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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9261
GMAT 1: 760 Q51 V42
GPA: 3.82
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[GMAT math practice question]

(exponent) m and n are positive numbers. (2^m)^n = 512. What is the minimum value of m + n?

A. 3
B. 4
C. 5
D. 6
E. 8

=>

(2^m)^n = 2^{mn} = 512 = 2^9.
Thus mn = 9.
Recall that if mn is a fixed constant, and m and n are positive numbers, then m + n has its minimum value when m = n.
Thus, we must have m = n = 3 to obtain the minimum value of m + n = 3 + 3 = 6.

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[GMAT math practice question]

(number properties) Is pqr is a multiple of 5?

1) p, q and r are consecutive odd integers.
2) p, q and r are prime numbers.

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (p, q and r) 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):
3, 5 and 7 is the unique triplet of three consecutive odd integers which are prime numbers, and 3*5*7 is a multiple of 5.
Thus, 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 p = 3, q = 5 and r = 7, then pqr = 105 is a multiple of 5, and the answer is ‘yes’.
If p = 7, q = 9 and r = 11, then pqr = 693 is not a multiple of 5, and the answer is ‘no’.
Condition 1) is not sufficient since it doesn’t yield a unique answer.

Condition 2)
If p = 3, q = 5 and r = 7, then pqr = 105 is a multiple of 5, and the answer is ‘yes’.
If p = 3, q = 7 and r = 11, then pqr = 231 is not a multiple of 5, and the answer is ‘no’.
Condition 2) is not sufficient since it doesn’t 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.
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[GMAT math practice question]

(absolute value) Is x<y<z ?

1) |x+2|<y<z+2
2) |x-2|<y<z-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.

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)
Now, by condition 1), x < x + 2 ≤ | x + 2 | < y < z + 2, and x < y.
By condition 2), |x-2|<y<z-2 < z, and so y < z.
Thus, both conditions together are sufficient since they yield x < y < z, and the unique answer is ‘yes’.

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 x = 1, y = 4, and z = 7, then the answer is ‘yes’.
If x = 1, y = 4, and z = 3, then the answer is ‘no’.
Condition 1) is not sufficient since it doesn’t yield a unique answer.

Condition 2)
If x = 1, y = 4, and z = 7, then the answer is ‘yes’.
If x = 5, y = 4, and z = 7, then the answer is ‘no’.
Condition 2) is not sufficient since it doesn’t yield a unique answer.

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]

(function) If the symbol # represents one of addition, subtraction, multiplication, or division, what is the value of (2 # 1)?

1) 2 # 2 = 1
2) (–1/2) # (-1) = 1/2

=>

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.

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

Condition 1)
# is the division operation since 2 / 2 = 1.
So, 2#1 = 2/1 = 2.
Condition 1) is sufficient since it yields a unique answer.

Condition 2)
# could be the multiplication, subtraction or the division operation
since (-1/2) * (-1) = 1/2, (-1/2)-(-1) = (1/2) and (-1/2)/(-1) = 1/2.

Since 2 * 1 = 2 and 2 / 1 = 2, but 2-1 = 1, condition 2) does not yield a unique value for 2#1.
Condition 2) is not sufficient since it doesn’t yield a unique answer.

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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Two people run the circle with 10 rounds, one runs in the inside and the other runs in the outside. When the difference between the inside and the outside is 15 feet, what is the difference of between two people's distance after running 10 rounds, approximately?
A. 800 B. 850 C. 900 D. 950 E. 1,000
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The sequence An satisfies An = A_(n-1)/A_(n-2) , where n is a positive integer greater than 2. If A1=1 and A2=2, then A123=?

A. 1/2
B. 1/4
C. 1
D. 2
E. 4

=>

A1 = 1
A2 = 2
A3 = A2/A1 = 2/1 =2
A4 = A3/A2 = 2/2 =1
A5 = A4/A3 = 1/2
A6 = A5/A4 = (1/2)/1 = 1/2
A7 = A6/A5 = (1/2)/(1/2) =1
A8 = A7/A8 = 1/(1/2) = 2
Since A7 = 1 and A8 = 2, the sequence is periodic with period 6.
Thus, A123 = A3 = 2 since 123=6•20+3

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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.

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 x and y are positive integers, condition 1) tells us that either x = 1 and y = 2, or x = 2 and y= 1. Thus, x and y are consecutive integers.

Since x and y are positive integers, condition 2 tells us that either x = 1 and y = 2, or x = 2 and y= 1. Thus, x and y are consecutive integers.

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).
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[GMAT math practice question]

(absolute value) If |2x|>|3y|, is x >y?

1) x>0
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. We should simplify the conditions, if necessary.

Condition 1)
Since x > 0, 3x > 2x ≥ |2x|>|3y| ≥ 3y. It follows that x > y.
Thus, condition 1) is sufficient.

Condition 2)
If x = 3 and y = 1, then x > y, and the answer is “yes”.
If x = -3 and y = 1, then x < y, and the answer is “no”.
Thus, condition 2) is not sufficient since it doesn’t yield a unique answer.

_________________
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Joined: 16 Aug 2015
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[GMAT math practice question]

(number properties) a, b, c and d are integers. Is abcd + abc + ab + a an even number?

1) abc is an odd integer
2) bcd 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 should simplify the conditions if necessary.

Modifying the question:
For abcd + abc + ab + a = a(bcd+bc+b+1) to be even, either a must be even or bcd + bc + b + 1 must be even.

Condition 2):
If bcd is an odd integer, then b, c and d are odd integers. This implies that bc is odd, and bcd + bc + b + 1 is an even integer. Condition 2) is sufficient.

Condition 1)
If a = b = c = d = 1, then abcd + abc + ab + a = 1 + 1 + 1 + 1 = 4, which is an even integer, and the answer is ‘yes’.
If a = b = c = 1 and d = 2, then abcd + abc + ab + a = 2 + 1 + 1 + 1 = 5, which is an odd integer, and the answer is ‘no’.
Condition 1) is not sufficient since it doesn’t yield a unique solution.

_________________
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[GMAT math practice question]

(probability) How many arrangements of the letters A, C, C, E, N, T include at least one letter between the two C’s?

A. 60
B. 120
C. 240
D. 360
E. 720

=>

The easiest way to solve this problem is to find the number of arrangements satisfying the complementary condition that there is no letter between the two C’s and subtract this from the total number of arrangements of the letters. The best approach to solving probability or counting problems, including the words, ‘at least’, is to use complementary counting.

The total number of arrangements of the letters A, C, C, E, N and T is (6!)/(2!) = 360.
To count the number of arrangements with no letter between the two C’s, we consider CC to be one letter. Thus, the number of arrangements satisfying the complementary condition is 5! = 120.

Thus, the number of arrangements of the letters A, C, C, E, N, T with at least one letter between the two C’s is 360 – 120 = 240.

Attachments 6.17.png [ 3.26 KiB | Viewed 662 times ]

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[GMAT math practice question]

(arithmetic, equation) What is the value of 1 / (√2+√1) + 1 / (√3+√2) + 1/(√4+√3) + … + 1/(√25+√24)?

A.1
B. 2
C. 3
D. 4
E. 5

=>

We rationalize the denominator of each fraction to give
1 / (√2+ √1) + 1 / (√3+ √2) + 1/( √4+ √3) + … + 1/( √25+ √24)
= (√2-√1)/[(√2+√1)(√2-√1)] + (√3-√2)/[(√3+√2) (√3-√2)] + (√4-√3)/[(√4+√3) (√4-√3)] + … + (√25-√24)/[(√25+√24)(√25-√24)]
= (√2-√1)/[2-1] + (√3-√2)/[3-2] + (√4-√3)/[4-3] + … + (√25-√24)/[25 - 24]
= (√2-√1) + (√3-√2) + (√4-√3) + … + (√25-√24)
= √25 - √1 = 5 - 1 = 4

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[GMAT math practice question]

(number properties) m is an odd integer. If m^3n^4=432, what is the value of n?

1) n is positive.
2) n 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.

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)
Since m is an odd integer, n is a positive integer and 432 = 33*24, the unique solution pair is m = 3 and n = 2.
Thus, both conditions are sufficient, when applied together.

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)
m = 3 and n = 2 is a pair of solutions.
m = 1 and n = 4√432 is another pair of solutions.
Since condition 1) doesn’t yield a unique solution, it is not sufficient.

Condition 2)
m = 3 and n = 2 is a pair of solutions.
m = 3 and n = -2 is another pair of solutions.
Since condition 2) doesn’t 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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Joined: 16 Aug 2015
Posts: 9261
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[GMAT math practice question]

(work rate) A factory has two machines. Machine X can produce 200 pins per hour and machine Y can produce 500 pins per hour. At least one machine is in operation throughout the 8-hour work day. The factory must produce 5,000 pins every day. What is the least possible number of hours that machines X and Y must work together on each day?

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

=>

In order to minimize the number of hours that the two machines work together, the more efficient machine Y should work for the entire day.
Thus, Y works for 8 hours and produces 8 * 500 = 4,000 pins.
There are 1,000 pins remaining for machine X to produce. This takes 1,000/200 = 5 hours.

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[GMAT math practice question]

(inequality) -2 ≤ x ≤ -1 and 1 ≤ y ≤ 2. What is the greatest possible value of ( x + y ) / x?

A. -1
B. –1/2
C. 0
D. 1/2
E. 1

=>

(x+y)/x = x/x + y/x = 1 + y/x
Since -2 ≤ x ≤ -1 and 1 ≤ y ≤ 2, we have -2 ≤ y/x ≤ -1/2 and -1 ≤ 1 + y/x ≤ 1/2.
Thus, the greatest possible value of (x+y)/x is ½.

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Joined: 16 Aug 2015
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[GMAT math practice question]

(number properties) p and q are different positive integers. What is the remainder when p^2 + q^2 is divided by 4?

1) p and q are prime numbers.
2) p and q are not consecutive integers.

=>

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 (p and q) 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)
Since p and q are prime numbers, which are not consecutive integers, p and q are odd integers.
So, both p^2 and q^2 have remainder 1 when they are divided by 4.
Thus, p^2 + q^2 has remainder 2 when it is divided by 4.
Since conditions 1) & 2) yield a unique solution, when they are applied together, 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)
If p = 2 and q = 3, then p^2 + q^2 = 4 + 9 = 13, which has remainder 1 when it is divided by 4.
If p = 3 and q = 5, then p^2 + q^2 = 9 + 25 = 34, which has remainder 2 when it is divided by 4.

Condition 1) is not sufficient since it doesn’t yield a unique solution.

Condition 2)
If p = 3 and q = 5, then p^2 + q^2 = 9 + 25 = 34, which has remainder 2 when it is divided by 4.
If p = 3 and q = 6, then p^2 + q^2 = 9 + 36 = 45, which has remainder 1 when it is divided by 4.

Condition 2) is not sufficient since it doesn’t yield a unique solution.

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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Joined: 16 Aug 2015
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[GMAT math practice question]

(number properties) m and n are positive integers. Is m^2 + n^2 is divisible by 3?

1) m = 1234
2) n = 4321

=>

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 should simplify the conditions if necessary.

Recall that the remainder when an integer is divided by 3 is the same as the remainder when the sum of all of its digits is divided by 3.

The square of an integer k has remainder 0 or 1 when it is divided by 3.
If k is a multiple of 3, then k = 3a for some integer a, and k^2 = (3a)^2 = 3(3a^2) has remainder 0 when it is divided by 3.
If k has remainder 1 when it is divided by 3, then k = 3a + 1 for some integer a, and k^2 = (3a+1)^2 = 9a^2 +6a + 1 = 3(3a^2 +2a) + 1 has remainder 1 when it is divided by 3.
If k has remainder 2 when it is divided by 3, then k = 3a + 2 for some integer a, and k^2 = (3a+2)^2 = 9a^2 +12a + 4 = 3(3a^2 +4a+1) + 1 has remainder 1 when it is divided by 3.

Since m=1234 has remainder 1 when it is divided by 3, m^2 has remainder 1 when it is divided by 3. Since n^2 could have remainder 0 or 1 when it is divided by 3, m^2 + n^2 is never divisible by 3, regardless of the value of n. Condition 1) is sufficient, since it yields the unique answer, ‘no’.

Since n=4321 has remainder 1 when it is divided by 3, n^2 has remainder 1 when it is divided by 3. Since m^2 could have remainder 0 or 1 when it is divided by 3, m^2 + n^2 is never divisible by 3, regardless of the value of m. Condition 2) is sufficient, since it yields the unique answer, ‘no’.

Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, both conditions are sufficient.
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[GMAT math practice question]

(arithmetic) A palindrome, such as 12321, is a number that remains the same when its digits are reversed. The numbers x and x+32 are three-digit and four-digit palindromes, respectively. What is the sum of the digits of x?

A. 18
B. 24
C. 28
D. 32
E. 36

=>

Let x = ABA and x + 32 = CDDC.
Then CDDC = ABA + 32
Since CDDC has a thousands digit, we must have A = 9.
Then the units digit of CDDC is equal to the units digit of A + 2 = 11.
And, looking at the tens digits, we must have B + 3 + 1 ≥ 10.
Therefore, B ≥ 6.
The possible values for B are:
B = 6, 7, 8, 9.
Let’s check which value gives a palindrome for ABA + 32:
969 + 32 = 1001
979 + 32 = 1011
989 + 32 = 1021
999 + 32 = 1031
The only palindrome is 1001, so x = 969.
Thus, the sum of the digits of x is 9 + 6 + 9 = 24.

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[GMAT math practice question]

(probability) A palindrome, such as 12321, is a number that remains the same when its digits are reversed. How many 4-digit palindromes are divisible by 4?

A. 10
B. 20
C. 25
D. 28
E. 30

=>

Palindromes between 1000 and 10,000 have the form ABBA.
When we check divisibility by 4, we need only check the divisibility of BA by 4.
The values of BA which are divisible by 4 are 04, 08, 12, 16, 24, 28, 32, 36, 44, 48, 52, 56, 64, 68, 72, 76, 84, 88, 92, and 96 (note that 20, 40, 60, 80 and 00 are not possible values of BA as they do not give rise to 4-digit numbers ABBA). The possible values for ABBA are 4004, 8008, 2112, 6116, 4224, 8228, 2332, 6336, 4444, 8448, 2552, 6556, 4664, 8668, 2772, 6776, 4884, 8888, 2992 and 6996.
Thus, there are 20 4-digit palindromes.

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[GMAT math practice question]

(probability) The vertices of a regular pentagon are to be colored using five different colors. In how many ways can the pentagon’s vertices be colored if the 5 colors are to be chosen from a palette of 6 different colors?

A. 64
B. 96
C. 108
D. 144
E. 192

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The number of ways to choose 5 colors out of 6 colors is 6C5 = 6.
The number of circular permutations of the 5 colors is (5-1)! = 4! = 24.
Thus, 6*24 = 144 different colorings of the pentagon are possible.

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[GMAT math practice question]

(algebra) x, y, z are 3 consecutive positive integers. What is the value of x?

1) x<y<z
2) 12+13+14+15=x+y+z

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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.

Consecutive integers have two variables for the first number and the number of integers. Since the number of integers is 3, we need one more equation and D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
If x = 1, y = 2 and z = 3, then we have x = 1.
If x = 2, y = 3 and z = 4, then we have x = 2.
Since condition 1) doesn’t yield a unique solution, it is not sufficient.

Condition 2)
Since 12 + 13 + 14 + 15 = x + y + z, we have x + y + z = 54.
If x = 17, y = 18, z = 19, then we have x = 17.
If x = 19, y = 18, z = 17, then we have x = 19.
Since condition 2) doesn’t yield a unique solution, it is not sufficient.

Conditions 1) & 2)
Since y = x + 1 and z = x + 2 from condition 1), we have 12 + 13 + 14 + 15 = 54 = x + y + z = x + x + 1 + x + 2 = 3x + 3 or 3x = 51.
Thus we have x = 17.
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.

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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# Overview of GMAT Math Question Types and Patterns on the GMAT  