Free GRE Practice Questions with Solutions
Sample 2

Solutions and detailed explanations to questions and problems similar to the questions in the GRE test. sample 2

Solution to Question 1

Let \( w \) be the mean of \( a, b, c \) and \( d \) is written as \[ w = \frac{a + b + c + d}{4} \] Let W is the average of \( m(a + k) \), \( m(b + k) \), \( m(c + k) \), and \( m(d + k) \) is given by \[ W = \frac{m(a + k) + m(b + k) + m(c + k) + m(d + k)}{4} \] \[ W = \frac{m(a + b + c + d + 4k)}{4} = \frac{m(a + b + c + d)}{4} + mk \] \[ = m(w + k) \] 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 2

Let \( r \) 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 to \[ R = 3r \] Areas \( A_1 \) of the smaller and \( A_2 \) of the larger circles are given by \[ A_1 = \pi r^2 \] \[ A_2 = \pi R^2 = \pi (3r)^2 = 9\pi r^2 \] Ratio \( R \) of areas (larger / smaller) is equal to \[ R = \frac{9\pi r^2}{\pi r^2} = 9 \]

Solution to Question 3

Let \( S_1 \) be the total salary of the group of 20 employees. Hence \[ 35,000 = \frac{S_1}{20} \] \[ S_1 = 20 \times 35,000 = 700,000 \] Let \( S_2 \) be the total salary of the group of 30 employees. Hence \[ 40,000 = \frac{S_2}{30} \] \[ S_2 = 1,200,000 \] The average of all 50 employees is given by \[ \frac{700,000 + 1,200,000}{50} = 38,000 \]

Solution to Question 4

Rewrite the given expression using the fact that \( 48 = 3 \times 16 \) \[ \sqrt{48} = \sqrt{3 \times 16} \] Use the formula \( \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \) to rewrite \( \sqrt{3 \times 16} \) as \[ \sqrt{48} = \sqrt{3 \times 16} = \sqrt{3} \times \sqrt{16} \] \[ = 4 \sqrt{3} \]

Solution to Question 5

Since the three angles are in the ratio 2:4:3, their sizes may be written in the form \[ \text{Size of A} = 2k, \quad \text{Size of B} = 4k, \quad \text{Size of C} = 3k \] where \( k \) is a constant. The sum of the angles of any triangle is equal to 180°; hence \[ 2k + 4k + 3k = 180 \] Solve for \( k \): \[ 9k = 180, \quad k = 20 \] The smallest angle is A and its size is equal to \( 2k \): \[ 2k = 2 \times 20 = 40^\circ \]

Solution to Question 6

If \( n \) is even, it can be written as follows \[ n = 2k, \quad \text{where } k \text{ (lower case k) is an integer} \] If \( m \) is odd, it can be written as follows \[ m = 2K + 1, \quad \text{where } K \text{ (upper case K) is an integer} \] We now express \( n + m \) in terms of \( k \) and \( K \) \[ n + m = 2k + 2K + 1 = 2(k + K) + 1 \] \( n + m \) is odd We now express \( n - m \) in terms of \( k \) and \( K \) \[ n - m = 2k - (2K + 1) = 2k - 2K - 1 \] \[ n - m = 2(k - K) - 1 \] \( n - m \) is odd We now express \( n \times m \) in terms of \( k \) and \( K \) \[ n \times m = (2k)(2K + 1) = 2(k(2K + 1)) \] \( n \times m \) is even We now express \( n^2 + m^2 + 1 \) in terms of \( k \) and \( K \) \[ n^2 + m^2 + 1 = (2k)^2 + (2K + 1)^2 + 1 \] \[ = 4k^2 + 4K^2 + 4K + 1 + 1 \] \[ = 2(2k^2 + 2K^2 + 2K + 1) \] \( n^2 + m^2 + 1 \) is even and hence statement D is true.

Solution to Question 7

Use the facts that: \( 25 = 5^2 \) and \( 125 = 5^3 \) to rewrite the given expression as follows: \[ 5^{100} + 25^{50} + 3\left(\frac{125^{34}}{25}\right) = 5^{100} + \left(5^2\right)^{50} + 3\left(\frac{\left(5^3\right)^{34}}{5^2}\right) \] Use the formula for exponents to simplify: \[ = 5^{100} + 5^{100} + 3\left( \frac{5^{102}}{5^2} \right) \] \[ = 5^{100} + 5^{100} + 3\left( 5^{100} \right) \] \[ = 5 \cdot 5^{100} \] \[ = 5^{101} \]

Solution to Question 8

Factor numerator as follows: \[ 6x^{10} - 2x^{9} = 2x^{9} (3x - 1) \] Factor denominator as follows: \[ 9x^{2} - 1 = (3x - 1)(3x + 1) \] Substitute numerator and denominator by their factored forms and simplify the given expression: \[ \frac{6x^{10} - 2x^{9}}{9x^{2} - 1} = \frac{2x^{9} (3x - 1)}{(3x - 1)(3x + 1)} \] \[ = \frac{2x^{9}}{3x + 1} \]

Solution to Question 9

Expand by multiplication or using the identity \( (x + y)^2 = x^2 + 2xy + y^2 \). \[ (-2x + 6)^2 = (-2x)^2 + 2(-2x)(6) + 6^2 \] \[ = 4x^2 - 24x + 36 \]

Solution to Question 10

The sum of all interior angles of a polygon of \( n \) sides is given by. \[ (n - 2) \cdot 180 \] and is equal to \( 1800^\circ \). Hence \[ (n - 2) \cdot 180 = 1800 \] Solve for \( n \): \[ (n - 2) = 10 \] \[ n = 12 \]

Free GRE Practice Questions with Solutions Sample 2