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[GMAT math practice question]

(algebra) \(x\) and \(y\) are integers. If \(y≠3\), is \(x=4\)?

\(1) x+y=7\)
\(2) x^2+y^2=25\)
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[GMAT math practice question]

(exponent) If \(x\) and \(y\) are positive integers, \(\frac{x}{y}\)=?

\(1) 2^{x+y}3^{xy}=72\)
\(2) 2^x3^y=12\)
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[GMAT math practice question]

(algebra) \(x\) and \(y\) are integers. If \(y≠3\), is \(x=4\)?

\(1) x+y=7\)
\(2) x^2+y^2=25\)

=>

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 \(2\) variables (\(x\) and \(y\)) and \(0\) equations, C 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):
Since \(y = 7 – x, x^2+y^2=25\) is equivalent to \(x^2+(7-x)^2=25\) or \(2x^2 – 14x + 24 = 0.\)
This factors as \(2(x^2 – 7x + 12) = 0\) or \(2(x-3)(x-4) = 0.\)
So, \(x = 3\) and \(y = 4\), or \(x = 4\) and \(y = 3\).
Since \(y ≠3\), we must have \(x = 3\) and \(x\) can’t be \(4\).
Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, both conditions are sufficient, when used together.

Since this question is an integer 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)
Since \(y = 7 – x ≠ 3, x\) can’t be \(4\). So, we have a unique answer, which is “no”.

Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, condition 1) is sufficient.

Condition 2)

If \(x = 4\) and \(y = -3\), then \(x^2+y2 = 25\), and the answer is “yes”.
If \(x = 3\) and \(y = 4,\) then \(x^2+y2 = 25\), 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

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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[GMAT math practice question]

If \(xyz≠0\), is \(x^3y^4z^5>0?\)

\(1) xz>0\)
\(2) xyz>0\)
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[GMAT math practice question]

If \(xyz≠0\), is \(x^3y^4z^5>0?\)

\(1) xz>0\)
\(2) xyz>0\)

Given: xyz ≠ 0

Target question: Is x³y⁴z⁵ > 0?
This is a great candidate for rephrasing the target question.
Since we know that x² must be POSITIVE, we can safely take the inequality x³y⁴z⁵ > 0 and divide both sides by x² to get: xy⁴z⁵ > 0
Similarly, since y⁴ is POSITIVE, we can safely divide both sides by y⁴ to get: xz⁵ > 0
Finally, since z⁴ is POSITIVE, we can safely divide both sides by z⁴ to get: xz > 0
REPHRASED target question: Is xz > 0?

Statement 1: xz > 0
PERFECT!
The answer to the REPHRASED target question is YES, xz IS greater than 0
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: xyz > 0
There are several values of x, y and z that satisfy statement 2. Here are two:
Case a: x = 1, y = 1 and z = 1. In this case, xz = (1)(1) = 1. So, the answer to the REPHRASED target question is YES, xz IS greater than 0
Case b: x = 1, y = -1 and z = -1. In this case, xz = (1)(-1) = -1. So, the answer to the REPHRASED target question is NO, xz is NOT greater than 0
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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How many DS questions will be there in the test?
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How many DS questions will be there in the test?

There is no actual count on how much DS questons one might face on the test day. It might range from 13-15. Totally depends.
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[GMAT math practice question]

(exponent) If \(x\) and \(y\) are positive integers, \(\frac{x}{y}\)=?

\(1) 2^{x+y}3^{xy}=72\)
\(2) 2^x3^y=12\)

=>

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) tells us that \(2^x3^y=12 = 2^23^1\) and \(x = 2, y = 1.\)
Thus \(\frac{x}{y} = \frac{2}{1} = 2\), and condition 2) is sufficient.

Condition 1)
\(2^{x+y}3^{xy}=72 = 2^33^2\) yields the equations \(x+y=3\) and \(xy=2\).
So, \(x = 1\) and \(y = 2, or x =2\) and \(y = 1\).
Thus, \(\frac{x}{y} = \frac{1}{2}\) or \(\frac{x}{y} = \frac{2}{1} = 2\).
Condition 1) is not sufficient since it does not yield a unique solution.

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

(function) If \(f(x)=ax^2+bx+c,\) where \(a, b\) and \(c\) are integers, is \(b=0\)?

\(1) f(49)=f(-49)=0\)
\(2) f(0)f(49)=0\)
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[GMAT math practice question]

If \(xyz≠0\), is \(x^3y^4z^5>0?\)

\(1) xz>0\)
\(2) xyz>0\)

=>

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.

Asking if \(x^3y^4z^5>0\) is equivalent to asking if \(xz > 0\) since we can ignore even exponents in inequalities.
Thus, condition 1) is sufficient.

Condition 2)
If \(x = 1, y = 1\) and \(z = 1\), then \(x^3y^4z^5>0\), and the answer is “yes”.
If \(x = 1, y = -1\) and \(z = -1\), then \(x^3y^4z^5<0\), and the answer is “no”.
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]

(function) The point (\(p,q\)) lies in which quadrant of the \(x-y\) plane?

1) (\(p+1, q\)) lies in the 2nd quadrant
2) (\(q-1, p\)) lies in the 4th quadrant
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[GMAT math practice question]

(function) If \(f(x)=ax^2+bx+c,\) where \(a, b\) and \(c\) are integers, is \(b=0\)?

\(1) f(49)=f(-49)=0\)
\(2) f(0)f(49)=0\)

=>

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.

In order to have \(b =0, f(x)\) must be symmetric about the y-axis. Thus, condition 1) is sufficient.
By the factor theorem, condition 1) tells us that \(f(x) = a(x-49)(x+49) = a(x^2 – 49^2) = ax^2 – 49^2a\) and \(b = 0\).
It is sufficient.

Condition 2)
If \(f(0)f(49) = 0\) then either \(f(0)= 0\) or \(f(49) = 0\).
Note that \(f(0) = c\).
If \(a = 1, b = 0\) and \(c = 0\), then \(f(0) = 0,\) so \(f(0)f(49) = 0\), and the answer is ‘yes’.
If \(a = 1, b = 1\) and \(c = 0\), then \(f(0) = 0\), so \(f(0)f(49) = 0\), 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]

(function) The point (\(p,q\)) lies in which quadrant of the \(x-y\) plane?

1) (\(p+1, q\)) lies in the 2nd quadrant
2) (\(q-1, p\)) lies in the 4th quadrant

=>

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) tells us that \(p + 1 < 0\) and \(q > 0\), which is equivalent to \(p < -1 < 0\) and \(q > 0\). Thus, (\(p, q\)) is in the 2nd quadrant.
Condition 1) is sufficient.

Condition 2) tells us that \(q - 1 > 0\) and \(p < 0\), which is equivalent to \(p < 0\) and \(q > 1 > 0\). Thus, (\(p, q\)) is in the 2nd quadrant.
Condition 2) is sufficient.

Therefore, D is the answer.
Answer: D

FYI, Tip 1) of the VA method states that D is most likely to be the answer if conditions 1) and 2) provide the same information.
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[GMAT math practice question]

(number properties) Is \(\sqrt{15n}\) an integer?

1) \(n\) is a multiple of \(3\)
2) \(n\) is the square of an integer
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[GMAT math practice question]

(absolute value) \(xy=?\)

\(1) (x-2)^2 = -|y-3|\)
\(2) |x-2| + \sqrt{y-3} =0\)
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(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
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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
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