# Free Solutions and Explanations to GRE Quantitative MCQ (Multiple Answers) to Sample 1

Free Solutions and Explanations to Practice GRE Quantitative MCQ (Multiple Answers) to sample 1.
## Solution to Question 1For a number to be divisible by 6, it must be divisible by 2 and 3. All the given numbers above are divisible by 2. We need to select values of B such the numbers are divisible by 3. For a number to be divisible 3, the sum of its digits must be divisible by 3. Hence3 + B + 3 + 2 + 4 = 12 + B The values of B that make 12 + B divisible by 3 are 0, 3, 6 and 9 The answer to the above question is A , D , G , J ## Solution to Question 2A prime number is a number that has two factors only 1 and itself. Let us write the above numbers and list their factors, then decide which are not prime.Factors of 27 are 1 , 3 , 9 , 27 Factors of 13 are 1 , 13 Factors of 43 are 1 , 43 Factors of 49 are 1 , 7 , 49 Factors of 119 are 1 , 7 , 17 , 119 Factors of 1111 are 1 , 11 , 101 , 1111 According to the definition above, 27, 49, 119 and 1111 are prime numbers. The answer to the above questions is: A, D, E, F ## Solution to Question 3The average of a set of non equal numbers is always greater than the smallest data value in the set and smaller than the largest value in the set. Hence in this case the average of a, b, c and d is less than d and greater than a. HenceStatement A is always true Statement B is always false Statement C is always true Statement D is not always true. Explanation with counter example: Let a = 0, b = 1, c = 3 and d = 100. Find the mean m of a, b, c and d. m = (0 + 1 + 3 + 100) / 4 = 26. Three times the average = 3 * 26 = 78 and is not greater than d = 100. Answer to above question: A , C ## Solution to Question 4The distance on a number line between x and -1 is given by.|x - (-1)| = |x + 1| The values of x that make. |x + 1| = 5 are x = 4 and x = -6 The answer to the above question is. A , C ## Solution to Question 5We first reduce the fraction 6 / 86 / 8 = 3 / 4 We reduce the given fraction in order to find which is equivalent to 3 / 4 30 / 40 = 3 / 4 (divide numerator and denominator by 10) 4 / 5 cannot be further reduced 12 / 16 = 3 / 4 (divide numerator and denominator by 4) 3 / 4 cannot be further reduced 99 / 132 = 3 / 4 (divide numerator and denominator by 33) 4 / 3 cannot be further reduced Answer to given question A , C , D , E ## Solution to Question 6If a, b and c are the sides of a right triangle and c has the largest length, then a,b and c must satisfy Pythagorean theorem as followsa ^{2} + b^{2} = c^{2}We now check if each of the given sides above satisfy Pythagora's theorem 1 , 2 , 3: 1 ^{2} + 2^{2} = 3^{2} ; 1 + 2 = 9 ; Not Satisfied
6 , 8 , 10: 6 ^{2} + 8^{2} = 10^{2} ; 36 + 64 = 100 ; Satisfied
1.2 , 1.6 , 2.0: 1.2 ^{2} + 1.6^{2} = 2.0^{2} ; 1.44 + 2.56 = 4.0 ; Satisfied
3 , 5 , √34: 3 ^{2} + 5^{2} = (√34)^{2} ; 9 + 25 = 34 ; Satisfied
10 , 12 , 20: 10 ^{2} + 12^{2} = 20^{2} ; 100 + 144 = 400 ; Not Satisfied
The answer to the above question is B , C , D ## Solution to Question 7Triangle ABC has 2 sides of equal length is therefore isosceles and angles ABC and ACB are equal in size.Statement A is true The sum of all three angles ia 180°. Hence. size of ABC + size of ACB + 30 = 180 angles ABC and ACB are equal in size, hence. 2(size of ABC) + 30 = 180 size of ABC = (180 - 30) / 2 = 75° Statement B is false angles ABC and ACB are equal in size, hence. size of ACB = 75° , statement C is true The area of a triangle may be calculated using two sides and the angle between them as follows. (1/2) sin (30°) 10 * 10 = (1/2)(1/2)10*10 = 25 , statement D is true. The answer to the above question is. A , C , D ## Solution to Question 8For x = 0x ^{2} = 0 and | x | = 0 , and therefore both statement A and B are false.
The square of a real number is either positive or zero and if you add any positive number to a square such x ^{2} + 1 it becomes greater than 0.Statment C is true. If x = -1, then | x + 1| = 0, Statment D is false. The absolute value of a real number is either positive or zero and if you add any positive number to an absolute value such | x | + 1 it becomes greater than 0. Statment E is true. The answer to the given question is. C , E ## Solution to Question 9|- x - y| may be written as|- x - y| = |-(x + y)| Apply the fact that | - x | = | x | to simplify |- x - y| = |-(x + y)| = | x + y | , statement A is true If x = 6 and y = 9, then |x - y| = |6 - 9| = 3 and |x + y| = |6 + 9| = 15 , statement B in not true |y - x| may be written as |y - x| = |-(x - y) | = |x - y| , statement C is true |- x + y| may be written as |- x + y| = | -(x - y) | = |x - y| , statement D is true The answer to the above question is A , C , D ## Solution to Question 10Since a × b × c = 100, none of the numbers a, b or c is equal to 0.Statement A is false Since a × d × e = 0 and b × f × h = 0, d may be equal to 0 or f may be equal to 0. Statement B is true Since a × d × e = 0 and a not equal to 0, d × e must be equal to 0. Statement C is false Since b × f × h = 0 and b not equal to 0, then either f equals 0 or h equals or both. Statment D is true The answer to the above question is. B , C ,D ## More References and Links to Maths Practice TestsFree GRE Quantitative for PracticeFree Practice for GAMT Maths tests Free Compass Maths tests Practice Free Practice for SAT, ACT Maths tests Free AP Calculus Questions (AB and BC) with Answers |