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The Ultimate Q51 Guide [Expert Level]

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New post 15 Aug 2018, 18:19
[Math Revolution GMAT math practice question]

If a, b, and c are consecutive 2-digit positive integers in that order, and a+b+c is a multiple of 10, what is the value of c?

1) a is a prime number
2) c is a 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.

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

Since a = b – 1 and c = b + 1, we have a + b + c = ( b – 1 ) + b + ( b + 1 ) = 3b is a multiple of 10 and b is a multiple of 10. Thus, the possible triples ( a, b, c ) are ( 19, 20, 21 ), ( 29, 30, 31 ), ( 39, 40, 41 ), ( 49, 50, 51 ), ( 59, 60, 61 ), ( 69, 70, 71 ), ( 79, 80, 81 ) and ( 89, 90, 91 ). Since we have 3 variables and 2 equations ( a = b – 1, c = b + 1), D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
The triples in which a is a prime number are
( 19, 20, 21 ), ( 29, 30, 31 ), ( 59, 60, 61 ), ( 79, 80, 81 ) and ( 89, 90, 91 ).
Thus, c could be 21, 31, 61, 81 or 91.
Since we don’t have a unique solution, condition 1) is not sufficient.

Condition 2)
The triples in which c is a prime number are

( 29, 30, 31 ), ( 39, 40, 41 ), ( 59, 60, 61 ), ( 69, 70, 71 ) and ( 89, 90, 91 ).
Thus, c could be 31, 41, 61, 71 or 91.
Since we don’t have a unique solution, condition 2) is not sufficient.

Conditions 1) & 2):
The triples in which both a and c are prime numbers are:
( 29, 30, 31 ), ( 59, 60, 61 ) and ( 89, 90, 91 ).
Thus, c could be 31, 61 or 91.
Since we don’t have a unique solution, both conditions are not sufficient, when considered together.


Therefore, E is the answer.
Answer: E
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New post 16 Aug 2018, 18:10
[Math Revolution GMAT math practice question]

x ≠ 0. Is x^2>x^4?

1) x^2<1
2) x^2<2x

=>

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^2>x^4
=> x^4 - x^2 < 0
=> x^2 ( x^2 – 1 ) < 0
=> ( x^2 – 1 ) < 0
=> ( x + 1 )( x – 1 ) < 0
=> -1 < x < 1

This is equivalent to condition 1).
Thus, condition 1) is sufficient.

Condition 2)
x^2<2x
=> x^2 - 2x < 0
=> x(x – 2) < 0
=> 0 < x < 2

In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient
However, the solution set of the question -1 < x < 1 does not include the solution set of the condition, 0 < x < 2. Thus, condition 2) is not sufficient.


Therefore, A is the answer.
_________________

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New post 19 Aug 2018, 18:47
[Math Revolution GMAT math practice question]

If the average (arithmetic mean) of 3 integers x, y, z is 10, what is the value of z?

1) x=1
2) x=5-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.

The original condition gives ( x + y + z ) / 3 = 10 or x + y + z = 30. Rearranging yields z = 30 – ( x + y ). Condition 2) gives x + y = 5, so condition 2) is sufficient .


Condition 1) is not sufficient since it gives no information about the value of y.

Therefore, B is the answer.

Answer: B
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New post 21 Aug 2018, 03:42
[Math Revolution GMAT math practice question]

What is the remainder when 2^n is divided by 10?

1) n is a positive multiple of 2
2) n is a positive multiple of 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.

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

The remainder when 2^n is divided by 10 is the units digit of 2^n.

Now, 2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, and 2^5 = 32.
So, the units digits of 2^n have period 4:
They form the cycle 2 -> 4 -> 8 -> 6.
Thus, 2^n has the units digit of 6 when n is a multiple of 4.
Condition 2) is sufficient.

Condition 1)
If n = 4, then 2^n = 2^4 = 16 has the units digit of 6.
If n = 2, then 2^n = 2^2 = 4 has the units digit of 4.
Since we don’t have a unique solution, condition 1) is not sufficient.

Therefore, B is the answer.

Answer: B
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New post 21 Aug 2018, 18:03
[Math Revolution GMAT math practice question]

Is n an integer?

1) 3n is an integer
2) 4n is an 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.

Condition 1)
If n = 1, the answer is ‘yes’.
If n = 1/3, the answer is ‘no’.
Since we don’t have a unique solution, condition 1) is not sufficient.

Condition 2)
If n = 1, the answer is ‘yes’.
If n = 1/4, the answer is ‘no’.
Since we don’t have a unique solution, condition 2) is not sufficient.

Conditions 1) & 2)
Since n = 4n – 3n is a difference of two integers, n is also an integer.
Conditions 1) and 2) are sufficient, when considered together.

Therefore, C is the answer.

Answer: C

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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New post 22 Aug 2018, 18:36
1
[Math Revolution GMAT math practice question]

If the greatest common divisor of (n-1)!, n!, and (n+1)! is 5040, what is the value of n?

A. 4
B. 5
C. 6
D. 7
E. 8

=>

Since n! and (n+1)! are multiples of (n-1)!, (n-1)! is their gcd.
It follows that n – 1 = 7 or n = 8, since 7! = 5040.

Therefore, E is the answer.
Answer: E

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New post 23 Aug 2018, 19:14
[Math Revolution GMAT math practice question]

If n is a positive integer, which of the following can’t be the value of (n+1)^4-n^4?

A. 2465
B. 4641
C. 6096
D. 7825
E. 9855

=>


If n + 1 is an even number, then n is an odd number and (n+1)^4-n^4 must be an odd number.
If n + 1 is an odd number, then n is an even number and (n+1)^4-n^4 must be an odd number.
All answer choices except for C) are odd numbers.

Therefore, the answer is C.

Answer: C
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New post 26 Aug 2018, 18:29
[Math Revolution GMAT math practice question]

If (|x|-2)(x-1)=0, then x=?

1) x>0
2) -1x1

=>

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.

Original condition:
(|x|-2)(x-1)=0
=> |x| = 2 or x = 1
=> x = 2, x = -2 or x = 1

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
By condition 1) (x > 0), the possible solutions are x = 1 and x = 2.
Since we don’t have a unique solution, condition 1) is not sufficient.

Condition 2)
By condition 2) (-1 x1), the only possible solution is x = 1.
Since we have a unique solution, condition 2) is 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 27 Aug 2018, 18:17
[Math Revolution GMAT math practice question]

If x and y are integers, is x^3+3x-y an even number?

1) x=11
2) y=10

=>
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+3x is always even number, regardless of the value of x. If y is even, then x^3+3x-y is an even number, and if y is odd, then x^3+3x-y is odd.

Thus, condition 2) is sufficient.

Condition 1) is not sufficient since it tells us nothing about the value of y.

Therefore, B is the answer.
Answer: B
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New post 28 Aug 2018, 18:09
[Math Revolution GMAT math practice question]

What is the length of the sides of a given cube?

1) The surface area of the cube is 150
2) The volume of the cube is 125

=>

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 a cube is uniquely determined by its side length, we have only one variable. So, we should consider each of the conditions on their own first.

Condition 1)
Let x be the side-length of the cube. Then condition 1) tells us that 6x^2 = 150. Solving this equation yields x = 5.
Condition 1) is sufficient.

Condition 2)
Let x be the side-length of the cube. Then condition 2) tells us that x^3 = 125. Solving this equation yields x = 5.
Condition 2) is sufficient.

Therefore, D is the answer.
Answer: D

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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New post 29 Aug 2018, 20:15
[Math Revolution GMAT math practice question]

Let z be the harmonic mean of x and y. If 1/z=(1/2)((1/x)+(1/y)), which of the following is an expression for z, in terms of x and y?

A. 2xy / ( x + y )
B. 2( x + y ) / ( x – y )
C. 2( x – y ) / ( x + y )
D. 2( x + y ) / xy
E. xy / ( x + y )

=>

1/z = (1/2)(1/x + 1/y) = (1/2)( (x+y)/xy ) = (x+y)/(2xy)
Thus, z = 2xy / ( x + y ).

Therefore, A is the answer.
Answer: A
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New post 30 Aug 2018, 19:01
[Math Revolution GMAT math practice question]

Which of the following is the best approximation for 11^{11}-9^9?

A. 11^8
B. 11^9
C. 11^{10}
D. 11^{11}
E. 11^{12}

=>

9^9 is much smaller than 11^{11}.
Thus, 11^{11}-9^9 is closest to 11^{11}.

Therefore, the answer is D.
Answer: D
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New post 02 Sep 2018, 18:58
[Math Revolution GMAT math practice question]

If m and n are integers greater than 1, is m^n>500?

1) n>8
2) n>2m

=>

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.

If a question states “greater than”, then we should find the minimum value because all data are greater than the minimum. Considering condition 1), the minimum value is m^n=2^9=512>500, so the answer is ‘yes’ and condition 1) is sufficient.

Condition 2)
If m = 2, n = 100, then 2^{100} > 500 and the answer is ‘yes’.
If m = 2, n = 5, then 2^5 = 32 < 500 and the answer is ‘no’.
Thus, condition 2) is not sufficient, since we don’t have a unique solution.

Therefore, A is the answer.

Answer: A
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New post 03 Sep 2018, 18:58
1
[Math Revolution GMAT math practice question]

Does x^2+px+q = 0 have a root?

1) p<0
2) q<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 discriminant of the equation is p^2 – 4q. If the discriminant is greater than or equal to zero, then the quadratic equation has roots.
The question asks if p^2-4q ≥ 0 or not.

Since p^2 ≥ 0, if q < 0, then p^2-4q ≥ 0. Thus, condition 2) is sufficient.

Condition 1)
If p = -1 and q = 0, then the discriminant is positive and the equation has 2 roots, which are 0 and 1. So, the answer is ‘yes’.
If p = -1 and q = 1, then the discriminant is negative and the equation has no real roots. So, the answer is ‘no’.
Since we don’t have a unique solution, 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 03 Sep 2018, 19:47
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Which of the following is the best approximation for 11^{11}-9^9?

A. 11^8
B. 11^9
C. 11^{10}
D. 11^{11}
E. 11^{12}

=>

9^9 is much smaller than 11^{11}.
Thus, 11^{11}-9^9 is closest to 11^{11}.



Therefore, the answer is D.
Answer: D


MathRevolution
how did u deduce that 9^9 is much smaller than 11^{11}.

I think answer should be 11^10 or 11^9 because 9^9 is still large term in comparison to 11^11.
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The Ultimate Q51 Guide [Expert Level]  [#permalink]

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New post 03 Sep 2018, 21:46
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If a=2, b=-2, and c=3, what is the value of a^4^{b-3}c^0?

A. -2
B. -1
C. 0
D. 2
E. 3

=>

a^4b^{-3}c^0
= 2^4*(-2)^{-3}*3^0
= -(2^4*2^{-3}*1)
= -2

Therefore, the answer is A.

Answer: A


MathRevolution

is question stem correct ? Question stem doesn't match with answer solution. Is it 4*b^(-3) or4^{b-3}. I am not able to understand what has been asked and what are you explaining. It will be great if you can use typeset formulas to post questions.
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New post 04 Sep 2018, 18:26
[Math Revolution GMAT math practice question]

What is the range of 10 numbers?

1) The average (arithmetic mean) of the 10 numbers is 10
2) The greatest of the 10 numbers is 15

=>

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 Range = Max – Min, we have 3 variables and 1 equation, so C is most likely to be the answer. There is no equation relating the range, the average and the maximum, so the answer is most likely to be E).

Conditions 1) & 2):
If the data set is { 5, 5, 5, 5, 5, 15, 15, 15, 15, 15 }, then its range is 10.
If the data set is { 4, 5, 5, 5, 6, 15, 15, 15, 15, 15 }, then its range is 11.
Since we don’t have a unique solution, both conditions are not sufficient, when taken together.

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.
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New post 05 Sep 2018, 20:47
[Math Revolution GMAT math practice question]

If x and y are positive integers and xy=36, what is the smallest possible value of x+y?

A. 6
B. 8
C. 10
D. 12
E. 18


=>

When we know the product of two numbers, the sum of those numbers is a minimum when the two numbers are equal.

Since xy = 36, the minimum value of x + y occurs when x = y = 6. It is 6 + 6 = 12.

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 06 Sep 2018, 19:57
[Math Revolution GMAT math practice question]

What is the sum of all multiples of 3 between 1 and 200?

A. 3300
B. 6600
C. 6633
D. 10100
E. 20100

=>

We need to find the sum of 3, 6, …, 198.
The number of data values is (198 – 3)/3 + 1 = 195/3 + 1 = 65 + 1 = 66.
Thus the sum of 3, 6, …, 198 is 3 + 6 + … + 198 = 3+6+…+198=3(1+2+…+66)=3(66)(66+1)/2=3(33)(67)=6,633.

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

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New post 09 Sep 2018, 18:30
[Math Revolution GMAT math practice question]

Is xy<1?

1) x^2+y^2 < 1
2) x + y < 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.

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):

x^2 + y^2 < 1
=> x^2< 1 - y^2 ≤ 1 since y^2 ≥ 0
=> x^2< 1
=> -1 < x < 1
and
x^2 + y^2 < 1
=> y^2< 1 - x^2 ≤ 1 since x^2 ≥ 0
=> y^2< 1
=> -1 < y < 1

Combining these two inequalities yields -1 < xy < 1, so xy < 1. Both conditions are sufficient, when taken together.

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)

The argument showing that conditions 1 and 2 are sufficient, when taken together, only used condition 1. Therefore, condition 1 is sufficient.

Using the same argument as above,

x^2 + y^2 < 1
=> x^2< 1 - y^2 ≤ 1 since y^2 ≥ 0
=> x^2< 1
=> -1 < x < 1
and
x^2 + y^2 < 1
=> y^2< 1 - x^2 ≤ 1 since x^2 ≥ 0
=> y^2< 1
=> -1 < y < 1.

So, -1 < xy < 1, and condition 1) is sufficient.

Condition 2)

If x = 1/3 and y = 1/3, then xy = 1/9 < 1 and the answer is ‘yes’
If x = -2 and y = -2, then xy = 4 > 1 and the answer is ‘no’
Since we don’t have a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.

Answer: A

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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