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There are 100 black balls, 100 red balls and 100 green balls evenly mixed in a jar. What is the minimum number of balls to ensure that at least 15 beans selected are the same color?

A. 40 B. 41 C. 42 D. 43 E. 44

=> We can assume we have 14 black balls, 14 red balls and 14 green balls. After that, if we have just one more ball, it ensures we have 15 balls with a same color.

Thus we need 43 ( = (15-1) + (15-1) + (15-1) + 1 = 14 + 14 + 14 + 1 ) balls.40 balls ensure there are 15 balls that have a same color.

2) x = 1, y = 2 and x = 2, y = 5/2 We don’t have a unique solution.

1) & 2) We have the following cases. x = 1, y = 5 / x = -1, y = -5 / x = 5, y = 1 / x = 1, y = 5 All of them have |x-y| = 4. Thus we have a unique solution. Both conditions together are sufficient.

If a, b, and c are different positive integers, what is the value of a+b+c?

1) a2+b2+c2=14 2) ab+bc+ca=11

=>

Condition 1) We can assume a < b < c without loss of generality. The maximum value of c is 3 and c^2 = 9 a^2 + b^2 = 5. Then we have b = 2 and a = 1. a + b + c = 1 + 2 + 3 = 6

Condition 2) We can assume a < b < c without loss of generality. ab + bc + ca = (a+b)c + ab = 11 Since a + b >= 3, the maximum value of c = 3. If c = 3, ab + 3b + 3a = 11 or ab + 3a + 3b + 9 = 20. We have (a+3)(b+3) = 20. Then a = 1 and b = 2. Thus a + b + c = 1 + 2 + 3 = 6.

For a positive integer n, if 5^^n is a factor of 25!, but 5^{n+1} is not a factor of 25!, what is the value of n?

A. 5 B. 6 C. 7 D. 8 E. 9

=>

25! = 1 x 2 x … x 5 x … x 10 x … x 15 x … x 20 x … x 25 = 1 x … x 4 x 6 x … x 9 x 11 x … x 14 x 16 x … x 19 x 21 x … x 24 x 5 x 10 x 15 x 20 x 25 = 1 x … x 4 x 6 x … x 9 x 11 x … x 14 x 16 x … x 19 x 21 x … x 24 x ( 5 x 5 x 2 x 5 x 3 x 5 x 4 x 5^2 ) = 1 x … x 4 x 6 x … x 9 x 11 x … x 14 x 16 x … x 19 x 21 x … x 24 x (2 x 3 x 4 x 5^6 )

3. (statistics) What is the median value of 5 data? 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.

What is the median of 3 consecutive integers? 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)(n2 + 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 equal number of variables and independent equations ensures a solution.

There are 2 variables and 0 equation. The answer could be C most likely.

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

Therefore, E is the answer 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.

In the coordinate plane, a triangle ABC has 3 points A(-1,0), B(1,0), and C(m,n). What is the area of the triangle ABC?

1) m=2 2) n=2

=> 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. 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 consider AB as the base and the y-coordinate n of the point C as another point of the triangle, and |n| is its height. Thus, its area is (1/2)*2*|n| = |n|.

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.

Since we have 1 variable and 0 equation, D is most likely to be the answer. According to CMT4(B), if D is chosen as an answer too easily for key area questions, we should check the answer A or B. This question is about integers, which is one of the key question areas.

Condition 1) n^2 = 2*3*k for an integer. Since 2 and 3 are prime factors of n^2, 2 and 3 are prime factor of n. Thus, 2*3 = 6 is a factor of n. This is sufficient.

Condition 2) We have two cases as follows. n = 3 does not have 6 as a factor. n = 12 has 6 as a factor. This is not sufficient.

Therefore, the answer is A.

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. Thus, 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.

The number of club X’s members is increased by 20 percent every year. If the number of members of the club X in 2000 was m, how many number of members in 2005 are greater than that of members in 2004?

A. (0.2)(1.2)^4m B. (0.8)(1.2)^4m C. (0.2)(1.2)^5m D. (0.8)(1.2)^5m E. (0.8)4(1.2)^4m

=> 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.

Since we have 3 variables and 0 equation, E is most likely to be the answer.

Conditions 1) & 2)

From the condition 1) and 2), |x+y| = |x| + |y| is equivalent to xy ≥ 0 and |y+z| = |y| + |z| is equivalent to yz ≥ 0. Then we have xy2z ≥ 0 or xz ≥ 0. According to CMT 1, since “No” is also an answer, both conditions are sufficient.

The answer is C.

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. Thus, 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. Answer: A
_________________

If x is a 5-digit integer, abc,de, is it divisible by 4?

1) d is an even number. 2) e is an even number.

=> 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 check the divisibility by 4 with the last two digits. The last two digit number is 10d + e. Since we have 2 variables and 0 equation, C is most likely to be the answer. Condition 1) & 2) d = 2, e = 4 ➔ 24 is divisible by 4. d = 4, e = 4 ➔ 44 is not divisible by 4. Therefore, 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 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. Answer: E
_________________

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