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Math Revolution GMAT Instructor
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Re: The Ultimate Q51 Guide [Expert Level]
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11 Apr 2019, 19:23
[GMAT math practice question] A right triangle has hypotenuse 10. If its perimeter is 25, what is its area? A. 125/4 B. 125/2 C. 125 D. 225/4 E. 225/2 => Let a and b be the legs of the right triangle. Since the hypotenuse is 10, a^2+b^2=100. Since the triangle’s perimeter is 25, we have a + b + 10 = 25 and a + b = 15. Recall that (a+b)^2 = a^2 + 2ab + b^2. 2ab = (a+b)^2 – (a^2+b^2) = 225 – 100 = 125. Thus, the area of the triangle is (1/2)ab = 125 / 4. Therefore, A is the answer. Answer: A
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Re: The Ultimate Q51 Guide [Expert Level]
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14 Apr 2019, 18:40
[GMAT math practice question] If the median and average (arithmetic mean) of a set of 4 different numbers are both 10, what is the smallest number? 1) The range of the 4 numbers is 10 2) The sum of the smallest and the largest numbers is 20 => 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. Let a, b, c and d be the 4 numbers, and suppose a < b < c < d. Then ( a + b + c + d ) / 4 = 10 and ( b + c ) / 2 = 10. Since b + c = 20 and a + b + c + d = 40, we must have a + d = 20. Condition 1) Since d – a = 10 by condition 1), we can figure out the values of a and d. Thus, condition 1) is sufficient. Condition 2) a + d = 20 can be deduced from the original condition as shown above. So, condition 2) provides no additional information. If a = 1, b = 9, c = 11 and d = 19, then the smallest number is 1. If a = 2, b = 9, c =11 and d = 18, then the smallest number is 2. Condition 2) is not sufficient since it does not yield a unique answer. Therefore, A is the answer. Answer: A
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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7252
GPA: 3.82

Re: The Ultimate Q51 Guide [Expert Level]
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15 Apr 2019, 19:03
[GMAT math practice question] If m and n are positive integers, is m^2n^2 divisible by 4? 1) m^2+n^2 has remainder 2 when it is divided by 4 2) m*n 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. The statement “m^2n^2 is divisible by 4” means that (m+n)(mn) is divisible by 4. This is equivalent to the requirement that m and n are either both even integers or both odd integers. Since condition 2) tells us that both m and n are odd integers, condition 2) is sufficient. Condition 1) The square of an odd integer (2a+1)^2 = 4a^2 + 4a + 1 = 4(a^2 + a) + 1 has remainder 1 when it is divided by 4. The square of an even integer (2b)^2 = 4b^2 has remainder 0 when it is divided by 4. Thus, if “m^2+n^2 has remainder 2 when it is divided by 4”, both m and n must be odd integers. Condition 1) is sufficient. Therefore, D is the answer. Answer: D FYI, Tip 1) of the VA method states that D is most likely to be the answer if conditions 1) and 2) provide the same information.
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MathRevolution: Finish GMAT Quant Section with 10 minutes to spareThe oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $149 for 3 month Online Course""Free Resources30 day online access & Diagnostic Test""Unlimited Access to over 120 free video lessons  try it yourself"



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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GPA: 3.82

Re: The Ultimate Q51 Guide [Expert Level]
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16 Apr 2019, 18:02
[GMAT math practice question] (number properties) What is the greatest common divisor of positive integers m and n? 1) m and n are consecutive 2) m^2 – n^2 = m + n => 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. Condition 1) The gcd of two consecutive integers is always 1. Thus, condition 1) is sufficient. Condition 2) If m^2–n^2 = m+n, then (m+n)(mn)=m+n and mn = 1 since m+n ≠0. This implies that m and n are consecutive integers, and their gcd is 1. Condition 2) is sufficient since it yields a unique answer. Therefore, D is the answer. Answer: D FYI, Tip 1) of the VA method states that D is most likely to be the answer if conditions 1) and 2) provide the same information.
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MathRevolution: Finish GMAT Quant Section with 10 minutes to spareThe oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $149 for 3 month Online Course""Free Resources30 day online access & Diagnostic Test""Unlimited Access to over 120 free video lessons  try it yourself"



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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GPA: 3.82

Re: The Ultimate Q51 Guide [Expert Level]
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17 Apr 2019, 18:02
[GMAT math practice question] (number property) If p and q are prime numbers, what is the number of factors of 6pq? 1) p and q are different 2) p<3<q => 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. Recall that if n = p^aq^br^c, where p, q and r are different prime numbers, and a, b and c are nonnegative integers, then n has (a+1)(b+1)(c+1) factors. Condition 2) We must have p = 2 since p is prime and p < 3. The prime factorization of 6pq is 2*3*p*q = 2^2*3*q since q is prime and q > 3. The number of factors of 2^2*3*q is (2+1)(1+1)(1+1) = 12. Condition 2) is sufficient since it yields a unique answer. Condition 1) If p = 2 and q = 3, then 6pq = 2^2*3^2 and the number of factors is (2+1)(2+1)= 9. If p = 5 and q = 7, then 6pq = 2*3*5*7 and the number of factors is (1+1)(1+1)(1+1)(1+1) = 8. Condition 1) is not sufficient since it does not yield a unique solution. Therefore, B is the answer. Answer: B
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Re: The Ultimate Q51 Guide [Expert Level]
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18 Apr 2019, 18:24
[GMAT math practice question] (probability) How many rectangles are found in the lattice below? Attachment:
4.9.png [ 1.72 KiB  Viewed 80 times ]
A. 90 B. 100 C. 120 D. 150 E. 180 => Each rectangle is uniquely determined by the intersections between two vertical lines and two horizontal lines. Since we have 6 vertical lines and 5 horizontal lines, the number of rectangles is 6C 2* 5C 2 = 15*10 = 150. Therefore, D is the answer. Answer: D
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Math Revolution GMAT Instructor
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Re: The Ultimate Q51 Guide [Expert Level]
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21 Apr 2019, 18:20
[GMAT math practice question] (number properties) m and n are positive integers greater than 6. What is the value of m + n? 1) m*n = 504 2) m and n are multiples of 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 2 variables (m and n) 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) Condition 1) allows us to write m = 6a and n = 6b, where a and b are integers greater than 1. So, m*n = 6a*6b = 36*a*b = 504. This yields ab = 14. So, a = 2 and b = 7 or a = 7 and b = 2. Thus, m=12 and n=42 or m=42 and n=12, and we obtain the unique answer m+n = 54. 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) m*n = 504 = 2^3*3^2*7 If m = 12 and n = 42, then m + n = 54. If m = 8 and n = 63, then m + n = 71. Condition 1) is not sufficient since it does not yield a unique answer. Condition 2) If m = 12 and n = 12, then m + n = 24. If m = 12 and n = 24, then m + n = 36. Condition 2) is not sufficient since it does not yield a unique answer. Therefore, C is the answer. Answer: C 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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Re: The Ultimate Q51 Guide [Expert Level]
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22 Apr 2019, 18:50
[GMAT math practice question] (number properties) m * n = 2145, where m and n are positive integers. What is the value of m + n? 1) m and n are twodigit integers. 2) m and n both have remainder 3 when they are divided by 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. Since we have 2 variables (m and n) and 1 equation (mn=2145), D is most likely to be the answer. Condition 1) m * n = 2145 = 3*5*11*13. Since m and n are twodigit numbers, there are four cases to consider: i) m=33 (= 3*11) and n=65 (= 5*13) ii) m=39 (=3*13) and n=55 (=5*11) iii) m=65 (=5*13) and n=33 (= 3*11) iv) m=55 (=5*11) and n=39 (=3*13) So, there are two possible values of m + n, which are 98 and 94. Condition 1) is not sufficient since it does not yield a unique answer. Condition 2) If m = 39 and n = 55, then m + n = 94. If m = 3 and n = 715(=5*11*13), then m + n = 718. Condition 2) is not sufficient since it does not yield a unique answer. Conditions 1) & 2) m * n = 2145 = 3*5*11*13. Condition 1) gives rise to the following four cases for the values of m and n: i) m=33 (= 3*11) and n=65 (= 5*13) ii) m=39 (=3*13) and n=55 (=5*11) iii) m=65 (=5*13) and n=33 (= 3*11) iv) m=55 (=5*11) and n=39 (=3*13) Of these, only m=39 and n=55, and m=55 and n=39 satisfy condition 2) So, we have a unique answer m+n=94. Thus, both conditions together are sufficient. Therefore, C is the answer. Answer: C 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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MathRevolution: Finish GMAT Quant Section with 10 minutes to spareThe oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $149 for 3 month Online Course""Free Resources30 day online access & Diagnostic Test""Unlimited Access to over 120 free video lessons  try it yourself"



Math Revolution GMAT Instructor
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Posts: 7252
GPA: 3.82

Re: The Ultimate Q51 Guide [Expert Level]
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25 Apr 2019, 01:29
[GMAT math practice question] (function) The parabola y=f(x)=a(xh)^2+k lies in the xy plane. What is the value of k? 1) y=f(x) passes through (1,0) and (3,0). 2) y=f(x) passes through (2,1) and no yvalue is greater than 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. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. Attachment:
4.18.png [ 10.53 KiB  Viewed 16 times ]
The question asks for the minimum or maximum value for the function. Condition 2 is sufficient since it provides the maximum value for the function. Condition 1) Attachment:
4.18...png [ 10.05 KiB  Viewed 16 times ]
The parabolas drawn above both pass through the points (1,0) and (3,0). It is obvious that we don’t have a unique maximum or minimum function value. Condition 1) is not sufficient. Therefore, B is the answer. Answer: B
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MathRevolution: Finish GMAT Quant Section with 10 minutes to spareThe oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $149 for 3 month Online Course""Free Resources30 day online access & Diagnostic Test""Unlimited Access to over 120 free video lessons  try it yourself"



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7252
GPA: 3.82

Re: The Ultimate Q51 Guide [Expert Level]
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25 Apr 2019, 18:29
[GMAT math practice question] (number properties) What is the largest multiple of 24 that can be written using each of the digits 0, 1, 2, … , 9 exactly once? A. 9876543210 B. 9876543120 C. 987654102 D. 1234567890 E. 1234567980 => A number is a multiple of 24 precisely when it is a multiple of both 3 and 8. The sum of 0, 1, 2, … , 9, “0 + 1 + 2 + … + 9 = 45” is a multiple of 3. Thus, a number which uses each of the digits 0, 1, 2, … , 9 exactly once is a multiple of 3. In order for a number to be a multiple of 8, the last three digits of the number must be a multiple of 8. And the smallest multiple of 8 that can be formed using three of the digits 0, 1, 2, … , 9 is 120. Thus, the biggest possible multiple of 24 is 9876543120. Therefore, B is the answer. Answer: B
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MathRevolution: Finish GMAT Quant Section with 10 minutes to spareThe oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $149 for 3 month Online Course""Free Resources30 day online access & Diagnostic Test""Unlimited Access to over 120 free video lessons  try it yourself"




Re: The Ultimate Q51 Guide [Expert Level]
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