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Nsp10
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02 Aug 2023, 11:11
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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02 Aug 2023, 12:59
Nsp10
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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09 Jan 2024, 02:07
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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09 Jan 2024, 02:30
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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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09 Jan 2024, 02:49
Bunuel
Roop14S
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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10 Jul 2024, 23:54
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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31 Aug 2024, 08:39
­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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14 Jan 2025, 00:02
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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14 Jan 2025, 11:53
KumarRishav
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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08 Feb 2025, 22:21
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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09 Feb 2025, 00:28
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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09 Feb 2025, 01:41
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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21 Nov 2025, 19:23
Hey
Could anybody provide the solution for Q2?
Is xy<=1/2?
(1) x^2+y^2=1
(2) x^2-y^2=0
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21 Nov 2025, 23:33
sanyagoyal812
Hey
Could anybody provide the solution for Q2?
Is xy<=1/2?
(1) x^2+y^2=1
(2) x^2-y^2=0
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Please, read the whole thread: https://gmatclub.com/forum/the-discreet ... l#p1039634

P.S. Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:

  • Word Problems
  • Work Problems
  • Distance Problems
  • Mixture Problems
  • Percent and Interest Problems
  • Overlapping Sets Problems
  • Statistics Problems
  • Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check GMAT Syllabus for Focus Edition

You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.

So, you can ignore this question.

Hope it helps.­

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