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Sandeepanisha
(integer) [x] is the greatest integer less than or equal to x, is x-[x]>0?

1) x is positive
2) x is not integer

[x] is the integer part of x.
If x = n + h where n is an integer and 0 ≤ h < 1, then [x] = n.
For example, if x = 1.2 = 1+0.2, then [x] = 1.

Thus x - [x] = n + h - n = h is the decimal part of x.
x - [x] > 0 means h > 0 and x is not an integer.
Thus condition 2) is sufficient.

Condition 1)
If x = 1.2, then x - [x] = 1.2 - 1 = 0.2 > 0 and the answer is 'yes'.
If x = 1, then x - [x] = 1 - 1 = 0 and answer is 'no'.
Since we don't have a unique answer, condition 1) is not sufficient.

Therefore, the answer is B.
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(ratio) All of the households used a total of x watts for 24 hours and one household used y watts per hour. What is the total number of all the households, in terms of x and y?

A. x/y B. 24x/y C. x/24y D. 24xy E. 24/xy

Let n be the number of households.
Then we have 24y*n = x, which is the total waats they use.
Thus n = x/(24y).

Therefore, then answer is C.
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[GMAT math practice question]

(absolute values) If \(|2x|>|3y|\), is \(x >y\)?

\(1) x>0\)
\(2) y>0\)
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MathRevolution
[GMAT math practice question]

(number properties) If \(x, y\) are integers, is \((x-y)(x+y)(x^2+y^2)\) an odd number?

1) \(x\) is an odd number
2) \(x-y\) is an odd number
fskilnik,
Sir, is it possible to solve this question by rotations style?
Thanks__

Hi, AsadAbu .

Thank you for your interest in my opinions, but (by contract) I must limit myself to post questions/solutions (and some brief explanations about our solutions, when asked).

I am not here to judge or compare GMATH´s method to other experts/companies approaches.

Thank you for your understanding.

Regards and success in your studies,
Fabio.
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MathRevolution
[GMAT math practice question]

(number properties) If \(x, y\) are integers, is \((x-y)(x+y)(x^2+y^2)\) an odd number?

1) \(x\) is an odd number
2) \(x-y\) is an odd number
fskilnik,
Sir, is it possible to solve this question by rotations style?
Thanks__

Hi, AsadAbu .

Thank you for your interest in my opinions, but (by contract) I must limit myself to post questions/solutions (and some brief explanations about our solutions, when asked).

I am not here to judge or compare GMATH´s method to other experts/companies approaches.

Thank you for your understanding.

Regards and success in your studies,
Fabio.
Yep, I get it.
Thanks__
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[GMAT math practice question]

(absolute value) Is \(x<y<z\) ?

\(1) |x+1|<y<z+1\)
\(2) |x-1|<y<z-1\)

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have \(3\) variables (\(x, y\) and \(z\)) and \(0\) equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
By condition 1), \(x < y\) since \(x < x + 1 ≤ | x + 1 | < y\).
By condition 2), \(y < z\) since \(y < z – 1 < z.\)
Therefore, \(x < y < z\).
Thus, both conditions 1) & 2) together are sufficient.

Since this question is an absolute value question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If \(x = 1, y = 3\), and \(z = 5,\) then the answer is ‘yes’.
If \(x = 1, y = 3,\) and \(z = 2.9\), then the answer is ‘no’ since z < y.
Thus, condition 1) is not sufficient, since it does not yield a unique solution.

Condition 2)
If \(x = 1, y = 3,\) and \(z = 5\), then the answer is ‘yes’.
If \(x = 3.1, y = 3,\) and \(z = 5\), then the answer is ‘no’ since \(x > y\).
Thus, condition 2) is not sufficient, since it does not yield a unique solution.

Therefore, C is the answer.
Answer: C

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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[GMAT math practice question]

\(m\) and \(n\) are positive integers. Are \(m\) and \(n\) consecutive integers?

\(1) m^2+n^2 = 5\)
\(2) m-n = 1\)
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[GMAT math practice question]

(absolute values) If \(|2x|>|3y|\), is \(x >y\)?

\(1) x>0\)
\(2) y>0\)

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

From condition 1), we have \(3x > 2x = |2x| > |3y| ≥ 3y\) since \(|x| = x.\) So, \(x > y\) and the answer is ‘yes’.
Thus, condition 1) is sufficient.

Condition 2)
If \(x = 10\), and \(y = 1\), then \(x > y\) and the answer is ‘yes’.
If \(x = -10\), and \(y = 1,\) then \(x < y\) and the answer is ‘no’.
Thus, condition 2) is not sufficient, since it does not yield a unique solution.

Therefore, A is the answer.
Answer: A
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[GMAT math practice question]

(number properties) \(m\) and \(n\) are integers. Is \(m + n\) an odd number?

\(1) m – n = 2\)
\(2) m^2n^2 = 225\)
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[GMAT math practice question]

\(m\) and \(n\) are positive integers. Are \(m\) and \(n\) consecutive integers?

\(1) m^2+n^2 = 5\)
\(2) m-n = 1\)

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Condition 1)
The only positive integers satisfying \(m^2+n^2 = 5\) are \(m = 1\) and \(n = 2\), or \(m=2\) and \(n = 1.\)
These are consecutive integers, so condition 1) is sufficient.

Condition 2)
Since the difference between \(m\) and \(n\) is \(1\), they are consecutive integers. Condition 2) is sufficient.

Therefore, D is the answer.
Answer: D
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[GMAT math practice question]

(inequality) Is \(x < 0\)?

\(1) x^3 + 1 < 0\)
\(2) x^3 + x + 1 < 0\)
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[GMAT math practice question]

(number properties) \(m\) and \(n\) are integers. Is \(m + n\) an odd number?

\(1) m – n = 2\)
\(2) m^2n^2 = 225\)

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Condition 2)
Since \(m^2n^2 = 225\) is an odd number, both \(m\) and \(n\) must be odd numbers, and \(m + n\) is an even number.
Thus, the answer is ‘no’, and condition 2) is sufficient by CMT (Common Mistake Type) 1.

Condition 1)
Since \(m – n = 2\), both \(m\) and \(n\) are odd numbers or even numbers.
Since \(m\) and \(n\) have the same parity, \(m + n\) is an even number.
Thus, the answer is ‘no’, and condition 1) is sufficient by CMT (Common Mistake Type) 1.

Therefore, D is the answer.
Answer: D
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MathRevolution
[GMAT math practice question]

(inequality) Is \(x < 0\)?

\(1) x^3 + 1 < 0\)
\(2) x^3 + x + 1 < 0\)

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (\(x\)) and \(0\) equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
\(x^3 + 1 < 0\)
\(=> (x+1)(x^2-x+1) < 0\)
\(=> x + 1 < 0\) since \(x^2-x+1 > 0\)
\(=> x < -1 < 0\)
Thus, condition 1) is sufficient, and the answer is ‘yes’.

Condition 2)
\(x^3 + x + 1 < 0\)
\(=> x^3 + x < -1\)
\(=> x(x^2 + 1) < -1\)
\(=> x < \frac{-1}{(x^2 + 1)}\) since \(x^2 + 1 > 0\)
\(=> x <\frac{-1}{(x^2 + 1)} < 0\) since \(x^2 + 1 > 0\)
Thus, condition 2) is sufficient, and the answer is ‘yes’.

Therefore, D is the answer.
Answer: D
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[GMAT math practice question]

(statistics) If the median and average (arithmetic mean) of a set of 4 different numbers are both 10, what is the smallest number?

1) The range of the 4 numbers is 10
2) The sum of the smallest and the largest numbers is 20
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[GMAT math practice question]

(statistics) \(x, y\) and \(z\) are different integers. Is their average equal to their median?

1) Their range is \(11\).
2) Their median is \(11\).
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[GMAT math practice question]

(statistics) If the median and average (arithmetic mean) of a set of 4 different numbers are both 10, what is the smallest number?

1) The range of the 4 numbers is 10
2) The sum of the smallest and the largest numbers is 20

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Let \(a, b, c\) and \(d\) be the \(4\) numbers, and suppose \(a < b < c < d.\)

Then \(\frac{( a + b + c + d )}{4} = 10\) and \(\frac{( b + c )}{2} = 10.\)

Since \(b + c = 20\) and \(a + b + c + d = 40\), we must have \(a + d = 20.\)

Condition 1)

Since \(d – a = 10\) by condition 1), we can figure out the values of \(a\) and \(d\). Thus, condition 1) is sufficient.

Condition 2)

\(a + d = 20\) can be deduced from the original condition as shown above.

So, condition 2) provides no additional information.

If \(a = 1, b = 9, c = 11\) and \(d = 19\), then the smallest number is \(1\).

If \(a = 2, b = 9, c =11\) and \(d = 18\), then the smallest number is \(2\).

Condition 2) is not sufficient since it does not yield a unique answer.

Therefore, A is the answer.
Answer: A
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[GMAT math practice question]

(number properties) If \(m\) and \(n\) are positive integers, is \(m^2-n^2\) divisible by \(4\)?

1) \(m^2+n^2\) has remainder \(2\) when it is divided by \(4\)

2) \(m*n\) is an odd integer
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