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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
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[GMAT math practice question]

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

=>

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.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

If x, y, z are positive integers, is xyz64?

1) xy≥yz≥zx≥16
2) x+y+z=64

=>

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.

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.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

3x+4y=?

1) 2|x|+3|y|=0
2) 3|x|+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.

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.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

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

=>

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.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

If (x-7)^2=-|y-5|, xy=?

A. 5
B. 7
C. 12
D. 35
E. cannot be determined

=>

(x-7)^2=-|y-5|
=> (x-7)^2+|y-5| = 0
=> x = 7 and y = 5
This yields xy = 35.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

What is the area of a right triangle?

1) The length of one side is 3
2) The length of one side is 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.

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 (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 √(4^2 – 3^2) = √7, and the area of the triangle is (1/2)*3*√7 =(3√7)/2.
Thus, both conditions together are not sufficient, since they do not yield a unique solution.

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: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

x^3-y^3=?

1) x-y=0
2) |x|=|y|

=>

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.

x^3-y^3 = (x-y)(x^2+xy+y^2)

Condition 1)
If x – y = 0, then x^3-y^3 = (x-y)(x^2+xy+y^2) = 0.
Thus, condition 1) is sufficient.

Condition 2)
If x = 1 and y = 1, then x^3-y^3 = 1 – 1 = 0.
If x = 1 and y = -1, then x^3-y^3 = 1 –(-1) = 2.
Condition 2) is not sufficient since it does not yield a unique solution.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

How many numbers between 1 and 200, inclusive, have a 2 in the units place and are divisible by 4?

A. 8
B. 9
C. 10
D. 11
E. 12

=>

In order for a number with 2 in the units place to be divisible by 4, the last two digits must be one of 12, 32, 52, 72 and 92.
Thus, the numbers between 1 and 200, inclusive, satisfying this property are
12, 32, 52, 72, 92, 112, 132, 152 and 172.

There are 8 such numbers. Therefore, A is the answer.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

How many integers between 18 and 3399 are multiples of 17?

A. 198
B. 199
C. 200
D. 201
E. 202

=>

18 = 17*1 + 1, 34 = 17*2 is the least multiple of 17 greater than 18.
3399 = 17*199 + 16, 3383 = 17*199 is the greatest multiple of 17 less than 3399.
Thus, the multiples of 17 between 18 and 3399 are 34=17*2, … , 3383 = 17*199. There are 198 = (199 – 2) + 1 of these numbers as they are in one-to-one correspondence with the integers between 2 and 199, inclusive.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

What is the value of (2^{2020} + 2^{2018}) / (2^{2020}-2^{2018})?

A. 3/5
B. 5/3
C. 2^{2018}
D. 2^{2019}
E. 2^{2020}

=>

(2^{2020} + 2^{2018}) / (2^{2020}-2^{2018})
= (2^22^{2018} + 2^{2018}) / (2^22^{2018}-2^{2018})
= (2^{2018})(2^2+1) / (2^{2018})(2^2-1) = (2^2+1) / (2^2-1)
= 5/3

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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If a and b are integers, and x and y are positive integers, is ax+by >0?

1) ax+y>0
2) bx+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.

Since we have 4 variables (a, b, x and y) 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):
If a = 1, b = 1, x = 1 and y = 1, then a^x+b^y = 2, and the answer is ‘yes’.
If a = -1, b = -1, x = 1 and y = 1, then a^x+b^y = -2, and the answer is ‘no’.

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

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: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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If a and b are positive integers, is a+b divisible by 7?

1) a – b is divisible by 7
2) a+b+7 is divisible by 91

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 a + b + 7 = 91k for some integer k, and so a + b = 91k – 7 = 7(13k-1). Therefore, a + b is a multiple of 7.
Thus, condition 2) is sufficient.

Condition 1)
If a = 14 and b = 7, then a + b = 21 is a multiple of 7, and the answer is ‘yes’.
If a = 8 and b = 1, then a + b = 9 is not a multiple of 7, and the answer is ‘no’.
Thus, condition 1) is not sufficient since it does not yield a unique solution.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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a, b, and c are positive integers. Is a+b+c an odd number?

1) ab is an odd number
2) c is an odd 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.

Since we have 4 variables (a, b, x and y) 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 with the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Condition 1), ab is an odd integer, is equivalent to the statement that both a and b are odd numbers.
As c is also an odd number, a + b + c is an odd number since it is the sum of three odd numbers.

Both conditions together 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)
If a = 1, b = 1, and c = 1, then a + b + c = 3 is an odd number, and the answer is ‘yes’.
If a = 1, b = 1, and c = 2, then a + b + c = 4 is not an odd number, and the answer is ‘no’.
Thus, condition 1) is not sufficient.

Condition 2)
If a = 1, b = 1, and c = 1, then a + b + c = 3 is an odd number, and the answer is ‘yes’.
If a = 2, b = 1, and c = 1, then we have a + b + c = 4 is not an odd number, and the answer is ‘no’.
Thus, condition 2) is 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: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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A box contains 1 red ball, 3 green balls, 5 yellow balls, 7 blue balls, 9 white balls, and 11 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 7 balls of the same color are drawn?

A. 7
B. 14
C. 15
D. 27
E. 28

The maximum number of draws without 7 balls of a single color is 1 + 3 + 5 + 6 + 6 + 6 = 27, obtained by drawing 1 red ball, 3 green balls, 5 yellow balls, 6 blue balls, 6 white balls and 6 black balls. If we draw one more ball, then we must have 7 balls of one color.
Thus, we need to draw 28 balls to ensure that 7 balls of the same color are drawn. Therefore, E is the answer.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

What is the largest digit n for which the number 123,45n is divisible by 3?

A. 3
B. 5
C. 6
D. 7
E. 9

=>

Recall that a number is divisible by 3 if and only if the sum of its digits is divisible by 3. The sum of the digits of 123,45n is 1 + 2 + 3 + 4 + 5 + n = n + 15. This is divisible by 3 exactly when n is divisible by 3.
So, n must be a multiple of 3.
Thus, the largest digit, n is 9.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

n is a positive integer. What is the remainder when n is divided by 3?

1) n^2 has remainder 1 when it is divided by 3
2) n has remainder 7 when it is divided by 9

=>

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 (n) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
n could be any of the integers 1, 2, 4, 5, 7, 8, …
If n is one of 1, 4, 7, then n has a remainder 1 when it is divided by 3.
If n is one of 2, 5, 8, then n has a remainder 2 when it is divided by 3.
Thus, condition 1) is not sufficient, since it does not yield a unique solution.

Condition 2)
n = 9k +7 can be expressed as n = 9k + 7 = 9k + 6 + 1 = 3(3k+2)+1. Therefore, n has remainder 1 when it is divided by 3.
Thus, condition 2) is sufficient.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

Is x = y?

1) x ≤ y
2) |x| ≥ |y|

=>

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)
If x =1, and y = 1, then x and y satisfy both conditions, and the answer is “yes” since x = y.
If x = -2, and y = 1, then x and y satisfy both conditions, but the answer is “no” since x ≠ y.

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

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

(b-a)/ab = 1/a – 1/b. What is the value of 1/2 + 1/6 + 1/12 + … + 1/90?

A. 4/5
B. 5/6
C. 8/9
D. 9/10
E. 11/12

=>

1/2 + 1/6 + 1/12 + … + 1/90
= 1/(1*2) + 1/(2*3) + 1/(3*4) + … + 1(9*10)
= (2-1)/(1*2) + (3-2)/(2*3) + (4-3)/(3*4) + … + (10-9)/(9*10)
= (1/1 – 1/2) + (1/2 – 1/3) + (1/3-1/4) + … + (1/9 – 1/10)
= 1/1 – 1/10
= 9/10

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

How many whole numbers between 100 and 400, inclusive, contain the digit 2?

A. 100
B. 125
C. 138
D. 145
E. 150

=>

There are 19 such numbers with hundreds digit 1: 102, 112, 120, 121, …, 129, 132, 142, …, 192.
There are 100 such numbers with hundreds digit 2: 200, 201, 202, 203, …, 299.
There are 19 such numbers with hundreds digit 3: 302, 312, 320, 321, …, 329, 332, 342, …, 392.

Thus, there are a total of 19 + 100 + 19 = 138 such numbers.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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[GMAT math practice question]

When 32 is divided by k, the remainder is k-3. What is the value of k?

1) k>20
2) k<40

=>

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 says that 32 = k*q + ( k – 3 ) or 35 = k*q + k = k(q+1).
That is, k is a factor of 35.
So, k = 1, 5, 7 or 35.

Since condition 1) “k>20” yields the unique solution “k=35”, condition 1) is sufficient.

Condition 2) yields k = 1, 5, 7 or 35. Since it does not yield a unique solution, condition 2) is not sufficient.

_________________ Re: The Ultimate Q51 Guide [Expert Level]   [#permalink] 10 Mar 2019, 17:23

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# The Ultimate Q51 Guide [Expert Level]

Moderator: DisciplinedPrep  