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The Ultimate Q51 Guide [Expert Level]

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Math Revolution GMAT Instructor
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GMAT 1: 760 Q51 V42
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New post 20 Jan 2019, 18:43
[Math Revolution GMAT math practice question]

Is a > b - c?

1) |c-b| < a
2) a, b, and c are the lengths of the 3 sides of a triangle.

=>

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) is sufficient since it yields b – c ≤ | b - c | = | c – b | < a.

Condition 2)
Since the sum of the lengths of two sides of a triangle is greater than the length of the third side, we must have a + c > b. Thus, 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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New post 21 Jan 2019, 19:26
[Math Revolution GMAT math practice question]

If x and y are integers, is x2-y2 an even integer?

1) x^3-y^3 is an even integer
2) x+y is an even integer

=>

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 whether x^2-y^2 is even is equivalent to asking whether (x+y)(x-y) is an even integer.

Condition 2) is sufficient since x^2-y^2 = (x+y)(x-y) is an even integer if x + y is an even integer.

Condition 1)
In order for x^3-y^3 to be an even integer, x and y must both have the same parity.
There are only two cases to consider: both x and y are even integers or both are odd integers.
Since x and y have the same parity, x – y is always an even integer.
Thus, x^2-y^2 is an even integer. Condition 1) is sufficient.

Therefore, D is the answer.
Answer: D
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New post 22 Jan 2019, 18:17
[Math Revolution GMAT math practice question]

A sequence An satisfies An+3=An+8. What is the remainder when A99 is divided by 8?

1) A1=1
2) A2=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.

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

The initial condition, An+3=An+8, tells us that every third term has the same remainder when it is divided by 8.
By condition 1), A1 = 1, A4 = 9, A7 = 17, … , A97 = 257, A100 = 265 …. Each of these terms has remainder 1 when it is divided by 8.
By condition 2), A2 = 2, A5 = 10, A8 = 18, … , A98 = 258, A101 = 266 …. Each of these terms has remainder 2 when it is divided by 8.

A99 does not appear in either of these lists. Thus, both conditions together do not provide enough information to find the remainder when A99 is divided by 8. Therefore, E is the answer.

Note: in order to find the remainder when A99 is divided by 8, we need the value of A3. Thus, we need the values of 3 variables, and we are given only 2, so E should be the answer.

Therefore, the correct answer is E.
Answer: E
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New post 23 Jan 2019, 18:07
[Math Revolution GMAT math practice question]

Is ab<0?

1) |a+b| = - ( a + b )
2) |a+b| + 1 = |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.

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

You should remember that the inequality |x+y| < |x| + |y| is equivalent to the inequality xy < 0.

Condition 2) tells us that |a+b| + 1 = |a| + |b|. Thus, |a + b| < |a| + |b| and ab < 0.
Thus, condition 2) is sufficient.

Condition 1)
If a = -2 and b = 1, then the answer is ‘yes’.
If a = -1 and b = -1, then the answer is ‘no’.
Since it does not yield a unique solution, condition 1) is not sufficient.

Therefore, the answer is B.
Answer: B
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New post 24 Jan 2019, 18:55
[Math Revolution GMAT math practice question]

A rectangular wall has dimensions of 800 by 300. If the wall is tiled using tiles measuring 12 x 25 of the same units, how many tiles are used?

A. 360
B. 400
C. 500
D. 600
E. 800

=>

Now, 800 / 25 = 32 and 300 / 12 = 25. Thus, the number of tiles required is 32 * 25 = 800.

Therefore, the answer is E.
Answer: E

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New post 27 Jan 2019, 18:29
[Math Revolution GMAT math practice question]

n is a 3 digit integer of the form ab6. Is n divisible by 4?

1) a+b is an even integer
2) ab is an odd integer.

=>

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.

We can determine whether a number is divisible by 4 from its final two digits.
Numbers with the final digits 16, 36, 56, 76 and 96 are divisible by 4 and those with final digits 06, 26, 46, 66 and 88 are not divisible by 4. Thus, asking whether n is divisible by 4 is equivalent to asking whether b is odd.

Since it implies that both a and b are odd integers, condition 2) is sufficient.

Condition 1)
There are two cases to consider.
If a is an even integer and b is an odd integer, the answer is ‘yes’.
If a is an odd integer and b is an even integer, the answer is ‘no’.
Since it does not yield a unique solution, condition 1) is not sufficient.

Therefore, B is the answer.
Answer: B
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New post 28 Jan 2019, 18:21
[Math Revolution GMAT math practice question]

x=?

1) x^3-x=0
2) x=-x

=>

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

Condition 1)

x^3-x = 0
=> x(x^2-1) = 0
=> x(x+1)(x-1) = 0
=> x = 0, x = -1 or x = 1.
Since it does not yield a unique solution, condition 1) is not sufficient.

Condition 2)
x = -x
=> 2x = 0
=> x = 0.
Since it gives a unique solution, condition 2) is sufficient.

Therefore, B is the answer.
Answer: B
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New post 29 Jan 2019, 21:42
[Math Revolution GMAT math practice question]

If x and y are non-zero numbers and x≠±y, then ( x^2 + y^2 ) / ( x^2 - y^2 )=?

1) |x/y|=1/3
2) y=-3x

=>

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 question asks for the value of ( x^2 + y^2 ) / ( x^2 - y^2 )= ( (x/y)^2 + 1 ) / (x/y)^2 – 1 ).

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.

Condition 1)

Since |x/y|=1/3, x/y = ±(1/3), and ( x^2 + y^2 ) / ( x^2 - y^2 )= ( (x/y)^2 + 1 ) / ( (x/y)^2 – 1 ) = ( (1/3)^2 + 1 ) / ( (1/3)^2 – 1) = (1/9 + 1)/(1/9-1) = (10/9)/(-8/9) = -10/8 = -5/4.
Condition 1) is sufficient since it gives a unique solution.


Condition 2)
Since y = -3x, x/y = -1/3, and ( x^2 + y^2 ) / ( x^2 - y^2 )= ( (x/y)^2 + 1 ) / ( (x/y)^2 – 1 ) = ( (-1/3)^2 + 1 ) / ( (-1/3)^2 – 1) = (1/9 + 1)/(1/9-1) = (10/9)/(-8/9) = -10/8 = -5/4.
Condition 2) is sufficient since it gives a unique solution.

Therefore, the answer is D.
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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New post 30 Jan 2019, 18:27
[Math Revolution GMAT math practice question]

If x-2y=4, then x^2-4xy+4y^2-x+2y=?

A. 12
B. 16
C. 20
D. 24
E. 28

=>

x^2-4xy+4y^2-x+2y
=(x-2y)^2 – (x-2y)
=4^2 – 4 = 16 – 4 = 12.

Therefore, the answer is A.
Answer: A
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New post 31 Jan 2019, 17:29
[Math Revolution GMAT math practice question]

A = (2-3+4)^{11}, and B = (-2+3-4)^{11}. What is the value of 2^{A+B}?

A. 1/2048
B. 1/1024
C. 1
D. 1024
E. 2048

=>

A + B
= (2-3+4)^{11}+ (-2+3-4)^{11}
= 3^{11}+ (-3)^{11}
= 3^{11}- 3^{11}
= 0.
Therefore,
2^{A+B} = 2^0 = 1.

Therefore, the answer is C.
Answer: C

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New post 04 Feb 2019, 21:12
[GMAT math practice question]

If n is positive integer, is 4^n+n^2+1 divisible by 2?

1) n is a multiple of 4
2) n is a multiple of 6

=>

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 4^n is a multiple of 2, we only need to look at n^2+1.
If n is an odd number, 4^n+n^2+1 is divisible by 2.
If n is an even number, 4^n+n^2+1 is not divisible by 2.
The question asks if n is an odd number.
Thus, each of conditions is sufficient.

Therefore, D is the answer.
Answer: D
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Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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New post 04 Feb 2019, 22:01
MathRevolution wrote:
[GMAT math practice question]

If n is positive integer, is 4^n+n^2+1 divisible by 2?

1) n is a multiple of 4
2) n is a multiple of 6

=>

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 4^n is a multiple of 2, we only need to look at n^2+1.
If n is an odd number, 4^n+n^2+1 is divisible by 2.
If n is an even number, 4^n+n^2+1 is not divisible by 2.
The question asks if n is an odd number.
Thus, each of conditions is sufficient.

Therefore, D is the answer.
Answer: D


0 is also a multiple of 4 and 6. If we take n as 0, then 4^n+n^2+1 = 4^0+0^2+1 = 2, which is divisible by 4 and 6, but if we take n as 4 or 6, then the expression is not divisible by both 4 and 6. Hence Shouldn't E be the answer?
chetan2u sir, Am I wrong?
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Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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New post 04 Feb 2019, 23:46
1
Afc0892 wrote:
MathRevolution wrote:
[GMAT math practice question]

If n is positive integer, is 4^n+n^2+1 divisible by 2?

1) n is a multiple of 4
2) n is a multiple of 6

=>

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 4^n is a multiple of 2, we only need to look at n^2+1.
If n is an odd number, 4^n+n^2+1 is divisible by 2.
If n is an even number, 4^n+n^2+1 is not divisible by 2.
The question asks if n is an odd number.
Thus, each of conditions is sufficient.

Therefore, D is the answer.
Answer: D


0 is also a multiple of 4 and 6. If we take n as 0, then 4^n+n^2+1 = 4^0+0^2+1 = 2, which is divisible by 4 and 6, but if we take n as 4 or 6, then the expression is not divisible by both 4 and 6. Hence Shouldn't E be the answer?
chetan2u sir, Am I wrong?


Afc0892, you have missed out on word positive integer.
n is a positive integer, so n cannot be 0
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New post 05 Feb 2019, 18:42
[GMAT math practice question]

Five data values are 11, 14, 16, 18 and x. What is the value of x?

1) The mode of the 5 data values is 11
2) The average (arithmetic mean) of the 5 data values is 14

=>

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

Condition 1)
The data values include four different values and x.
Since the mode of the five data values is 11, x must equal 11.
Condition 1) is sufficient.

Condition 2)
Calculating the mean of the five data values yields
( 11 + 14 + 16 + 18 + x ) / 5 = 14.
Solving for x gives
11 + 14 + 16 + 18 + x = 70
59 + x = 70
x = 11
Condition 2) is also sufficient.

Therefore, D is the answer.
Answer: D
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New post 06 Feb 2019, 18:20
[GMAT math practice question]

Is 3 a factor of x?

1) x-3 is divisible by 6
2) x+3 is divisible by 6

=>

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

To solve remainder questions, plugging in numbers is recommended.

Condition 1)
If we plug in x = 9, then x – 3 = 6 is divisible by 6 and x is a multiple of 3. Condition 1) is sufficient.

Condition 2)
If we plug in x = 9, then x + 3 = 12 is divisible by 6 and x is a multiple of 3. Condition 2) is sufficient.

Therefore, the answer is D.
Answer: D
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New post 07 Feb 2019, 18:27
[GMAT math practice question]

If p, x, and y are integers, x^p/x^q=?

1) p=q+4
2) x^q=16

=>

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 and z) 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 x = 2, p = 8 and q = 4, then x^p/x^q = x^{p-q} = x^4 = 2^4 = 16.
If x = 16, p = 5 and q = 1, then x^p/x^q = x^{p-q} = x^4 = 16^4 = 2^{16} = 65536.
Since they do not yield a unique solution, both conditions are not sufficient, when considered together.

Therefore, the answer is E.
Answer: E

Note: This question is related to finding a hidden 1.
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.
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New post 10 Feb 2019, 18:16
[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.

Therefore, B is the answer.
Answer: B
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Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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New post 11 Feb 2019, 18:26
[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.

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.
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Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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New post 12 Feb 2019, 18:27
[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.

Therefore, the answer is D.
Answer: D
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Re: The Ultimate Q51 Guide [Expert Level]  [#permalink]

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New post 13 Feb 2019, 18:21
[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.

Therefore, the answer is A.
Answer: A
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Re: The Ultimate Q51 Guide [Expert Level]   [#permalink] 13 Feb 2019, 18:21

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