Math Revolution DS Expert - Ask Me Anything about GMAT DS

Hi, Archit3110 !

Please check the parts in red.

Regards,
Fabio.

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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.

Since \((x-y)(x+y)(x^2+y^2) = x^4-y^4\), the question asks if \(x\) and \(y\) have different parities.

By Condition 2), \(x\) and \(y\) must have different parities since \(x – y\) is an odd number.
Condition 2) is sufficient.

Condition 1)
Since we don’t know whether \(y\) is even or odd, condition 1) is not sufficient.

Therefore, B is the answer.
Answer: B

[GMAT math practice question]

(inequality) If \(x, y, z\) are positive integers, is \(xyz ≥ 64\)?

\(1) xy ≥ yz ≥ zx ≥ 16\)
\(2) x+y+z=64\)

Fabio
understood thanks..

x,y,z are +ve integers

#1
\(1) xy ≥ yz ≥ zx ≥ 16\)

x=y=z; then only we can have \(1) xy ≥ yz ≥ zx ≥ 16\)
so x=y=z=4 atleast
so \(xyz ≥ 64\); sufficient
#2

\(2) x+y+z=64\)
since x,y,z are + ve integers
so x= 62, y=z=1
so insufficient
IMO A

I am glad to know it, Archit3110!

I found your post by coincidence... if you want to "mention" me, so that I know I have a post of yours like this one, do not forget to use my "nickname" fskilnik , please!

("Fabio" is related to someone with no posts, who left this forum in 2009... I have checked that one minute ago.)

Regards and success in your studies!
Fabio.

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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.
The original condition \(a>b>c>d>0\) is equivalent to \(0 < \frac{1}{a} < \frac{1}{b} < \frac{1}{c} < \frac{1}{d}\). The question asks if \(d < 4\). This is equivalent to asking if \(\frac{1}{d} > \frac{1}{4}\).

By condition \(1, \frac{1}{d} > \frac{1}{4}\) since \(\frac{1}{c} < \frac{1}{d}\). So, \(d < 4\). Condition 1) is sufficient.

Condition 2)
Since \(0 < \frac{1}{a} < \frac{1}{b} < \frac{1}{c} < \frac{1}{d}\) and \((\frac{1}{a})+(\frac{1}{b})+(\frac{1}{c})+(\frac{1}{d})=1\), we have \(\frac{1}{a}<\frac{1}{d}\), \(\frac{1}{b}<\frac{1}{d}, \frac{1}{c}<\frac{1}{d}\) and \(\frac{1}{a} +\frac{1}{b} + \frac{1}{c} + \frac{1}{d} < \frac{1}{d} +\frac{1}{d} +\frac{1}{d} + \frac{1}{d} = \frac{4}{d}.\)
Therefore, \(1 < \frac{4}{d}\) and \(d < 4\).
Condition 2) is sufficient.

Therefore, D is the answer.
Answer: D

Note: This question is a CMT4(B) question: condition 1) is easy to work with and condition 2) is hard. For CMT4(B) questions, D is most likely to be the answer.

[GMAT math practice question]

(absolute value) \(3x+4y=?\)

\(1) 2|x|+3|y|=0\)
\(2) 3|x|+2|y|=0\)

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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 (\(x, y, z\)) and \(0\) equations, E is most likely to be the answer. So, we should consider each condition on its own first. As condition 1) includes 3 equations, we should consider it first.

Condition 1)
Since \(xy ≥ 16, yz ≥ 16\), and \(zx ≥ 16\), we have \((xyz)^2 ≥ 16^3\) or \(xyz ≥ 64\).
Condition 1) is sufficient.

Condition 2)
If \(x = 20, y = 20\) and \(z = 24\), then \(xyz = 9600 ≥ 64\) and the answer is ‘yes’.
If \(x = 1, y = 1\) and \(z = 62\), then \(xyz = 62 < 64\) and the answer is ‘no’.
Since condition 2) does not yield a unique answer, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A

This is a CMT(Common Mistake Type) 4(A) question. If a question is from one of the key question areas and C should be the answer, CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

[GMAT math practice question]

(function) Does \(f(x)=ax^4+bx^3+cx^2+dx+e\) have (\(x-2\)) as a factor?

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

IMO D

Modulus of any variable is a positive number.

if sum of 2 modulus variables is zero then it means the variable has a value zero

therefore 3x+4y = 0

Hence D

#1 :
\(1) 2|x|+3|y|=0\)
whenx=y=0 sufficient
#2
again at x=y=0. sufficient
IMO D

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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)
Since \(|x| ≥ 0, |y| ≥ 0\) and \(2|x| + 3|y| = 0,\) we have \(x = y = 0.\)
Therefore, \(3x + 4y = 0.\)
Condition 1) is sufficient.

Condition 2)
Since \(|x| ≥ 0, |y| ≥ 0\) and \(3|x| + 2|y| = 0\), we have \(x = y = 0.\)
Therefore, \(3x + 4y = 0\). Condition 2) is sufficient.

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.

Therefore, the answer is D.
Answer: D

=>

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.
The statement that \(f(x)=ax^4+bx^3+cx^2+dx+e\) has \((x-2)\) as a factor is equivalent to saying \(f(x) = (x-2)g(x)\) for some function \(g(x)\). This is equivalent to the requirement that \(f(2) = 0\).

Thus condition 1) is sufficient.

Condition 2)
If \(f(x) = (x-2)(x+2)(x-1)(x+1)\), then \(f(x)\) has \(x-2\) as a factor, The answer is ‘yes’.
If \(f(x) = x(x+2)(x+1)(x-1)\), then \(f(x)\) doesn’t have \(x-2\) as a factor. The answer is ‘no’.
Since condition 2) does not yield a unique answer, condition 2) is not sufficient.

Therefore, the answer is A.
Answer: A

[GMAT math practice question]

(geometry) What is the area of a right triangle?

1) The length of one side is \(3\)
2) The length of one side is \(4\)

[GMAT math practice question]

(inequality) Is \(3(a-b)>0\)?

\(1) a^3>b^3\)
\(2) a>b\)

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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 questions about triangles involve three variables, E is most likely to be the answer.

Conditions 1) & 2)
If the sides given are legs, then the area of the right triangle is \((\frac{1}{2})*3*4 = 6.\)
If one leg has length \(3\) and the hypotenuse has length \(4\), then, by Pythagoras’ theorem, the length of the other leg is \(\sqrt{(4^2 – 3^2)} = √7\), and the area of the triangle is \((\frac{1}{2})*3*√7 =\frac{(3√7)}{2}.\)
Thus, both conditions together are not sufficient, since they do not yield a unique solution.

Therefore, E is the answer.
Answer: E

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.