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I. 1/2,2/3,3/4,4/5,5/6 II. 1/2,1/3,1/4,1/5,1/6 III. 1/6,2/6,3/6,4/6,5/6

For which of the above lists is the average of the numbers less than the median of numbers?

A. I only B. II only C. III only D. II and III E. I, II and III

=>

I. The differences between the consecutive terms are 1/6 (= 2/3 – 1/2 ), 1/12 (= 3/4 – 2/3 ), 1/20 (= 4/5 – 3/4 ), and 1/30 (= 5/6 – 4/5). As these are decreasing, the average is smaller than the median.

II. The differences between consecutive terms are - 1/6(= 1/3 – 1/2),-1/12(= 1/4 – 1/3 ), - 1/20(= 1/5 – 1/4), and -1/30(= 1/6 – 1/5). As these are increasing, the average is larger than the median.

III. As the data are symmetric, the average and the median are equal.

A. |x-3|<4 B. |x-3|<5 C. |x-3|<6 D. |x-3|>4 E. |x-3|>5

=>

|x-a|<b ⟺ -b < x – a < b ⟺ a - b < x < a + b So, a is the midpoint of the end points a – b and a + b, and b is half the distance between the end points a – b and a + b.

The midpoint of -3 and 9 is [(-3)+9]/2 = 6/2 = 3. So, a = 3. Since the distance between -3 and 9 is 12, b = 122 = 6. Thus, |x-3| < 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.

Condition 1) x + 1 > 0 x > -1 Since the range of the question does not include the range of the condition, this is not sufficient.

Condition 2) |x+4| < |x-2|⟺|x+4|^2 < |x-2|^2⟺(x+4)^2 < (x-2)^2⟺x^2 + 8x + 16 < x^2 -4x + 4⟺8x + 16 < -4x + 4⟺12x < -12⟺x < -1 The answer is ‘no’. By CMT (Common Mistake Type) 1, ‘no’ is also an answer. This is sufficient.

Therefore, the answer is 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.

A group ticket for admission to art gallery A costs $p for up to 10 people. Each additional person is charged an admission fee of $q. A group ticket for admission to museum M costs $r for up to 15 people. Each additional person is charged an admission fee of $s. If a group of 20 people visit both the art gallery A and the museum M, is the total cost of their admission to the art gallery less than the total cost of their admission to the museum?

1) p<r 2) q<s

=> 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 question asks if p + (20 – 10)q = p + 10q is smaller than r + ( 20 – 15 )s = r + 5s.

Since we have 4 variables (p, q, r, and s) and 0 equations, E is most likely to be the answer. Therefore, we should consider both conditions together first.

Conditions 1) & 2) If p = 10, q = 20, r = 100, and s = 200, then p+10q=210<1,100=r+5s. The answer is ‘yes’. If p = 10, q = 20, r = 11, and s = 21, then p+10q=210>116=r+5s. The answer is ‘no’. So, conditions 1) & 2) are NOT sufficient when taken together.

Therefore, the answer is E, as expected.

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.

What is the value of x in the list {36, 47, 51, 65, x}?

1) The median of the 5 values is 51. 2) The range of the 5 values is 29.

=>

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.

Condition 1) Since the median is 51, x could be any value greater than or equal to 51. This is NOT sufficient.

Condition 2) Since the range is 29=65-36, x could be any value between 36 and 65, inclusive. This is NOT sufficient.

Conditions 1) and 2) Condition 1) tells us that x could be any value greater than or equal to 51, while condition 2) tells us that x could be any value between 36 and 65, inclusive. Combining these ranges tells us that x could be any value between 51 and 65, inclusive. Therefore, the two conditions are NOT sufficient, when considered together. The answer is E.

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.

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.

00 lies outside the scope of the GMAT exam.

Condition 1) There are two possible solutions: x = 1 and x = -1. Since the solution is not unique, this condition is not sufficient.

Condition 2) We have the unique solution x = 1. Since this solution is unique, this condition is sufficient.

Therefore, the answer is 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. Answer: B
_________________

For which value of x will y=ax^2+100x+b have a maximum in the xy-plane?

1) b=100 2) a=-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, and then recheck the numbers of variables and equations.

We can modify the original condition and question as follows:

If a < 0, the function will have a maximum at x = (-100)/(2a) = (-50)/a. If a > 0, the function has no maximum. So, to answer the question, we need to find the value of a. Thus, condition 2) is sufficient.

Note: condition 1) cannot be sufficient as it provides no information about the value of a.

For positive integers x and y, x@y is defined by x@y=(x+y)/xy. If a, b, and c are positive integers, what is the value of 1/a@1/(1/b@1/c)?

A. a+b+c B. 1/abc C. 1/(a+b+c) D. 1/(ab+bc+ca) E. 3/abc

=>

We can simplify the definition of the operation in the following way: x@y=(x+y)/xy = x/(xy) + y/(xy) = 1/x+1/y. So, 1/a@1/(1/b@1/c) = a + (1/b @1/ c) = a + ( b + c ).

Therefore, the answer is A. Answer: A
_________________

When 2 fair dice are tossed, what is the probability that the difference between the 2 numbers that land face up will be 3?

A. 1/6 B. 1/3 C. 1/2 D. 2/3 E. 5/6

=>

6 pairs of numbers with this property can appear on the dice: (1,4), (4,1), (2,5), (5,2), (3,6) and (6,3). The total number of outcomes from rolling two dice is 36. Thus, the probability that the two numbers will have a difference of 3 is 6/36 = 1/6. Therefore, the answer is A.

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.

Condition 1) & 2) Since x and y are different integers, xy = 4 has the following four solutions: x = 1, y = 4 x = 4, y = 1 x = -1, y = -4 x = -4, y = -1 In all cases, |x-y| = 3. Thus, both conditions 1) and 2) together are sufficient.

Therefore, C is the answer.

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. Answer: C
_________________

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 method is to modify the original condition and the question, and then recheck the question.

Condition 1) If a > 0, then a^b>0. The answer is ‘yes’. So, this condition is sufficient.

Condition 2) For a = 2, b = 2, we have a^b = 2^2 = 4 > 0, and the answer is ‘no’. For a = -2, b = 3, we have a^b = (-2)^2 = -8 < 0, and the answer is ‘yes’. So, condition 2) is not sufficient. Therefore, the answer is A. Normally, in problems which require 2 or more additional 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).

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 VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations.

The question asks if 0<x<1. This question has the unique answer, ‘yes’, if we require both x>0 and x<1. Thus, both conditions together are sufficient.

Therefore, C is the answer.

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. Answer: C

If x is an integer and |x|+(x/3)<5, what is the value of x?

1) x>-12 2) x<-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 VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations.

Modifying the original condition: There are two cases to consider. Case 1) x ≥ 0 |x|+(x/3)<5 ⇔ x + x/3 < 5 ⇔ (4/3)x < 5 ⇔ x < 15/4 ⇔ x = 0, 1, 2, 3

Case 2) x < 0 |x|+(x/3)<5 ⇔ -x + x/3 < 5 ⇔ -(2/3)x <5 ⇔ x > -15/2 ⇔ x > -7.5 ⇔ x = -7, -6, -5, -4, -3, -2 or -1

The question asks for the value of x if x is one of -7, -6, -5, …, 0, 1, 2, 3. Since we have 1 variable (x), D is most likely to be the answer.

Condition 1) All of the possible values of x are greater than -12. Therefore, we do not have a unique solution, and this condition is not sufficient.

Condition 2) The only possible value of x is -7. Since we have a unique solution, condition 2) is sufficient.

=> 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, and then recheck the question.

If x and y are integers, is x^3-y^3 an odd integer?

1) x is an odd number 2) x and y are 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.

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

We can modify the original condition and question as follows. There are two different ways in which x^3-y^3 can be odd: x is even and y is odd. x is odd and y is even.

Since condition 2) tells us that x and y are consecutive integers, one of them must be odd, and the other must be even. In both cases, the answer is ‘yes’. Therefore, condition 2) is sufficient.

As condition 1) gives us no information about y, it is not sufficient.

Therefore, B is the answer.

Normally, in problems which require 2 or more additional 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.

I, …, E, …, M, …, S, …, K, …, W, … Z The frequencies of the appearance of the letters from ‘A’ to ‘Z’ in a book are listed in decreasing order as shown above. ‘I’ is the most frequently occurring letter. The book contains a total of 100,000 alphabetic characters. Is the probability of randomly selecting an ‘S’ from the alphabetic characters appearing in the book greater than 1/27?

1) The probability of selecting ‘M’ is greater than 1/27. 2) The probability of selecting ‘K’ is greater than 1/26.

=>

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 which are the probability of S, P(S), that of M, P(M) and that of K, P(K) and we have 1 equation which is P(M) > P(S) > P(K), C is most likely to be the answer and so we should consider both conditions 1) & 2) together first.

Conditions 1) & 2) Since P(M) > P(S) > P(K) > 1/26 > 1/27, both conditions together are sufficient.

Since this is an inequality question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Condition 1) If P(M) > P(S) = 1/27, the answer is ‘no’. If P(M) = 1/25 and P(S) = 1/26 > 1/27, the answer is ‘yes’. Thus, condition 1) is not sufficient.

Condition 2) Since P(S) > P(K) > 1/26 > 1/27, condition 2) is sufficient.

Therefore, B is the answer.

Normally, in problems which require 2 or more additional 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.