,
Am I wrong in this case ?



Bunuel
4. How many numbers of 5 consecutive positive integers is divisible by 4?

(1) The median of these numbers is odd --> the median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.

(2) The average (arithmetic mean) of these numbers is a prime number --> in any evenly spaced set the arithmetic mean (average) is equal to the median --> mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.

Answer: D.
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{5, 6, 7, 8, 9}
{9, 10, 11, 12, 13}
{11, 12, 13, 14, 15}
{15, 16, 17, 18, 19}
...
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The Discreet Charm of the DS 09 Jan 2024, 03:07
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Bunuel
2. Is xy<=1/2?

(1) x^2+y^2=1. Recall that \((x-y)^2\geq{0}\) (square of any number is more than or equal to zero) --> \(x^2-2xy+y^2\geq{0}\) --> since \(x^2+y^2=1\) then: \(1-2xy\geq{0}\) --> \(xy\leq{\frac{1}{2}}\). Sufficient.

(2) x^2-y^2=0 --> \(|x|=|y|\). Clearly insufficient.

Answer: A.
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Hello,
in point 1 if we take (x+y)^2 >= 0, does that work as well. As it will prove that the inequality does not hold true for the given statements.
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The Discreet Charm of the DS 09 Jan 2024, 03:30
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Bunuel
2. Is xy<=1/2?

(1) x^2+y^2=1. Recall that \((x-y)^2\geq{0}\) (square of any number is more than or equal to zero) --> \(x^2-2xy+y^2\geq{0}\) --> since \(x^2+y^2=1\) then: \(1-2xy\geq{0}\) --> \(xy\leq{\frac{1}{2}}\). Sufficient.

(2) x^2-y^2=0 --> \(|x|=|y|\). Clearly insufficient.

Answer: A.

Hello,
in point 1 if we take (x+y)^2 >= 0, does that work as well. As it will prove that the inequality does not hold true for the given statements.
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The key is that if you look at \((x+y)^2 \ge 0\), you'll find \(xy\geq{-\frac{1}{2} }\). But this isn't really helpful. It's true that both \(xy\geq{-\frac{1}{2} }\) (from \((x+y)^2 \ge 0\)) and \(xy \le \frac{1}{2}\) (from \((x-y)^2 \ge 0\)) are correct. So, this means \(-\frac{1}{2} \leq{xy} \leq{\frac{1}{2} }\). But only the method shown in the original solution gets us the answer we want. The other way just gives us an inequality that doesn't help much.

Hope it's clear.
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The Discreet Charm of the DS 09 Jan 2024, 03:49
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Bunuel
2. Is xy<=1/2?

(1) x^2+y^2=1. Recall that \((x-y)^2\geq{0}\) (square of any number is more than or equal to zero) --> \(x^2-2xy+y^2\geq{0}\) --> since \(x^2+y^2=1\) then: \(1-2xy\geq{0}\) --> \(xy\leq{\frac{1}{2}}\). Sufficient.

(2) x^2-y^2=0 --> \(|x|=|y|\). Clearly insufficient.

Answer: A.

Hello,
in point 1 if we take (x+y)^2 >= 0, does that work as well. As it will prove that the inequality does not hold true for the given statements.

The key is that if you look at \((x+y)^2 \ge 0\), you'll find \(xy\geq{-\frac{1}{2} }\). But this isn't really helpful. It's true that both \(xy\geq{-\frac{1}{2} }\) (from \((x+y)^2 \ge 0\)) and \(xy \le \frac{1}{2}\) (from \((x-y)^2 \ge 0\)) are correct. So, this means \(-\frac{1}{2} \leq{xy} \leq{\frac{1}{2} }\). But only the method shown in the original solution gets us the answer we want. The other way just gives us an inequality that doesn't help much.

Hope it's clear.
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Yes, thanks this helps. Just got confused between different approaches.
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The Discreet Charm of the DS 11 Jul 2024, 00:54
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Bunuel Hi.. shouldn't statement (2) be Bonnie and Clyde complete the painting of the car together at 10:30am

1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y?
(1) x^2+y^2<12
(2) Bonnie and Clyde complete the painting of the car at 10:30am

The question stem states they are working simultaneously and independently. Then the 2nd statement states Bonnie and Clyde complete the painting at 10:30am. At first glance, it appears that they both complete the work simultaneously and independently.
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The Discreet Charm of the DS 31 Aug 2024, 09:39
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­For Question 2

Can we use AM >= GM?

(x*x + y*y)/2 >= √(x*x)(y*y)
=> Statement 1 : 1/2 >= xy; which is required.
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The Discreet Charm of the DS 14 Jan 2025, 01:02
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Bunuel
2. Is \(xy \le \frac{1}{2}\)?

(1) \(x^2+y^2=1\).

Recall that \((x-y)^2 \ge 0\) (since the square of any number is always non-negative). If we expand this, we get \(x^2-2xy+y^2 \ge 0\). Given that \(x^2+y^2=1\), we can substitute this into our equation, resulting in \(1-2xy \ge 0\). From this, we can deduce that \(xy \le \frac{1}{2}\). Sufficient.

(2) \(x^2-y^2=0\).

If we rearrange this and take the square root of both sides, we get \(|x|=|y|\). However, this does not provide us with enough information to answer whether \(xy \le \frac{1}{2}\). Not sufficient.

Answer: A.
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If you solve it by (x+y)^2 >= 0, then answer is coming different.
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The Discreet Charm of the DS 14 Jan 2025, 12:53
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Bunuel
2. Is \(xy \le \frac{1}{2}\)?

(1) \(x^2+y^2=1\).

Recall that \((x-y)^2 \ge 0\) (since the square of any number is always non-negative). If we expand this, we get \(x^2-2xy+y^2 \ge 0\). Given that \(x^2+y^2=1\), we can substitute this into our equation, resulting in \(1-2xy \ge 0\). From this, we can deduce that \(xy \le \frac{1}{2}\). Sufficient.

(2) \(x^2-y^2=0\).

If we rearrange this and take the square root of both sides, we get \(|x|=|y|\). However, this does not provide us with enough information to answer whether \(xy \le \frac{1}{2}\). Not sufficient.

Answer: A.
If you solve it by (x+y)^2 >= 0, then answer is coming different.
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Please read this: https://gmatclub.com/forum/the-discreet ... l#p3332721
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The Discreet Charm of the DS 08 Feb 2025, 23:21
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Bunuel the algebra done here in the 2nd part is incorrect. It should be -6
Bunuel
5. What is the value of integer \(x\)?

(1) \(2x^2+9 < 9x\).

First, rearrange the inequality: \(2x^2 - 9x + 9 < 0\). Now, factor the quadratic: \((2x - 3)(x - 3) < 0\). The roots are \(\frac{3}{2}\) and 3. Since the inequality sign is "\( < \)", the solution lies between the roots: \(1.5 < x < 3\). The only integer in this range is 2, so \(x=2\). Sufficient.

(2) \(|x+10|=2x+8\).

The left-hand side (LHS) is an absolute value, which is always non-negative. Thus, the right-hand side (RHS) must also be non-negative: \(2x+8 \ge 0\), which gives us \(x \ge -4\). For this range, \(x+10\) is positive, so \(|x+10|=x+10\). Therefore, \(|x+10|=2x+8\) can be rewritten as \(x+10=2x+8\). Solving for \(x\) gives \(x=2\). Sufficient.

Answer: D.

Check this for more on solving inequalities like the one in the first statement:
https://gmatclub.com/forum/x2-4x-94661.html#p731476
https://gmatclub.com/forum/inequalities ... 91482.html
https://gmatclub.com/forum/everything-i ... me#p868863
https://gmatclub.com/forum/xy-plane-714 ... ic#p841486

Hope it helps.
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The Discreet Charm of the DS 09 Feb 2025, 01:28
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yrao1000
Bunuel the algebra done here in the 2nd part is incorrect. It should be -6
Bunuel
5. What is the value of integer \(x\)?

(1) \(2x^2+9 < 9x\).

First, rearrange the inequality: \(2x^2 - 9x + 9 < 0\). Now, factor the quadratic: \((2x - 3)(x - 3) < 0\). The roots are \(\frac{3}{2}\) and 3. Since the inequality sign is "\( < \)", the solution lies between the roots: \(1.5 < x < 3\). The only integer in this range is 2, so \(x=2\). Sufficient.

(2) \(|x+10|=2x+8\).

The left-hand side (LHS) is an absolute value, which is always non-negative. Thus, the right-hand side (RHS) must also be non-negative: \(2x+8 \ge 0\), which gives us \(x \ge -4\). For this range, \(x+10\) is positive, so \(|x+10|=x+10\). Therefore, \(|x+10|=2x+8\) can be rewritten as \(x+10=2x+8\). Solving for \(x\) gives \(x=2\). Sufficient.

Answer: D.

Check this for more on solving inequalities like the one in the first statement:
https://gmatclub.com/forum/x2-4x-94661.html#p731476
https://gmatclub.com/forum/inequalities ... 91482.html
https://gmatclub.com/forum/everything-i ... me#p868863
https://gmatclub.com/forum/xy-plane-714 ... ic#p841486

Hope it helps.
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x = -6 does not satisfy |x + 10| = 2x + 8. Substituting, we get 4 = -4, which is clearly incorrect. However, for x = 2, we obtain 12 = 12, which is correct. Thus, the solution is accurate. You would benefit from investing more time in reviewing the provided solutions carefully and double-checking your reasoning with simple number plugging.
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The Discreet Charm of the DS 09 Feb 2025, 02:41
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Thanks, appreciate the feedback
Bunuel
yrao1000
Bunuel the algebra done here in the 2nd part is incorrect. It should be -6
Bunuel
5. What is the value of integer \(x\)?

(1) \(2x^2+9 < 9x\).

First, rearrange the inequality: \(2x^2 - 9x + 9 < 0\). Now, factor the quadratic: \((2x - 3)(x - 3) < 0\). The roots are \(\frac{3}{2}\) and 3. Since the inequality sign is "\( < \)", the solution lies between the roots: \(1.5 < x < 3\). The only integer in this range is 2, so \(x=2\). Sufficient.

(2) \(|x+10|=2x+8\).

The left-hand side (LHS) is an absolute value, which is always non-negative. Thus, the right-hand side (RHS) must also be non-negative: \(2x+8 \ge 0\), which gives us \(x \ge -4\). For this range, \(x+10\) is positive, so \(|x+10|=x+10\). Therefore, \(|x+10|=2x+8\) can be rewritten as \(x+10=2x+8\). Solving for \(x\) gives \(x=2\). Sufficient.

Answer: D.

Check this for more on solving inequalities like the one in the first statement:
https://gmatclub.com/forum/x2-4x-94661.html#p731476
https://gmatclub.com/forum/inequalities ... 91482.html
https://gmatclub.com/forum/everything-i ... me#p868863
https://gmatclub.com/forum/xy-plane-714 ... ic#p841486

Hope it helps.

x = -6 does not satisfy |x + 10| = 2x + 8. Substituting, we get 4 = -4, which is clearly incorrect. However, for x = 2, we obtain 12 = 12, which is correct. Thus, the solution is accurate. You would benefit from investing more time in reviewing the provided solutions carefully and double-checking your reasoning with simple number plugging.
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