#1
29th February 2016, 03:07 PM
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GRE Math Practice
hii sir, I am preparing for the GRE math test will you please provide me the asample question paper with solution for the mathematics ?
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#2
29th February 2016, 03:49 PM
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Re: GRE Math Practice
As you asking for the GRE mathematics sample question papwer with solution that is as follow : Question Type: Multiple Answer Questions (Choose all that apply) Concept: Absolute Value/Algebra Level: 145 – 150 What are all the possible solutions of | |x – 2| – 2| = 5? -5 -3 -1 7 9 Answers: A, E. If we focus just on the delim{|}{x-2}{|}, we can see that the result must be positive. Stepping back and looking at the entire equation we substitute u for delim{|}{x-2}{|}, to get delim{|}{u-2}{|}=5. Solving for absolute value, we get the following: u-2=5 u-2=-5 Thus, u=7 and u=-3. Because u must be positive, we discount the second result. Next, we have to find x in the original delim{|}{x-2}{|}, which we had substituted with u. Replacing u with 7 we get: delim{|}{x-2}{|}=7 and delim{|}{x-2}{|}=-2 x=9 and x = -5 . A faster way is to plug in the answer choices to see which ones work. Concept: Symbolic Reasoning/Exponents Level: 165 – 170 If (a^2)(b) is an integer which of the following must be an integer? a b ab b^2 None of the above Answer: E. Let’s choose numbers to disprove each case. By the way, the word disprove is very important here – the question says ‘must’ so by picking numbers that prove the case, we are not necessarily proving that an answer choice must always be an integer. For A. I can choose sqrt{3}=a, and b is any integer. Because a is not an integer, A. is not correct. For B. it’s a bit tricky. However, if you keep in mind that there are no constraints in the problem stating that a cannot equal b, we can make a=3^{1/4} and b=sqrt{3}. For C. we can choose the same numbers to show that ab is not an integer. For D. if b=3^{1/4} and a=3^{3/8} (a^2)(b) equals an integer, but b^2 does not. Concept: Prime Numbers/Factors Level: 150 – 155 How many positive integers less than 100 are the product of three distinct primes? [5] Answer: 5 Let’s write out some primes: 2, 3, 5, 7, 11, 13, and 17. I’m stopping at 17 because the smallest distinct primes, 2 and 3, when multiplied. by 17 give us 102. Therefore 13 is the greatest prime conforming to the question. Here is one instance. 2*5*13 is greater than 100 so we can discount it. Working in this fashion we can add the following instances: 2*3*5 2*3*7 2*3*11 2*5*7 3*5*7 is too great 2*3*13=78. Therefore, there are five instances. Concept: Exponents and Fractions Level: 155 – 160 -1<x<100 Column A Column B x^3 x^6 The quantity in Column A is greater The quantity in Column B is greater The two quantities are equal relationship cannot be determined from the information given Answer: D. If x is less than 0 the answer is B. If x is 0 <x<1, the answer is A. Therefore, the answer is D. Gre Math practise question paper What are all the possible solutions of | |x – 2| – 2| = 5? 1. -5 2. -3 3. -1 4. 7 5. 9 Answers: A, E. If we focus just on the , we can see that the result must be positive. Stepping back and looking at the entire equation we substitute u for , to get . Solving for absolute value, we get the following: Thus, and . Because u must be positive, we discount the second result. Next, we have to find in the original , which we had substituted with u. Replacing u with 7 we get: and and . A faster way is to plug in the answer choices to see which ones work. Question Type: Multiple Choice Concept: Symbolic Reasoning/Exponents Level: 165 – 170 If is an integer which of the following must be an integer? 1. 2. 3. 4. 5. None of the above Answer: E. Let’s choose numbers to disprove each case. By the way, the word disprove is very important here – the question says ‘must’ so by picking numbers that prove the case, we are not necessarily proving that an answer choice must always be an integer. For A. I can choose , and b is any integer. Because a is not an integer, A. is not correct. For B. it’s a bit tricky. However, if you keep in mind that there are no constraints in the problem stating that a cannot equal b, we can make and . For C. we can choose the same numbers to show that ab is not an integer. For D. if and equals an integer, but does not. Question Type: Numeric Entry Concept: Prime Numbers/Factors Level: 150 – 155 How many positive integers less than 100 are the product of three distinct primes? [5] Answer: 5 Let’s write out some primes: 2, 3, 5, 7, 11, 13, and 17. I’m stopping at 17 because the smallest distinct primes, 2 and 3, when multiplied. by 17 give us 102. Therefore 13 is the greatest prime conforming to the question. Here is one instance. is greater than 100 so we can discount it. Working in this fashion we can add the following instances: . Therefore, there are five instances. Question Type: Quantitative Comparison Concept: Exponents and Fractions Level: 155 – 160 Column A Column B 1. The quantity in Column A is greater 2. The quantity in Column B is greater 3. The two quantities are equal 4. The relationship cannot be determined from the information given Answer: D. If x is less than 0 the answer is B. If x is , the answer is A. Therefore, the answer is D. Question Type: Multiple Choice Concept: Geometry/Variables in Answer Choices Level: 160 – 165 A square garden is surrounded by a path of uniform width. If the path and the garden both have an area of x, then what is the width of the path in terms of x? (160 – 165) 1. 2. 3. 4. 5. Answer: E If the area of the small square is x, then each side is √x. The area of the large square is 2x (you want to add the area of the small square to that of the path), leaving us with sides of √2x. If we subtract the length of a side of the small square from a side of the large square, that leaves us with √2x – √x. Remember that there are two parts of the path, so we have to divide by 2: √2x/2 – √x/2, which is (E). |
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