Math Revolution DS Expert - Ask Me Anything about GMAT DS

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. A standard deviation is the square root of an average of squares minus the square of the average. Thus, C is the answer.

The following is more detailed reasoning.

Assume \(m\) is the average of x1, x2, …, xn and p is the average of x1^2, x2^2, …, xn^2.
Then we have (x1 + x2 + … + xn) / n = m or (x1 + x2 + … + xn) = mn, and we have
(x1^2+x2^2+⋯+xn^2)/n= p or (x1^2+x2^2+⋯+xn^2)= pn.

Then we have
(x1-m)^2+(x2-m)^2+⋯+(xn-m)^2/n
= x1^2 + x2^2 +⋯+ xn^2- 2m(x1 + x2 + ⋯ + xn) + nm^2/n
= np - 2m∙nm + nm^2/n = p - m^2
The standard deviation is \(\sqrt{p-m^2}.\)
Since we have m = 1 and p = 5 from both conditions 1) and 2), we have the standard deviation \(\sqrt{5-1^2}=\sqrt{4}=2.\)
Since both conditions together yield a unique solution, they are sufficient.

Therefore, C is the answer.
Answer: C
_________________

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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

Since we have \(2\) variables (\(x\) and \(y\)) and \(0\) equations, C is most likely 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)

We have \(x = 1, y = 2\) or \(x = 2, y = 1.\)

Then we have \(x^5 + y^5 = 33\) for both cases.

Since both conditions together yield a unique solution, they are sufficient.

Therefore, C is the answer.
Answer: C

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.
_________________

[GMAT math practice question]

(Algebra) \(x, y,\) and \(z\) are real numbers. What is the value of \(\frac{y}{x}\)?

1) \(x\) and \(y\) are positive numbers.

2) \(\frac{x}{y}=\frac{2y}{x-z}=\frac{2x+y}{z}\)
_________________

[GMAT math practice question]

(Number Properties) \(x\) in an integer. Is \(x\) a perfect square?

1) \(x\) is one greater than the product of \(4\) consecutive positive integers.

2) \(x\) is the sum of five consecutive odd numbers.
_________________

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have \(3\) variables (\(x, y,\) and \(z\)) and \(0\) equations, E is most likely 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)

We have \(x^2 - zx = 2y^2\) when we cross multiply \(\frac{x}{y}=\frac{2y}{x-z}.\)

We have \(2xy + y^2 = zx\) when we cross multiply \(\frac{x}{y}=\frac{2x+y}{z}\).

When we subtract the second equation from the first equation, we have
\(x^2 – zx – (2xy + y^2) = 2y^2 – zx\)

\(x^2 – zx – 2xy – y^2 = 2y^2 – zx\) (multiplying \(-1\) through the bracket)

\(x^2 – zx – 2xy – y^2 – 2y^2 + zx = 0\) (moving all terms to the left side of the equation)

\(x^2 - 2xy - 3y^2 = 0\) (gathering like terms)

\((x - 3y)(x + y) = 0 \)(factoring)

Since \(x > 0\) and \(y > 0\) from condition 1), we have \(x + y ≠ 0\) and \(x = 3y.\)

Thus \(\frac{y}{x} = \frac{y}{3y} = \frac{1}{3}.\)

Since both conditions together yield a unique solution, they are sufficient.

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.
_________________

[GMAT math practice question]

(Number Properties) Is \(111…11 – 222…22\) a perfect square, where \(111…11\) and \(222…22\) are \(n\) and \(m\) digit numbers, respectively?

1) \(n = 2m.\)

2) \(m = 5.\)
_________________

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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

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

Condition 1)

Assume \(x = k(k + 1)(k + 2)(k + 3) + 1.\)

Then \(x = (k^2 + 3k)(k^2 + 3k + 2) + 1 = A(A + 2) + 1 = A^2 + 2A + 1 = (A + 1)^2\) for \(A = k^2 + 3k\) for an integer \(k\).

Thus, \(x\) is a perfect integer, and the answer is ‘yes’.

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)

If \(x = 1 + 3 + 5 + 7 + 9 = 25\), then \(x\) is a perfect integer and the answer is ‘yes’.
If \(x = 3 + 5 + 7 + 9 + 11 = 35\), then \(x\) is not a perfect integer and the answer is ‘no’.

Since condition 2) does not yield a unique solution, it is not sufficient.

Therefore, A is the answer.
Answer: A

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations,” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A, B, C, or E.
_________________

[GMAT math practice question]

(Algebra) What is the value of \(x^2(x - y) + y^2(y - x)\)?

1) x\( + y = 3.\)

2) \(xy = 1. \)
_________________

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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

\(111…11 – 222…22 = 999…99 / 9 – 2(999…99/9) = (10^n - 1)/9 – 2(10m - 1) / 9 \)

\(= [(10^n - 1) - 2(10^m - 1)] / 9\)

\(= [(10^n – 1 + 2*10^m + 2)] / 9\)

\(= [(10^{2m} – 2*10^m + 1)] / 9\)

\(= (10^m – 1)^2 / 9 \)

\(= [(10^m – 1) / 3]^2\), if \(n = 2m\) from condition 1).

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)

Since we don’t have any information about n, condition 2) does not yield a unique solution, and it is not sufficient.

Therefore, A is the answer.
Answer: A

Since both conditions together are trivial, C is not an answer. If one condition includes a ratio and the other condition just gives a number, the condition, including the ratio is most likely to be sufficient by Tip 4. This tells us that A is most likely to be the answer to this question.
_________________

[GMAT math practice question]

(Number Properties) \(x^3 + y^3 = A(x + y), x^2 + y^2 = B,\) where \(x\) and \(y\) are positive integers. What is the value of \(xy\)?

1) \(A - B = 36.\)

2) \(B = 97.\)
_________________

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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

The expression in the question \(x^2(x - y) + y^2(y - x)\) is equivalent to \((x - y)^2(x + y)\) for the following reason:

\(x^2(x - y) + y^2(y - x) \)

\(= x^2(x - y) - y^2(x - y)\) (taking out a common factor of \(-1\) from the second bracket)

\(= (x - y)(x^2 - y^2)\) (taking out a common factor of (\(x – y\)))

\(= (x - y)(x - y)(x + y)\) (factoring \((x^2 – y^2)\) using a difference of squares)

\(= (x - y)^2(x + y)\) (putting like terms together)

\((x - y)^2\)

\(= x^2 - 2xy + y^2\) (foiling \((x - y)(x – y)\))

\(= x^2 + 2xy + y^2 – 4xy\) (since \(2xy – 4xy = -2xy \)in the previous equation)

\(= (x + y)^2 – 4xy\) (factoring \(x^2 + 2xy + y^2\) using trinomial factoring)

\(= 3^2- 4*1 = 9 – 4 = 5\) when we have \(x + y = 3\) and \(xy = 1\) from both conditions 1) & 2).

Then we have \((x - y)^2(x + y) = 5*3 = 15.\)

Therefore, C is the answer.
Answer: C
_________________

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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

We can simplify the equation \(x^3 + y^3 = A(x + y)\) found in the original equation as follows:
\(x^3 + y^3 = A(x + y)\)

\(=> (x + y)(x^2 – xy + y^2) = A(x + y)\) (factoring using the sum of cubes)

\(=> \frac{[(x + y)(x^2 – xy + y^2)] }{ (x + y)} = A\) (dividing both sides by (\(x + y\)))

\(=> (x^2 – xy + y^2) = A\) (simplifying)

Since \(xy = x^2+xy+y^2 – (x^2+y^2) = \frac{(x^3+y^3)}{(x+y)} – (x^2+y^2) = A – B = 36\), condition 1) is sufficient.

Condition 2)
Since \(x^2 + y^2 = 97\), the possible pairs of \((x, y)\) are \((4, 9)\) and \((9, 4).\)

If \(x = 4, y = 9\), then we have\( xy = 36.\)

If \(x = 9, y = 4,\) then we have \(xy = 36.\)

Since condition 2) yields a unique solution, it is sufficient.

Therefore, D is the answer.
Answer: D

This question is a CMT 4(B) question: condition 1) is easy to work with, and condition 2) is difficult to work with. For CMT 4(B) questions, D is most likely the answer, since this question is an integer question.
_________________

[GMAT math practice question]

(Algebra) \(xyz≠0\). What is the value of \(x^2 + y^2 + z^2\)?

1) \(x + y + z = 3.\)

2) \(x^2(\frac{1}{y}+\frac{1}{z})+y^2(\frac{1}{z}+\frac{1}{x})+z^2(\frac{1}{x}+\frac{1}{y})=-3.\)
_________________

[GMAT math practice question]

(Algebra) What is \(\frac{1}{a+1}+\frac{1}{b+1}\)?

1) \(a=\frac{1}{(\sqrt{3}+\sqrt{2})^3}\)

2) \(b=\frac{1}{(\sqrt{3}-\sqrt{2})^3}\)
_________________

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have \(3\) variables (\(x, y,\) and \(z\)) and \(0\) equations, E is most likely 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)

\(x^2(\frac{1}{y}+\frac{1}{z})+y^2(\frac{1}{z}+\frac{1}{x})+z^2(\frac{1}{x}+\frac{1}{y}) = -3\)

\(=> x^2(\frac{1}{y}+\frac{1}{z})+y^2(\frac{1}{z}+\frac{1}{x})+z^2(\frac{1}{x}+\frac{1}{y}) + 3 = 0\)

\(=> x^2(\frac{1}{y}+\frac{1}{z})+y^2(\frac{1}{z}+\frac{1}{x})+z^2(\frac{1}{x}+\frac{1}{y}) + (x + y + z) = 0\)

\(=> x^2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+y^2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+z^2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) = 0\)

\(=> (x^2+y^2+z^2)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) = 0\)

Then we have \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\), since \(x^2+y^2+z^2≠0.\)

We have \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+zx}{xyz}=0\) or \(xy + yz + zx = 0.\)

Then \(x^2 + y^2 + z^2 = (x + y + z)^2 – 2(xy + yz + zx) = 3^2 – 0 = 9.\)

Since both conditions together yield a unique solution, they are sufficient.

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.
_________________

[GMAT math practice question]

(Absolute Value) What is the value of \(x + y\)?

1) \(|x - 2| = 4.\)

2) \(|x – y + 3| = 4.\)
_________________

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have \(2\) variables (\(a\) and \(b\)) and \(0\) equations, C is most likely 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)

We have \(\frac{1}{a+1}+\frac{1}{b+1}= \frac{b+1+a+1}{(a+1)(b+1)}=\frac{a+b+2}{ab+a+b+1}=\frac{a+b+2}{a+b+2}=1\), since we have ab=\(\frac{1}{(\sqrt{3}+\sqrt{2})^3}*\frac{1}{(\sqrt{3}-\sqrt{2})^3}=\frac{1}{{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}^3}=1.\)

Since both conditions together yield a unique solution, they are sufficient.

Therefore, C is the answer.
Answer: C

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.
_________________

[GMAT math practice question]

(Algebra) \(abc ≠ 0\). What is the value of \(a^2+b^2+c^2\)?

1) \(a+b+c=3.\)

2) \(a^3+b^3+c^3=27.\)
_________________

=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have \(2\) variables (\(x\) and \(y\)) and \(0\) equations, C is most likely 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)

We have \(x = 6\) or \(x = -2\), since \(|x - 2| = 4 ⇔ x - 2 = ±4 ⇔ x = 2 ± 4 ⇔ x = 6\) or \(x = -2\) from condition 1).

We have \(x – y = 1\) or \(x – y = -7\), since \(|x – y + 3| = 4 ⇔ x – y + 3 = ±4 ⇔ x - y = -3 ± 4 ⇔ x - y = 1\) or \(x – y = -7\) from condition 2).

If \(x = 6\) and \(x – y = 1\), then we have \(x = 6, y = 5\) and \(x + y = 11.\)

If \(x = 6\) and \(x – y = -7,\) then we have \(x = 6, y = 13\) and \(x + y = 19.\)

Since both conditions together do not yield a unique solution, they are not sufficient.

Therefore, E is the answer.
Answer: E

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.
_________________