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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42 GPA: 3.82

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[GMAT math practice question]

(Equation) $$2ax - 3b = a - bx$$ is an equation in terms of $$x$$. What is its solution?

1) $$\frac{-3}{2}$$ is a solution of $$(b-a)x - (2a-3b) = 0$$

2) $$a = 3b$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(algebra) $$a≠b$$ and $$b≠c$$ are given. Is it true that $$a=c$$?

1) $$(a-b)(b-c)(c-a) = 0$$

2) $$\frac{(a^2+3a)}{(a+1)} = \frac{(b^2+3b)}{(b+1)} = \frac{(c^2+3c)}{(c+1)}$$

=>

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.

Condition 1) tells that $$a = b$$ or $$b =c$$ or $$c = a$$. However, we have $$a≠b$$ and $$b≠c$$ from the original condition, so we have $$a = c$$.

Thus, condition 1) is sufficient.

Condition 2)
Assume $$\frac{(a^2+3a)}{(a+1)} = \frac{(b^2+3b)}{(b+1)} = \frac{(c^2+3c)}{(c+1)} = k$$

We have $$(a^2+3a) = k(a+1), (b^2+3b) = k(b+1)$$ and $$(c^2+3c) = k(c+1)$$

When we subtract the first two equations, we have
$$(a^2+3a) - (b^2+3b) = k(a+1) - k(b+1)$$

=> $$(a^2-b^2+3a-3b) = ka+k-kb-k$$

=>$$(a^2-b^2) + 3(a-b) = ka-kb$$

=> $$(a^2-b^2) + 3(a-b) = k(a-b)$$

=> $$(a+b)(a-b) + 3(a-b) = k(a-b)$$

=> $$(a+b+3)(a-b) = k(a-b)$$

=> $$a+b+3 = k$$ since $$a≠b$$

When we subtract the last two equations, we have
$$(b^2+3b) - (c^2+3c) = k(b+1) - k(c+1)$$

=> $$b^2-c^2+3b-3c = kb+k-kc-k$$

=> $$(b^2-c^2) + 3(b-c) = kb-kc$$

=> $$(b^2-c^2) + 3(b-c) = k(b-c)$$

=> $$(b+c)(b-c) + 3(b-c) = k(b-c)$$

=> $$(b+c+3)(b-c) = k(b-c)$$

=> $$b+c+3 = k$$ since $$b≠c$$

Since we have $$a+b+3 = k$$ and $$b+c+3 = k$$, we have $$a = c.$$
Thus condition 1) is sufficient.

When a question asks for a ratio, 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. This tells us that D is most likely to be the answer to this question, since each condition includes a ratio.

Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).

This question is a CMT4(B) question: condition 2) is easy to work with, and condition 1) is difficult to work with. For CMT4(B) questions, D is most likely to be the answer.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Equation) $$2ax - 3b = a - bx$$ is an equation in terms of $$x$$. What is its solution?

1) $$\frac{-3}{2}$$ is a solution of $$(b-a)x - (2a-3b) = 0$$

2) $$a = 3b$$

=>

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 question asks the value of $$\frac{(a+3b)}{(2a+b)}$$ for the following reason.
$$2ax - 3b = a - bx$$

=> $$2ax + bx = a + 3b$$

=> $$x(2a+b) = a+3b$$

=> $$x = \frac{(a+3b)}{(2a+b)}$$

Since we have $$a = 3b$$ from condition 2), we have $$x = \frac{(a+3b)}{(2a+b)} = \frac{(3b+3b)}{(6b+b)} = \frac{(6b)}{(7b)} = \frac{6}{7}.$$

Thus, condition 2) is sufficient.

Condition 1)
When we substitute $$\frac{-3}{2}$$ for $$x,$$ we have $$(b-a)(\frac{-3}{2})- (2a-3b) = 0$$ or $$(-3)(b-a) = 2(2a-3b)$$. We have $$-3b+3a = 4a-6b$$ or $$a = 3b.$$

Condition 1) is equivalent to condition 2), and it is also sufficient.

When a question asks for a ratio, 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. This tells us that D is most likely to be the answer to this question, since each condition includes a ratio.

Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).

This question is a CMT4(B) question: condition 2) is easy to work with, and condition 1) is difficult to work with. For CMT4(B) questions, D is most likely to be the answer.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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[GMAT math practice question]

(number properties) $$p$$ and $$q$$ are positive integers and relative primes. Is $$p$$ divisible by $$1979$$?

1) $$p$$ is a multiple of $$1979.$$

2) $$\frac{p}{q} = 1 - (\frac{1}{2}) + (\frac{1}{3}) - (\frac{1}{4}) +…- (\frac{1}{1318}) + (\frac{1}{1319}).$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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[GMAT math practice question]

(arithmetic) What is the maximum value of $$2^a + 2^b + 2^c$$?

1) $$a + b + c = 5$$

2) for any $$a, b ≥ 0, 2^a + 2^b ≤ 1 + 2^{a+b}$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(number properties) $$p$$ and $$q$$ are positive integers and relative primes. Is $$p$$ divisible by $$1979$$?

1) $$p$$ is a multiple of $$1979.$$

2) $$\frac{p}{q} = 1 - (\frac{1}{2}) + (\frac{1}{3}) - (\frac{1}{4}) +…- (\frac{1}{1318}) + (\frac{1}{1319}).$$

=>

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 question “is $$p$$ divisible by $$1979$$” is equivalent to condition 1) “$$p$$ is a multiple of $$1979$$”.

Condition 2)
Remember that $$\frac{-1}{2k} = \frac{1}{2k} – \frac{1}{k}$$ for $$k = 1, 2, 3, …, 659.$$

$$\frac{-1}{2} = \frac{1}{2} – \frac{1}{1}$$

$$\frac{-1}{4} = \frac{1}{4} – \frac{1}{2}$$

$$\frac{-1}{6} = \frac{1}{6} – \frac{1}{3}$$

$$\frac{-1}{1318} = \frac{1}{1318} – \frac{1}{659}$$

$$\frac{p}{q} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + … + \frac{1}{1317} + \frac{1}{1318} + \frac{1}{1319} – 2(\frac{1}{2} + \frac{1}{4} + … + \frac{1}{1318})$$

$$= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + … + \frac{1}{1317} + \frac{1}{1318} + \frac{1}{1319} – (\frac{1}{1} + \frac{1}{2} + … + \frac{1}{659})$$

$$= \frac{1}{660} + \frac{1}{661} + … + \frac{1}{1318} + \frac{1}{1319}$$

$$= (\frac{1}{660} + \frac{1}{1319}) + (\frac{1}{661} + \frac{1}{1318}) + … + (\frac{1}{989} + \frac{1}{990})$$

$$= \frac{1979}{(660*1319)} + \frac{1979}{(661*1318)} + … + \frac{1979}{(989*990)}$$

$$= \frac{(1979*k)}{(660*661*…*1318*1319)}$$

Then, we have $$p(660*661*…*1318*1319) = q(1979*k).$$

Since $$1979$$ is a prime number, $$p$$ is a multiple of $$1979$$.

This question is a CMT4 (B) question: condition 1) is easy to work with, and condition 2) is difficult to work with. For CMT4 (B) questions, D is most likely to be the answer.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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[GMAT math practice question]

(algebra) Is $$y^x = x^y$$?

1) $$x = y$$
2) $$x = (1 + (\frac{1}{n}))^n, y = (1 + (\frac{1}{n}))^{n+1}$$ for a positive integer $$n$$.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(arithmetic) What is the maximum value of $$2^a + 2^b + 2^c$$?

1) $$a + b + c = 5$$

2) for any $$a, b ≥ 0, 2^a + 2^b ≤ 1 + 2^{a+b}$$

=>

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 $$3$$ variables ($$a, b$$, and $$c$$) 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)
$$2^a + 2^b + 2^c$$

$$≤1 + 2^{a+b} + 2^c$$

$$≤1 + 1 + 2^{a+b+c},$$ since $$2^{a+b} + 2^c ≤ 1 + 2^{a+b+c}$$

$$≤1 + 1 + 2^5 = 34$$

So, the maximum value of $$2^a + 2^b + 2^c$$ is $$34.$$

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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[GMAT math practice question]

(Inequalities) Is $$\frac{m}{n} > \frac{(m+n)}{mn}$$?

1) $$m > n$$

2) $$m$$ and $$n$$ are integers greater than $$1$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(algebra) Is $$y^x = x^y$$?

1) $$x = y$$
2) $$x = (1 + (\frac{1}{n}))^n, y = (1 + (\frac{1}{n}))^{n+1}$$ for a positive integer $$n$$.

=>

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.

Condition 1) is sufficient, obviously.

Condition 2)
Since $$y^x = ((1 + (\frac{1}{n}))^{n+1})$$^$$(1 + (\frac{1}{n}))^n$$ $$= ((1 + (\frac{1}{n})))^{(n + 1)}*(1 + (\frac{1}{n}))^n$$, the exponent of $$y^x$$ with the base $$(1 + (\frac{1}{n}))$$ is $$(n+1)*(1+(\frac{1}{n}))^n = n(1+\frac{1}{n})* (1+(\frac{1}{n}))^n = n(1+(\frac{1}{n}))^{n+1}$$

Since $$x^y = ((1 + (1/n))^n)$$^$$(1 + (\frac{1}{n}))^{n+1} = ((1 + (\frac{1}{n})))^n(1 + (\frac{1}{n}))^n$$, the exponent of $$x^y$$ with the base $$(1 + (\frac{1}{n}))$$ is $$n(1 + (\frac{1}{n}))^n$$ is equal to the exponent of $$y^x$$.

So, we have $$y^x = x^y.$$

This question is a CMT4 (B) question: condition 1) is easy to work with, and condition 2) is difficult to work with. For CMT4 (B) questions, D is most likely to be the answer.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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[GMAT math practice question]

(geometry) $$a, b,$$ and $$c$$ are the dimensions of a rectangular box. What is the value $$a + b + c$$?

1) the volume of the rectangular box is $$1000m^3$$

2) the surface area is $$720m^2$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Inequalities) Is $$\frac{m}{n} > \frac{(m+n)}{mn}$$?

1) $$m > n$$

2) $$m$$ and $$n$$ are integers greater than $$1$$

=>

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.

Since we have $$2$$ variables ($$m$$ and $$n$$) 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)
Since $$m$$ and $$n$$ are integers greater than $$1$$,

The question $$\frac{m}{n} > \frac{(m+n)}{mn}$$ is equivalent to $$m(m – 2) + (m – n) > 0$$ for the following reason

$$\frac{m}{n} > \frac{(m+n)}{mn}$$

=> $$m^2 > m+n$$, by multiplying both sides by $$mn$$

=> $$m^2 – m – n > 0$$

=> $$m^2 – 2m + m – n > 0$$

=> $$m(m – 2) + (m – n) > 0$$

We have $$m(m - 2) ≥ 0$$, since $$m$$ is an integer greater than or equal to $$2$$ from condition 2).

We have $$m – n > 0$$ from condition 1)

So we have $$m(m – 2) + (m – n) > 0$$ and the answer is ‘yes’.

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

Since this question is an inequality 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 $$m = 4$$ and $$n = 2$$, then we have $$\frac{m}{n} = \frac{4}{2} = 2, \frac{(m+n)}{mn} = \frac{6}{8}$$ and $$\frac{m}{n} > \frac{(m+n)}{mn},$$ which means the answer is ‘yes’.

If $$m = 4$$ and $$n = -2$$, then we have $$\frac{m}{n} = \frac{4}{(-2)} = -2, \frac{(m+n)}{mn} = \frac{2}{(-8)} = \frac{-1}{4}$$ and $$\frac{m}{n} < \frac{(m+n)}{mn},$$ which means the answer is ‘no’.

Since condition 1) does not yield a unique solution, it is not sufficient

Condition 2)
If $$m = 4$$ and $$n = 2$$, then we have $$\frac{m}{n} = \frac{4}{2} = 2, \frac{(m+n)}{mn} = \frac{6}{8}$$ and $$\frac{m}{n} > \frac{(m+n)}{mn}$$, which means the answer is ‘yes’.

If $$m = 2$$ and $$n = 4$$, then we have $$\frac{m}{n} = \frac{2}{4} = \frac{1}{2}, \frac{(m+n)}{mn} = \frac{6}{8} = \frac{3}{4}$$ and $$\frac{m}{n} < \frac{(m+n)}{mn}$$, which means the answer is ‘no’.

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

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(geometry) $$a, b,$$ and $$c$$ are the dimensions of a rectangular box. What is the value $$a + b + c$$?

1) the volume of the rectangular box is $$1000m^3$$

2) the surface area is $$720m^2$$

=>

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 ($$a, b$$, and $$c$$) 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 $$abc = 100$$ and $$2(ab + bc + ca) = 720$$ or $$ab + bc + ca = 360.$$

If we have $$a = 10,$$ then we have $$bc = 100$$ and $$ab + bc + ca = 10(b+c) + 100 = 720$$ or $$b + c = 62$$, which yields $$a + b + c = 72.$$

If we have $$a = 5,$$ then we have $$bc = 200$$ and $$ab + bc + ca = 5(b+c) + 200 = 720$$ or $$b + c = 104,$$ which yields $$a + b + c = 5 + 104 = 109.$$

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

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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[GMAT math practice question]

(Number Properties) Adam is a teenage boy and his age is $$x$$ and his father’s age is $$y$$. We have a four-digit number $$S$$ by putting $$x$$ after $$y$$. What is the value of $$x$$?

1) $$y - x = 27$$

2) $$S( - (y - x)) = 4289$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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[GMAT math practice question]

(Number Properties) What is the solution to $$4x + 3y = 13$$?

1) $$x$$ and, $$y$$ are positive integers
2) $$x$$ and, $$y$$ are real numbers
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8245
GMAT 1: 760 Q51 V42 GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Number Properties) Adam is a teenage boy and his age is $$x$$ and his father’s age is $$y$$. We have a four-digit number $$S$$ by putting $$x$$ after $$y$$. What is the value of $$x$$?

1) $$y - x = 27$$

2) $$S( - (y - x)) = 4289$$

=>

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 put $$x$$ after $$y$$, we have $$S = 100y + x.$$

And we have $$13 ≤ x ≤ 9$$ and $$13 ≤ x ≤ y ≤ 99$$ since $$x$$ and $$y$$ are two-digit numbers.

Since we have $$3$$ variables ($$x, y$$, and $$S$$) and $$1$$ equation, 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 can rearrange the equation $$S - (y - x) = 4289$$ from condition 1) to get $$S = 4289 + (y - x).$$ Substituting $$y - x = 27$$ from condition 2) we get $$S = 4289 + 27 = 4316.$$ Then $$y = 43$$ and $$x = 16$$.

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

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)
We have many possible pairs of $$x$$ and $$y$$ from $$y – x = 27.$$

Since condition 1) does not yield a unique solution, it is not sufficient

Condition 2)
Substituting $$S = 100y + x$$, into $$S – (y-x)$$ we get $$100y + x – (y - x) = 99y + 2x = 4289.$$

If we make $$y = 10a + b$$ and $$x = 10 + c$$ where $$1 ≤ a ≤ 9, 0 ≤ b ≤ 9$$ and $$3 ≤ c ≤ 9.$$

Then we have $$S = 99y + 2x = 990a + 99b + 20 + 2c = 4289$$ or $$990a + 99b + 2c = 4269.$$

When we substitute integers between $$1$$ and $$9$$ for the variable a one by one, we notice $$a = 4$$ is the unique value in order to have units digits $$b$$ and $$c$$.

Then we have $$99b + 2c = 4269 – 3960 = 309.$$

When we substitute integers between $$1$$ and $$9$$ for the variable $$b$$ one by one, we notice $$b = 3$$ is the unique value in order to have a units digit $$c$$.
Then we have $$c = 6.$$

Thus the boy’s age is $$16$$ and his father’s age is $$43.$$

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

This question is a CMT 4 (A) question: When we easily get C as an answer, consider A and B as an answer.
If the question has both C and B as its answer, then B is an answer rather than C by the definition of DS questions. Also this question is a 50/51 level question and can be solved by using the relationship between the Variable Approach and Common Mistake Type 3 and 4 (A or B).

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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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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[GMAT math practice question]

(Equation) What is the value of $$x + 2y$$?

1) The system of equations with $$ax + by + c = 0$$ and $$bx + 2cy + 4a = 0 (abc ≠ 0)$$ has infinitely many solutions.

2) $$x + 2y$$ is negative and $$|x + 2y|$$ is an even prime number.
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MathRevolution wrote:
[GMAT math practice question]

(Number Properties) What is the solution to $$4x + 3y = 13$$?

1) $$x$$ and, $$y$$ are positive integers
2) $$x$$ and, $$y$$ are real numbers

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

Review the following property.
When we have solve an equation $$ax + by = c$$ in terms of $$x$$ and $$y$$, we can presume this equation has a unique solution under the following conditions.
1) $$x$$ and $$y$$ are positive integers.
2) coefficients $$a$$ and $$b$$ are relative primes.
3) $$c$$ is not a big number.

Since condition 1) tells us that $$x$$ and $$y$$ are positive integers. The coefficients $$4$$ and $$3$$ are relative primes and the constant term $$c$$ is not a big number.
Thus we can presume this equation has a unique solution.

The logical reason is as follows.
We can substitute positive integers from $$1$$ into the variable $$x$$ of the equation $$3y = 13 – 4x$$ and we should notice that $$13 – 4x$$ must be a multiple of $$3$$ since we have $$3y$$ on the left hand side.
The unique possible value of $$x$$ is $$1$$ in order to have a positive value of $$y$$ from the equation.
Then we have a unique solution $$x = 1$$ and $$y = 3.$$
Since condition 1) yields a unique solution, it is sufficient.

Condition 2)
Since condition 2) allows all real numbers, there many possibilities for solutions to this equation $$4x + 3y = 13$$.
Since condition 2) does not yield a unique solution, it is not sufficient.

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]

(Algebra) What is the difference between $$x$$ and $$y$$?

1) $$|x - y|$$ is the first odd prime number

2) $$x$$ and $$y$$ are positive integers such that $$3x + 5y = 23$$
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Math Revolution GMAT Instructor V
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GMAT 1: 760 Q51 V42 GPA: 3.82

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MathRevolution wrote:
[GMAT math practice question]

(Equation) What is the value of $$x + 2y$$?

1) The system of equations with $$ax + by + c = 0$$ and $$bx + 2cy + 4a = 0 (abc ≠ 0)$$ has infinitely many solutions.

2) $$x + 2y$$ is negative and $$|x + 2y|$$ is an even prime number.

=>

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.

Even though we have $$2$$ variables ($$x$$ and $$y$$) and $$0$$ equations, D is most likely the answer, since each condition has $$2$$ equations. So, we should consider each condition on its own first.

Condition 1)

Since $$ax + by + c = 0$$ and $$bx + 2cy + 4a = 0 (abc ≠ 0)$$ has infinitely many solutions, we have $$\frac{a}{b} = \frac{b}{2c} = \frac{3}{4a}$$. Assume $$\frac{a}{b} = \frac{b}{2c} = \frac{3}{4a} = k.$$

Then we have $$b = ak, 2c = bk$$ and $$4a = ck$$. When we multiply those equations together, we have $$8abc = abdk^3$$. Since $$abc$$ is not equal to $$0$$, we have $$k^3 = 8$$ or $$k = 2.$$

Then, $$b = 2a, 2c = 2b, 4a = 2c.$$ The second equation reduces to $$c = b,$$ so we have $$b = c = 2a.$$

Thus both equations in the system of equations are $$x + 2y + 2 = 0.$$

We have $$x + 2y = -2.$$

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

Condition 2)
Since the unique even prime number is $$2$$, we have $$|x + 2y| = 2$$ which yields $$x + 2y = 2$$ or $$x + 2y = -2.$$

Since $$x + 2y$$ is negative, we have $$x + 2y = -2.$$

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

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

Note: Tip 1) of the VA method states that D is most likely the answer if condition 1) gives the same information as condition 2).

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.
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