Sample 2

Solutions and detailed explanations to questions and problems similar to the questions in the GRE test. sample 2## Solution to Question 1w is the mean of a, b, c and d is written asw = (a + b + c + d) / 4 w is the average W of m(a + k), m(b + k), m(c + k) and m(d + k) is given by W = [ m(a + k) + m(b + k) + m(c + k) + m(d + k) ] / 4 W = m [ a + b + c + d + 4 k] / 4 = m [a + b + c + d ] / 4 + m k = m (w + k) If w is the average of a, b, c and d, then the average W of m(a + k), m(b + k), m(c + k) and m(d + k) is given by W = m (w + k) ## Solution to Question 2Let r and R be the radii of the smaller and larger circles respectively. The radius of the larger circle is three times the radius of the smaller circle leads toR = 3r Areas A1 of smaller and A2 of larger circles are given by A1 = Pi r ^{2}A2 = Pi R ^{2} = Pi (3r)^{2} = 9 Pi r^{2}ratio R of areas larger / smaller is equal to R = 9 Pi r ^{2} / Pi r^{2} = 9
## Solution to Question 3Let S1 be the total salary of the group of 20 employess. Hence35,000 = S1 / 20 S1 = 20 * 35,000 = $700,000 Let S2 be the total salary of the group of 30 employess. Hence 40,000 = S2 / 30 S2 = $1,200,000 The average of all 50 employers is given by (700,000 + 1,200,000) / 50 = $38,000 ## Solution to Question 4Rewrite the given expression using the fact that 48 = 3 × 16√48 = √(3 × 16) Use the formula √(a × b) = √a × √a to rewite √(3 × 16) as √48 = √(3 × 16) = √3 × √16 = 4 √3 ## Solution to Question 5Since the three angles are in the ration 2:4:3, their sizes they may be written in the formSize of A = 2 k , Size of B = 4 k and size of C = 3 k , where k is a constant. The sum of the angles of a ny triangle is equal to 180°; hence 2 k + 4 k + 3 k = 180 Solve for k 9 k = 180 , k = 20 The smallest angle is A and its size is equal to 2 k 2 k = 2 × 20 = 40° ## Solution to Question 6If n is even, it can be written as followsn = 2 k , where k is an integer If m is odd, it can be written as follows m = 2 K + 1 , where K is an integer We now express n + m in terms of k and K n + m = 2 k + 2 K + 1 = 2(k + K) + 1 n + m is odd We now express n - m in terms of k and K n - m = 2 k - (2 K + 1) = 2 k - 2 K - 1 n - m = 2 (k - K) - 1 n - m is odd We now express n * m in terms of k and K n * m = (2 k)(2 K + 1) = 2( k(2K + 1) ) n * m is even We now express n ^{2} + m^{2} + 1 in terms of k and K
n ^{2} + m^{2} + 1 = (2 k)^{2} + (2 K + 1)^{2} + 1
= 4 k^{2} + 4 K^{2} + 4 K + 1 + 1
= 2 ( 2 k ^{2} + 2 K^{2} + 2 K + 1)
n ^{2} + m^{2} + 1 is even
Statement D is true. ## Solution to Question 7Use the facts that 25 = 5^{2} and 125 = 5^{3} to rewrite the given expression as follows
5 ^{100} + 25^{50} + 3(125^{34} / 25) = 5^{100} + (5^{2})^{50} + 3( (5^{3})^{34} / (5^{2}))
Use formula for exponents to simplify = 5 ^{100} + 5^{100} + 3( 5^{102} / 5^{2})
= 5 ^{100} + 5^{100} + 3( 5^{100})
= 5 * 5 ^{100}= 5 ^{101}## Solution to Question 8Factor numerator as follows6x ^{10} - 2x^{9} = 2x^{9} (3x - 1)
Factor denominator as follows 9x ^{2} - 1 = (3x - 1)(3x + 1)
Substitute numerator and denominator by their factored forms and simplify the given expression [ 6x ^{10} - 2x^{9} ] / (9x^{2} - 1) = [2x^{9} (3x - 1)
] / [(3x - 1)(3x + 1)]
= 2x ^{9} / (3x + 1)
## Solution to Question 9Expand by multiplication or using the identity (x + y)^{2} = x^{2} + 2 x y + y^{2}.
(- 2x + 6) ^{2} = (-2x)^{2} + 2 (-2x)(6) + 6^{2}= 4 x ^{2} - 24 x + 36
## Solution to Question 10The sum of all interior angles of a polygon of n sides is given by.(n - 2) * 180 and is equal to 1800°. Hence (n - 2) * 180 = 1800 Solve for n (n - 2) = 10 n = 12 ## More References and Links to More Maths Practice TestsFree GRE Quantitative for Practice Free 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 |