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When a positive integer n is divided by 4, what is the remainder?

1) The remainder is 5 when n is divided by 6. 2) The remainder is 7 when n is divided by 9.

=>

Condition 1) The divisor 6 of the condition 1) is not a multiple of 4, which is the divisor of the question. This is not sufficient.

Condition 2) The divisor 9 of the condition 2) is not a multiple of 4, which is the divisor of the question. This is not sufficient.

Condition 1) & 2) The LCM of 6 and 9 which are divisors of both conditions is 18. 18 is not a multiple of 4. Thus both conditions together are not sufficient.

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

Using VA (Variable Approach) method, since we have 2 variables and 0 equation, C is the answer most likely.

Condition 1) Since we don’t know values of x and y, this is not sufficient.

Condition 2) x = 1, y =5 ⇒ | x – y | = | 1 – 5 | = | - 4 | = 4 x = 1/2, y = 10 ⇒ | x – y | = | 1/2 – 10 | = | - 19/10 | = 19/10 This is not sufficient.

Condition 1) & 2) Since x and y are integers with xy = 5, we have ( x, y ) = (1, 5), (5, 1), (-1, -5) or (-5, -1). | x – y | of all cases are 4. Both conditions together are sufficient.

Therefore, the answer is C as expected.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore, C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

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

Using VA (Variable Approach) method, since we have 2 variables and 0 equation, C could be the answer most likely.

Condition 1) x^2-4xy=-4y^2 x^2-4xy+4y^2= 0 (x-2y)^2 = 0 x – 2y = 0 x = 2y

( x + y ) / ( x – y ) = ( 2y + y ) / ( 2y – y ) = 3y / y = 3

This is sufficient.

Condition 2)

(x^2+1)(x-2y)=0 x^2 + 1 = 0 or x – 2y = 0. We can take x – 2y = 0, since x^2 + 1 ≠ 0. Thus, we have x = 2y.

( x + y ) / ( x – y ) = ( 2y + y ) / ( 2y – y ) = 3y / y = 3

This is sufficient too.

Therefore, the answer is D unlike our expectation..

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore, C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

If x, y, and z are positive integers, is x+y divisible by 5?

1) x+z is divisible by 5 2) y+z is divisible by 5

=>

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

There are 3 variables and 0 equation. Thus the answer E is most likely.

Condition 1) & 2) x = 2, y = 7, z = 3 ➔ x + y = 9 is not divisible by 5. x = 5, y = 10, z = 15 ➔ x + y = 15 is divisible by 5.

Therefore, E is the answer as expected.

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Since E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously there may be cases where the answer is A, B, C or D.

=> Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. In the original condition, there is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.

Condition 1) We have 2 possible values, which are x = 1 and x = -1. This is not sufficient.

Condition 2) The only possible value is x = 1.

Therefore, the answer is B unlike our expectation. For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E. Answer: B
_________________

If the operation @ is defined by x@y=ax+by (a and b are constants), what is the value of 24?

1) 1@2=5 2) 1@1=6

=>

Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures 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 is what the value of 2a + 4b is.

Condition 1) a + 2b = 5

2a + 4b = 2(a+2b) = 2*5 = 10 This is sufficient.

Condition 2) a + b = 6

a = 1, b = 5 ➔ 2a + 4b = 2 + 20 = 22 a = 3, b = 3 ➔ 2a + 4b = 6 + 12 = 18 This is not sufficient.

Therefore, A is the answer.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore, C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

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

There are 2 variables and 0 equation. Therefore, C is most likely to be the answer.

Conditin 1) & 2) x = 1, y = 2 ➔ xy = 2 < 15 : Yes x = -4, y = -4 ➔ xy = 16 > 16 : No

Therefore, unlike our expectation, E is the answer.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both con 1) and con 2). Therefore, C has a high chance of being the answer, which is why we attempt to solve the question using con 1) and con 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using con 1) and con 2) together. (It saves us time). Obviously, there may be cases where the answer is A, B, D or E.

1) The first 3 smallest data are 10, 12, 15. 2) The biggest data is 20

=> Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures a solution. We can assume that a1 ≤ a2 ≤ a3 ≤ a4 ≤a5. The question asks what the value of a3 is. In the original condition, there is 5 variables and 0 equation. Therefore, E is most likely to be the answer.

Condition 1) a1 = 10, a2 = 12, a3 = 15. Thus, the median is 15 and this is sufficient.

Condition 2) a5 = 20. This is not sufficient.

Therefore, unlike our expectation, E is the answer. Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both con 1) and con 2). Therefore, C has a high chance of being the answer, which is why we attempt to solve the question using con 1) and con 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using con 1) and con 2) together. (It saves us time). Obviously, there may be cases where the answer is A, B, D or E.

If n is an integer, is (n-1)(n+1) a multiple of 24?

1) n is odd. 2) n is not divisible by 3.

=>

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures a solution. We have 1 variable and 0 equation from the original condition. Therefore, D is most likely to be the answer.

Condition 1) Since n is odd, n-1 and n+1 are consecutive even integers. One of two consecutive even integers must be a multiple of 4. For example, (2,4), (6,8), (8,10) and (10,12). Thus (n-1)(n+1) is a multiple of 4, but its divisibility by 3 is not identified. Counterexamples are n = 3 and n = 5. (3-1)(3+1) = 2*4 = 8, which is not a multiple of 24. (5-1)(5+1) = 4*6 = 24, which is a multiple of 24. This is not sufficient.

Condition 2) Since n is not a multiple of 3, n = 3k +1 or n = 3k + 2 for some integer k. For the case, n = 3k +1, (n-1)(n+1) = (3k+1-1)(3k+1+1) = 3k(3k+2) is a multiple of 3. However, we can’t identify if n is a multiple of 24. Counterexamples are n = 4 for which (n-1)(n+1) = 3*5 = 15 is not a multiple of 24 and n = 7 for which (n-1)(n+1) = 6*8 = 48 is a multiple of 24. For the case, n = 3k +2, (n-1)(n+1) = (3k+2-1)(3k+2+1) = (3k+1)(3k+3) = 3(3k+1)(k+1) is a multiple of 3. However, we can’t identify if n is a multiple of 24. Counterexamples are n = 5 for which (n-1)(n+1) = 4*6 = 15 is a multiple of 24 and n = 6 for which (n-1)(n+1) = 5*7 = 35 is a not multiple of 24.

Condition 1) & 2) From the condition 1), (n-1)(n+1) is a multiple of 8 and from the condition 2), (n-1)(n+1) is a multiple of 3. Thus, (n-1)(n+1) is a multiple of 24. Therefore, unlike our expectation, C is the answer.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both con 1) and con 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using con 1) and con 2) separately according to DS definition. Obviously, there may be cases where the answer is A, B, C or E.

1) The product of the integers is 0 2) The sum of the integers and the product of the integers are the same

=>

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures a solution. We assume that we have 3 consecutive integers, n, n+1 and n+2. We have 1 variable and 0 equation from the original condition. Therefore, D is most likely to be the answer.

Condition 1) n(n+1)(n+2) = 2. We have n = 0, n = -1 or n = -2. This is not sufficient.

Condition 2) n + ( n + 1 ) + ( n + 2 ) = n(n+1)(n+2) 3n + 3 = n(n+1)(n+2) 3(n + 1) = n(n+1)(n+2) n(n+1)(n+2) - 3(n + 1) = 0 (n+1){ n(n+2) – 3 } = 0 (n+1)(n^2 + 2n – 3) = 0 (n+1)(n-1)(n+3) = 0 n = -1, n = 1 or n = -3 This is not sufficient. Condition 1) & 2) From the condition 1), we have n = 0, n = -1 or n = -2. From the condition 2), we have n = 1, n = -1 or n = -3. Thus, n = -1. Both conditions together are sufficient.

Therefore, unlike our expectation, C is the answer.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both con 1) and con 2). Therefore, C has a high chance of being the answer, which is why we attempt to solve the question using con 1) and con 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using con 1) and con 2) together. (It saves us time). Obviously, there may be cases where the answer is A, B, D or E.

Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures 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.

Under the condition ab > 0, the question ab2 > 0 is equivalent to a > 0.

The condition 1) is equivalent to the question a > 0. This is sufficient. And under the original condition ab > 0, the condition 2) b > 0 is equivalent to a > 0. This is sufficient too.

Therefore, the answer is D.

By the tip 1) of VA method, since two conditions are same, we can choose D as the answer.

If the average (arithmetic mean) of 5 numbers is 30, what is the standard deviation of the numbers?

1) The smallest number of them is 30 2) The greatest number of them is 30

=> Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures 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. We can encounter this kind of question in GMAT quant exam these days. If we have the mean same as one of the maximum and the minimum or the range ( = Max – Min ) is zero, the standard deviation is zero.

Condition 1) All data are greater than or equal to 30, and their average is 30. It means all data are 30. Thus, their standard deviation is 0, since all data are same. This is sufficient.

Condition 2) All data are less than or equal to 30, and their average is 30. It means all data are 30. Thus, their standard deviation is 0, since all data are same. This is sufficient too.

Therefore, the answer is D.

By the tip 1) of VA method, since two conditions are same, we can choose D as the answer.

If n is a positive integer, is 3^n+3^8 divisible by 4?

1) n is a multiple of 3. 2) 3^2 +3^n+2 is divisible by 4.

=> Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures 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.

Since 38 has a remainder 1 when it is divided by 4, the question asks if 3n has a remainder 3 if it is divided by 4. 3^1 = 3 has the remainder 3, when it is divided by 4. 3^2 = 9 has the remainder 1, when it is divided by 4. 3^3 = 27 has the remainder 3, when it is divided by 4. 3^4 = 81 has the remainder 1, when it is divided by 4. ... Thus, the question asks if n is an odd number.

Condition 1) Since n is a multiple of 3, it is unknown if n is odd or even. This is not sufficient.

Condition 2) “3^2 +3^n+2 is divisible by 4” means 3^n+2 has a remainder of 3 when divided by 4. It means n + 2 and n are odd integers. Thus, this condition is sufficient. Therefore, the answer is B.