GRE Algebra Theory & Practice Questions

GRE Algebra – Word Problems

January 25, 2013 By K S Baskar Leave a Comment

When four is added to six times a number and the result squared, the result obtained is four times the square of the sum of the number and its next multiple. What is the number?Explanation:The question can be directly converted to an equation as follows.Let ‘n’ be the number.(4 + 6n)2= 4(n + 2n)222(2 + 3n)2 = 4(3n)24 + 12n + 9n2= 9n2n = … [Read more...] about GRE Algebra – Word Problems

Algebra – Inequalities and Exponents

December 31, 2012 By K S Baskar Leave a Comment

Solve: (a) 7^(4-16x) > 1 (b) 2/(x-1) < 1 (a) For any a^b > 1, if a > 1 then b should be greater than 0. Hence 4 – 16x > 0 Therefore, x<1/4. (b) 2/(x-1) < 1 2/(x-1) - 1< 0 [2-(x-1)]/(x-1) < 0 (3-x)/(x-1) < 0 (x-3)/(x-1) > 0 Hence ‘x’ lies outside the roots ‘1’ and ‘3’. Therefore, x € (-∞ , 1) U (3, ∞) … [Read more...] about Algebra – Inequalities and Exponents

Algbera Practice: Functions

October 8, 2012 By K S Baskar Leave a Comment

A function is defined as f(x+2) = 3 + f(x) when x is even f(x+2) = x + f(x) when x is odd f(1) = 4 and f(2) = 3 Find f(f(f(1))) + f(f(f(2))) Answer: 17 Explanation: f(f(f(1))) + f(f(f(2))) = f(f(4)) + f(f(3)) Now f(4) can be found by using the definition of the function when x is even. Take x = 2. f(2+2) = 3 + f(2) f(4) = 3 + 3 f(4) = 6 Likewise, f(3) can be … [Read more...] about Algbera Practice: Functions

Inequalities and Absolute Values

August 17, 2012 By K S Baskar Leave a Comment

The solution set for the inequality |x+2| < |4x+1| (a) -3/5 < x < 1/3 (b) -1/3 < x < 3/5 (c) 0 < x < 1/3 (d) -1/3 < x < 1/3 (e) None of these Correct Answer: (E) Explanation Since the expressions on both sides of the inequality are under modulus, you can square on both sides without disturbing the inequality. (x+2)2 < … [Read more...] about Inequalities and Absolute Values